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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
779 | A | Pupils Redistribution | PROGRAMMING | 1,000 | [
"constructive algorithms",
"math"
] | null | null | In Berland each high school student is characterized by academic performance — integer value between 1 and 5.
In high school 0xFF there are two groups of pupils: the group *A* and the group *B*. Each group consists of exactly *n* students. An academic performance of each student is known — integer value between 1 and 5.
The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1, the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.
To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class *A* and one student of class *B*. After that, they both change their groups.
Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance. | The first line of the input contains integer number *n* (1<=≤<=*n*<=≤<=100) — number of students in both groups.
The second line contains sequence of integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=5), where *a**i* is academic performance of the *i*-th student of the group *A*.
The third line contains sequence of integer numbers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=5), where *b**i* is academic performance of the *i*-th student of the group *B*. | Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained. | [
"4\n5 4 4 4\n5 5 4 5\n",
"6\n1 1 1 1 1 1\n5 5 5 5 5 5\n",
"1\n5\n3\n",
"9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1\n"
] | [
"1\n",
"3\n",
"-1\n",
"4\n"
] | none | 500 | [
{
"input": "4\n5 4 4 4\n5 5 4 5",
"output": "1"
},
{
"input": "6\n1 1 1 1 1 1\n5 5 5 5 5 5",
"output": "3"
},
{
"input": "1\n5\n3",
"output": "-1"
},
{
"input": "9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1",
"output": "4"
},
{
"input": "1\n1\n2",
"output": "-1"
},
{
"input": "1\n1\n1",
"output": "0"
},
{
"input": "8\n1 1 2 2 3 3 4 4\n4 4 5 5 1 1 1 1",
"output": "2"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1\n2 2 2 2 2 2 2 2 2 2",
"output": "5"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "2\n1 1\n1 1",
"output": "0"
},
{
"input": "2\n1 2\n1 1",
"output": "-1"
},
{
"input": "2\n2 2\n1 1",
"output": "1"
},
{
"input": "2\n1 2\n2 1",
"output": "0"
},
{
"input": "2\n1 1\n2 2",
"output": "1"
},
{
"input": "5\n5 5 5 5 5\n5 5 5 5 5",
"output": "0"
},
{
"input": "5\n5 5 5 3 5\n5 3 5 5 5",
"output": "0"
},
{
"input": "5\n2 3 2 3 3\n2 3 2 2 2",
"output": "1"
},
{
"input": "5\n4 4 1 4 2\n1 2 4 2 2",
"output": "1"
},
{
"input": "50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "50\n1 3 1 3 3 3 1 3 3 3 3 1 1 1 3 3 3 1 3 1 1 1 3 1 3 1 3 3 3 1 3 1 1 3 3 3 1 1 1 1 3 3 1 1 1 3 3 1 1 1\n1 3 1 3 3 1 1 3 1 3 3 1 1 1 1 3 3 1 3 1 1 3 1 1 3 1 1 1 1 3 3 1 3 3 3 3 1 3 3 3 3 3 1 1 3 3 1 1 3 1",
"output": "0"
},
{
"input": "50\n1 1 1 4 1 1 4 1 4 1 1 4 1 1 4 1 1 4 1 1 4 1 4 4 4 1 1 4 1 4 4 4 4 4 4 4 1 4 1 1 1 1 4 1 4 4 1 1 1 4\n1 4 4 1 1 4 1 4 4 1 1 4 1 4 1 1 4 1 1 1 4 4 1 1 4 1 4 1 1 4 4 4 4 1 1 4 4 1 1 1 4 1 4 1 4 1 1 1 4 4",
"output": "0"
},
{
"input": "50\n3 5 1 3 3 4 3 4 2 5 2 1 2 2 5 5 4 5 4 2 1 3 4 2 3 3 3 2 4 3 5 5 5 5 5 5 2 5 2 2 5 4 4 1 5 3 4 2 1 3\n3 5 3 2 5 3 4 4 5 2 3 4 4 4 2 2 4 4 4 3 3 5 5 4 3 1 4 4 5 5 4 1 2 5 5 4 1 2 3 4 5 5 3 2 3 4 3 5 1 1",
"output": "3"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "100\n1 1 3 1 3 1 1 3 1 1 3 1 3 1 1 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 1 1 1 3 1 1 1 3 1 1 3 3 1 3 3 1 3 1 3 3 3 3 1 1 3 3 3 1 1 3 1 3 3 3 1 3 3 3 3 3 1 3 3 3 3 1 3 1 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 1 1 3 1 1 1\n1 1 1 3 3 3 3 3 3 3 1 3 3 3 1 3 3 3 3 3 3 1 3 3 1 3 3 1 1 1 3 3 3 3 3 3 3 1 1 3 3 3 1 1 3 3 1 1 1 3 3 3 1 1 3 1 1 3 3 1 1 3 3 3 3 3 3 1 3 3 3 1 1 3 3 3 1 1 3 3 1 3 1 3 3 1 1 3 3 1 1 3 1 3 3 3 1 3 1 3",
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{
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"output": "0"
},
{
"input": "100\n3 3 4 3 3 4 3 1 4 2 1 3 1 1 2 4 4 4 4 1 1 4 1 4 4 1 1 2 3 3 3 2 4 2 3 3 3 1 3 4 2 2 1 3 4 4 3 2 2 2 4 2 1 2 1 2 2 1 1 4 2 1 3 2 4 4 4 2 3 1 3 1 3 2 2 2 2 4 4 1 3 1 1 4 2 3 3 4 4 2 4 4 2 4 3 3 1 3 2 4\n3 1 4 4 2 1 1 1 1 1 1 3 1 1 3 4 3 2 2 4 2 1 4 4 4 4 1 2 3 4 2 3 3 4 3 3 2 4 2 2 2 1 2 4 4 4 2 1 3 4 3 3 4 2 4 4 3 2 4 2 4 2 4 4 1 4 3 1 4 3 3 3 3 1 2 2 2 2 4 1 2 1 3 4 3 1 3 3 4 2 3 3 2 1 3 4 2 1 1 2",
"output": "0"
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{
"input": "100\n2 4 5 2 1 5 5 2 1 5 1 5 1 1 1 3 4 5 1 1 2 3 3 1 5 5 4 4 4 1 1 1 5 2 3 5 1 2 2 1 1 1 2 2 1 2 4 4 5 1 3 2 5 3 5 5 3 2 2 2 1 3 4 4 4 4 4 5 3 1 4 1 5 4 4 5 4 5 2 4 4 3 1 2 1 4 5 3 3 3 3 2 2 2 3 5 3 1 3 4\n3 2 5 1 5 4 4 3 5 5 5 2 1 4 4 3 2 3 3 5 5 4 5 5 2 1 2 4 4 3 5 1 1 5 1 3 2 5 2 4 4 2 4 2 4 2 3 2 5 1 4 4 1 1 1 5 3 5 1 1 4 5 1 1 2 2 5 3 5 1 1 1 2 3 3 2 3 2 4 4 5 4 2 1 3 4 1 1 2 4 1 5 3 1 2 1 3 4 1 3",
"output": "0"
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{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "100\n1 4 4 1 4 4 1 1 4 1 1 1 1 4 4 4 4 1 1 1 1 1 1 4 4 4 1 1 4 4 1 1 1 1 4 4 4 4 4 1 1 4 4 1 1 1 4 1 1 1 1 4 4 4 4 4 4 1 4 4 4 4 1 1 1 4 1 4 1 1 1 1 4 1 1 1 4 4 4 1 4 4 1 4 4 4 4 4 1 4 1 1 4 1 4 1 1 1 4 4\n4 1 1 4 4 4 1 4 4 4 1 1 4 1 1 4 1 4 4 4 1 1 4 1 4 1 1 1 4 4 1 4 1 4 1 4 4 1 1 4 1 4 1 1 1 4 1 4 4 4 1 4 1 4 4 4 4 1 4 1 1 4 1 1 4 4 4 1 4 1 4 1 4 4 4 1 1 4 1 4 4 4 4 1 1 1 1 1 4 4 1 4 1 4 1 1 1 4 4 1",
"output": "1"
},
{
"input": "100\n5 2 5 2 2 3 3 2 5 3 2 5 3 3 3 5 2 2 5 5 3 3 5 3 2 2 2 3 2 2 2 2 3 5 3 3 2 3 2 5 3 3 5 3 2 2 5 5 5 5 5 2 3 2 2 2 2 3 2 5 2 2 2 3 5 5 5 3 2 2 2 3 5 3 2 5 5 3 5 5 5 3 2 5 2 3 5 3 2 5 5 3 5 2 3 3 2 2 2 2\n5 3 5 3 3 5 2 5 3 2 3 3 5 2 5 2 2 5 2 5 2 5 3 3 5 3 2 2 2 3 5 3 2 2 3 2 2 5 5 2 3 2 3 3 5 3 2 5 2 2 2 3 3 5 3 3 5 2 2 2 3 3 2 2 3 5 3 5 5 3 3 2 5 3 5 2 3 2 5 5 3 2 5 5 2 2 2 2 3 2 2 5 2 5 2 2 3 3 2 5",
"output": "1"
},
{
"input": "100\n4 4 5 4 3 5 5 2 4 5 5 5 3 4 4 2 5 2 5 3 3 3 3 5 3 2 2 2 4 4 4 4 3 3 4 5 3 2 2 2 4 4 5 3 4 5 4 5 5 2 4 2 5 2 3 4 4 5 2 2 4 4 5 5 5 3 5 4 5 5 5 4 3 3 2 4 3 5 5 5 2 4 2 5 4 3 5 3 2 3 5 2 5 2 2 5 4 5 4 3\n5 4 2 4 3 5 2 5 5 3 4 5 4 5 3 3 5 5 2 3 4 2 3 5 2 2 2 4 2 5 2 4 4 5 2 2 4 4 5 5 2 3 4 2 4 5 2 5 2 2 4 5 5 3 5 5 5 4 3 4 4 3 5 5 3 4 5 3 2 3 4 3 4 4 2 5 3 4 5 5 3 5 3 3 4 3 5 3 2 2 4 5 4 5 5 2 3 4 3 5",
"output": "1"
},
{
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"output": "1"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\n5 2 2 2 5 2 5 5 5 2 5 2 5 5 5 5 5 5 2 2 2 5 5 2 5 2 2 5 2 5 5 2 5 2 5 2 5 5 5 5 5 2 2 2 2 5 5 2 5 5 5 2 5 5 5 2 5 5 5 2 2 2 5 2 2 2 5 5 2 5 5 5 2 5 2 2 5 2 2 2 5 5 5 5 2 5 2 5 2 2 5 2 5 2 2 2 2 5 5 2\n5 5 2 2 5 5 2 5 2 2 5 5 5 5 2 5 5 2 5 2 2 5 2 2 5 2 5 2 2 5 2 5 2 5 5 2 2 5 5 5 2 5 5 2 5 5 5 2 2 5 5 5 2 5 5 5 2 2 2 5 5 5 2 2 5 5 2 2 2 5 2 5 5 2 5 2 5 2 2 5 5 2 2 5 5 2 2 5 2 2 5 2 2 2 5 5 2 2 2 5",
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{
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"output": "1"
},
{
"input": "100\n5 3 3 2 5 3 2 4 2 3 3 5 3 4 5 4 3 3 4 3 2 3 3 4 5 4 2 4 2 4 5 3 3 4 5 3 5 3 5 3 3 2 5 3 4 5 2 5 2 2 4 2 2 2 2 5 4 5 4 3 5 4 2 5 5 3 4 5 2 3 2 2 2 5 3 2 2 2 3 3 5 2 3 2 4 5 3 3 3 5 2 3 3 3 5 4 5 5 5 2\n4 4 4 5 5 3 5 5 4 3 5 4 3 4 3 3 5 3 5 5 3 3 3 5 5 4 4 3 2 5 4 3 3 4 5 3 5 2 4 2 2 2 5 3 5 2 5 5 3 3 2 3 3 4 2 5 2 5 2 4 2 4 2 3 3 4 2 2 2 4 4 3 3 3 4 3 3 3 5 5 3 4 2 2 3 5 5 2 3 4 5 4 5 3 4 2 5 3 2 4",
"output": "3"
},
{
"input": "100\n5 3 4 4 2 5 1 1 4 4 3 5 5 1 4 4 2 5 3 2 1 1 3 2 4 4 4 2 5 2 2 3 1 4 1 4 4 5 3 5 1 4 1 4 1 5 5 3 5 5 1 5 3 5 1 3 3 4 5 3 2 2 4 5 2 5 4 2 4 4 1 1 4 2 4 1 2 2 4 3 4 1 1 1 4 3 5 1 2 1 4 5 4 4 2 1 4 1 3 2\n1 1 1 1 4 2 1 4 1 1 3 5 4 3 5 2 2 4 2 2 4 1 3 4 4 5 1 1 2 2 2 1 4 1 4 4 1 5 5 2 3 5 1 5 4 2 3 2 2 5 4 1 1 4 5 2 4 5 4 4 3 3 2 4 3 4 5 5 4 2 4 2 1 2 3 2 2 5 5 3 1 3 4 3 4 4 5 3 1 1 3 5 1 4 4 2 2 1 4 5",
"output": "2"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\n3 3 4 3 3 4 3 3 4 4 3 3 3 4 3 4 3 4 4 3 3 3 3 3 3 4 3 3 4 3 3 3 3 4 3 3 3 4 4 4 3 3 4 4 4 3 4 4 3 3 4 3 3 3 4 4 4 3 4 3 3 3 3 3 3 3 4 4 3 3 3 3 4 3 3 3 3 3 4 4 3 3 3 3 3 4 3 4 4 4 4 3 4 3 4 4 4 4 3 3\n4 3 3 3 3 4 4 3 4 4 4 3 3 4 4 3 4 4 4 4 3 4 3 3 3 4 4 4 3 4 3 4 4 3 3 4 3 3 3 3 3 4 3 3 3 3 4 4 4 3 3 4 3 4 4 4 4 3 4 4 3 3 4 3 3 4 3 4 3 4 4 4 4 3 3 4 3 4 4 4 3 3 4 4 4 4 4 3 3 3 4 3 3 4 3 3 3 3 3 3",
"output": "5"
},
{
"input": "100\n4 2 5 2 5 4 2 5 5 4 4 2 4 4 2 4 4 5 2 5 5 2 2 4 4 5 4 5 5 5 2 2 2 2 4 4 5 2 4 4 4 2 2 5 5 4 5 4 4 2 4 5 4 2 4 5 4 2 4 5 4 4 4 4 4 5 4 2 5 2 5 5 5 5 4 2 5 5 4 4 2 5 2 5 2 5 4 2 4 2 4 5 2 5 2 4 2 4 2 4\n5 4 5 4 5 2 2 4 5 2 5 5 5 5 5 4 4 4 4 5 4 5 5 2 4 4 4 4 5 2 4 4 5 5 2 5 2 5 5 4 4 5 2 5 2 5 2 5 4 5 2 5 2 5 2 4 4 5 4 2 5 5 4 2 2 2 5 4 2 2 4 4 4 5 5 2 5 2 2 4 4 4 2 5 4 5 2 2 5 4 4 5 5 4 5 5 4 5 2 5",
"output": "5"
},
{
"input": "100\n3 4 5 3 5 4 5 4 4 4 2 4 5 4 3 2 3 4 3 5 2 5 2 5 4 3 4 2 5 2 5 3 4 5 2 5 4 2 4 5 4 3 2 4 4 5 2 5 5 3 3 5 2 4 4 2 3 3 2 5 5 5 2 4 5 5 4 2 2 5 3 3 2 4 4 2 4 5 5 2 5 5 3 2 5 2 4 4 3 3 5 4 5 5 2 5 4 5 4 3\n4 3 5 5 2 4 2 4 5 5 5 2 3 3 3 3 5 5 5 5 3 5 2 3 5 2 3 2 2 5 5 3 5 3 4 2 2 5 3 3 3 3 5 2 4 5 3 5 3 4 4 4 5 5 3 4 4 2 2 4 4 5 3 2 4 5 5 4 5 2 2 3 5 4 5 5 2 5 4 3 2 3 2 5 4 5 3 4 5 5 3 5 2 2 4 4 3 2 5 2",
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},
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},
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"output": "0"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 1 4 4 4 4 4 4 4 4 4 4\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 1 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "1"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 4 3 3 3 3 3 3 3 3 3 3 1 3 1 3 3 3 3 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3\n3 3 3 4 3 3 3 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 1 3 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 3 3 3 3 1 3 3 3 3 3 3 3 3 3 3",
"output": "1"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "50"
},
{
"input": "100\n3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5\n3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1",
"output": "25"
},
{
"input": "100\n3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5 3 5\n2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4",
"output": "50"
},
{
"input": "100\n1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "40"
},
{
"input": "100\n1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5\n2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3",
"output": "30"
},
{
"input": "5\n4 4 4 4 5\n4 5 5 5 5",
"output": "-1"
},
{
"input": "4\n1 1 1 1\n3 3 3 3",
"output": "2"
},
{
"input": "6\n1 1 2 2 3 4\n1 2 3 3 4 4",
"output": "-1"
},
{
"input": "4\n1 1 1 2\n3 3 3 3",
"output": "-1"
},
{
"input": "3\n2 2 2\n4 4 4",
"output": "-1"
},
{
"input": "2\n1 2\n3 4",
"output": "-1"
},
{
"input": "6\n1 1 1 3 3 3\n2 2 2 4 4 4",
"output": "-1"
},
{
"input": "5\n1 2 2 2 2\n1 1 1 1 3",
"output": "-1"
},
{
"input": "2\n1 3\n2 2",
"output": "-1"
},
{
"input": "2\n1 3\n4 5",
"output": "-1"
},
{
"input": "4\n1 2 3 4\n5 5 5 5",
"output": "-1"
},
{
"input": "2\n1 3\n2 4",
"output": "-1"
},
{
"input": "2\n1 2\n4 4",
"output": "-1"
},
{
"input": "2\n1 2\n3 3",
"output": "-1"
},
{
"input": "10\n4 4 4 4 2 3 3 3 3 1\n2 2 2 2 4 1 1 1 1 3",
"output": "-1"
},
{
"input": "6\n1 2 3 3 4 4\n1 1 2 2 3 4",
"output": "-1"
},
{
"input": "5\n3 3 3 3 1\n1 1 1 1 3",
"output": "-1"
},
{
"input": "2\n1 1\n2 3",
"output": "-1"
},
{
"input": "8\n1 1 2 2 3 3 3 3\n2 2 2 2 1 1 1 1",
"output": "2"
},
{
"input": "5\n1 1 1 3 3\n1 1 1 1 2",
"output": "-1"
},
{
"input": "6\n2 2 3 3 4 4\n2 3 4 5 5 5",
"output": "-1"
},
{
"input": "6\n1 1 2 2 3 4\n3 3 4 4 1 2",
"output": "-1"
},
{
"input": "4\n1 2 3 3\n3 3 3 3",
"output": "-1"
},
{
"input": "3\n1 2 3\n3 3 3",
"output": "-1"
},
{
"input": "5\n3 3 3 2 2\n2 2 2 3 3",
"output": "-1"
},
{
"input": "10\n1 2 3 4 1 2 3 4 1 2\n1 2 3 4 1 2 3 4 3 4",
"output": "-1"
},
{
"input": "2\n2 2\n1 3",
"output": "-1"
},
{
"input": "3\n1 2 3\n1 1 4",
"output": "-1"
},
{
"input": "4\n3 4 4 4\n3 3 4 4",
"output": "-1"
}
] | 1,668,895,847 | 2,147,483,647 | PyPy 3 | OK | TESTS | 81 | 92 | 0 | def solve(s1,s2):
#dc={"r": 0, "g" : 0, "b" : 0, "w": 0, "b": 0}
dc1={1: 0, 2 : 0, 3 : 0, 4 : 0, 5: 0}
dc2={1: 0, 2 : 0, 3 : 0, 4 : 0, 5: 0}
ln=len(s1)
for i in range(ln):
dc1[s1[i]]+=1
for i in range(ln):
dc2[s2[i]]+=1
ans=0
a=False
for i in range(1,6):
if (dc1[i] + dc2[i]) &1:
a=True
ans+=abs(dc1[i]-dc2[i])
if ans%4!=0 or a==True:
return -1
return ans//4
n=int(input())
s1=list(map(int,input().split()))
s2=list(map(int,input().split()))
print(solve(s1,s2))
| Title: Pupils Redistribution
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland each high school student is characterized by academic performance — integer value between 1 and 5.
In high school 0xFF there are two groups of pupils: the group *A* and the group *B*. Each group consists of exactly *n* students. An academic performance of each student is known — integer value between 1 and 5.
The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1, the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.
To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class *A* and one student of class *B*. After that, they both change their groups.
Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance.
Input Specification:
The first line of the input contains integer number *n* (1<=≤<=*n*<=≤<=100) — number of students in both groups.
The second line contains sequence of integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=5), where *a**i* is academic performance of the *i*-th student of the group *A*.
The third line contains sequence of integer numbers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=5), where *b**i* is academic performance of the *i*-th student of the group *B*.
Output Specification:
Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained.
Demo Input:
['4\n5 4 4 4\n5 5 4 5\n', '6\n1 1 1 1 1 1\n5 5 5 5 5 5\n', '1\n5\n3\n', '9\n3 2 5 5 2 3 3 3 2\n4 1 4 1 1 2 4 4 1\n']
Demo Output:
['1\n', '3\n', '-1\n', '4\n']
Note:
none | ```python
def solve(s1,s2):
#dc={"r": 0, "g" : 0, "b" : 0, "w": 0, "b": 0}
dc1={1: 0, 2 : 0, 3 : 0, 4 : 0, 5: 0}
dc2={1: 0, 2 : 0, 3 : 0, 4 : 0, 5: 0}
ln=len(s1)
for i in range(ln):
dc1[s1[i]]+=1
for i in range(ln):
dc2[s2[i]]+=1
ans=0
a=False
for i in range(1,6):
if (dc1[i] + dc2[i]) &1:
a=True
ans+=abs(dc1[i]-dc2[i])
if ans%4!=0 or a==True:
return -1
return ans//4
n=int(input())
s1=list(map(int,input().split()))
s2=list(map(int,input().split()))
print(solve(s1,s2))
``` | 3 | |
202 | A | LLPS | PROGRAMMING | 800 | [
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null | This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z". | The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10. | Print the lexicographically largest palindromic subsequence of string *s*. | [
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] | [
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] | Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | 500 | [
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
{
"input": "helloworld",
"output": "w"
},
{
"input": "zzzzzzzazz",
"output": "zzzzzzzzz"
},
{
"input": "testcase",
"output": "tt"
},
{
"input": "hahahahaha",
"output": "hhhhh"
},
{
"input": "abbbbbbbbb",
"output": "bbbbbbbbb"
},
{
"input": "zaz",
"output": "zz"
},
{
"input": "aza",
"output": "z"
},
{
"input": "dcbaedcba",
"output": "e"
},
{
"input": "abcdeabcd",
"output": "e"
},
{
"input": "edcbabcde",
"output": "ee"
},
{
"input": "aaaaaaaaab",
"output": "b"
},
{
"input": "testzzzzzz",
"output": "zzzzzz"
},
{
"input": "zzzzzzwait",
"output": "zzzzzz"
},
{
"input": "rrrrrqponm",
"output": "rrrrr"
},
{
"input": "zzyzyy",
"output": "zzz"
},
{
"input": "aababb",
"output": "bbb"
},
{
"input": "zanzibar",
"output": "zz"
},
{
"input": "hhgfedcbaa",
"output": "hh"
},
{
"input": "aabcdefghh",
"output": "hh"
},
{
"input": "aruaru",
"output": "uu"
},
{
"input": "uraura",
"output": "uu"
},
{
"input": "aru",
"output": "u"
},
{
"input": "aburvabur",
"output": "v"
},
{
"input": "ura",
"output": "u"
},
{
"input": "eurottat",
"output": "u"
},
{
"input": "referee",
"output": "rr"
},
{
"input": "joking",
"output": "o"
},
{
"input": "seriously",
"output": "y"
},
{
"input": "sets",
"output": "t"
},
{
"input": "test",
"output": "tt"
},
{
"input": "klmgameklm",
"output": "mmm"
},
{
"input": "dfkjafdkdd",
"output": "kk"
},
{
"input": "zzzzzzzzzz",
"output": "zzzzzzzzzz"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaaaa"
},
{
"input": "buzz",
"output": "zz"
},
{
"input": "b",
"output": "b"
},
{
"input": "y",
"output": "y"
},
{
"input": "yy",
"output": "yy"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "bb",
"output": "bb"
},
{
"input": "aa",
"output": "aa"
},
{
"input": "a",
"output": "a"
},
{
"input": "z",
"output": "z"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "zzazazzzaz",
"output": "zzzzzzz"
},
{
"input": "hellhellhe",
"output": "llll"
},
{
"input": "hellohello",
"output": "oo"
},
{
"input": "refer",
"output": "rr"
}
] | 1,602,847,257 | 2,147,483,647 | PyPy 3 | OK | TESTS | 54 | 310 | 0 | s=input().strip()
res=[]
mx=s[0]
for i in range(1,len(s)):
mx=max(mx,s[i])
for i in range(len(s)):
if s[i]==mx:
res.append(s[i])
print(''.join(res)) | Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | ```python
s=input().strip()
res=[]
mx=s[0]
for i in range(1,len(s)):
mx=max(mx,s[i])
for i in range(len(s)):
if s[i]==mx:
res.append(s[i])
print(''.join(res))
``` | 3 | |
400 | C | Inna and Huge Candy Matrix | PROGRAMMING | 1,500 | [
"implementation",
"math"
] | null | null | Inna and Dima decided to surprise Sereja. They brought a really huge candy matrix, it's big even for Sereja! Let's number the rows of the giant matrix from 1 to *n* from top to bottom and the columns — from 1 to *m*, from left to right. We'll represent the cell on the intersection of the *i*-th row and *j*-th column as (*i*,<=*j*). Just as is expected, some cells of the giant candy matrix contain candies. Overall the matrix has *p* candies: the *k*-th candy is at cell (*x**k*,<=*y**k*).
The time moved closer to dinner and Inna was already going to eat *p* of her favourite sweets from the matrix, when suddenly Sereja (for the reason he didn't share with anyone) rotated the matrix *x* times clockwise by 90 degrees. Then he performed the horizontal rotate of the matrix *y* times. And then he rotated the matrix *z* times counterclockwise by 90 degrees. The figure below shows how the rotates of the matrix looks like.
Inna got really upset, but Duma suddenly understood two things: the candies didn't get damaged and he remembered which cells contained Inna's favourite sweets before Sereja's strange actions. Help guys to find the new coordinates in the candy matrix after the transformation Sereja made! | The first line of the input contains fix integers *n*, *m*, *x*, *y*, *z*, *p* (1<=≤<=*n*,<=*m*<=≤<=109; 0<=≤<=*x*,<=*y*,<=*z*<=≤<=109; 1<=≤<=*p*<=≤<=105).
Each of the following *p* lines contains two integers *x**k*, *y**k* (1<=≤<=*x**k*<=≤<=*n*; 1<=≤<=*y**k*<=≤<=*m*) — the initial coordinates of the *k*-th candy. Two candies can lie on the same cell. | For each of the *p* candies, print on a single line its space-separated new coordinates. | [
"3 3 3 1 1 9\n1 1\n1 2\n1 3\n2 1\n2 2\n2 3\n3 1\n3 2\n3 3\n"
] | [
"1 3\n1 2\n1 1\n2 3\n2 2\n2 1\n3 3\n3 2\n3 1\n"
] | Just for clarity. Horizontal rotating is like a mirroring of the matrix. For matrix: | 1,500 | [
{
"input": "3 3 3 1 1 9\n1 1\n1 2\n1 3\n2 1\n2 2\n2 3\n3 1\n3 2\n3 3",
"output": "1 3\n1 2\n1 1\n2 3\n2 2\n2 1\n3 3\n3 2\n3 1"
},
{
"input": "5 5 0 0 0 1\n1 4",
"output": "1 4"
},
{
"input": "14 76 376219315 550904689 16684615 24\n11 21\n1 65\n5 25\n14 63\n11 30\n1 19\n5 7\n9 51\n2 49\n13 75\n9 9\n3 63\n8 49\n5 1\n1 67\n13 31\n9 35\n3 53\n13 73\n5 71\n1 32\n5 49\n1 41\n14 69",
"output": "4 21\n14 65\n10 25\n1 63\n4 30\n14 19\n10 7\n6 51\n13 49\n2 75\n6 9\n12 63\n7 49\n10 1\n14 67\n2 31\n6 35\n12 53\n2 73\n10 71\n14 32\n10 49\n14 41\n1 69"
},
{
"input": "63 67 18046757 61758841 85367218 68\n22 30\n25 40\n56 58\n29 11\n34 63\n28 66\n51 5\n39 64\n1 23\n24 61\n19 47\n10 31\n55 28\n52 26\n38 7\n28 31\n13 27\n37 42\n10 52\n19 33\n7 36\n13 1\n46 40\n21 41\n1 1\n6 35\n10 4\n46 9\n21 57\n1 49\n34 14\n14 35\n43 4\n1 41\n25 22\n18 25\n27 23\n43 17\n34 23\n29 4\n50 40\n43 67\n55 37\n4 60\n35 32\n22 58\n22 12\n9 2\n42 44\n20 57\n5 37\n22 48\n26 8\n33 1\n61 28\n55 18\n21 1\n1 2\n36 29\n45 65\n1 41\n22 46\n25 67\n25 41\n36 42\n8 66\n52 60\n28 50",
"output": "38 42\n28 39\n10 8\n57 35\n5 30\n2 36\n63 13\n4 25\n45 63\n7 40\n21 45\n37 54\n40 9\n42 12\n61 26\n37 36\n41 51\n26 27\n16 54\n35 45\n32 57\n67 51\n28 18\n27 43\n67 63\n33 58\n64 54\n59 18\n11 43\n19 63\n54 30\n33 50\n64 21\n27 63\n46 39\n43 46\n45 37\n51 21\n45 30\n64 35\n28 14\n1 21\n31 9\n8 60\n36 29\n10 42\n56 42\n66 55\n24 22\n11 44\n31 59\n20 42\n60 38\n67 31\n40 3\n50 9\n67 43\n66 63\n39 28\n3 19\n27 63\n22 42\n1 39\n27 39\n26 28\n2 56\n8 12\n18 36"
},
{
"input": "75 18 163006189 147424057 443319537 71\n56 7\n1 5\n17 4\n67 13\n45 1\n55 9\n46 14\n23 10\n10 1\n1 1\n14 9\n18 16\n25 9\n22 4\n73 13\n51 7\n43 13\n59 1\n62 15\n37 6\n43 11\n66 17\n61 13\n45 1\n16 7\n46 7\n25 1\n52 13\n74 7\n16 17\n34 11\n37 16\n24 5\n10 11\n20 5\n74 1\n57 7\n72 10\n21 11\n66 13\n46 1\n46 13\n65 1\n68 11\n14 13\n72 11\n58 1\n16 15\n49 1\n53 10\n30 1\n75 1\n45 4\n42 13\n52 10\n25 1\n31 1\n26 1\n21 7\n27 4\n55 10\n61 1\n37 3\n13 18\n24 1\n11 3\n14 17\n34 5\n49 4\n56 13\n19 11",
"output": "20 7\n75 5\n59 4\n9 13\n31 1\n21 9\n30 14\n53 10\n66 1\n75 1\n62 9\n58 16\n51 9\n54 4\n3 13\n25 7\n33 13\n17 1\n14 15\n39 6\n33 11\n10 17\n15 13\n31 1\n60 7\n30 7\n51 1\n24 13\n2 7\n60 17\n42 11\n39 16\n52 5\n66 11\n56 5\n2 1\n19 7\n4 10\n55 11\n10 13\n30 1\n30 13\n11 1\n8 11\n62 13\n4 11\n18 1\n60 15\n27 1\n23 10\n46 1\n1 1\n31 4\n34 13\n24 10\n51 1\n45 1\n50 1\n55 7\n49 4\n21 10\n15 1\n39 3\n63 18\n52 1\n65 3\n62 17\n42 5\n27 4\n20 13\n57 11"
},
{
"input": "99 65 100328801 11658361 60379320 41\n46 61\n92 23\n46 16\n60 56\n50 42\n24 19\n43 54\n40 1\n41 16\n19 34\n57 59\n84 20\n33 3\n82 59\n74 53\n26 65\n83 30\n76 14\n73 55\n58 33\n97 62\n10 18\n70 1\n56 27\n64 25\n25 57\n28 21\n96 2\n10 41\n99 59\n25 15\n1 3\n46 27\n38 65\n34 25\n64 55\n37 53\n78 43\n70 64\n64 49\n4 12",
"output": "61 46\n23 92\n16 46\n56 60\n42 50\n19 24\n54 43\n1 40\n16 41\n34 19\n59 57\n20 84\n3 33\n59 82\n53 74\n65 26\n30 83\n14 76\n55 73\n33 58\n62 97\n18 10\n1 70\n27 56\n25 64\n57 25\n21 28\n2 96\n41 10\n59 99\n15 25\n3 1\n27 46\n65 38\n25 34\n55 64\n53 37\n43 78\n64 70\n49 64\n12 4"
},
{
"input": "60 1 884622497 447787585 45746569 5\n41 1\n3 1\n57 1\n1 1\n28 1",
"output": "20 1\n58 1\n4 1\n60 1\n33 1"
},
{
"input": "29 9 101222353 522378781 221562741 21\n8 1\n12 8\n21 7\n29 2\n12 3\n1 4\n18 9\n28 6\n2 3\n10 8\n16 4\n3 9\n14 4\n15 3\n16 6\n28 7\n18 1\n12 1\n23 1\n11 1\n18 4",
"output": "22 1\n18 8\n9 7\n1 2\n18 3\n29 4\n12 9\n2 6\n28 3\n20 8\n14 4\n27 9\n16 4\n15 3\n14 6\n2 7\n12 1\n18 1\n7 1\n19 1\n12 4"
},
{
"input": "14 33 331499150 82809609 266661996 75\n9 10\n1 1\n8 8\n13 26\n3 1\n5 1\n8 13\n3 19\n1 13\n1 6\n13 1\n12 19\n5 25\n3 10\n6 19\n6 23\n7 1\n11 7\n11 16\n7 32\n8 30\n1 2\n11 2\n13 25\n8 7\n9 33\n9 1\n1 7\n1 30\n14 32\n9 10\n11 7\n12 5\n11 31\n7 10\n7 21\n9 28\n3 23\n11 31\n9 12\n5 14\n9 7\n10 11\n5 14\n5 14\n4 16\n3 32\n3 16\n13 28\n5 10\n2 8\n4 11\n8 4\n11 15\n1 12\n5 17\n14 10\n13 12\n7 7\n2 32\n3 25\n4 5\n4 31\n10 23\n10 28\n5 8\n5 31\n4 25\n3 25\n13 7\n1 26\n6 4\n9 33\n5 4\n1 14",
"output": "6 10\n14 1\n7 8\n2 26\n12 1\n10 1\n7 13\n12 19\n14 13\n14 6\n2 1\n3 19\n10 25\n12 10\n9 19\n9 23\n8 1\n4 7\n4 16\n8 32\n7 30\n14 2\n4 2\n2 25\n7 7\n6 33\n6 1\n14 7\n14 30\n1 32\n6 10\n4 7\n3 5\n4 31\n8 10\n8 21\n6 28\n12 23\n4 31\n6 12\n10 14\n6 7\n5 11\n10 14\n10 14\n11 16\n12 32\n12 16\n2 28\n10 10\n13 8\n11 11\n7 4\n4 15\n14 12\n10 17\n1 10\n2 12\n8 7\n13 32\n12 25\n11 5\n11 31\n5 23\n5 28\n10 8\n10 31\n11 25\n12 25\n2 7\n14 26\n9 4\n6 33\n10 4\n14 14"
},
{
"input": "26 89 146819986 242756320 184308201 43\n20 71\n12 22\n3 73\n9 48\n1 32\n5 20\n1 18\n19 57\n23 77\n1 4\n17 86\n1 13\n16 64\n1 56\n7 63\n18 38\n17 82\n21 43\n5 16\n9 39\n7 23\n5 53\n19 8\n25 10\n11 69\n11 7\n16 47\n25 48\n20 87\n14 16\n1 16\n14 43\n22 43\n11 89\n7 3\n1 57\n5 43\n21 1\n1 21\n3 85\n5 7\n19 16\n7 15",
"output": "71 7\n22 15\n73 24\n48 18\n32 26\n20 22\n18 26\n57 8\n77 4\n4 26\n86 10\n13 26\n64 11\n56 26\n63 20\n38 9\n82 10\n43 6\n16 22\n39 18\n23 20\n53 22\n8 8\n10 2\n69 16\n7 16\n47 11\n48 2\n87 7\n16 13\n16 26\n43 13\n43 5\n89 16\n3 20\n57 26\n43 22\n1 6\n21 26\n85 24\n7 22\n16 8\n15 20"
},
{
"input": "57 62 402127657 5834146 166754152 26\n55 15\n3 10\n10 21\n25 45\n28 50\n54 39\n1 57\n5 11\n13 54\n52 17\n52 9\n28 3\n37 25\n29 15\n55 33\n23 25\n28 1\n46 7\n39 25\n20 43\n33 49\n52 47\n22 11\n37 37\n52 48\n25 53",
"output": "15 3\n10 55\n21 48\n45 33\n50 30\n39 4\n57 57\n11 53\n54 45\n17 6\n9 6\n3 30\n25 21\n15 29\n33 3\n25 35\n1 30\n7 12\n25 19\n43 38\n49 25\n47 6\n11 36\n37 21\n48 6\n53 33"
},
{
"input": "83 53 263444877 330109611 453128994 25\n47 7\n40 13\n47 53\n23 37\n57 23\n4 38\n39 25\n42 41\n61 23\n74 6\n48 5\n56 53\n48 37\n13 37\n34 32\n49 4\n43 32\n14 1\n75 15\n59 18\n25 14\n46 23\n47 48\n72 3\n55 17",
"output": "47 37\n41 44\n1 37\n17 61\n31 27\n16 80\n29 45\n13 42\n31 23\n48 10\n49 36\n1 28\n17 36\n17 71\n22 50\n50 35\n22 41\n53 70\n39 9\n36 25\n40 59\n31 38\n6 37\n51 12\n37 29"
},
{
"input": "65 66 68528825 50348481 104442753 7\n1 49\n54 47\n16 37\n1 34\n51 29\n36 17\n11 16",
"output": "65 49\n12 47\n50 37\n65 34\n15 29\n30 17\n55 16"
},
{
"input": "63 5 311153546 666957619 681867949 3\n14 3\n22 1\n14 2",
"output": "3 50\n5 42\n4 50"
},
{
"input": "1 9 549924215 115901887 855235569 1\n1 6",
"output": "1 4"
},
{
"input": "85 26 48272945 423830401 423026164 3\n35 1\n50 17\n55 2",
"output": "1 35\n17 50\n2 55"
},
{
"input": "67 61 443905131 226973811 158369983 1\n52 51",
"output": "16 51"
},
{
"input": "1 13 348638338 31146449 81215464 3\n1 4\n1 10\n1 6",
"output": "1 4\n1 10\n1 6"
},
{
"input": "68 45 637151929 93583345 392834373 10\n52 11\n28 1\n67 21\n33 29\n43 28\n30 17\n39 30\n52 37\n11 7\n11 26",
"output": "17 11\n41 1\n2 21\n36 29\n26 28\n39 17\n30 30\n17 37\n58 7\n58 26"
},
{
"input": "25 45 20761261 857816695 7926985 1\n1 26",
"output": "25 26"
},
{
"input": "1 1 0 0 0 2\n1 1\n1 1",
"output": "1 1\n1 1"
},
{
"input": "2 2 1 0 0 4\n1 1\n1 2\n2 2\n2 1",
"output": "1 2\n2 2\n2 1\n1 1"
},
{
"input": "2 2 0 1 0 4\n1 1\n1 2\n2 2\n2 1",
"output": "1 2\n1 1\n2 1\n2 2"
},
{
"input": "2 2 0 0 1 4\n1 1\n1 2\n2 2\n2 1",
"output": "2 1\n1 1\n1 2\n2 2"
},
{
"input": "2 1 1 0 0 2\n1 1\n2 1",
"output": "1 2\n1 1"
},
{
"input": "2 1 0 1 0 2\n1 1\n2 1",
"output": "1 1\n2 1"
},
{
"input": "2 1 0 0 1 2\n1 1\n2 1",
"output": "1 1\n1 2"
},
{
"input": "1 1 0 1 0 2\n1 1\n1 1",
"output": "1 1\n1 1"
}
] | 1,689,259,191 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1689259191.485174")# 1689259191.4851892 | Title: Inna and Huge Candy Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Inna and Dima decided to surprise Sereja. They brought a really huge candy matrix, it's big even for Sereja! Let's number the rows of the giant matrix from 1 to *n* from top to bottom and the columns — from 1 to *m*, from left to right. We'll represent the cell on the intersection of the *i*-th row and *j*-th column as (*i*,<=*j*). Just as is expected, some cells of the giant candy matrix contain candies. Overall the matrix has *p* candies: the *k*-th candy is at cell (*x**k*,<=*y**k*).
The time moved closer to dinner and Inna was already going to eat *p* of her favourite sweets from the matrix, when suddenly Sereja (for the reason he didn't share with anyone) rotated the matrix *x* times clockwise by 90 degrees. Then he performed the horizontal rotate of the matrix *y* times. And then he rotated the matrix *z* times counterclockwise by 90 degrees. The figure below shows how the rotates of the matrix looks like.
Inna got really upset, but Duma suddenly understood two things: the candies didn't get damaged and he remembered which cells contained Inna's favourite sweets before Sereja's strange actions. Help guys to find the new coordinates in the candy matrix after the transformation Sereja made!
Input Specification:
The first line of the input contains fix integers *n*, *m*, *x*, *y*, *z*, *p* (1<=≤<=*n*,<=*m*<=≤<=109; 0<=≤<=*x*,<=*y*,<=*z*<=≤<=109; 1<=≤<=*p*<=≤<=105).
Each of the following *p* lines contains two integers *x**k*, *y**k* (1<=≤<=*x**k*<=≤<=*n*; 1<=≤<=*y**k*<=≤<=*m*) — the initial coordinates of the *k*-th candy. Two candies can lie on the same cell.
Output Specification:
For each of the *p* candies, print on a single line its space-separated new coordinates.
Demo Input:
['3 3 3 1 1 9\n1 1\n1 2\n1 3\n2 1\n2 2\n2 3\n3 1\n3 2\n3 3\n']
Demo Output:
['1 3\n1 2\n1 1\n2 3\n2 2\n2 1\n3 3\n3 2\n3 1\n']
Note:
Just for clarity. Horizontal rotating is like a mirroring of the matrix. For matrix: | ```python
print("_RANDOM_GUESS_1689259191.485174")# 1689259191.4851892
``` | 0 | |
88 | B | Keyboard | PROGRAMMING | 1,500 | [
"implementation"
] | B. Keyboard | 1 | 256 | Vasya learns to type. He has an unusual keyboard at his disposal: it is rectangular and it has *n* rows of keys containing *m* keys in each row. Besides, the keys are of two types. Some of the keys have lowercase Latin letters on them and some of the keys work like the "Shift" key on standard keyboards, that is, they make lowercase letters uppercase.
Vasya can press one or two keys with one hand. However, he can only press two keys if the Euclidean distance between the centers of the keys does not exceed *x*. The keys are considered as squares with a side equal to 1. There are no empty spaces between neighbouring keys.
Vasya is a very lazy boy, that's why he tries to type with one hand as he eats chips with his other one. However, it is possible that some symbol can't be typed with one hand only, because the distance between it and the closest "Shift" key is strictly larger than *x*. In this case he will have to use his other hand. Having typed the symbol, Vasya returns other hand back to the chips.
You are given Vasya's keyboard and the text. Count the minimum number of times Vasya will have to use the other hand. | The first line contains three integers *n*, *m*, *x* (1<=≤<=*n*,<=*m*<=≤<=30,<=1<=≤<=*x*<=≤<=50).
Next *n* lines contain descriptions of all the keyboard keys. Each line contains the descriptions of exactly *m* keys, without spaces. The letter keys are marked with the corresponding lowercase letters. The "Shift" keys are marked with the "S" symbol.
Then follow the length of the text *q* (1<=≤<=*q*<=≤<=5·105). The last line contains the text *T*, which consists of *q* symbols, which are uppercase and lowercase Latin letters. | If Vasya can type the text, then print the minimum number of times he will have to use his other hand. Otherwise, print "-1" (without the quotes). | [
"2 2 1\nab\ncd\n1\nA\n",
"2 2 1\nab\ncd\n1\ne\n",
"2 2 1\nab\ncS\n5\nabcBA\n",
"3 9 4\nqwertyuio\nasdfghjkl\nSzxcvbnmS\n35\nTheQuIcKbRoWnFOXjummsovertHeLazYDOG\n"
] | [
"-1\n",
"-1\n",
"1\n",
"2\n"
] | In the first sample the symbol "A" is impossible to print as there's no "Shift" key on the keyboard.
In the second sample the symbol "e" is impossible to print as there's no such key on the keyboard.
In the fourth sample the symbols "T", "G" are impossible to print with one hand. The other letters that are on the keyboard can be printed. Those symbols come up in the text twice, thus, the answer is 2. | 1,000 | [
{
"input": "2 2 1\nab\ncd\n1\nA",
"output": "-1"
},
{
"input": "2 2 1\nab\ncd\n1\ne",
"output": "-1"
},
{
"input": "2 2 1\nab\ncS\n5\nabcBA",
"output": "1"
},
{
"input": "3 9 4\nqwertyuio\nasdfghjkl\nSzxcvbnmS\n35\nTheQuIcKbRoWnFOXjummsovertHeLazYDOG",
"output": "2"
},
{
"input": "10 9 3\noboxlgpey\nyxcuwkkmp\njuqeflhwq\nsfnxqtjqS\nkkudcnyjl\nhgjlcrkjq\njnofqksxn\nqbhsnuguv\nlvahnifao\nebwnnlrwe\n35\nCodeforcesBetaRoundproblemAtestfive",
"output": "4"
},
{
"input": "2 7 4\niuqtieo\nysxcgmS\n2\nsQ",
"output": "1"
},
{
"input": "1 2 4\nbS\n8\nbBbbbBbb",
"output": "0"
},
{
"input": "7 8 5\nfqiubjpm\nqbshcsyk\ncjbxpbef\nptwpmapx\nryazscbm\nqnvsgzrf\nhtardzkz\n9\nuxrmwkayy",
"output": "0"
},
{
"input": "8 6 4\nefvmov\nkeofnw\npwajpe\nknptky\nSibruu\nrgdukk\nbsxosd\nhovgSe\n10\nECreruXmsC",
"output": "-1"
},
{
"input": "10 3 2\nukk\neqt\nfex\nqSh\ntvz\nfjn\niol\nehd\nnte\ngyx\n5\ncgQxI",
"output": "-1"
},
{
"input": "10 10 19\nowqjcaSpqn\nvgrhboqahn\nbzziocjmbu\npurqsmiSop\nxcsifctjhy\nycyytwoamk\nrnjfxsxowl\nnkgcywcdff\nbazljrisqv\nkcakigSekq\n100\nzewpATtssQVicNrlRrcoifTutTAfFMUEfDFKoNyQbSrSYxTGMadNkRpmJvoEqUsqPYgAdQreaUrwDKMNFWiwdRRCcJBPorfMVMoK",
"output": "0"
},
{
"input": "10 10 26\nwxmssptheb\nzpxbxsyxsy\nqbjkpaywqp\nfwhnuzjcgq\nycgaanzedz\njrycrbzqfs\ngswwakybus\nfhtxhljedz\noSepmyjosv\ndwviycevdn\n100\nyapwUfnyPzgZyFvAHGKWVbXQHkuhJDoUTvCAtdMMCQmKchxKkilUTECOqYJFUSHPqKiRKhDXZgHxwApDWlShdwakmVCgaeKCLOMX",
"output": "0"
},
{
"input": "10 10 3\nrvouufmnqu\nbyukrnmnhr\nzjggwxgvkz\ntcagkSitiw\nhryajgtpwc\njragfhqoks\nkgroxxkuvp\nbpgrkqiyns\njbuhjjkziw\nomjmbaggsw\n100\nCpRzrPqPngYvrVJFCWRPMRwrpXcbtiwfoFcAkRaNjzpMMKOQAzBxSrxGbIHaYgmSqhhxhZTmhFttKnhFzRfKxYXshUZRvtKJIzZq",
"output": "12"
},
{
"input": "10 10 2\nfriuxvShvg\nerslojqtgu\nzeqsmdewry\nwvhbeeyeSu\ngkofbjaavr\ntwkcdxugps\nnzlylSmafu\nstamkpxnzt\nuwxwximkrm\nmzxyboazbl\n100\nmRIfAtrLKmztpVkAmojDCiIgseBwlUilBIixDQhqNhNAqVLLIobuCIretLdSvixNNdCiouFMXtwHZFlObCeaygmIiFBfaCirbmCa",
"output": "19"
},
{
"input": "10 10 2\nbddahSqkmk\npxbocxayjs\nottvdazstk\nlaxuidqlqb\nkfjwdpdfat\nxlipuubkgv\niqyomzfktm\niwbgidmwyu\nrngqkeupsf\nbqndtekryw\n100\nMNQgWFLhHycqwjSsbTkbgMYAIHFYARRmOsinYMFjOxxnLjiKfeiBbMpoeTdzUMORPaAxRNfvdAPFaKkPdxdAjjJgGCxkDzmSasqq",
"output": "37"
},
{
"input": "10 10 2\nnxcwdrsmrv\nSyjahsosvp\nvkrqbxhgbv\nwkxywavtnn\nepkyoviqbi\nsfmpvhuwwq\nnlsostrotx\ntcdguorhny\nimixrqzSdu\nxzhdhdwibt\n100\nUzzaWiRFYbAqxIDMrRBBDoGQhSzSqSLEddAiJsZcxbemdeuddamNYdWOvzlYSCuHIRpnuxdNxAsnZMiLXBYwnrMcrbNeLrUYhZOB",
"output": "17"
},
{
"input": "10 10 23\nhtyvouoiqi\nvySvsfqadv\nxvqyqjyutq\npjcrrphzbk\nhlqfyoqfmo\nezcSwleoew\nxkwqrajxyg\nngSiftgoso\njyndgicccr\nlgjvokydhp\n100\nJzVVfotldIRcyjhTNRcFlTxFeZKRwavZxYcvdDOQyUvTmryFRuRBcRvmscegtspkPuchqlFEKbrfpTOSlSFOARsbbvSenMwNmaRj",
"output": "0"
},
{
"input": "10 10 7\nifcwalsdbj\njpykymrbei\nrylzgkyefh\noilvvexpjp\niptgodpfim\ndSrqejaixu\npksxlsniwa\nmoSenxtfbc\noqssptcenz\nqdhmouvyas\n100\nqtMDVUXJpSEFgPsLKyRJVRbfVoYaCKJDnQDLFVngVjSPzzVyMnMyuyahMRiBJuNhKtgpVqvukUolLvYEmidvXotgQUJukYwIweUW",
"output": "0"
},
{
"input": "10 10 1\nmdxafehbkr\nyuhenybjps\ntvfwmiwcoh\njmzrepzjvx\nnqyorkSnuk\ntSmztmwidv\ncmmajnlqrw\nfiqewpdwax\nuesmkdcplt\nlgkomdcqbo\n100\nmcEQmAvFqKYMXLHQUDeIulkmAMRkIUtbKihTFJwJYQfcAelNrZWSAwHunwZTrdHaRWokgCyLqbubOpEHuZiDVoFHjvkMSoBPyGOI",
"output": "39"
},
{
"input": "10 10 2\nnhfafdwqhh\neyvitpcthk\nrpiotuoqzh\nnxxnhuaxee\nyevrtirzwf\nkbtSsamyel\nfeenjvxsmo\nkqpenxjmde\nlqsamthlwp\njdyyqsbtbk\n100\nUHucxPWDaKonVpXEctuqYUAQnrFEZaTYxhoacNbHIMevlbDejXjitEzyVrTfcfBHWRMdJvaTkbkqccyHjtzpTbKmRAXwlXCtFKNX",
"output": "29"
},
{
"input": "10 10 1\nsufnxxpdnx\nvttibpllhv\nlvbrjmfdjx\ngmtexvrnfh\nygsqrsSwxd\nkxbbjxgbzs\nedutwocmzd\nfebjgknyai\nvcvquagvrs\ndrdoarhgoc\n100\nZoZJXhUWyaLgBTpgbznABKHuyFcKzJmGaMhoKkKfyOGacLwBspaKtAEdwMZJFYiZUFNDxdDIDgKSCRvsbGUOXRqalbpuEqkduYpW",
"output": "44"
},
{
"input": "10 10 2\ncstcrltzsl\nblotmquzvj\nuiitiytlgx\nwumpfdaprd\ntfxohqpztn\nvfrpsccddo\nneegusrkxw\niijfjozqjq\nioegbvuhew\npjjpqdxvqu\n100\nkPCBONfZLkeXzWVuSgvinPENazcnRoBcUHXwRzPyvNIiDlDSeKOYmiUmjooXuzTCtIRxKDAYeTLgjsenxHoymVazMALUADQpjVjV",
"output": "-1"
},
{
"input": "10 10 1\nqztnjglyrc\nnukswgzajl\nqnpbdwjvbb\nliiakzcrlz\nnolwfzzvxd\nmqvhiySttx\nqwuizSjuto\nqbgwiwjukx\nkomyvblgkc\ntkzlxzgsru\n100\nYUzTZDzLFkMUhjQWbwljJCRyZGFzgJcozvROiwPktRGxkMKiPyiTvhDrtusPYhMgVAOFIjAvlpzcrUvrMrMbhkpUiyAytKfYOGTF",
"output": "37"
},
{
"input": "10 10 1\nmgziiihbkq\niobjknuogh\nvntwahSopu\nsjsxjpaqvm\nwqgrodizst\nselzugktoi\nvbhfzvgjfn\nliqlfdcjhf\nbpbtpmimxb\npksfiydpfw\n100\nwAVZXEhKTuajdCauVTIwgnfbxWuUSmtXkjHZtNVcfTsiSAPLdpdEFdTJLZRjptUcRhAmrNjKMXmuDGatAQoaIpbddnzRGHsJrhoq",
"output": "39"
},
{
"input": "10 10 2\nshbqxycvfm\notydudkttw\nqhatsxsngz\nixvyujtyjc\nsbvqhnjbak\neggcguuuka\nxcydfgjzeb\nytpdkcdrsq\nefqlpywggu\nfcnfrhnouo\n100\nHPqtuVckdUOhsnuhnbpekWvWKUnAEaOCihpeEvmaOKOPcIZiMixGJGEuXAaOxuUNyrIesmldLEqGnvyDKPDvFkCbRebCORHmUgeV",
"output": "-1"
},
{
"input": "1 1 50\nS\n29\nargjhoaiogjiSjqfhjksdvjkSvcvn",
"output": "-1"
},
{
"input": "1 1 50\nS\n1\nS",
"output": "-1"
},
{
"input": "1 1 50\na\n29\nargjhoaiogjiSjqfhjksdvjkSvcvn",
"output": "-1"
},
{
"input": "1 1 50\nz\n29\nargjhoaiogjiSjqfhjksdvjkSvcvn",
"output": "-1"
},
{
"input": "2 1 2\nS\nc\n4\nCSSA",
"output": "-1"
}
] | 1,617,004,029 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 46 | 204,800 | # link: https://codeforces.com/contest/88/problem/B
if __name__ == "__main__":
n,m,x = map(int, input().split())
kb = []
nn = n
while nn:
kb.append(input())
nn -= 1
l = int(input())
word = input()
shift = []
reachable = set()
for i in range(n):
if 'S' in kb[i]:
for j in range(m):
if kb[i][j] == 'S':
shift.append([i,j])
if len(shift) == 0:
flag = True
for i in range(l):
w = word[i]
if w.isupper():
flag = False
break
else:
for ii in range(n):
if w.lower() in kb[ii]:
flag = True
break
if flag:
print(0)
else:
print(-1)
else:
for i in range(n):
for j in range(m):
if kb[i][j] != 'S' and kb[i][j] not in reachable:
# max euclidean distance is x
for k in shift:
f = 0
distance = ((k[0] - i)**2 + (k[1] - j)**2) ** 0.5
if distance <= x:
reachable.add(kb[i][j])
break
hands = 0
for i in range(l):
w = word[i]
if w.isupper() and w.lower() not in reachable:
hands += 1
else:
found = False
for ii in range(n):
if w.lower() in kb[ii]:
found = True
break
else:
continue
if not found:
print(-1)
exit(0)
print(hands)
| Title: Keyboard
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya learns to type. He has an unusual keyboard at his disposal: it is rectangular and it has *n* rows of keys containing *m* keys in each row. Besides, the keys are of two types. Some of the keys have lowercase Latin letters on them and some of the keys work like the "Shift" key on standard keyboards, that is, they make lowercase letters uppercase.
Vasya can press one or two keys with one hand. However, he can only press two keys if the Euclidean distance between the centers of the keys does not exceed *x*. The keys are considered as squares with a side equal to 1. There are no empty spaces between neighbouring keys.
Vasya is a very lazy boy, that's why he tries to type with one hand as he eats chips with his other one. However, it is possible that some symbol can't be typed with one hand only, because the distance between it and the closest "Shift" key is strictly larger than *x*. In this case he will have to use his other hand. Having typed the symbol, Vasya returns other hand back to the chips.
You are given Vasya's keyboard and the text. Count the minimum number of times Vasya will have to use the other hand.
Input Specification:
The first line contains three integers *n*, *m*, *x* (1<=≤<=*n*,<=*m*<=≤<=30,<=1<=≤<=*x*<=≤<=50).
Next *n* lines contain descriptions of all the keyboard keys. Each line contains the descriptions of exactly *m* keys, without spaces. The letter keys are marked with the corresponding lowercase letters. The "Shift" keys are marked with the "S" symbol.
Then follow the length of the text *q* (1<=≤<=*q*<=≤<=5·105). The last line contains the text *T*, which consists of *q* symbols, which are uppercase and lowercase Latin letters.
Output Specification:
If Vasya can type the text, then print the minimum number of times he will have to use his other hand. Otherwise, print "-1" (without the quotes).
Demo Input:
['2 2 1\nab\ncd\n1\nA\n', '2 2 1\nab\ncd\n1\ne\n', '2 2 1\nab\ncS\n5\nabcBA\n', '3 9 4\nqwertyuio\nasdfghjkl\nSzxcvbnmS\n35\nTheQuIcKbRoWnFOXjummsovertHeLazYDOG\n']
Demo Output:
['-1\n', '-1\n', '1\n', '2\n']
Note:
In the first sample the symbol "A" is impossible to print as there's no "Shift" key on the keyboard.
In the second sample the symbol "e" is impossible to print as there's no such key on the keyboard.
In the fourth sample the symbols "T", "G" are impossible to print with one hand. The other letters that are on the keyboard can be printed. Those symbols come up in the text twice, thus, the answer is 2. | ```python
# link: https://codeforces.com/contest/88/problem/B
if __name__ == "__main__":
n,m,x = map(int, input().split())
kb = []
nn = n
while nn:
kb.append(input())
nn -= 1
l = int(input())
word = input()
shift = []
reachable = set()
for i in range(n):
if 'S' in kb[i]:
for j in range(m):
if kb[i][j] == 'S':
shift.append([i,j])
if len(shift) == 0:
flag = True
for i in range(l):
w = word[i]
if w.isupper():
flag = False
break
else:
for ii in range(n):
if w.lower() in kb[ii]:
flag = True
break
if flag:
print(0)
else:
print(-1)
else:
for i in range(n):
for j in range(m):
if kb[i][j] != 'S' and kb[i][j] not in reachable:
# max euclidean distance is x
for k in shift:
f = 0
distance = ((k[0] - i)**2 + (k[1] - j)**2) ** 0.5
if distance <= x:
reachable.add(kb[i][j])
break
hands = 0
for i in range(l):
w = word[i]
if w.isupper() and w.lower() not in reachable:
hands += 1
else:
found = False
for ii in range(n):
if w.lower() in kb[ii]:
found = True
break
else:
continue
if not found:
print(-1)
exit(0)
print(hands)
``` | 0 |
999 | A | Mishka and Contest | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$.
Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.
Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list.
How many problems can Mishka solve? | The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill.
The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list. | Print one integer — the maximum number of problems Mishka can solve. | [
"8 4\n4 2 3 1 5 1 6 4\n",
"5 2\n3 1 2 1 3\n",
"5 100\n12 34 55 43 21\n"
] | [
"5\n",
"0\n",
"5\n"
] | In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$.
In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$.
In the third example, Mishka's solving skill is so amazing that he can solve all the problems. | 0 | [
{
"input": "8 4\n4 2 3 1 5 1 6 4",
"output": "5"
},
{
"input": "5 2\n3 1 2 1 3",
"output": "0"
},
{
"input": "5 100\n12 34 55 43 21",
"output": "5"
},
{
"input": "100 100\n44 47 36 83 76 94 86 69 31 2 22 77 37 51 10 19 25 78 53 25 1 29 48 95 35 53 22 72 49 86 60 38 13 91 89 18 54 19 71 2 25 33 65 49 53 5 95 90 100 68 25 5 87 48 45 72 34 14 100 44 94 75 80 26 25 7 57 82 49 73 55 43 42 60 34 8 51 11 71 41 81 23 20 89 12 72 68 26 96 92 32 63 13 47 19 9 35 56 79 62",
"output": "100"
},
{
"input": "100 99\n84 82 43 4 71 3 30 92 15 47 76 43 2 17 76 4 1 33 24 96 44 98 75 99 59 11 73 27 67 17 8 88 69 41 44 22 91 48 4 46 42 21 21 67 85 51 57 84 11 100 100 59 39 72 89 82 74 19 98 14 37 97 20 78 38 52 44 83 19 83 69 32 56 6 93 13 98 80 80 2 33 71 11 15 55 51 98 58 16 91 39 32 83 58 77 79 88 81 17 98",
"output": "98"
},
{
"input": "100 69\n80 31 12 89 16 35 8 28 39 12 32 51 42 67 64 53 17 88 63 97 29 41 57 28 51 33 82 75 93 79 57 86 32 100 83 82 99 33 1 27 86 22 65 15 60 100 42 37 38 85 26 43 90 62 91 13 1 92 16 20 100 19 28 30 23 6 5 69 24 22 9 1 10 14 28 14 25 9 32 8 67 4 39 7 10 57 15 7 8 35 62 6 53 59 62 13 24 7 53 2",
"output": "39"
},
{
"input": "100 2\n2 2 2 2 1 1 1 2 1 2 2 2 1 2 2 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 2 1 2 1 2 1 1 2 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 2 1 1 1 2 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 16",
"output": "99"
},
{
"input": "100 3\n86 53 82 40 2 20 59 2 46 63 75 49 24 81 70 22 9 9 93 72 47 23 29 77 78 51 17 59 19 71 35 3 20 60 70 9 11 96 71 94 91 19 88 93 50 49 72 19 53 30 38 67 62 71 81 86 5 26 5 32 63 98 1 97 22 32 87 65 96 55 43 85 56 37 56 67 12 100 98 58 77 54 18 20 33 53 21 66 24 64 42 71 59 32 51 69 49 79 10 1",
"output": "1"
},
{
"input": "13 7\n1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "13"
},
{
"input": "1 5\n4",
"output": "1"
},
{
"input": "3 2\n1 4 1",
"output": "2"
},
{
"input": "1 2\n100",
"output": "0"
},
{
"input": "7 4\n4 2 3 4 4 2 3",
"output": "7"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "1 2\n15",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "2"
},
{
"input": "5 3\n3 4 3 2 1",
"output": "4"
},
{
"input": "1 1\n2",
"output": "0"
},
{
"input": "1 5\n1",
"output": "1"
},
{
"input": "6 6\n7 1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n6 5 5 5 5",
"output": "4"
},
{
"input": "1 4\n2",
"output": "1"
},
{
"input": "9 4\n1 2 1 2 4 2 1 2 1",
"output": "9"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 10\n5",
"output": "1"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "100 10\n2 5 1 10 10 2 7 7 9 4 1 8 1 1 8 4 7 9 10 5 7 9 5 6 7 2 7 5 3 2 1 82 4 80 9 8 6 1 10 7 5 7 1 5 6 7 19 4 2 4 6 2 1 8 31 6 2 2 57 42 3 2 7 1 9 5 10 8 5 4 10 8 3 5 8 7 2 7 6 5 3 3 4 10 6 7 10 8 7 10 7 2 4 6 8 10 10 2 6 4",
"output": "71"
},
{
"input": "100 90\n17 16 5 51 17 62 24 45 49 41 90 30 19 78 67 66 59 34 28 47 42 8 33 77 90 41 61 16 86 33 43 71 90 95 23 9 56 41 24 90 31 12 77 36 90 67 47 15 92 50 79 88 42 19 21 79 86 60 41 26 47 4 70 62 44 90 82 89 84 91 54 16 90 53 29 69 21 44 18 28 88 74 56 43 12 76 10 22 34 24 27 52 28 76 90 75 5 29 50 90",
"output": "63"
},
{
"input": "100 10\n6 4 8 4 1 9 4 8 5 2 2 5 2 6 10 2 2 5 3 5 2 3 10 5 2 9 1 1 6 1 5 9 16 42 33 49 26 31 81 27 53 63 81 90 55 97 70 51 87 21 79 62 60 91 54 95 26 26 30 61 87 79 47 11 59 34 40 82 37 40 81 2 7 1 8 4 10 7 1 10 8 7 3 5 2 8 3 3 9 2 1 1 5 7 8 7 1 10 9 8",
"output": "61"
},
{
"input": "100 90\n45 57 52 69 17 81 85 60 59 39 55 14 87 90 90 31 41 57 35 89 74 20 53 4 33 49 71 11 46 90 71 41 71 90 63 74 51 13 99 92 99 91 100 97 93 40 93 96 100 99 100 92 98 96 78 91 91 91 91 100 94 97 95 97 96 95 17 13 45 35 54 26 2 74 6 51 20 3 73 90 90 42 66 43 86 28 84 70 37 27 90 30 55 80 6 58 57 51 10 22",
"output": "72"
},
{
"input": "100 10\n10 2 10 10 10 10 10 10 10 7 10 10 10 10 10 10 9 10 10 10 10 10 10 10 10 7 9 10 10 10 37 10 4 10 10 10 59 5 95 10 10 10 10 39 10 10 10 10 10 10 10 5 10 10 10 10 10 10 10 10 10 10 10 10 66 10 10 10 10 10 5 10 10 10 10 10 10 44 10 10 10 10 10 10 10 10 10 10 10 7 10 10 10 10 10 10 10 10 10 2",
"output": "52"
},
{
"input": "100 90\n57 90 90 90 90 90 90 90 81 90 3 90 39 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 92 90 90 90 90 90 90 90 90 98 90 90 90 90 90 90 90 90 90 90 90 90 90 54 90 90 90 90 90 62 90 90 91 90 90 90 90 90 90 91 90 90 90 90 90 90 90 3 90 90 90 90 90 90 90 2 90 90 90 90 90 90 90 90 90 2 90 90 90 90 90",
"output": "60"
},
{
"input": "100 10\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 78 90 61 40 87 39 91 50 64 30 10 24 10 55 28 11 28 35 26 26 10 57 45 67 14 99 96 51 67 79 59 11 21 55 70 33 10 16 92 70 38 50 66 52 5 10 10 10 2 4 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 10 10 10 10 8 10 10 10 10 10",
"output": "56"
},
{
"input": "100 90\n90 90 90 90 90 90 55 21 90 90 90 90 90 90 90 90 90 90 69 83 90 90 90 90 90 90 90 90 93 95 92 98 92 97 91 92 92 91 91 95 94 95 100 100 96 97 94 93 90 90 95 95 97 99 90 95 98 91 94 96 99 99 94 95 95 97 99 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 12 90 3 90 90 90 90 90 90 90",
"output": "61"
},
{
"input": "100 49\n71 25 14 36 36 48 36 49 28 40 49 49 49 38 40 49 33 22 49 49 14 46 8 44 49 11 37 49 40 49 2 49 3 49 37 49 49 11 25 49 49 32 49 11 49 30 16 21 49 49 23 24 30 49 49 49 49 49 49 27 49 42 49 49 20 32 30 29 35 49 30 49 9 49 27 25 5 49 49 42 49 20 49 35 49 22 15 49 49 49 19 49 29 28 13 49 22 7 6 24",
"output": "99"
},
{
"input": "100 50\n38 68 9 6 50 18 19 50 50 20 33 34 43 50 24 50 50 2 50 50 50 50 50 21 30 50 41 40 50 50 50 50 50 7 50 21 19 23 1 50 24 50 50 50 25 50 50 50 50 50 50 50 7 24 28 18 50 5 43 50 20 50 13 50 50 16 50 3 2 24 50 50 18 5 50 4 50 50 38 50 33 49 12 33 11 14 50 50 50 33 50 50 50 50 50 50 7 4 50 50",
"output": "99"
},
{
"input": "100 48\n8 6 23 47 29 48 48 48 48 48 48 26 24 48 48 48 3 48 27 28 41 45 9 29 48 48 48 48 48 48 48 48 48 48 47 23 48 48 48 5 48 22 40 48 48 48 20 48 48 57 48 32 19 48 33 2 4 19 48 48 39 48 16 48 48 44 48 48 48 48 29 14 25 43 46 7 48 19 30 48 18 8 39 48 30 47 35 18 48 45 48 48 30 13 48 48 48 17 9 48",
"output": "99"
},
{
"input": "100 57\n57 9 57 4 43 57 57 57 57 26 57 18 57 57 57 57 57 57 57 47 33 57 57 43 57 57 55 57 14 57 57 4 1 57 57 57 57 57 46 26 57 57 57 57 57 57 57 39 57 57 57 5 57 12 11 57 57 57 25 37 34 57 54 18 29 57 39 57 5 57 56 34 57 24 7 57 57 57 2 57 57 57 57 1 55 39 19 57 57 57 57 21 3 40 13 3 57 57 62 57",
"output": "99"
},
{
"input": "100 51\n51 51 38 51 51 45 51 51 51 18 51 36 51 19 51 26 37 51 11 51 45 34 51 21 51 51 33 51 6 51 51 51 21 47 51 13 51 51 30 29 50 51 51 51 51 51 51 45 14 51 2 51 51 23 9 51 50 23 51 29 34 51 40 32 1 36 31 51 11 51 51 47 51 51 51 51 51 51 51 50 39 51 14 4 4 12 3 11 51 51 51 51 41 51 51 51 49 37 5 93",
"output": "99"
},
{
"input": "100 50\n87 91 95 73 50 50 16 97 39 24 58 50 33 89 42 37 50 50 12 71 3 55 50 50 80 10 76 50 52 36 88 44 66 69 86 71 77 50 72 50 21 55 50 50 78 61 75 89 65 2 50 69 62 47 11 92 97 77 41 31 55 29 35 51 36 48 50 91 92 86 50 36 50 94 51 74 4 27 55 63 50 36 87 50 67 7 65 75 20 96 88 50 41 73 35 51 66 21 29 33",
"output": "3"
},
{
"input": "100 50\n50 37 28 92 7 76 50 50 50 76 100 57 50 50 50 32 76 50 8 72 14 8 50 91 67 50 55 82 50 50 24 97 88 50 59 61 68 86 44 15 61 67 88 50 40 50 36 99 1 23 63 50 88 59 76 82 99 76 68 50 50 30 31 68 57 98 71 12 15 60 35 79 90 6 67 50 50 50 50 68 13 6 50 50 16 87 84 50 67 67 50 64 50 58 50 50 77 51 50 51",
"output": "3"
},
{
"input": "100 50\n43 50 50 91 97 67 6 50 86 50 76 60 50 59 4 56 11 38 49 50 37 50 50 20 60 47 33 54 95 58 22 50 77 77 72 9 57 40 81 57 95 50 81 63 62 76 13 87 50 39 74 69 50 99 63 1 11 62 84 31 97 99 56 73 70 36 45 100 28 91 93 9 19 52 73 50 83 58 84 52 86 12 50 44 64 52 97 50 12 71 97 52 87 66 83 66 86 50 9 49",
"output": "6"
},
{
"input": "88 10\n10 8 1 10 10 1 3 7 10 5 8 8 10 2 7 10 10 10 10 10 1 10 10 10 10 1 2 9 10 9 10 10 10 64 100 25 10 12 9 52 13 8 10 56 10 4 10 7 10 3 10 79 74 8 73 10 10 10 9 10 3 5 10 10 10 5 1 10 10 4 3 10 10 10 4 10 6 4 10 10 10 10 3 3 8 5 6 8",
"output": "66"
},
{
"input": "100 50\n80 39 33 69 75 50 23 88 50 50 67 90 87 50 29 15 55 32 60 50 50 50 38 95 62 50 50 88 8 97 45 50 42 12 22 93 49 50 24 50 50 71 60 4 50 72 57 57 50 50 50 83 69 17 1 31 72 55 50 11 50 80 93 41 91 94 20 60 50 50 51 48 53 56 76 73 50 72 19 98 50 50 50 50 50 28 48 45 62 11 16 67 93 88 63 50 50 66 48 95",
"output": "0"
},
{
"input": "100 50\n70 50 38 50 38 50 32 30 50 31 26 42 50 33 34 50 50 50 28 21 50 44 50 47 50 50 9 40 50 50 50 50 50 42 50 50 16 50 50 3 24 50 50 50 4 26 50 2 50 50 33 1 27 50 50 50 8 29 50 23 33 50 6 29 50 50 15 50 50 50 32 50 43 50 50 50 31 50 4 50 50 31 50 50 31 16 50 17 50 17 31 13 25 16 50 10 50 47 50 66",
"output": "0"
},
{
"input": "2 8\n8 8",
"output": "2"
},
{
"input": "1 6\n3",
"output": "1"
},
{
"input": "1 5\n5",
"output": "1"
}
] | 1,624,888,137 | 2,147,483,647 | PyPy 3 | OK | TESTS | 48 | 108 | 0 | n, k = list(map(int, input().split()))
a = list(map(int, input().split()))
left = 0
right = n - 1
res1 = 0
res2 = 0
while left < n and a[left] <= k:
left += 1
res1 += 1
while right >= 0 and a[right] <= k:
right -= 1
res2 += 1
if res1 == n:
print(res1)
else:
print(res1 + res2)
| Title: Mishka and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$.
Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.
Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list.
How many problems can Mishka solve?
Input Specification:
The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill.
The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.
Output Specification:
Print one integer — the maximum number of problems Mishka can solve.
Demo Input:
['8 4\n4 2 3 1 5 1 6 4\n', '5 2\n3 1 2 1 3\n', '5 100\n12 34 55 43 21\n']
Demo Output:
['5\n', '0\n', '5\n']
Note:
In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$.
In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$.
In the third example, Mishka's solving skill is so amazing that he can solve all the problems. | ```python
n, k = list(map(int, input().split()))
a = list(map(int, input().split()))
left = 0
right = n - 1
res1 = 0
res2 = 0
while left < n and a[left] <= k:
left += 1
res1 += 1
while right >= 0 and a[right] <= k:
right -= 1
res2 += 1
if res1 == n:
print(res1)
else:
print(res1 + res2)
``` | 3 | |
812 | B | Sagheer, the Hausmeister | PROGRAMMING | 1,600 | [
"bitmasks",
"brute force",
"dp"
] | null | null | Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of *n* floors with stairs at the left and the right sides. Each floor has *m* rooms on the same line with a corridor that connects the left and right stairs passing by all the rooms. In other words, the building can be represented as a rectangle with *n* rows and *m*<=+<=2 columns, where the first and the last columns represent the stairs, and the *m* columns in the middle represent rooms.
Sagheer is standing at the ground floor at the left stairs. He wants to turn all the lights off in such a way that he will not go upstairs until all lights in the floor he is standing at are off. Of course, Sagheer must visit a room to turn the light there off. It takes one minute for Sagheer to go to the next floor using stairs or to move from the current room/stairs to a neighboring room/stairs on the same floor. It takes no time for him to switch the light off in the room he is currently standing in. Help Sagheer find the minimum total time to turn off all the lights.
Note that Sagheer does not have to go back to his starting position, and he does not have to visit rooms where the light is already switched off. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=15 and 1<=≤<=*m*<=≤<=100) — the number of floors and the number of rooms in each floor, respectively.
The next *n* lines contains the building description. Each line contains a binary string of length *m*<=+<=2 representing a floor (the left stairs, then *m* rooms, then the right stairs) where 0 indicates that the light is off and 1 indicates that the light is on. The floors are listed from top to bottom, so that the last line represents the ground floor.
The first and last characters of each string represent the left and the right stairs, respectively, so they are always 0. | Print a single integer — the minimum total time needed to turn off all the lights. | [
"2 2\n0010\n0100\n",
"3 4\n001000\n000010\n000010\n",
"4 3\n01110\n01110\n01110\n01110\n"
] | [
"5\n",
"12\n",
"18\n"
] | In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to the second room in the last floor.
In the third example, he will walk through the whole corridor alternating between the left and right stairs at each floor. | 1,000 | [
{
"input": "2 2\n0010\n0100",
"output": "5"
},
{
"input": "3 4\n001000\n000010\n000010",
"output": "12"
},
{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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"output": "228"
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{
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{
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{
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{
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{
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{
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"output": "100"
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{
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"output": "7"
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] | 1,644,786,499 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 8 | 140 | 4,710,400 | from math import inf
from collections import *
import math, os, sys, heapq, bisect, random
from functools import lru_cache
from itertools import *
def inp(): return sys.stdin.readline().rstrip("\r\n")
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def inpu(): return int(inp())
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
M,M1=1000000007,998244353
def main():
how_much_noob_I_am = 1
#how_much_noob_I_am = inpu()
for __ in range(how_much_noob_I_am):
n ,m = sep()
l=[inp() for i in range(n)]
l.reverse()
last=n
for i in range(len(l)-1,-1,-1):
if l[i].count("1")==0:
last=i
if last==0:
print(0)
continue
m=m+2
#l.reverse()
#print(last)
def fun(ind,j,last):
#print(ind,j)
if ind==last-1:
if j==0:
return l[ind].rfind("1")
else:
return m-l[ind].find("1")-1
if j==0:
p = fun(ind+1,m,last) + m
q = fun(ind+1,0,last) + l[ind].rfind("1")*2 + 1
#print(p,q)
return min(p,q)
if j==m:
p = fun(ind+1,0,last)+m
q = fun(ind+1,m,last) + (m-l[ind].find("1")-1)*2+1
return min(p,q)
print(fun(0,0,last))
if __name__ == '__main__':
main()
| Title: Sagheer, the Hausmeister
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of *n* floors with stairs at the left and the right sides. Each floor has *m* rooms on the same line with a corridor that connects the left and right stairs passing by all the rooms. In other words, the building can be represented as a rectangle with *n* rows and *m*<=+<=2 columns, where the first and the last columns represent the stairs, and the *m* columns in the middle represent rooms.
Sagheer is standing at the ground floor at the left stairs. He wants to turn all the lights off in such a way that he will not go upstairs until all lights in the floor he is standing at are off. Of course, Sagheer must visit a room to turn the light there off. It takes one minute for Sagheer to go to the next floor using stairs or to move from the current room/stairs to a neighboring room/stairs on the same floor. It takes no time for him to switch the light off in the room he is currently standing in. Help Sagheer find the minimum total time to turn off all the lights.
Note that Sagheer does not have to go back to his starting position, and he does not have to visit rooms where the light is already switched off.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=15 and 1<=≤<=*m*<=≤<=100) — the number of floors and the number of rooms in each floor, respectively.
The next *n* lines contains the building description. Each line contains a binary string of length *m*<=+<=2 representing a floor (the left stairs, then *m* rooms, then the right stairs) where 0 indicates that the light is off and 1 indicates that the light is on. The floors are listed from top to bottom, so that the last line represents the ground floor.
The first and last characters of each string represent the left and the right stairs, respectively, so they are always 0.
Output Specification:
Print a single integer — the minimum total time needed to turn off all the lights.
Demo Input:
['2 2\n0010\n0100\n', '3 4\n001000\n000010\n000010\n', '4 3\n01110\n01110\n01110\n01110\n']
Demo Output:
['5\n', '12\n', '18\n']
Note:
In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to the second room in the last floor.
In the third example, he will walk through the whole corridor alternating between the left and right stairs at each floor. | ```python
from math import inf
from collections import *
import math, os, sys, heapq, bisect, random
from functools import lru_cache
from itertools import *
def inp(): return sys.stdin.readline().rstrip("\r\n")
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def inpu(): return int(inp())
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
M,M1=1000000007,998244353
def main():
how_much_noob_I_am = 1
#how_much_noob_I_am = inpu()
for __ in range(how_much_noob_I_am):
n ,m = sep()
l=[inp() for i in range(n)]
l.reverse()
last=n
for i in range(len(l)-1,-1,-1):
if l[i].count("1")==0:
last=i
if last==0:
print(0)
continue
m=m+2
#l.reverse()
#print(last)
def fun(ind,j,last):
#print(ind,j)
if ind==last-1:
if j==0:
return l[ind].rfind("1")
else:
return m-l[ind].find("1")-1
if j==0:
p = fun(ind+1,m,last) + m
q = fun(ind+1,0,last) + l[ind].rfind("1")*2 + 1
#print(p,q)
return min(p,q)
if j==m:
p = fun(ind+1,0,last)+m
q = fun(ind+1,m,last) + (m-l[ind].find("1")-1)*2+1
return min(p,q)
print(fun(0,0,last))
if __name__ == '__main__':
main()
``` | 0 | |
166 | A | Rank List | PROGRAMMING | 1,100 | [
"binary search",
"implementation",
"sortings"
] | null | null | Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place.
You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*.
It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place.
Your task is to count what number of teams from the given list shared the *k*-th place. | The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=≤<=*p**i*,<=*t**i*<=≤<=50) — the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces. | In the only line print the sought number of teams that got the *k*-th place in the final results' table. | [
"7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n",
"5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n"
] | [
"3\n",
"4\n"
] | The final results' table for the first sample is:
- 1-3 places — 4 solved problems, the penalty time equals 10 - 4 place — 3 solved problems, the penalty time equals 20 - 5-6 places — 2 solved problems, the penalty time equals 1 - 7 place — 1 solved problem, the penalty time equals 10
The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams.
The final table for the second sample is:
- 1 place — 5 solved problems, the penalty time equals 3 - 2-5 places — 3 solved problems, the penalty time equals 1
The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams. | 500 | [
{
"input": "7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10",
"output": "3"
},
{
"input": "5 4\n3 1\n3 1\n5 3\n3 1\n3 1",
"output": "4"
},
{
"input": "5 1\n2 2\n1 1\n1 1\n1 1\n2 2",
"output": "2"
},
{
"input": "6 3\n2 2\n3 1\n2 2\n4 5\n2 2\n4 5",
"output": "1"
},
{
"input": "5 5\n3 1\n10 2\n2 2\n1 10\n10 2",
"output": "1"
},
{
"input": "3 2\n3 3\n3 3\n3 3",
"output": "3"
},
{
"input": "4 3\n10 3\n6 10\n5 2\n5 2",
"output": "2"
},
{
"input": "5 3\n10 10\n10 10\n1 1\n10 10\n4 3",
"output": "3"
},
{
"input": "3 1\n2 1\n1 1\n1 2",
"output": "1"
},
{
"input": "1 1\n28 28",
"output": "1"
},
{
"input": "2 2\n1 2\n1 2",
"output": "2"
},
{
"input": "5 3\n2 3\n4 2\n5 3\n2 4\n3 5",
"output": "1"
},
{
"input": "50 22\n4 9\n8 1\n3 7\n1 2\n3 8\n9 8\n8 5\n2 10\n5 8\n1 3\n1 8\n2 3\n7 9\n10 2\n9 9\n7 3\n8 6\n10 6\n5 4\n8 1\n1 5\n6 8\n9 5\n9 5\n3 2\n3 3\n3 8\n7 5\n4 5\n8 10\n8 2\n3 5\n3 2\n1 1\n7 2\n2 7\n6 8\n10 4\n7 5\n1 7\n6 5\n3 1\n4 9\n2 3\n3 6\n5 8\n4 10\n10 7\n7 10\n9 8",
"output": "1"
},
{
"input": "50 6\n11 20\n18 13\n1 13\n3 11\n4 17\n15 10\n15 8\n9 16\n11 17\n16 3\n3 20\n14 13\n12 15\n9 10\n14 2\n12 12\n13 17\n6 10\n20 9\n2 8\n13 7\n7 20\n15 3\n1 20\n2 13\n2 5\n14 7\n10 13\n15 12\n15 5\n17 6\n9 11\n18 5\n10 1\n15 14\n3 16\n6 12\n4 1\n14 9\n7 14\n8 17\n17 13\n4 6\n19 16\n5 6\n3 15\n4 19\n15 20\n2 10\n20 10",
"output": "1"
},
{
"input": "50 12\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "50"
},
{
"input": "50 28\n2 2\n1 1\n2 1\n1 2\n1 1\n1 1\n1 1\n2 2\n2 2\n2 2\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 1\n2 2\n1 2\n2 2\n2 2\n2 1\n1 1\n1 2\n1 2\n1 1\n1 1\n1 1\n2 2\n2 1\n2 1\n2 2\n1 2\n1 2\n1 2\n1 1\n2 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n2 1\n1 1\n2 2\n2 2\n2 2\n2 2",
"output": "13"
},
{
"input": "50 40\n2 3\n3 1\n2 1\n2 1\n2 1\n3 1\n1 1\n1 2\n2 3\n1 3\n1 3\n2 1\n3 1\n1 1\n3 1\n3 1\n2 2\n1 1\n3 3\n3 1\n3 2\n2 3\n3 3\n3 1\n1 3\n2 3\n2 1\n3 2\n3 3\n3 1\n2 1\n2 2\n1 3\n3 3\n1 1\n3 2\n1 2\n2 3\n2 1\n2 2\n3 2\n1 3\n3 1\n1 1\n3 3\n2 3\n2 1\n2 3\n2 3\n1 2",
"output": "5"
},
{
"input": "50 16\n2 1\n3 2\n5 2\n2 2\n3 4\n4 4\n3 3\n4 1\n2 3\n1 5\n4 1\n2 2\n1 5\n3 2\n2 1\n5 4\n5 2\n5 4\n1 1\n3 5\n2 1\n4 5\n5 1\n5 5\n5 4\n2 4\n1 2\n5 5\n4 4\n1 5\n4 2\n5 1\n2 4\n2 5\n2 2\n3 4\n3 1\n1 1\n5 5\n2 2\n3 4\n2 4\n5 2\n4 1\n3 1\n1 1\n4 1\n4 4\n1 4\n1 3",
"output": "1"
},
{
"input": "50 32\n6 6\n4 2\n5 5\n1 1\n2 4\n6 5\n2 3\n6 5\n2 3\n6 3\n1 4\n1 6\n3 3\n2 4\n3 2\n6 2\n4 1\n3 3\n3 1\n5 5\n1 2\n2 1\n5 4\n3 1\n4 4\n5 6\n4 1\n2 5\n3 1\n4 6\n2 3\n1 1\n6 5\n2 6\n3 3\n2 6\n2 3\n2 6\n3 4\n2 6\n4 5\n5 4\n1 6\n3 2\n5 1\n4 1\n4 6\n4 2\n1 2\n5 2",
"output": "1"
},
{
"input": "50 48\n5 1\n6 4\n3 2\n2 1\n4 7\n3 6\n7 1\n7 5\n6 5\n5 6\n4 7\n5 7\n5 7\n5 5\n7 3\n3 5\n4 3\n5 4\n6 2\n1 6\n6 3\n6 5\n5 2\n4 2\n3 1\n1 1\n5 6\n1 3\n6 5\n3 7\n1 5\n7 5\n6 5\n3 6\n2 7\n5 3\n5 3\n4 7\n5 2\n6 5\n5 7\n7 1\n2 3\n5 5\n2 6\n4 1\n6 2\n6 5\n3 3\n1 6",
"output": "1"
},
{
"input": "50 8\n5 3\n7 3\n4 3\n7 4\n2 2\n4 4\n5 4\n1 1\n7 7\n4 8\n1 1\n6 3\n1 5\n7 3\n6 5\n4 5\n8 6\n3 6\n2 1\n3 2\n2 5\n7 6\n5 8\n1 3\n5 5\n8 4\n4 5\n4 4\n8 8\n7 2\n7 2\n3 6\n2 8\n8 3\n3 2\n4 5\n8 1\n3 2\n8 7\n6 3\n2 3\n5 1\n3 4\n7 2\n6 3\n7 3\n3 3\n6 4\n2 2\n5 1",
"output": "3"
},
{
"input": "20 16\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "20"
},
{
"input": "20 20\n1 2\n2 2\n1 1\n2 1\n2 2\n1 1\n1 1\n2 1\n1 1\n1 2\n2 2\n1 2\n1 2\n2 2\n2 2\n1 2\n2 1\n2 1\n1 2\n2 2",
"output": "6"
},
{
"input": "30 16\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "30"
},
{
"input": "30 22\n2 1\n1 2\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n2 2\n1 2\n2 2\n1 2\n1 2\n2 1\n1 2\n2 2\n2 2\n1 2\n2 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 2\n2 2\n1 2\n2 2\n2 1\n1 1",
"output": "13"
},
{
"input": "30 22\n1 1\n1 3\n2 3\n3 1\n2 3\n3 1\n1 2\n3 3\n2 1\n2 1\n2 2\n3 1\n3 2\n2 3\n3 1\n1 3\n2 3\n3 1\n1 2\n1 2\n2 3\n2 1\n3 3\n3 2\n1 3\n3 3\n3 3\n3 3\n3 3\n3 1",
"output": "5"
},
{
"input": "50 16\n2 1\n3 2\n5 2\n2 2\n3 4\n4 4\n3 3\n4 1\n2 3\n1 5\n4 1\n2 2\n1 5\n3 2\n2 1\n5 4\n5 2\n5 4\n1 1\n3 5\n2 1\n4 5\n5 1\n5 5\n5 4\n2 4\n1 2\n5 5\n4 4\n1 5\n4 2\n5 1\n2 4\n2 5\n2 2\n3 4\n3 1\n1 1\n5 5\n2 2\n3 4\n2 4\n5 2\n4 1\n3 1\n1 1\n4 1\n4 4\n1 4\n1 3",
"output": "1"
},
{
"input": "50 22\n4 9\n8 1\n3 7\n1 2\n3 8\n9 8\n8 5\n2 10\n5 8\n1 3\n1 8\n2 3\n7 9\n10 2\n9 9\n7 3\n8 6\n10 6\n5 4\n8 1\n1 5\n6 8\n9 5\n9 5\n3 2\n3 3\n3 8\n7 5\n4 5\n8 10\n8 2\n3 5\n3 2\n1 1\n7 2\n2 7\n6 8\n10 4\n7 5\n1 7\n6 5\n3 1\n4 9\n2 3\n3 6\n5 8\n4 10\n10 7\n7 10\n9 8",
"output": "1"
},
{
"input": "50 22\n29 15\n18 10\n6 23\n38 28\n34 40\n40 1\n16 26\n22 33\n14 30\n26 7\n15 16\n22 40\n14 15\n6 28\n32 27\n33 3\n38 22\n40 17\n16 27\n21 27\n34 26\n5 15\n34 9\n38 23\n7 36\n17 6\n19 37\n40 1\n10 28\n9 14\n8 31\n40 8\n14 2\n24 16\n38 33\n3 37\n2 9\n21 21\n40 26\n28 33\n24 31\n10 12\n27 27\n17 4\n38 5\n21 31\n5 12\n29 7\n39 12\n26 14",
"output": "1"
},
{
"input": "50 14\n4 20\n37 50\n46 19\n20 25\n47 10\n6 34\n12 41\n47 9\n22 28\n41 34\n47 40\n12 42\n9 4\n15 15\n27 8\n38 9\n4 17\n8 13\n47 7\n9 38\n30 48\n50 7\n41 34\n23 11\n16 37\n2 32\n18 46\n37 48\n47 41\n13 9\n24 50\n46 14\n33 49\n9 50\n35 30\n49 44\n42 49\n39 15\n33 42\n3 18\n44 15\n44 28\n9 17\n16 4\n10 36\n4 22\n47 17\n24 12\n2 31\n6 30",
"output": "2"
},
{
"input": "2 1\n50 50\n50 50",
"output": "2"
},
{
"input": "2 2\n50 50\n50 50",
"output": "2"
},
{
"input": "2 1\n50 50\n50 49",
"output": "1"
},
{
"input": "2 2\n50 50\n50 49",
"output": "1"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50",
"output": "50"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n49 50",
"output": "1"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 49",
"output": "49"
},
{
"input": "10 1\n3 1\n3 2\n3 2\n2 3\n2 3\n2 3\n2 4\n2 4\n2 4\n2 4",
"output": "1"
},
{
"input": "4 3\n1 1\n1 2\n2 1\n2 2",
"output": "1"
},
{
"input": "4 2\n1 2\n1 3\n1 5\n1 2",
"output": "2"
},
{
"input": "6 1\n4 3\n4 2\n5 1\n5 2\n5 3\n5 3",
"output": "1"
},
{
"input": "5 3\n2 1\n1 3\n1 2\n1 1\n1 1",
"output": "2"
},
{
"input": "3 1\n1 1\n1 2\n1 3",
"output": "1"
},
{
"input": "10 10\n3 1\n3 2\n3 2\n2 3\n2 3\n2 3\n2 4\n2 4\n2 4\n2 4",
"output": "4"
},
{
"input": "5 1\n5 5\n5 6\n5 7\n5 8\n5 5",
"output": "2"
},
{
"input": "4 1\n1 1\n1 1\n1 2\n1 3",
"output": "2"
}
] | 1,691,661,606 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | # LUOGU_RID: 120110648
n, k = map(int, input().split())
cong = []
for i in range(n):
a, b = map(int, input().split())
cong.append((a, -b))
cong = sorted(cong)
print(cong.count(cong[k - 1])) | Title: Rank List
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place.
You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*.
It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place.
Your task is to count what number of teams from the given list shared the *k*-th place.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=≤<=*p**i*,<=*t**i*<=≤<=50) — the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces.
Output Specification:
In the only line print the sought number of teams that got the *k*-th place in the final results' table.
Demo Input:
['7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n', '5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n']
Demo Output:
['3\n', '4\n']
Note:
The final results' table for the first sample is:
- 1-3 places — 4 solved problems, the penalty time equals 10 - 4 place — 3 solved problems, the penalty time equals 20 - 5-6 places — 2 solved problems, the penalty time equals 1 - 7 place — 1 solved problem, the penalty time equals 10
The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams.
The final table for the second sample is:
- 1 place — 5 solved problems, the penalty time equals 3 - 2-5 places — 3 solved problems, the penalty time equals 1
The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams. | ```python
# LUOGU_RID: 120110648
n, k = map(int, input().split())
cong = []
for i in range(n):
a, b = map(int, input().split())
cong.append((a, -b))
cong = sorted(cong)
print(cong.count(cong[k - 1]))
``` | 0 | |
281 | A | Word Capitalization | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged. | A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103. | Output the given word after capitalization. | [
"ApPLe\n",
"konjac\n"
] | [
"ApPLe\n",
"Konjac\n"
] | none | 500 | [
{
"input": "ApPLe",
"output": "ApPLe"
},
{
"input": "konjac",
"output": "Konjac"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "A"
},
{
"input": "z",
"output": "Z"
},
{
"input": "ABACABA",
"output": "ABACABA"
},
{
"input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX",
"output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX"
},
{
"input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO",
"output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO"
},
{
"input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD",
"output": "HDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD"
},
{
"input": "qUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW",
"output": "QUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW"
},
{
"input": "kHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL",
"output": "KHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL"
},
{
"input": "aHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC",
"output": "AHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC"
},
{
"input": "sLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN",
"output": "SLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN"
},
{
"input": "iRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE",
"output": "IRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE"
},
{
"input": "cRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP",
"output": "CRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP"
},
{
"input": "wVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG",
"output": "WVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG"
},
{
"input": "kEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV",
"output": "KEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV"
},
{
"input": "eIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI",
"output": "EIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI"
},
{
"input": "uOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX",
"output": "UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX"
},
{
"input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ",
"output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ"
},
{
"input": "P",
"output": "P"
},
{
"input": "Xyzzy",
"output": "Xyzzy"
},
{
"input": "Zzz",
"output": "Zzz"
},
{
"input": "Zp",
"output": "Zp"
}
] | 1,696,239,114 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 6 | 92 | 0 | word = input()
output = ""
for letter in word:
if letter == word[0]:
output+=letter.upper()
else:
output+=letter
print(output)
| Title: Word Capitalization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
Input Specification:
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output Specification:
Output the given word after capitalization.
Demo Input:
['ApPLe\n', 'konjac\n']
Demo Output:
['ApPLe\n', 'Konjac\n']
Note:
none | ```python
word = input()
output = ""
for letter in word:
if letter == word[0]:
output+=letter.upper()
else:
output+=letter
print(output)
``` | 0 | |
490 | C | Hacking Cypher | PROGRAMMING | 1,700 | [
"brute force",
"math",
"number theory",
"strings"
] | null | null | Polycarpus participates in a competition for hacking into a new secure messenger. He's almost won.
Having carefully studied the interaction protocol, Polycarpus came to the conclusion that the secret key can be obtained if he properly cuts the public key of the application into two parts. The public key is a long integer which may consist of even a million digits!
Polycarpus needs to find such a way to cut the public key into two nonempty parts, that the first (left) part is divisible by *a* as a separate number, and the second (right) part is divisible by *b* as a separate number. Both parts should be positive integers that have no leading zeros. Polycarpus knows values *a* and *b*.
Help Polycarpus and find any suitable method to cut the public key. | The first line of the input contains the public key of the messenger — an integer without leading zeroes, its length is in range from 1 to 106 digits. The second line contains a pair of space-separated positive integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=108). | In the first line print "YES" (without the quotes), if the method satisfying conditions above exists. In this case, next print two lines — the left and right parts after the cut. These two parts, being concatenated, must be exactly identical to the public key. The left part must be divisible by *a*, and the right part must be divisible by *b*. The two parts must be positive integers having no leading zeros. If there are several answers, print any of them.
If there is no answer, print in a single line "NO" (without the quotes). | [
"116401024\n97 1024\n",
"284254589153928171911281811000\n1009 1000\n",
"120\n12 1\n"
] | [
"YES\n11640\n1024\n",
"YES\n2842545891539\n28171911281811000\n",
"NO\n"
] | none | 1,500 | [
{
"input": "116401024\n97 1024",
"output": "YES\n11640\n1024"
},
{
"input": "284254589153928171911281811000\n1009 1000",
"output": "YES\n2842545891539\n28171911281811000"
},
{
"input": "120\n12 1",
"output": "NO"
},
{
"input": "604\n6 4",
"output": "YES\n60\n4"
},
{
"input": "2108\n7 8",
"output": "YES\n210\n8"
},
{
"input": "7208\n10 1",
"output": "YES\n720\n8"
},
{
"input": "97502821\n25 91",
"output": "YES\n9750\n2821"
},
{
"input": "803405634\n309 313",
"output": "YES\n80340\n5634"
},
{
"input": "15203400\n38 129",
"output": "NO"
},
{
"input": "8552104774\n973 76",
"output": "NO"
},
{
"input": "2368009434\n320 106",
"output": "YES\n236800\n9434"
},
{
"input": "425392502895812\n4363 2452",
"output": "YES\n42539250\n2895812"
},
{
"input": "142222201649130\n4854 7853",
"output": "YES\n14222220\n1649130"
},
{
"input": "137871307228140\n9375 9092",
"output": "NO"
},
{
"input": "8784054131798916\n9 61794291",
"output": "YES\n87840\n54131798916"
},
{
"input": "24450015102786098\n75 55729838",
"output": "YES\n244500\n15102786098"
},
{
"input": "100890056766780885\n177 88010513",
"output": "YES\n1008900\n56766780885"
},
{
"input": "2460708054301924950\n9428 85246350",
"output": "YES\n24607080\n54301924950"
},
{
"input": "39915186055525904358\n90102 63169402",
"output": "YES\n399151860\n55525904358"
},
{
"input": "199510140021146591389\n458644 28692797",
"output": "YES\n1995101400\n21146591389"
},
{
"input": "4802711808015050898224\n8381696 51544172",
"output": "YES\n48027118080\n15050898224"
},
{
"input": "6450225349035040017740\n8872387 56607460",
"output": "YES\n64502253490\n35040017740"
},
{
"input": "4530228043401488\n71454701 8",
"output": "YES\n453022804340\n1488"
},
{
"input": "18769213650033200\n56876405 100",
"output": "YES\n187692136500\n33200"
},
{
"input": "389744672208415\n17019418 765",
"output": "YES\n38974467220\n8415"
},
{
"input": "1256363256202133560\n26228878 7460",
"output": "YES\n125636325620\n2133560"
},
{
"input": "10213094404080691512\n64639838 83359",
"output": "YES\n102130944040\n80691512"
},
{
"input": "14525757302059286788\n44151238 152801",
"output": "YES\n145257573020\n59286788"
},
{
"input": "443852406270256089240\n54194433 423288",
"output": "YES\n443852406270\n256089240"
},
{
"input": "6450225349035040017740\n8872387 56607460",
"output": "YES\n64502253490\n35040017740"
},
{
"input": "16375289070073689\n33903290 216",
"output": "NO"
},
{
"input": "3415280033041307294\n15179 79809921",
"output": "NO"
},
{
"input": "4261508098904115227\n52546339 6430",
"output": "NO"
},
{
"input": "15016\n15 16",
"output": "YES\n150\n16"
},
{
"input": "120007\n120 7",
"output": "YES\n12000\n7"
},
{
"input": "23\n2 3",
"output": "YES\n2\n3"
}
] | 1,683,052,299 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 35 | 1,000 | 3,072,000 | s=input()
a,b=map(int,input().split());j=len(s);flag=True
while j>-1:
if j!= 0 and int(s[:j])%a==0 and j!=len(s) and int(s[j:])!=0 and int(s[j:])%b==0:
print("YES")
print(s[:j])
print(s[j:])
flag=False
break
else:
j-=1
if flag:
print("NO") | Title: Hacking Cypher
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus participates in a competition for hacking into a new secure messenger. He's almost won.
Having carefully studied the interaction protocol, Polycarpus came to the conclusion that the secret key can be obtained if he properly cuts the public key of the application into two parts. The public key is a long integer which may consist of even a million digits!
Polycarpus needs to find such a way to cut the public key into two nonempty parts, that the first (left) part is divisible by *a* as a separate number, and the second (right) part is divisible by *b* as a separate number. Both parts should be positive integers that have no leading zeros. Polycarpus knows values *a* and *b*.
Help Polycarpus and find any suitable method to cut the public key.
Input Specification:
The first line of the input contains the public key of the messenger — an integer without leading zeroes, its length is in range from 1 to 106 digits. The second line contains a pair of space-separated positive integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=108).
Output Specification:
In the first line print "YES" (without the quotes), if the method satisfying conditions above exists. In this case, next print two lines — the left and right parts after the cut. These two parts, being concatenated, must be exactly identical to the public key. The left part must be divisible by *a*, and the right part must be divisible by *b*. The two parts must be positive integers having no leading zeros. If there are several answers, print any of them.
If there is no answer, print in a single line "NO" (without the quotes).
Demo Input:
['116401024\n97 1024\n', '284254589153928171911281811000\n1009 1000\n', '120\n12 1\n']
Demo Output:
['YES\n11640\n1024\n', 'YES\n2842545891539\n28171911281811000\n', 'NO\n']
Note:
none | ```python
s=input()
a,b=map(int,input().split());j=len(s);flag=True
while j>-1:
if j!= 0 and int(s[:j])%a==0 and j!=len(s) and int(s[j:])!=0 and int(s[j:])%b==0:
print("YES")
print(s[:j])
print(s[j:])
flag=False
break
else:
j-=1
if flag:
print("NO")
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | For an array $b$ of length $m$ we define the function $f$ as
where $\oplus$ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
For example, $f(1,2,4,8)=f(1\oplus2,2\oplus4,4\oplus8)=f(3,6,12)=f(3\oplus6,6\oplus12)=f(5,10)=f(5\oplus10)=f(15)=15$
You are given an array $a$ and a few queries. Each query is represented as two integers $l$ and $r$. The answer is the maximum value of $f$ on all continuous subsegments of the array $a_l, a_{l+1}, \ldots, a_r$. | The first line contains a single integer $n$ ($1 \le n \le 5000$) — the length of $a$.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 2^{30}-1$) — the elements of the array.
The third line contains a single integer $q$ ($1 \le q \le 100\,000$) — the number of queries.
Each of the next $q$ lines contains a query represented as two integers $l$, $r$ ($1 \le l \le r \le n$). | Print $q$ lines — the answers for the queries. | [
"3\n8 4 1\n2\n2 3\n1 2\n",
"6\n1 2 4 8 16 32\n4\n1 6\n2 5\n3 4\n1 2\n"
] | [
"5\n12\n",
"60\n30\n12\n3\n"
] | In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.
In second sample, optimal segment for first query are $[3,6]$, for second query — $[2,5]$, for third — $[3,4]$, for fourth — $[1,2]$. | 0 | [
{
"input": "3\n8 4 1\n2\n2 3\n1 2",
"output": "5\n12"
},
{
"input": "6\n1 2 4 8 16 32\n4\n1 6\n2 5\n3 4\n1 2",
"output": "60\n30\n12\n3"
}
] | 1,621,995,324 | 2,147,483,647 | PyPy 3 | OK | TESTS | 51 | 1,809 | 110,080,000 | n = int(input())
A = list(map(int,input().split(" ")))
matrix = []
for i in range(n):
matrix.append([0]*n)
for i in range(n):
matrix[i][i] = A[i]
for i in reversed(range(0,n)):
for j in range(i+1,n):
matrix[i][j] = matrix[i][j-1]^matrix[i+1][j]
for i in reversed(range(0,n)):
for j in range(i+1,n):
matrix[i][j] = max(matrix[i][j],matrix[i][j-1],matrix[i+1][j])
q = int(input())
for i in range(q):
parameters = list(map(int,input().split(" ")))
l,r = parameters[0]-1,parameters[1]-1
print(matrix[l][r])
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For an array $b$ of length $m$ we define the function $f$ as
where $\oplus$ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
For example, $f(1,2,4,8)=f(1\oplus2,2\oplus4,4\oplus8)=f(3,6,12)=f(3\oplus6,6\oplus12)=f(5,10)=f(5\oplus10)=f(15)=15$
You are given an array $a$ and a few queries. Each query is represented as two integers $l$ and $r$. The answer is the maximum value of $f$ on all continuous subsegments of the array $a_l, a_{l+1}, \ldots, a_r$.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 5000$) — the length of $a$.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 2^{30}-1$) — the elements of the array.
The third line contains a single integer $q$ ($1 \le q \le 100\,000$) — the number of queries.
Each of the next $q$ lines contains a query represented as two integers $l$, $r$ ($1 \le l \le r \le n$).
Output Specification:
Print $q$ lines — the answers for the queries.
Demo Input:
['3\n8 4 1\n2\n2 3\n1 2\n', '6\n1 2 4 8 16 32\n4\n1 6\n2 5\n3 4\n1 2\n']
Demo Output:
['5\n12\n', '60\n30\n12\n3\n']
Note:
In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.
In second sample, optimal segment for first query are $[3,6]$, for second query — $[2,5]$, for third — $[3,4]$, for fourth — $[1,2]$. | ```python
n = int(input())
A = list(map(int,input().split(" ")))
matrix = []
for i in range(n):
matrix.append([0]*n)
for i in range(n):
matrix[i][i] = A[i]
for i in reversed(range(0,n)):
for j in range(i+1,n):
matrix[i][j] = matrix[i][j-1]^matrix[i+1][j]
for i in reversed(range(0,n)):
for j in range(i+1,n):
matrix[i][j] = max(matrix[i][j],matrix[i][j-1],matrix[i+1][j])
q = int(input())
for i in range(q):
parameters = list(map(int,input().split(" ")))
l,r = parameters[0]-1,parameters[1]-1
print(matrix[l][r])
``` | 3 | |
604 | A | Uncowed Forces | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.
Codeforces scores are computed as follows: If the maximum point value of a problem is *x*, and Kevin submitted correctly at minute *m* but made *w* wrong submissions, then his score on that problem is . His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points for each unsuccessful hack.
All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer. | The first line of the input contains five space-separated integers *m*1, *m*2, *m*3, *m*4, *m*5, where *m**i* (0<=≤<=*m**i*<=≤<=119) is the time of Kevin's last submission for problem *i*. His last submission is always correct and gets accepted.
The second line contains five space-separated integers *w*1, *w*2, *w*3, *w*4, *w*5, where *w**i* (0<=≤<=*w**i*<=≤<=10) is Kevin's number of wrong submissions on problem *i*.
The last line contains two space-separated integers *h**s* and *h**u* (0<=≤<=*h**s*,<=*h**u*<=≤<=20), denoting the Kevin's numbers of successful and unsuccessful hacks, respectively. | Print a single integer, the value of Kevin's final score. | [
"20 40 60 80 100\n0 1 2 3 4\n1 0\n",
"119 119 119 119 119\n0 0 0 0 0\n10 0\n"
] | [
"4900\n",
"4930\n"
] | In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/42158dc2bc78cd21fa679530ae9ef8b9ea298d15.png" style="max-width: 100.0%;max-height: 100.0%;"/> of the points on each problem. So his score from solving problems is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/fdf392d8508500b57f8057ac0c4c892ab5f925a2.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Adding in 10·100 = 1000 points from hacks, his total score becomes 3930 + 1000 = 4930. | 500 | [
{
"input": "20 40 60 80 100\n0 1 2 3 4\n1 0",
"output": "4900"
},
{
"input": "119 119 119 119 119\n0 0 0 0 0\n10 0",
"output": "4930"
},
{
"input": "3 6 13 38 60\n6 10 10 3 8\n9 9",
"output": "5088"
},
{
"input": "21 44 11 68 75\n6 2 4 8 4\n2 8",
"output": "4522"
},
{
"input": "16 112 50 114 68\n1 4 8 4 9\n19 11",
"output": "5178"
},
{
"input": "55 66 75 44 47\n6 0 6 6 10\n19 0",
"output": "6414"
},
{
"input": "47 11 88 5 110\n6 10 4 2 3\n10 6",
"output": "5188"
},
{
"input": "5 44 61 103 92\n9 0 10 4 8\n15 7",
"output": "4914"
},
{
"input": "115 53 96 62 110\n7 8 1 7 9\n7 16",
"output": "3416"
},
{
"input": "102 83 26 6 11\n3 4 1 8 3\n17 14",
"output": "6704"
},
{
"input": "36 102 73 101 19\n5 9 2 2 6\n4 13",
"output": "4292"
},
{
"input": "40 115 93 107 113\n5 7 2 6 8\n6 17",
"output": "2876"
},
{
"input": "53 34 53 107 81\n4 3 1 10 8\n7 7",
"output": "4324"
},
{
"input": "113 37 4 84 66\n2 0 10 3 0\n20 19",
"output": "6070"
},
{
"input": "10 53 101 62 1\n8 0 9 7 9\n0 11",
"output": "4032"
},
{
"input": "45 45 75 36 76\n6 2 2 0 0\n8 17",
"output": "5222"
},
{
"input": "47 16 44 78 111\n7 9 8 0 2\n1 19",
"output": "3288"
},
{
"input": "7 54 39 102 31\n6 0 2 10 1\n18 3",
"output": "6610"
},
{
"input": "0 46 86 72 40\n1 5 5 5 9\n6 5",
"output": "4924"
},
{
"input": "114 4 45 78 113\n0 4 8 10 2\n10 12",
"output": "4432"
},
{
"input": "56 56 96 105 107\n4 9 10 4 8\n2 1",
"output": "3104"
},
{
"input": "113 107 59 50 56\n3 7 10 6 3\n10 12",
"output": "4586"
},
{
"input": "96 104 9 94 84\n6 10 7 8 3\n14 11",
"output": "4754"
},
{
"input": "98 15 116 43 55\n4 3 0 9 3\n10 7",
"output": "5400"
},
{
"input": "0 26 99 108 35\n0 4 3 0 10\n9 5",
"output": "5388"
},
{
"input": "89 24 51 49 84\n5 6 2 2 9\n2 14",
"output": "4066"
},
{
"input": "57 51 76 45 96\n1 0 4 3 6\n12 15",
"output": "5156"
},
{
"input": "79 112 37 36 116\n2 8 4 7 5\n4 12",
"output": "3872"
},
{
"input": "71 42 60 20 7\n7 1 1 10 6\n1 7",
"output": "5242"
},
{
"input": "86 10 66 80 55\n0 2 5 10 5\n15 6",
"output": "5802"
},
{
"input": "66 109 22 22 62\n3 1 5 4 5\n10 5",
"output": "5854"
},
{
"input": "97 17 43 84 58\n2 8 3 8 6\n10 7",
"output": "5028"
},
{
"input": "109 83 5 114 104\n6 0 3 9 5\n5 2",
"output": "4386"
},
{
"input": "94 18 24 91 105\n2 0 7 10 3\n1 4",
"output": "4118"
},
{
"input": "64 17 86 59 45\n8 0 10 2 2\n4 4",
"output": "5144"
},
{
"input": "70 84 31 57 2\n7 0 0 2 7\n12 5",
"output": "6652"
},
{
"input": "98 118 117 86 4\n2 10 9 7 5\n11 15",
"output": "4476"
},
{
"input": "103 110 101 97 70\n4 2 1 0 5\n7 5",
"output": "4678"
},
{
"input": "78 96 6 97 62\n7 7 9 2 9\n10 3",
"output": "4868"
},
{
"input": "95 28 3 31 115\n1 9 0 7 3\n10 13",
"output": "5132"
},
{
"input": "45 17 116 58 3\n8 8 7 6 4\n3 19",
"output": "3992"
},
{
"input": "19 12 0 113 77\n3 0 10 9 2\n8 6",
"output": "5040"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0",
"output": "7500"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n20 0",
"output": "9500"
},
{
"input": "119 119 119 119 119\n10 10 10 10 10\n0 20",
"output": "1310"
},
{
"input": "0 0 0 0 0\n10 10 10 10 10\n0 20",
"output": "4150"
},
{
"input": "119 0 0 0 0\n10 0 0 0 0\n5 5",
"output": "7400"
},
{
"input": "0 119 0 0 0\n0 10 0 0 0\n5 5",
"output": "7050"
},
{
"input": "0 0 119 0 0\n0 0 10 0 0\n0 0",
"output": "6450"
},
{
"input": "0 0 0 119 0\n0 0 0 10 0\n5 5",
"output": "6350"
},
{
"input": "0 0 0 0 119\n0 0 0 0 10\n5 5",
"output": "6060"
},
{
"input": "119 0 0 0 0\n2 0 0 0 0\n5 5",
"output": "7412"
},
{
"input": "0 119 0 0 0\n0 2 0 0 0\n5 5",
"output": "7174"
},
{
"input": "0 0 119 0 0\n0 0 2 0 0\n5 5",
"output": "6936"
},
{
"input": "0 0 0 119 0\n0 0 0 2 0\n5 5",
"output": "6698"
},
{
"input": "0 0 0 0 119\n0 0 0 0 2\n5 5",
"output": "6460"
},
{
"input": "119 0 0 0 0\n0 0 0 0 0\n4 9",
"output": "7212"
}
] | 1,517,999,087 | 2,147,483,647 | Python 3 | OK | TESTS | 57 | 62 | 5,632,000 | p = [500, 1000, 1500, 2000, 2500]
r = lambda: list(map(int,input().split()))
score = 0
for x,m,w in zip(p,r(),r()):
score += max(0.3*x,(1-m/250)*x-50*w)
hack, hack_wrong = r()
score += 100*hack - 50*hack_wrong
print (int(score)) | Title: Uncowed Forces
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.
Codeforces scores are computed as follows: If the maximum point value of a problem is *x*, and Kevin submitted correctly at minute *m* but made *w* wrong submissions, then his score on that problem is . His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points for each unsuccessful hack.
All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer.
Input Specification:
The first line of the input contains five space-separated integers *m*1, *m*2, *m*3, *m*4, *m*5, where *m**i* (0<=≤<=*m**i*<=≤<=119) is the time of Kevin's last submission for problem *i*. His last submission is always correct and gets accepted.
The second line contains five space-separated integers *w*1, *w*2, *w*3, *w*4, *w*5, where *w**i* (0<=≤<=*w**i*<=≤<=10) is Kevin's number of wrong submissions on problem *i*.
The last line contains two space-separated integers *h**s* and *h**u* (0<=≤<=*h**s*,<=*h**u*<=≤<=20), denoting the Kevin's numbers of successful and unsuccessful hacks, respectively.
Output Specification:
Print a single integer, the value of Kevin's final score.
Demo Input:
['20 40 60 80 100\n0 1 2 3 4\n1 0\n', '119 119 119 119 119\n0 0 0 0 0\n10 0\n']
Demo Output:
['4900\n', '4930\n']
Note:
In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/42158dc2bc78cd21fa679530ae9ef8b9ea298d15.png" style="max-width: 100.0%;max-height: 100.0%;"/> of the points on each problem. So his score from solving problems is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/fdf392d8508500b57f8057ac0c4c892ab5f925a2.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Adding in 10·100 = 1000 points from hacks, his total score becomes 3930 + 1000 = 4930. | ```python
p = [500, 1000, 1500, 2000, 2500]
r = lambda: list(map(int,input().split()))
score = 0
for x,m,w in zip(p,r(),r()):
score += max(0.3*x,(1-m/250)*x-50*w)
hack, hack_wrong = r()
score += 100*hack - 50*hack_wrong
print (int(score))
``` | 3 | |
171 | B | Star | PROGRAMMING | 1,300 | [
"*special",
"combinatorics"
] | null | null | The input contains a single integer *a* (1<=≤<=*a*<=≤<=18257). | Print a single integer *output* (1<=≤<=*output*<=≤<=2·109). | [
"2\n"
] | [
"13"
] | none | 0 | [
{
"input": "2",
"output": "13"
},
{
"input": "1",
"output": "1"
},
{
"input": "3",
"output": "37"
},
{
"input": "4",
"output": "73"
},
{
"input": "5",
"output": "121"
},
{
"input": "6",
"output": "181"
},
{
"input": "7",
"output": "253"
},
{
"input": "8",
"output": "337"
},
{
"input": "9",
"output": "433"
},
{
"input": "15000",
"output": "1349910001"
},
{
"input": "4845",
"output": "140815081"
},
{
"input": "6914",
"output": "286778893"
},
{
"input": "3994",
"output": "95688253"
},
{
"input": "12504",
"output": "938025073"
},
{
"input": "13170",
"output": "1040614381"
},
{
"input": "427",
"output": "1091413"
},
{
"input": "11877",
"output": "846307513"
},
{
"input": "3202",
"output": "61497613"
},
{
"input": "5689",
"output": "194154193"
},
{
"input": "15302",
"output": "1404815413"
},
{
"input": "17042",
"output": "1742476333"
},
{
"input": "1481",
"output": "13151281"
},
{
"input": "15592",
"output": "1458569233"
},
{
"input": "16344",
"output": "1602659953"
},
{
"input": "4222",
"output": "106926373"
},
{
"input": "11808",
"output": "836502337"
},
{
"input": "13366",
"output": "1071819541"
},
{
"input": "3823",
"output": "87669037"
},
{
"input": "581",
"output": "2021881"
},
{
"input": "15479",
"output": "1437503773"
},
{
"input": "6543",
"output": "256825837"
},
{
"input": "11136",
"output": "743996161"
},
{
"input": "16331",
"output": "1600111381"
},
{
"input": "8543",
"output": "437845837"
},
{
"input": "7530",
"output": "340160221"
},
{
"input": "3154",
"output": "59667373"
},
{
"input": "11501",
"output": "793569001"
},
{
"input": "12038",
"output": "869408437"
},
{
"input": "13082",
"output": "1026753853"
},
{
"input": "18257",
"output": "1999798753"
}
] | 1,647,253,869 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 62 | 0 | n = int(input())
print (6*n*(n-1)+1) | Title: Star
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Input Specification:
The input contains a single integer *a* (1<=≤<=*a*<=≤<=18257).
Output Specification:
Print a single integer *output* (1<=≤<=*output*<=≤<=2·109).
Demo Input:
['2\n']
Demo Output:
['13']
Note:
none | ```python
n = int(input())
print (6*n*(n-1)+1)
``` | 3 | ||
667 | A | Pouring Rain | PROGRAMMING | 1,100 | [
"geometry",
"math"
] | null | null | A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do.
Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation.
Thus, your cup is a cylinder with diameter equals *d* centimeters. Initial level of water in cup equals *h* centimeters from the bottom.
You drink a water with a speed equals *v* milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on *e* centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously.
Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 104 seconds.
Note one milliliter equals to one cubic centimeter. | The only line of the input contains four integer numbers *d*,<=*h*,<=*v*,<=*e* (1<=≤<=*d*,<=*h*,<=*v*,<=*e*<=≤<=104), where:
- *d* — the diameter of your cylindrical cup, - *h* — the initial level of water in the cup, - *v* — the speed of drinking process from the cup in milliliters per second, - *e* — the growth of water because of rain if you do not drink from the cup. | If it is impossible to make the cup empty, print "NO" (without quotes).
Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=4. It is guaranteed that if the answer exists, it doesn't exceed 104. | [
"1 2 3 100\n",
"1 1 1 1\n"
] | [
"NO\n",
"YES\n3.659792366325\n"
] | In the first example the water fills the cup faster than you can drink from it.
In the second example area of the cup's bottom equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/419dc74dcd7bc392019c9fe748fe1fdb08ab521a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, thus we can conclude that you decrease the level of water by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8edb237e1f805fe83c2f47e48d3a9d03f2ee304.png" style="max-width: 100.0%;max-height: 100.0%;"/> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9dae615d7e2c5c7c03cb478848fb06aba1a8942e.png" style="max-width: 100.0%;max-height: 100.0%;"/> seconds. | 500 | [
{
"input": "1 2 3 100",
"output": "NO"
},
{
"input": "1 1 1 1",
"output": "YES\n3.659792366325"
},
{
"input": "48 7946 7992 72",
"output": "NO"
},
{
"input": "72 6791 8546 46",
"output": "NO"
},
{
"input": "100 5635 9099 23",
"output": "NO"
},
{
"input": "20 287 3845 5",
"output": "YES\n39.646277165210"
},
{
"input": "48 6428 9807 83",
"output": "NO"
},
{
"input": "72 5272 4552 64",
"output": "NO"
},
{
"input": "100 4117 5106 34",
"output": "NO"
},
{
"input": "20 2961 9852 15",
"output": "YES\n180.991437129723"
},
{
"input": "48 1805 3109 93",
"output": "NO"
},
{
"input": "72 8534 7042 65",
"output": "NO"
},
{
"input": "1 47 80 68",
"output": "YES\n1.388102806810"
},
{
"input": "4 495 8813 1",
"output": "YES\n0.706823517575"
},
{
"input": "5 2797 5925 9",
"output": "YES\n9.553973511669"
},
{
"input": "1 8324 4362 23",
"output": "YES\n1.505007106354"
},
{
"input": "6 1976 8455 3",
"output": "YES\n6.674898722265"
},
{
"input": "7 2644 8080 5",
"output": "YES\n12.900417790197"
},
{
"input": "3 4183 5491 98",
"output": "YES\n6.162185601824"
},
{
"input": "2 8591 320 101",
"output": "YES\n9999.259991757254"
},
{
"input": "10000 10000 10000 10000",
"output": "NO"
},
{
"input": "2 5000 12 3",
"output": "YES\n6099.653943875812"
},
{
"input": "10 1000 100 1",
"output": "YES\n3659.792366325487"
}
] | 1,526,050,351 | 2,147,483,647 | Python 3 | OK | TESTS | 23 | 78 | 7,372,800 | from math import pi
d, h, v, e = input().split()
d, h, v, e = int(d), int(h), int(v), int(e)
e1 = v /( pi * (d/2)**2)
if e1 - e > 0:
print("YES")
print(h/(e1-e))
else:
print("NO")
| Title: Pouring Rain
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do.
Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation.
Thus, your cup is a cylinder with diameter equals *d* centimeters. Initial level of water in cup equals *h* centimeters from the bottom.
You drink a water with a speed equals *v* milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on *e* centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously.
Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 104 seconds.
Note one milliliter equals to one cubic centimeter.
Input Specification:
The only line of the input contains four integer numbers *d*,<=*h*,<=*v*,<=*e* (1<=≤<=*d*,<=*h*,<=*v*,<=*e*<=≤<=104), where:
- *d* — the diameter of your cylindrical cup, - *h* — the initial level of water in the cup, - *v* — the speed of drinking process from the cup in milliliters per second, - *e* — the growth of water because of rain if you do not drink from the cup.
Output Specification:
If it is impossible to make the cup empty, print "NO" (without quotes).
Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=4. It is guaranteed that if the answer exists, it doesn't exceed 104.
Demo Input:
['1 2 3 100\n', '1 1 1 1\n']
Demo Output:
['NO\n', 'YES\n3.659792366325\n']
Note:
In the first example the water fills the cup faster than you can drink from it.
In the second example area of the cup's bottom equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/419dc74dcd7bc392019c9fe748fe1fdb08ab521a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, thus we can conclude that you decrease the level of water by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8edb237e1f805fe83c2f47e48d3a9d03f2ee304.png" style="max-width: 100.0%;max-height: 100.0%;"/> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9dae615d7e2c5c7c03cb478848fb06aba1a8942e.png" style="max-width: 100.0%;max-height: 100.0%;"/> seconds. | ```python
from math import pi
d, h, v, e = input().split()
d, h, v, e = int(d), int(h), int(v), int(e)
e1 = v /( pi * (d/2)**2)
if e1 - e > 0:
print("YES")
print(h/(e1-e))
else:
print("NO")
``` | 3 | |
389 | A | Fox and Number Game | PROGRAMMING | 1,000 | [
"greedy",
"math"
] | null | null | Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum. | The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100). | Output a single integer — the required minimal sum. | [
"2\n1 2\n",
"3\n2 4 6\n",
"2\n12 18\n",
"5\n45 12 27 30 18\n"
] | [
"2\n",
"6\n",
"12\n",
"15\n"
] | In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>. | 500 | [
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n2 4 6",
"output": "6"
},
{
"input": "2\n12 18",
"output": "12"
},
{
"input": "5\n45 12 27 30 18",
"output": "15"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "2\n100 100",
"output": "200"
},
{
"input": "2\n87 58",
"output": "58"
},
{
"input": "39\n52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52",
"output": "2028"
},
{
"input": "59\n96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96",
"output": "5664"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "10000"
},
{
"input": "100\n70 70 77 42 98 84 56 91 35 21 7 70 77 77 56 63 14 84 56 14 77 77 63 70 14 7 28 91 63 49 21 84 98 56 77 98 98 84 98 14 7 56 49 28 91 98 7 56 14 91 14 98 49 28 98 14 98 98 14 70 35 28 63 28 49 63 63 56 91 98 35 42 42 35 63 35 42 14 63 21 77 56 42 77 35 91 56 21 28 84 56 70 70 91 98 70 84 63 21 98",
"output": "700"
},
{
"input": "39\n63 21 21 42 21 63 21 84 42 21 84 63 42 63 84 84 84 42 42 84 21 63 42 63 42 42 63 42 42 63 84 42 21 84 21 63 42 21 42",
"output": "819"
},
{
"input": "59\n70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70",
"output": "4130"
},
{
"input": "87\n44 88 88 88 88 66 88 22 22 88 88 44 88 22 22 22 88 88 88 88 66 22 88 88 88 88 66 66 44 88 44 44 66 22 88 88 22 44 66 44 88 66 66 22 22 22 22 88 22 22 44 66 88 22 22 88 66 66 88 22 66 88 66 88 66 44 88 44 22 44 44 22 44 88 44 44 44 44 22 88 88 88 66 66 88 44 22",
"output": "1914"
},
{
"input": "15\n63 63 63 63 63 63 63 63 63 63 63 63 63 63 63",
"output": "945"
},
{
"input": "39\n63 77 21 14 14 35 21 21 70 42 21 70 28 77 28 77 7 42 63 7 98 49 98 84 35 70 70 91 14 42 98 7 42 7 98 42 56 35 91",
"output": "273"
},
{
"input": "18\n18 18 18 36 36 36 54 72 54 36 72 54 36 36 36 36 18 36",
"output": "324"
},
{
"input": "46\n71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71",
"output": "3266"
},
{
"input": "70\n66 11 66 11 44 11 44 99 55 22 88 11 11 22 55 44 22 77 44 77 77 22 44 55 88 11 99 99 88 22 77 77 66 11 11 66 99 55 55 44 66 44 77 44 44 55 33 55 44 88 77 77 22 66 33 44 11 22 55 44 22 66 77 33 33 44 44 44 22 33",
"output": "770"
},
{
"input": "10\n60 12 96 48 60 24 60 36 60 60",
"output": "120"
},
{
"input": "20\n51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51",
"output": "1020"
},
{
"input": "50\n58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58",
"output": "2900"
},
{
"input": "98\n70 60 100 30 70 20 30 50 50 30 90 40 30 40 60 80 60 60 80 50 10 80 20 10 20 10 50 70 30 80 30 50 60 90 90 100 60 30 90 20 30 60 90 80 60 60 10 90 10 50 40 40 80 90 100 40 70 40 30 50 60 50 60 30 40 20 90 60 20 20 20 70 60 70 50 100 90 50 20 40 80 60 10 60 50 40 40 10 50 10 40 10 80 100 100 90 10 90",
"output": "980"
},
{
"input": "100\n82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82",
"output": "8200"
},
{
"input": "100\n11 87 77 93 3 54 21 93 9 71 37 23 69 85 74 3 48 99 51 31 56 19 21 96 39 6 4 4 29 69 100 42 1 22 81 53 48 49 81 61 10 7 40 61 7 71 51 59 79 44 50 35 95 80 83 8 98 40 18 94 84 49 52 74 66 69 39 37 100 44 38 62 2 80 46 31 35 53 5 60 21 49 63 55 20 53 80 53 66 34 23 92 77 50 86 63 65 24 12 70",
"output": "100"
},
{
"input": "2\n100 1",
"output": "2"
},
{
"input": "2\n18 30",
"output": "12"
},
{
"input": "2\n3 5",
"output": "2"
},
{
"input": "2\n1 10",
"output": "2"
},
{
"input": "2\n8 5",
"output": "2"
},
{
"input": "5\n2 3 5 8 18",
"output": "5"
},
{
"input": "5\n2 4 1 6 8",
"output": "5"
},
{
"input": "3\n12 10 5",
"output": "3"
},
{
"input": "3\n6 10 15",
"output": "3"
}
] | 1,614,964,788 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 77 | 0 | n=int(input())
l=list(map(int,input().split()))
while len(set(l))!=1:
l.sort()
for i in range(1,len(l)+1):
for j in range(1,len(l)+1):
if l[-i]>l[-j]:
l[-i]=l[-i]-l[-j]
print(l)
continue
print(sum(l)) | Title: Fox and Number Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100).
Output Specification:
Output a single integer — the required minimal sum.
Demo Input:
['2\n1 2\n', '3\n2 4 6\n', '2\n12 18\n', '5\n45 12 27 30 18\n']
Demo Output:
['2\n', '6\n', '12\n', '15\n']
Note:
In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>. | ```python
n=int(input())
l=list(map(int,input().split()))
while len(set(l))!=1:
l.sort()
for i in range(1,len(l)+1):
for j in range(1,len(l)+1):
if l[-i]>l[-j]:
l[-i]=l[-i]-l[-j]
print(l)
continue
print(sum(l))
``` | 0 | |
82 | A | Double Cola | PROGRAMMING | 1,100 | [
"implementation",
"math"
] | A. Double Cola | 1 | 256 | Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon. | The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers. | Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially. | [
"1\n",
"6\n",
"1802\n"
] | [
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] | none | 500 | [
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,578,301,308 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 93 | 0 | """
Author: evangeloskarountzos
Date: 2020-01-06
Project: ELO 1100
File: 82A - Double Cola
IDE: PyCharm
"""
x = int(input())
line = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
if x < 5:
print(line[x-1])
else:
print(line[x%5])
| Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none | ```python
"""
Author: evangeloskarountzos
Date: 2020-01-06
Project: ELO 1100
File: 82A - Double Cola
IDE: PyCharm
"""
x = int(input())
line = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
if x < 5:
print(line[x-1])
else:
print(line[x%5])
``` | 0 |
888 | G | Xor-MST | PROGRAMMING | 2,300 | [
"bitmasks",
"constructive algorithms",
"data structures"
] | null | null | You are given a complete undirected graph with *n* vertices. A number *a**i* is assigned to each vertex, and the weight of an edge between vertices *i* and *j* is equal to *a**i*<=*xor*<=*a**j*.
Calculate the weight of the minimum spanning tree in this graph. | The first line contains *n* (1<=≤<=*n*<=≤<=200000) — the number of vertices in the graph.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (0<=≤<=*a**i*<=<<=230) — the numbers assigned to the vertices. | Print one number — the weight of the minimum spanning tree in the graph. | [
"5\n1 2 3 4 5\n",
"4\n1 2 3 4\n"
] | [
"8\n",
"8\n"
] | none | 0 | [
{
"input": "5\n1 2 3 4 5",
"output": "8"
},
{
"input": "4\n1 2 3 4",
"output": "8"
},
{
"input": "1\n1",
"output": "0"
}
] | 1,692,110,358 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1692110358.445972")# 1692110358.445988 | Title: Xor-MST
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a complete undirected graph with *n* vertices. A number *a**i* is assigned to each vertex, and the weight of an edge between vertices *i* and *j* is equal to *a**i*<=*xor*<=*a**j*.
Calculate the weight of the minimum spanning tree in this graph.
Input Specification:
The first line contains *n* (1<=≤<=*n*<=≤<=200000) — the number of vertices in the graph.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (0<=≤<=*a**i*<=<<=230) — the numbers assigned to the vertices.
Output Specification:
Print one number — the weight of the minimum spanning tree in the graph.
Demo Input:
['5\n1 2 3 4 5\n', '4\n1 2 3 4\n']
Demo Output:
['8\n', '8\n']
Note:
none | ```python
print("_RANDOM_GUESS_1692110358.445972")# 1692110358.445988
``` | 0 | |
382 | A | Ksenia and Pan Scales | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | Ksenia has ordinary pan scales and several weights of an equal mass. Ksenia has already put some weights on the scales, while other weights are untouched. Ksenia is now wondering whether it is possible to put all the remaining weights on the scales so that the scales were in equilibrium.
The scales is in equilibrium if the total sum of weights on the left pan is equal to the total sum of weights on the right pan. | The first line has a non-empty sequence of characters describing the scales. In this sequence, an uppercase English letter indicates a weight, and the symbol "|" indicates the delimiter (the character occurs in the sequence exactly once). All weights that are recorded in the sequence before the delimiter are initially on the left pan of the scale. All weights that are recorded in the sequence after the delimiter are initially on the right pan of the scale.
The second line contains a non-empty sequence containing uppercase English letters. Each letter indicates a weight which is not used yet.
It is guaranteed that all the English letters in the input data are different. It is guaranteed that the input does not contain any extra characters. | If you cannot put all the weights on the scales so that the scales were in equilibrium, print string "Impossible". Otherwise, print the description of the resulting scales, copy the format of the input.
If there are multiple answers, print any of them. | [
"AC|T\nL\n",
"|ABC\nXYZ\n",
"W|T\nF\n",
"ABC|\nD\n"
] | [
"AC|TL\n",
"XYZ|ABC\n",
"Impossible\n",
"Impossible\n"
] | none | 500 | [
{
"input": "AC|T\nL",
"output": "AC|TL"
},
{
"input": "|ABC\nXYZ",
"output": "XYZ|ABC"
},
{
"input": "W|T\nF",
"output": "Impossible"
},
{
"input": "ABC|\nD",
"output": "Impossible"
},
{
"input": "A|BC\nDEF",
"output": "ADF|BCE"
},
{
"input": "|\nABC",
"output": "Impossible"
},
{
"input": "|\nZXCVBANMIO",
"output": "XVAMO|ZCBNI"
},
{
"input": "|C\nA",
"output": "A|C"
},
{
"input": "|\nAB",
"output": "B|A"
},
{
"input": "A|XYZ\nUIOPL",
"output": "Impossible"
},
{
"input": "K|B\nY",
"output": "Impossible"
},
{
"input": "EQJWDOHKZRBISPLXUYVCMNFGT|\nA",
"output": "Impossible"
},
{
"input": "|MACKERIGZPVHNDYXJBUFLWSO\nQT",
"output": "Impossible"
},
{
"input": "ERACGIZOVPT|WXUYMDLJNQS\nKB",
"output": "ERACGIZOVPTB|WXUYMDLJNQSK"
},
{
"input": "CKQHRUZMISGE|FBVWPXDLTJYN\nOA",
"output": "CKQHRUZMISGEA|FBVWPXDLTJYNO"
},
{
"input": "V|CMOEUTAXBFWSK\nDLRZJGIYNQHP",
"output": "VDLRZJGIYNQHP|CMOEUTAXBFWSK"
},
{
"input": "QWHNMALDGKTJ|\nPBRYVXZUESCOIF",
"output": "QWHNMALDGKTJF|PBRYVXZUESCOI"
},
{
"input": "|\nFXCVMUEWZAHNDOSITPRLKQJYBG",
"output": "XVUWANOIPLQYG|FCMEZHDSTRKJB"
},
{
"input": "IB|PCGHZ\nFXWTJQNEKAUM",
"output": "Impossible"
},
{
"input": "EC|IWAXQ\nJUHSRKGZTOMYN",
"output": "ECJUHRGTMN|IWAXQSKZOY"
},
{
"input": "VDINYMA|UQKWBCLRHZJ\nXEGOF",
"output": "Impossible"
},
{
"input": "ZLTPSIQUBAR|XFDEMYC\nHNOJWG",
"output": "ZLTPSIQUBARG|XFDEMYCHNOJW"
},
{
"input": "R|FLZOTJNU\nGIYHKVX",
"output": "RGIYHKVX|FLZOTJNU"
},
{
"input": "W|TL\nQROFSADYPKHEJNMXBZVUCIG",
"output": "WQOSDPHJMBVCG|TLRFAYKENXZUI"
},
{
"input": "NRDFQSEKLAYMOT|ZH\nGUXIBJCVPW",
"output": "Impossible"
},
{
"input": "FGRT|\nAC",
"output": "Impossible"
},
{
"input": "|FGRT\nAC",
"output": "Impossible"
},
{
"input": "A|\nB",
"output": "A|B"
},
{
"input": "|A\nB",
"output": "B|A"
},
{
"input": "|\nA",
"output": "Impossible"
},
{
"input": "|\nQWERTYUIOPASDFGHJKLZXCVBNM",
"output": "WRYIPSFHKZCBM|QETUOADGJLXVN"
},
{
"input": "QWERTYUIOPASDFGHJKLZXCVBN|\nM",
"output": "Impossible"
},
{
"input": "QWERTY|VBN\nUIOPASDFGHJKLZXC",
"output": "Impossible"
},
{
"input": "ABC|D\nKSL",
"output": "Impossible"
},
{
"input": "A|BCDEF\nGH",
"output": "Impossible"
},
{
"input": "|ABC\nD",
"output": "Impossible"
},
{
"input": "A|BC\nDE",
"output": "Impossible"
},
{
"input": "|ASD\nX",
"output": "Impossible"
},
{
"input": "AB|CDEF\nXYZRT",
"output": "Impossible"
}
] | 1,389,975,641 | 3,041 | Python 3 | WRONG_ANSWER | PRETESTS | 0 | 61 | 307,200 | import sys
from sys import stdin
import heapq
# stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
a = stdin.readline().strip().split('|')
w = stdin.readline()
if((len(a[0]) + len(a[1]) + len(w)) % 2 == 0):
balance = int((len(a[0]) + len(a[1]) + len(w)) / 2)
n = balance - len(a[0])
a[0] = a[0] + w[:n]
a[1] = a[1] + w[n:]
if (len(a[1]) == len(a[0])):
print('|'.join(a))
else:
print('Impossible')
else:
print('Impossible')
# stdin.close()
# sys.stdout.close()
| Title: Ksenia and Pan Scales
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ksenia has ordinary pan scales and several weights of an equal mass. Ksenia has already put some weights on the scales, while other weights are untouched. Ksenia is now wondering whether it is possible to put all the remaining weights on the scales so that the scales were in equilibrium.
The scales is in equilibrium if the total sum of weights on the left pan is equal to the total sum of weights on the right pan.
Input Specification:
The first line has a non-empty sequence of characters describing the scales. In this sequence, an uppercase English letter indicates a weight, and the symbol "|" indicates the delimiter (the character occurs in the sequence exactly once). All weights that are recorded in the sequence before the delimiter are initially on the left pan of the scale. All weights that are recorded in the sequence after the delimiter are initially on the right pan of the scale.
The second line contains a non-empty sequence containing uppercase English letters. Each letter indicates a weight which is not used yet.
It is guaranteed that all the English letters in the input data are different. It is guaranteed that the input does not contain any extra characters.
Output Specification:
If you cannot put all the weights on the scales so that the scales were in equilibrium, print string "Impossible". Otherwise, print the description of the resulting scales, copy the format of the input.
If there are multiple answers, print any of them.
Demo Input:
['AC|T\nL\n', '|ABC\nXYZ\n', 'W|T\nF\n', 'ABC|\nD\n']
Demo Output:
['AC|TL\n', 'XYZ|ABC\n', 'Impossible\n', 'Impossible\n']
Note:
none | ```python
import sys
from sys import stdin
import heapq
# stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
a = stdin.readline().strip().split('|')
w = stdin.readline()
if((len(a[0]) + len(a[1]) + len(w)) % 2 == 0):
balance = int((len(a[0]) + len(a[1]) + len(w)) / 2)
n = balance - len(a[0])
a[0] = a[0] + w[:n]
a[1] = a[1] + w[n:]
if (len(a[1]) == len(a[0])):
print('|'.join(a))
else:
print('Impossible')
else:
print('Impossible')
# stdin.close()
# sys.stdout.close()
``` | 0 | |
605 | A | Sorting Railway Cars | PROGRAMMING | 1,600 | [
"constructive algorithms",
"greedy"
] | null | null | An infinitely long railway has a train consisting of *n* cars, numbered from 1 to *n* (the numbers of all the cars are distinct) and positioned in arbitrary order. David Blaine wants to sort the railway cars in the order of increasing numbers. In one move he can make one of the cars disappear from its place and teleport it either to the beginning of the train, or to the end of the train, at his desire. What is the minimum number of actions David Blaine needs to perform in order to sort the train? | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cars in the train.
The second line contains *n* integers *p**i* (1<=≤<=*p**i*<=≤<=*n*, *p**i*<=≠<=*p**j* if *i*<=≠<=*j*) — the sequence of the numbers of the cars in the train. | Print a single integer — the minimum number of actions needed to sort the railway cars. | [
"5\n4 1 2 5 3\n",
"4\n4 1 3 2\n"
] | [
"2\n",
"2\n"
] | In the first sample you need first to teleport the 4-th car, and then the 5-th car to the end of the train. | 500 | [
{
"input": "5\n4 1 2 5 3",
"output": "2"
},
{
"input": "4\n4 1 3 2",
"output": "2"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "6\n5 3 6 1 4 2",
"output": "4"
},
{
"input": "7\n1 2 3 6 7 4 5",
"output": "2"
},
{
"input": "8\n6 2 1 8 5 7 3 4",
"output": "5"
},
{
"input": "3\n1 2 3",
"output": "0"
},
{
"input": "3\n1 3 2",
"output": "1"
},
{
"input": "3\n2 1 3",
"output": "1"
},
{
"input": "3\n2 3 1",
"output": "1"
},
{
"input": "3\n3 1 2",
"output": "1"
},
{
"input": "3\n3 2 1",
"output": "2"
},
{
"input": "7\n1 3 5 7 2 4 6",
"output": "5"
},
{
"input": "7\n1 5 2 6 3 7 4",
"output": "3"
},
{
"input": "5\n1 4 2 3 5",
"output": "2"
},
{
"input": "9\n1 6 4 5 9 8 7 3 2",
"output": "7"
},
{
"input": "10\n5 1 6 2 8 3 4 10 9 7",
"output": "6"
},
{
"input": "50\n39 8 41 9 45 1 5 18 38 31 28 7 12 49 33 19 26 6 42 13 37 27 2 21 20 22 14 16 48 47 32 50 25 17 35 24 36 4 29 15 43 10 11 30 40 46 3 23 44 34",
"output": "46"
},
{
"input": "50\n43 15 10 33 32 31 13 7 5 22 36 1 25 14 38 19 8 6 24 42 28 21 44 35 4 3 49 30 27 46 2 9 17 37 45 41 18 39 12 11 16 20 50 26 29 34 40 47 48 23",
"output": "47"
},
{
"input": "50\n10 40 34 43 50 17 15 13 9 2 32 18 11 46 27 24 36 16 29 45 42 4 47 19 48 37 41 5 21 26 22 25 44 31 35 49 20 8 12 23 6 38 14 1 7 28 3 33 39 30",
"output": "46"
},
{
"input": "50\n10 37 3 46 45 29 36 13 21 25 35 5 18 33 12 19 50 16 30 47 20 42 39 28 2 6 38 8 7 31 22 27 26 9 15 14 34 48 4 32 40 43 44 24 11 1 23 17 49 41",
"output": "46"
},
{
"input": "50\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 42 37 38 39 40 41 36 43 44 45 46 47 48 49 50",
"output": "14"
},
{
"input": "50\n1 2 3 4 5 6 7 8 43 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 50 33 34 35 36 37 38 39 40 41 42 9 44 45 46 47 48 49 32",
"output": "27"
},
{
"input": "50\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 49 40 41 47 43 44 45 46 42 50 39 48",
"output": "11"
},
{
"input": "50\n1 2 3 4 27 6 7 8 9 10 30 12 13 14 15 16 17 18 19 20 21 22 23 24 28 26 5 25 29 11 31 32 33 34 38 36 37 35 39 40 41 42 43 44 45 46 47 48 49 50",
"output": "36"
},
{
"input": "50\n1 2 3 4 5 6 7 49 9 10 17 12 13 14 15 16 11 18 19 20 21 22 23 24 25 26 27 38 29 36 30 32 33 34 35 31 37 28 39 40 41 42 43 44 45 46 47 48 8 50",
"output": "38"
},
{
"input": "50\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 31 18 19 20 21 23 22 24 25 26 27 28 29 49 17 32 33 34 39 36 37 38 47 44 41 42 43 40 45 46 35 48 30 50",
"output": "33"
},
{
"input": "50\n1 2 15 4 5 6 7 8 9 10 11 12 13 14 3 16 17 18 19 32 21 22 36 28 23 26 27 24 29 30 31 20 33 34 37 25 35 38 40 39 41 42 43 44 45 46 47 48 49 50",
"output": "39"
},
{
"input": "5\n4 3 1 2 5",
"output": "3"
},
{
"input": "6\n1 3 5 6 4 2",
"output": "4"
},
{
"input": "10\n2 1 4 3 6 5 8 7 10 9",
"output": "8"
},
{
"input": "5\n1 2 4 5 3",
"output": "2"
},
{
"input": "7\n1 4 2 3 7 6 5",
"output": "4"
},
{
"input": "4\n3 1 2 4",
"output": "2"
},
{
"input": "6\n2 5 4 3 6 1",
"output": "4"
},
{
"input": "5\n1 3 4 5 2",
"output": "2"
},
{
"input": "6\n2 4 6 5 1 3",
"output": "4"
},
{
"input": "6\n1 2 4 5 6 3",
"output": "3"
},
{
"input": "9\n9 8 7 4 5 6 3 2 1",
"output": "6"
},
{
"input": "7\n4 1 2 3 6 5 7",
"output": "4"
}
] | 1,479,502,599 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 62 | 0 | n = int(input())
L = input().split()
L = [10000000000] + L + [0]
def max(a, b):
if a > b:
return a
return b
mx = 0
act = 0
for i in range(1, n + 1):
x, y = int(L[i]), int(L[i+1])
if y > x:
act += 1
mx = max(mx, act)
else:
act = 0
print(n - mx - 1)
| Title: Sorting Railway Cars
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An infinitely long railway has a train consisting of *n* cars, numbered from 1 to *n* (the numbers of all the cars are distinct) and positioned in arbitrary order. David Blaine wants to sort the railway cars in the order of increasing numbers. In one move he can make one of the cars disappear from its place and teleport it either to the beginning of the train, or to the end of the train, at his desire. What is the minimum number of actions David Blaine needs to perform in order to sort the train?
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cars in the train.
The second line contains *n* integers *p**i* (1<=≤<=*p**i*<=≤<=*n*, *p**i*<=≠<=*p**j* if *i*<=≠<=*j*) — the sequence of the numbers of the cars in the train.
Output Specification:
Print a single integer — the minimum number of actions needed to sort the railway cars.
Demo Input:
['5\n4 1 2 5 3\n', '4\n4 1 3 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample you need first to teleport the 4-th car, and then the 5-th car to the end of the train. | ```python
n = int(input())
L = input().split()
L = [10000000000] + L + [0]
def max(a, b):
if a > b:
return a
return b
mx = 0
act = 0
for i in range(1, n + 1):
x, y = int(L[i]), int(L[i+1])
if y > x:
act += 1
mx = max(mx, act)
else:
act = 0
print(n - mx - 1)
``` | 0 | |
938 | A | Word Correction | PROGRAMMING | 800 | [
"implementation"
] | null | null | Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.
Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.
You are given a word *s*. Can you predict what will it become after correction?
In this problem letters a, e, i, o, u and y are considered to be vowels. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of letters in word *s* before the correction.
The second line contains a string *s* consisting of exactly *n* lowercase Latin letters — the word before the correction. | Output the word *s* after the correction. | [
"5\nweird\n",
"4\nword\n",
"5\naaeaa\n"
] | [
"werd\n",
"word\n",
"a\n"
] | Explanations of the examples:
1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a. | 0 | [
{
"input": "5\nweird",
"output": "werd"
},
{
"input": "4\nword",
"output": "word"
},
{
"input": "5\naaeaa",
"output": "a"
},
{
"input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyw",
"output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyw"
},
{
"input": "69\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "12\nmmmmmmmmmmmm",
"output": "mmmmmmmmmmmm"
},
{
"input": "18\nyaywptqwuyiqypwoyw",
"output": "ywptqwuqypwow"
},
{
"input": "85\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "13\nmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmm"
},
{
"input": "10\nmmmmmmmmmm",
"output": "mmmmmmmmmm"
},
{
"input": "11\nmmmmmmmmmmm",
"output": "mmmmmmmmmmm"
},
{
"input": "15\nmmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmmm"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "14\nmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmm"
},
{
"input": "33\nmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm"
},
{
"input": "79\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "90\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "2\naa",
"output": "a"
},
{
"input": "18\niuiuqpyyaoaetiwliu",
"output": "iqpytiwli"
},
{
"input": "5\nxxxxx",
"output": "xxxxx"
},
{
"input": "6\nxxxahg",
"output": "xxxahg"
},
{
"input": "3\nzcv",
"output": "zcv"
},
{
"input": "4\naepo",
"output": "apo"
},
{
"input": "5\nqqqqq",
"output": "qqqqq"
},
{
"input": "6\naaaaaa",
"output": "a"
},
{
"input": "4\naeta",
"output": "ata"
},
{
"input": "20\nttyttlwaoieulyiluuri",
"output": "ttyttlwalyluri"
},
{
"input": "1\nb",
"output": "b"
},
{
"input": "3\nanc",
"output": "anc"
},
{
"input": "1\ne",
"output": "e"
},
{
"input": "3\naie",
"output": "a"
},
{
"input": "3\nvio",
"output": "vi"
},
{
"input": "2\nea",
"output": "e"
},
{
"input": "3\nuas",
"output": "us"
},
{
"input": "2\nba",
"output": "ba"
},
{
"input": "2\naq",
"output": "aq"
},
{
"input": "2\nya",
"output": "y"
},
{
"input": "2\nou",
"output": "o"
},
{
"input": "2\nbb",
"output": "bb"
},
{
"input": "7\nayylmao",
"output": "alma"
},
{
"input": "2\nab",
"output": "ab"
},
{
"input": "19\nyuouiyaoiiweqrryqqp",
"output": "yweqrryqqp"
},
{
"input": "25\niqypwqpriiioetiuqqqttouei",
"output": "iqypwqpritiqqqtto"
},
{
"input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyz",
"output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyz"
},
{
"input": "17\naccccccccccccccca",
"output": "accccccccccccccca"
},
{
"input": "5\nababa",
"output": "ababa"
},
{
"input": "10\naaaaaaaaaa",
"output": "a"
},
{
"input": "22\naaaaabbbbboyoyoyoyoyac",
"output": "abbbbboc"
},
{
"input": "7\nmahmoud",
"output": "mahmod"
}
] | 1,519,400,896 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 5,632,000 | n=int(input())
name=list(input())
vowels=['a','e','o','u','y','i']
for i in range(n-1) :
if name[i] in vowels and name[i+1] in vowels :
name[i]='~'
for i in name :
if i!='~' :
print(i,end='')
| Title: Word Correction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.
Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.
You are given a word *s*. Can you predict what will it become after correction?
In this problem letters a, e, i, o, u and y are considered to be vowels.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of letters in word *s* before the correction.
The second line contains a string *s* consisting of exactly *n* lowercase Latin letters — the word before the correction.
Output Specification:
Output the word *s* after the correction.
Demo Input:
['5\nweird\n', '4\nword\n', '5\naaeaa\n']
Demo Output:
['werd\n', 'word\n', 'a\n']
Note:
Explanations of the examples:
1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a. | ```python
n=int(input())
name=list(input())
vowels=['a','e','o','u','y','i']
for i in range(n-1) :
if name[i] in vowels and name[i+1] in vowels :
name[i]='~'
for i in name :
if i!='~' :
print(i,end='')
``` | 0 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,620,476,871 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 154 | 20,172,800 | import sys
def get_ints():
return map(int, sys.stdin.readline().strip().split())
M,N = get_ints()
if (M % 2 == 0):
ans = N * (M/2)
sys.stdout.write(str(int(ans)))
else:
ans = N * (M/2) + N/2
sys.stdout.write(str(int(ans)))
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
import sys
def get_ints():
return map(int, sys.stdin.readline().strip().split())
M,N = get_ints()
if (M % 2 == 0):
ans = N * (M/2)
sys.stdout.write(str(int(ans)))
else:
ans = N * (M/2) + N/2
sys.stdout.write(str(int(ans)))
``` | 0 |
812 | A | Sagheer and Crossroads | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (*l* — left, *s* — straight, *r* — right) and a light *p* for a pedestrian crossing.
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible. | The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers *l*, *s*, *r*, *p* — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light. | On a single line, print "YES" if an accident is possible, and "NO" otherwise. | [
"1 0 0 1\n0 1 0 0\n0 0 1 0\n0 0 0 1\n",
"0 1 1 0\n1 0 1 0\n1 1 0 0\n0 0 0 1\n",
"1 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 1 0\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur. | 500 | [
{
"input": "1 0 0 1\n0 1 0 0\n0 0 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 1 1 0\n1 0 1 0\n1 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "1 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 1 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "1 1 1 0\n0 1 0 1\n1 1 1 0\n1 1 1 1",
"output": "YES"
},
{
"input": "0 1 1 0\n0 1 0 0\n1 0 0 1\n1 0 0 0",
"output": "YES"
},
{
"input": "1 0 0 0\n0 1 0 0\n1 1 0 0\n0 1 1 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 1\n1 0 1 1\n1 1 1 0",
"output": "YES"
},
{
"input": "1 1 0 0\n0 1 0 1\n1 1 1 0\n0 0 1 1",
"output": "YES"
},
{
"input": "0 1 0 0\n0 0 0 0\n1 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 1 0\n0 0 0 0\n1 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 1 0\n0 1 0 1\n1 0 1 0\n0 0 1 0",
"output": "YES"
},
{
"input": "1 1 1 0\n0 1 0 1\n1 1 1 1\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 1 0\n0 0 0 0\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 1\n1 0 1 1\n0 0 0 1",
"output": "YES"
},
{
"input": "1 1 0 0\n0 1 0 0\n1 1 1 0\n1 0 1 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "1 0 1 0\n1 1 0 0\n1 1 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 1 0\n1 1 0 0\n1 0 1 0\n1 0 0 0",
"output": "NO"
},
{
"input": "0 0 1 0\n1 0 0 0\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "0 1 1 0\n1 1 0 1\n1 0 0 1\n1 1 1 0",
"output": "YES"
},
{
"input": "1 0 0 0\n1 1 0 0\n1 1 0 1\n0 0 1 0",
"output": "YES"
},
{
"input": "0 0 0 0\n1 1 0 0\n0 0 0 1\n0 0 1 0",
"output": "NO"
},
{
"input": "0 1 0 0\n0 0 0 1\n0 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 1 0 0\n1 1 0 1\n1 0 0 1\n1 1 0 1",
"output": "YES"
},
{
"input": "1 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n1 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 1 0 0\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 1 0\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n1 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 1 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 1 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n1 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n0 1 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 1 0",
"output": "YES"
},
{
"input": "1 0 0 0\n0 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 1 0 0\n0 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 1 0\n0 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n1 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 1 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 1 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 1\n1 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 1 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 1 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n0 1 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n0 0 1 0",
"output": "NO"
},
{
"input": "1 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 1 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 1 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n1 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 1 0\n0 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n1 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 1 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 1 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n1 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 1 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 1 0",
"output": "NO"
},
{
"input": "1 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 1 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 1 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n1 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 0\n0 0 0 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 1 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n1 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n1 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 1 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 1 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "1 1 1 1\n1 1 1 1\n1 1 1 1\n1 1 1 1",
"output": "YES"
},
{
"input": "1 0 0 0\n0 1 0 0\n0 0 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "1 1 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "1 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n1 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 1 0\n1 0 1 0\n1 1 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "1 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "1 1 1 0\n1 1 1 0\n1 1 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "1 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 1 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 1 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 1 1 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 1 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 1\n0 0 0 0\n0 1 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 1\n1 0 0 0\n0 0 0 0",
"output": "YES"
}
] | 1,496,333,299 | 6,799 | Python 3 | WRONG_ANSWER | PRETESTS | 0 | 30 | 0 | road1=input()
road2=input()
road3=input()
road4=input()
flag =0
def trafficlights (road):
road=road.split( )
def accident(road):
if road[3]=='1':
for i in range(0,3):
if road[i]=='1':
flag=1
return flag
else:
flag=0
return flag
trafficlights (road1)
trafficlights (road2)
trafficlights (road3)
trafficlights (road4)
flag1=accident (road1)
flag2=accident (road2)
flag3=accident (road3)
flag4=accident (road4)
if flag1==1 or flag2 ==1 or flag3==1 or flag4==1:
print('Yes')
if flag1==0 or flag2 ==0 or flag3==0 or flag4==0:
print('No')
| Title: Sagheer and Crossroads
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (*l* — left, *s* — straight, *r* — right) and a light *p* for a pedestrian crossing.
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input Specification:
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers *l*, *s*, *r*, *p* — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output Specification:
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Demo Input:
['1 0 0 1\n0 1 0 0\n0 0 1 0\n0 0 0 1\n', '0 1 1 0\n1 0 1 0\n1 1 0 0\n0 0 0 1\n', '1 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 1 0\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur. | ```python
road1=input()
road2=input()
road3=input()
road4=input()
flag =0
def trafficlights (road):
road=road.split( )
def accident(road):
if road[3]=='1':
for i in range(0,3):
if road[i]=='1':
flag=1
return flag
else:
flag=0
return flag
trafficlights (road1)
trafficlights (road2)
trafficlights (road3)
trafficlights (road4)
flag1=accident (road1)
flag2=accident (road2)
flag3=accident (road3)
flag4=accident (road4)
if flag1==1 or flag2 ==1 or flag3==1 or flag4==1:
print('Yes')
if flag1==0 or flag2 ==0 or flag3==0 or flag4==0:
print('No')
``` | 0 | |
99 | A | Help Far Away Kingdom | PROGRAMMING | 800 | [
"strings"
] | A. Help Far Away Kingdom | 2 | 256 | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Kingdom healers had already had three appointments with the merchants who were asked to sell, say, exactly 0.273549107 beer barrels. To deal with the problem somehow, the King issued an order obliging rounding up all numbers to the closest integer to simplify calculations. Specifically, the order went like this:
- If a number's integer part does not end with digit 9 and its fractional part is strictly less than 0.5, then the rounded up number coincides with the number’s integer part. - If a number's integer part does not end with digit 9 and its fractional part is not less than 0.5, the rounded up number is obtained if we add 1 to the last digit of the number’s integer part.- If the number’s integer part ends with digit 9, to round up the numbers one should go to Vasilisa the Wise. In the whole Kingdom she is the only one who can perform the tricky operation of carrying into the next position.
Merchants found the algorithm very sophisticated and they asked you (the ACMers) to help them. Can you write a program that would perform the rounding according to the King’s order? | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empty set of decimal digits). The number's length does not exceed 1000 characters, including the dot. There are no other characters in the input data. | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | [
"0.0\n",
"1.49\n",
"1.50\n",
"2.71828182845904523536\n",
"3.14159265358979323846\n",
"12345678901234567890.1\n",
"123456789123456789.999\n"
] | [
"0",
"1",
"2",
"3",
"3",
"12345678901234567890",
"GOTO Vasilisa."
] | none | 500 | [
{
"input": "0.0",
"output": "0"
},
{
"input": "1.49",
"output": "1"
},
{
"input": "1.50",
"output": "2"
},
{
"input": "2.71828182845904523536",
"output": "3"
},
{
"input": "3.14159265358979323846",
"output": "3"
},
{
"input": "12345678901234567890.1",
"output": "12345678901234567890"
},
{
"input": "123456789123456789.999",
"output": "GOTO Vasilisa."
},
{
"input": "12345678901234567890.9",
"output": "12345678901234567891"
},
{
"input": "123456789123456788.999",
"output": "123456789123456789"
},
{
"input": "9.000",
"output": "GOTO Vasilisa."
},
{
"input": "0.1",
"output": "0"
},
{
"input": "0.2",
"output": "0"
},
{
"input": "0.3",
"output": "0"
},
{
"input": "0.4",
"output": "0"
},
{
"input": "0.5",
"output": "1"
},
{
"input": "0.6",
"output": "1"
},
{
"input": "0.7",
"output": "1"
},
{
"input": "0.8",
"output": "1"
},
{
"input": "0.9",
"output": "1"
},
{
"input": "1.0",
"output": "1"
},
{
"input": "1.1",
"output": "1"
},
{
"input": "1.2",
"output": "1"
},
{
"input": "1.3",
"output": "1"
},
{
"input": "1.4",
"output": "1"
},
{
"input": "1.5",
"output": "2"
},
{
"input": "1.6",
"output": "2"
},
{
"input": "1.7",
"output": "2"
},
{
"input": "1.8",
"output": "2"
},
{
"input": "1.9",
"output": "2"
},
{
"input": "2.0",
"output": "2"
},
{
"input": "2.1",
"output": "2"
},
{
"input": "2.2",
"output": "2"
},
{
"input": "2.3",
"output": "2"
},
{
"input": "2.4",
"output": "2"
},
{
"input": "2.5",
"output": "3"
},
{
"input": "2.6",
"output": "3"
},
{
"input": "2.7",
"output": "3"
},
{
"input": "2.8",
"output": "3"
},
{
"input": "2.9",
"output": "3"
},
{
"input": "3.0",
"output": "3"
},
{
"input": "3.1",
"output": "3"
},
{
"input": "3.2",
"output": "3"
},
{
"input": "3.3",
"output": "3"
},
{
"input": "3.4",
"output": "3"
},
{
"input": "3.5",
"output": "4"
},
{
"input": "3.6",
"output": "4"
},
{
"input": "3.7",
"output": "4"
},
{
"input": "3.8",
"output": "4"
},
{
"input": "3.9",
"output": "4"
},
{
"input": "4.0",
"output": "4"
},
{
"input": "4.1",
"output": "4"
},
{
"input": "4.2",
"output": "4"
},
{
"input": "4.3",
"output": "4"
},
{
"input": "4.4",
"output": "4"
},
{
"input": "4.5",
"output": "5"
},
{
"input": "4.6",
"output": "5"
},
{
"input": "4.7",
"output": "5"
},
{
"input": "4.8",
"output": "5"
},
{
"input": "4.9",
"output": "5"
},
{
"input": "5.0",
"output": "5"
},
{
"input": "5.1",
"output": "5"
},
{
"input": "5.2",
"output": "5"
},
{
"input": "5.3",
"output": "5"
},
{
"input": "5.4",
"output": "5"
},
{
"input": "5.5",
"output": "6"
},
{
"input": "5.6",
"output": "6"
},
{
"input": "5.7",
"output": "6"
},
{
"input": "5.8",
"output": "6"
},
{
"input": "5.9",
"output": "6"
},
{
"input": "6.0",
"output": "6"
},
{
"input": "6.1",
"output": "6"
},
{
"input": "6.2",
"output": "6"
},
{
"input": "6.3",
"output": "6"
},
{
"input": "6.4",
"output": "6"
},
{
"input": "6.5",
"output": "7"
},
{
"input": "6.6",
"output": "7"
},
{
"input": "6.7",
"output": "7"
},
{
"input": "6.8",
"output": "7"
},
{
"input": "6.9",
"output": "7"
},
{
"input": "7.0",
"output": "7"
},
{
"input": "7.1",
"output": "7"
},
{
"input": "7.2",
"output": "7"
},
{
"input": "7.3",
"output": "7"
},
{
"input": "7.4",
"output": "7"
},
{
"input": "7.5",
"output": "8"
},
{
"input": "7.6",
"output": "8"
},
{
"input": "7.7",
"output": "8"
},
{
"input": "7.8",
"output": "8"
},
{
"input": "7.9",
"output": "8"
},
{
"input": "8.0",
"output": "8"
},
{
"input": "8.1",
"output": "8"
},
{
"input": "8.2",
"output": "8"
},
{
"input": "8.3",
"output": "8"
},
{
"input": "8.4",
"output": "8"
},
{
"input": "8.5",
"output": "9"
},
{
"input": "8.6",
"output": "9"
},
{
"input": "8.7",
"output": "9"
},
{
"input": "8.8",
"output": "9"
},
{
"input": "8.9",
"output": "9"
},
{
"input": "9.0",
"output": "GOTO Vasilisa."
},
{
"input": "9.1",
"output": "GOTO Vasilisa."
},
{
"input": "9.2",
"output": "GOTO Vasilisa."
},
{
"input": "9.3",
"output": "GOTO Vasilisa."
},
{
"input": "9.4",
"output": "GOTO Vasilisa."
},
{
"input": "9.5",
"output": "GOTO Vasilisa."
},
{
"input": "9.6",
"output": "GOTO Vasilisa."
},
{
"input": "9.7",
"output": "GOTO Vasilisa."
},
{
"input": "9.8",
"output": "GOTO Vasilisa."
},
{
"input": "9.9",
"output": "GOTO Vasilisa."
},
{
"input": "609942239104813108618306232517836377583566292129955473517174437591594761209877970062547641606473593416245554763832875919009472288995880898848455284062760160557686724163817329189799336769669146848904803188614226720978399787805489531837751080926098.1664915772983166314490532653577560222779830866949001942720729759794777105570672781798092416748052690224813237139640723361527601154465287615917169132637313918577673651098507390501962",
"output": "609942239104813108618306232517836377583566292129955473517174437591594761209877970062547641606473593416245554763832875919009472288995880898848455284062760160557686724163817329189799336769669146848904803188614226720978399787805489531837751080926098"
},
{
"input": "7002108534951820589946967018226114921984364117669853212254634761258884835434844673935047882480101006606512119541798298905598015607366335061012709906661245805358900665571472645463994925687210711492820804158354236327017974683658305043146543214454877759341394.20211856263503281388748282682120712214711232598021393495443628276945042110862480888110959179019986486690931930108026302665438087068150666835901617457150158918705186964935221768346957536540345814875615118637945520917367155931078965",
"output": "7002108534951820589946967018226114921984364117669853212254634761258884835434844673935047882480101006606512119541798298905598015607366335061012709906661245805358900665571472645463994925687210711492820804158354236327017974683658305043146543214454877759341394"
},
{
"input": "1950583094879039694852660558765931995628486712128191844305265555887022812284005463780616067.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1950583094879039694852660558765931995628486712128191844305265555887022812284005463780616068"
},
{
"input": "718130341896330596635811874410345440628950330.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "718130341896330596635811874410345440628950331"
},
{
"input": "927925904158088313481229162503626281882161630091489367140850985555900173018122871746924067186432044676083646964286435457446768031295712712803570690846298544912543439221596866052681116386179629036945370280722.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "927925904158088313481229162503626281882161630091489367140850985555900173018122871746924067186432044676083646964286435457446768031295712712803570690846298544912543439221596866052681116386179629036945370280723"
},
{
"input": "68289614863244584294178637364598054554769889.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "GOTO Vasilisa."
},
{
"input": "7536521504744364134984603189602839063535643888645969434165019366202558753840519.4999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999",
"output": "GOTO Vasilisa."
},
{
"input": "646188694587964249318078225173.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "646188694587964249318078225174"
},
{
"input": "7017472758334494297677068672222822550374695787171163207025095950296957261530.50",
"output": "7017472758334494297677068672222822550374695787171163207025095950296957261531"
},
{
"input": "259085737066615534998640212505663524594409165063310128108448186246980628179842202905722595400477937071746695941939306735605849342959111887834258250883469840846714848774368.50000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "259085737066615534998640212505663524594409165063310128108448186246980628179842202905722595400477937071746695941939306735605849342959111887834258250883469840846714848774369"
},
{
"input": "682500858233333594535201113441004740771119672961581796618069185960107115823662126812159957094407454522028503739299.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "GOTO Vasilisa."
},
{
"input": "41203422675619090661099806687619.49999999999999999999999999999999999999999999999999999999999999999999999999999999999999999",
"output": "GOTO Vasilisa."
}
] | 1,601,643,270 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 6 | 280 | 0 | i, j=map(str,input().split("."))
if j[0]>='5':
i = int(i)+1
print(i) | Title: Help Far Away Kingdom
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Kingdom healers had already had three appointments with the merchants who were asked to sell, say, exactly 0.273549107 beer barrels. To deal with the problem somehow, the King issued an order obliging rounding up all numbers to the closest integer to simplify calculations. Specifically, the order went like this:
- If a number's integer part does not end with digit 9 and its fractional part is strictly less than 0.5, then the rounded up number coincides with the number’s integer part. - If a number's integer part does not end with digit 9 and its fractional part is not less than 0.5, the rounded up number is obtained if we add 1 to the last digit of the number’s integer part.- If the number’s integer part ends with digit 9, to round up the numbers one should go to Vasilisa the Wise. In the whole Kingdom she is the only one who can perform the tricky operation of carrying into the next position.
Merchants found the algorithm very sophisticated and they asked you (the ACMers) to help them. Can you write a program that would perform the rounding according to the King’s order?
Input Specification:
The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empty set of decimal digits). The number's length does not exceed 1000 characters, including the dot. There are no other characters in the input data.
Output Specification:
If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes).
Demo Input:
['0.0\n', '1.49\n', '1.50\n', '2.71828182845904523536\n', '3.14159265358979323846\n', '12345678901234567890.1\n', '123456789123456789.999\n']
Demo Output:
['0', '1', '2', '3', '3', '12345678901234567890', 'GOTO Vasilisa.']
Note:
none | ```python
i, j=map(str,input().split("."))
if j[0]>='5':
i = int(i)+1
print(i)
``` | 0 |
710 | A | King Moves | PROGRAMMING | 800 | [
"implementation"
] | null | null | The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)). | The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'. | Print the only integer *x* — the number of moves permitted for the king. | [
"e4\n"
] | [
"8\n"
] | none | 0 | [
{
"input": "e4",
"output": "8"
},
{
"input": "a1",
"output": "3"
},
{
"input": "h8",
"output": "3"
},
{
"input": "a4",
"output": "5"
},
{
"input": "g7",
"output": "8"
},
{
"input": "e1",
"output": "5"
},
{
"input": "b2",
"output": "8"
},
{
"input": "c7",
"output": "8"
},
{
"input": "h6",
"output": "5"
},
{
"input": "c8",
"output": "5"
},
{
"input": "h2",
"output": "5"
},
{
"input": "h5",
"output": "5"
},
{
"input": "a8",
"output": "3"
},
{
"input": "f8",
"output": "5"
},
{
"input": "h1",
"output": "3"
},
{
"input": "f2",
"output": "8"
},
{
"input": "e8",
"output": "5"
},
{
"input": "h3",
"output": "5"
},
{
"input": "b8",
"output": "5"
},
{
"input": "g8",
"output": "5"
},
{
"input": "d8",
"output": "5"
},
{
"input": "h4",
"output": "5"
},
{
"input": "b1",
"output": "5"
},
{
"input": "a2",
"output": "5"
}
] | 1,472,054,203 | 2,147,483,647 | Python 3 | OK | TESTS | 24 | 62 | 4,608,000 | def main():
c, d = input()
ans = 0
if c > 'a' and c < 'h' and d > '1' and d < '8':
ans = 8
elif c > 'a' and c < 'h' and (d == '1' or d == '8'):
ans = 5
elif (c == 'a' or c == 'h') and d > '1' and d < '8':
ans = 5
elif (c == 'a' or c == 'h') and (d == '1' or d == '8'):
ans = 3
print(ans)
if __name__ == '__main__':
main()
| Title: King Moves
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)).
Input Specification:
The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'.
Output Specification:
Print the only integer *x* — the number of moves permitted for the king.
Demo Input:
['e4\n']
Demo Output:
['8\n']
Note:
none | ```python
def main():
c, d = input()
ans = 0
if c > 'a' and c < 'h' and d > '1' and d < '8':
ans = 8
elif c > 'a' and c < 'h' and (d == '1' or d == '8'):
ans = 5
elif (c == 'a' or c == 'h') and d > '1' and d < '8':
ans = 5
elif (c == 'a' or c == 'h') and (d == '1' or d == '8'):
ans = 3
print(ans)
if __name__ == '__main__':
main()
``` | 3 | |
5 | B | Center Alignment | PROGRAMMING | 1,200 | [
"implementation",
"strings"
] | B. Center Alignment | 1 | 64 | Almost every text editor has a built-in function of center text alignment. The developers of the popular in Berland text editor «Textpad» decided to introduce this functionality into the fourth release of the product.
You are to implement the alignment in the shortest possible time. Good luck! | The input file consists of one or more lines, each of the lines contains Latin letters, digits and/or spaces. The lines cannot start or end with a space. It is guaranteed that at least one of the lines has positive length. The length of each line and the total amount of the lines do not exceed 1000. | Format the given text, aligning it center. Frame the whole text with characters «*» of the minimum size. If a line cannot be aligned perfectly (for example, the line has even length, while the width of the block is uneven), you should place such lines rounding down the distance to the left or to the right edge and bringing them closer left or right alternatively (you should start with bringing left). Study the sample tests carefully to understand the output format better. | [
"This is\n\nCodeforces\nBeta\nRound\n5\n",
"welcome to the\nCodeforces\nBeta\nRound 5\n\nand\ngood luck\n"
] | [
"************\n* This is *\n* *\n*Codeforces*\n* Beta *\n* Round *\n* 5 *\n************\n",
"****************\n*welcome to the*\n* Codeforces *\n* Beta *\n* Round 5 *\n* *\n* and *\n* good luck *\n****************\n"
] | none | 0 | [
{
"input": "This is\n\nCodeforces\nBeta\nRound\n5",
"output": "************\n* This is *\n* *\n*Codeforces*\n* Beta *\n* Round *\n* 5 *\n************"
},
{
"input": "welcome to the\nCodeforces\nBeta\nRound 5\n\nand\ngood luck",
"output": "****************\n*welcome to the*\n* Codeforces *\n* Beta *\n* Round 5 *\n* *\n* and *\n* good luck *\n****************"
},
{
"input": "0\n2",
"output": "***\n*0*\n*2*\n***"
},
{
"input": "O\no\nd",
"output": "***\n*O*\n*o*\n*d*\n***"
},
{
"input": "0v uO M6Sy",
"output": "************\n*0v uO M6Sy*\n************"
},
{
"input": "fm v\nOL U W",
"output": "**********\n* fm v *\n*OL U W*\n**********"
},
{
"input": "vb\nJ\nyU\nZ",
"output": "****\n*vb*\n*J *\n*yU*\n* Z*\n****"
},
{
"input": "N\nSV\nEh\n6f\nX6\n9e",
"output": "****\n*N *\n*SV*\n*Eh*\n*6f*\n*X6*\n*9e*\n****"
},
{
"input": "Pj\nA\nFA\nP\nVJ\nU\nEb\nW",
"output": "****\n*Pj*\n*A *\n*FA*\n* P*\n*VJ*\n*U *\n*Eb*\n* W*\n****"
},
{
"input": "T\n7j\nS\nb\nq8\nVZ\nn\n4T\niZ\npA",
"output": "****\n*T *\n*7j*\n* S*\n*b *\n*q8*\n*VZ*\n* n*\n*4T*\n*iZ*\n*pA*\n****"
},
{
"input": "8\n\n\n\ny\nW\n\n\n\n3B\n\nw\nV\n\n\n\nL\nSr\n\n\nV\n\n5\n\nAq\n\n\n\nJ\nR\n\n04\nJ\nv\nhU\n\n\n\nY\nG\n4\n\nG\nb\n\n\n9\n\n6\nd\n\n2\n\n\nE\n7\n\nr\n\n\n\n\nKC\ns\nE\n\nab\n4\nx\n\n\n\n\n\nEe\n4\n\nl\n\np\n\nG\nM\n\n\nn\n\n\nm0\n\nw\n\n\nP\n\n\n\n0",
"output": "****\n*8 *\n* *\n* *\n* *\n* y*\n*W *\n* *\n* *\n* *\n*3B*\n* *\n* w*\n*V *\n* *\n* *\n* *\n* L*\n*Sr*\n* *\n* *\n*V *\n* *\n* 5*\n* *\n*Aq*\n* *\n* *\n* *\n*J *\n* R*\n* *\n*04*\n*J *\n* v*\n*hU*\n* *\n* *\n* *\n*Y *\n* G*\n*4 *\n* *\n* G*\n*b *\n* *\n* *\n* 9*\n* *\n*6 *\n* d*\n* *\n*2 *\n* *\n* *\n* E*\n*7 *\n* *\n* r*\n* *\n* *\n* *\n* *\n*KC*\n*s *\n* E*\n* *\n*ab*\n*4 *\n* x*\n* *\n* *\n* *\n* *\n* *\n*Ee*\n*4 *\n* *\n* l*\n* *\n*p *\n* *\n* G*\n*M *\n* *\n*..."
},
{
"input": "U",
"output": "***\n*U*\n***"
}
] | 1,514,688,437 | 497 | Python 3 | OK | TESTS | 19 | 124 | 6,451,200 | import sys
def main():
lines = [line.strip() for line in sys.stdin]
w = max(len(l) for l in lines) + 2
print('*' * w)
flag = True
for line in lines:
extra = w - 2 - len(line)
if extra % 2 != 0:
if flag:
left = extra // 2
right = extra - left
else:
right = extra // 2
left = extra - right
flag = not flag
else:
left = extra // 2
right = extra // 2
print('*' + ' ' * left + line + ' ' * right + '*')
print('*' * w)
main()
| Title: Center Alignment
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Almost every text editor has a built-in function of center text alignment. The developers of the popular in Berland text editor «Textpad» decided to introduce this functionality into the fourth release of the product.
You are to implement the alignment in the shortest possible time. Good luck!
Input Specification:
The input file consists of one or more lines, each of the lines contains Latin letters, digits and/or spaces. The lines cannot start or end with a space. It is guaranteed that at least one of the lines has positive length. The length of each line and the total amount of the lines do not exceed 1000.
Output Specification:
Format the given text, aligning it center. Frame the whole text with characters «*» of the minimum size. If a line cannot be aligned perfectly (for example, the line has even length, while the width of the block is uneven), you should place such lines rounding down the distance to the left or to the right edge and bringing them closer left or right alternatively (you should start with bringing left). Study the sample tests carefully to understand the output format better.
Demo Input:
['This is\n\nCodeforces\nBeta\nRound\n5\n', 'welcome to the\nCodeforces\nBeta\nRound 5\n\nand\ngood luck\n']
Demo Output:
['************\n* This is *\n* *\n*Codeforces*\n* Beta *\n* Round *\n* 5 *\n************\n', '****************\n*welcome to the*\n* Codeforces *\n* Beta *\n* Round 5 *\n* *\n* and *\n* good luck *\n****************\n']
Note:
none | ```python
import sys
def main():
lines = [line.strip() for line in sys.stdin]
w = max(len(l) for l in lines) + 2
print('*' * w)
flag = True
for line in lines:
extra = w - 2 - len(line)
if extra % 2 != 0:
if flag:
left = extra // 2
right = extra - left
else:
right = extra // 2
left = extra - right
flag = not flag
else:
left = extra // 2
right = extra // 2
print('*' + ' ' * left + line + ' ' * right + '*')
print('*' * w)
main()
``` | 3.889935 |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,679,650,183 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | s=input()
s1=""
for i in range(len(s)-1):
if((s[i]==s[i+1]=='l' and s1.count('l')==1) or s[i]!=s[i+1]):
s1=s1+s[i]
if "hello" in s1:
print("YES")
else:
print("NO") | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
s=input()
s1=""
for i in range(len(s)-1):
if((s[i]==s[i+1]=='l' and s1.count('l')==1) or s[i]!=s[i+1]):
s1=s1+s[i]
if "hello" in s1:
print("YES")
else:
print("NO")
``` | 0 |
729 | A | Interview with Oleg | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking! | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters. | Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences. | [
"7\naogogob\n",
"13\nogogmgogogogo\n",
"9\nogoogoogo\n"
] | [
"a***b\n",
"***gmg***\n",
"*********\n"
] | The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | 500 | [
{
"input": "7\naogogob",
"output": "a***b"
},
{
"input": "13\nogogmgogogogo",
"output": "***gmg***"
},
{
"input": "9\nogoogoogo",
"output": "*********"
},
{
"input": "32\nabcdefogoghijklmnogoopqrstuvwxyz",
"output": "abcdef***ghijklmn***opqrstuvwxyz"
},
{
"input": "100\nggogogoooggogooggoggogggggogoogoggooooggooggoooggogoooggoggoogggoogoggogggoooggoggoggogggogoogggoooo",
"output": "gg***oogg***oggoggoggggg******ggooooggooggooogg***ooggoggoogggo***ggogggoooggoggoggoggg***ogggoooo"
},
{
"input": "10\nogooggoggo",
"output": "***oggoggo"
},
{
"input": "20\nooggooogooogooogooog",
"output": "ooggoo***o***o***oog"
},
{
"input": "30\ngoggogoooggooggggoggoggoogoggo",
"output": "gogg***ooggooggggoggoggo***ggo"
},
{
"input": "40\nogggogooggoogoogggogooogogggoogggooggooo",
"output": "oggg***oggo***oggg***o***gggoogggooggooo"
},
{
"input": "50\noggggogoogggggggoogogggoooggooogoggogooogogggogooo",
"output": "ogggg***ogggggggo***gggoooggoo***gg***o***ggg***oo"
},
{
"input": "60\nggoooogoggogooogogooggoogggggogogogggggogggogooogogogggogooo",
"output": "ggooo***gg***o***oggooggggg***gggggoggg***o***ggg***oo"
},
{
"input": "70\ngogoooggggoggoggggggoggggoogooogogggggooogggogoogoogoggogggoggogoooooo",
"output": "g***ooggggoggoggggggoggggo***o***gggggoooggg*********ggogggogg***ooooo"
},
{
"input": "80\nooogoggoooggogogoggooooogoogogooogoggggogggggogoogggooogooooooggoggoggoggogoooog",
"output": "oo***ggooogg***ggoooo******o***ggggoggggg***ogggoo***oooooggoggoggogg***ooog"
},
{
"input": "90\nooogoggggooogoggggoooogggggooggoggoggooooooogggoggogggooggggoooooogoooogooggoooogggggooooo",
"output": "oo***ggggoo***ggggoooogggggooggoggoggooooooogggoggogggooggggooooo***oo***oggoooogggggooooo"
},
{
"input": "100\ngooogoggooggggoggoggooooggogoogggoogogggoogogoggogogogoggogggggogggggoogggooogogoggoooggogoooooogogg",
"output": "goo***ggooggggoggoggoooogg***ogggo***gggo***gg***ggogggggogggggoogggoo***ggooogg***oooo***gg"
},
{
"input": "100\ngoogoogggogoooooggoogooogoogoogogoooooogooogooggggoogoggogooogogogoogogooooggoggogoooogooooooggogogo",
"output": "go***oggg***ooooggo***o*********oooo***o***oggggo***gg***o******oooggogg***oo***ooooogg***"
},
{
"input": "100\ngoogoggggogggoooggoogoogogooggoggooggggggogogggogogggoogogggoogoggoggogooogogoooogooggggogggogggoooo",
"output": "go***ggggogggoooggo******oggoggoogggggg***ggg***gggo***gggo***ggogg***o***oo***oggggogggogggoooo"
},
{
"input": "100\nogogogogogoggogogogogogogoggogogogoogoggoggooggoggogoogoooogogoogggogogogogogoggogogogogogogogogogoe",
"output": "***gg***gg******ggoggooggogg******oo***oggg***gg***e"
},
{
"input": "5\nogoga",
"output": "***ga"
},
{
"input": "1\no",
"output": "o"
},
{
"input": "100\nogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogog",
"output": "***g"
},
{
"input": "99\nogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo",
"output": "***"
},
{
"input": "5\nggggg",
"output": "ggggg"
},
{
"input": "6\ngoogoo",
"output": "go***o"
},
{
"input": "7\nooogooo",
"output": "oo***oo"
},
{
"input": "8\ngggggggg",
"output": "gggggggg"
},
{
"input": "9\nogggogggg",
"output": "ogggogggg"
},
{
"input": "10\nogogoggogo",
"output": "***gg***"
},
{
"input": "11\noooggooggog",
"output": "oooggooggog"
},
{
"input": "12\nogggooooggog",
"output": "ogggooooggog"
},
{
"input": "13\nogoggogogooog",
"output": "***gg***oog"
},
{
"input": "15\nggooogoooggoggg",
"output": "ggoo***ooggoggg"
},
{
"input": "14\noogooooggooggo",
"output": "o***oooggooggo"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "1\ng",
"output": "g"
},
{
"input": "1\nz",
"output": "z"
},
{
"input": "46\nsdfkogogogogergergogogergerogogogogoergoegogog",
"output": "sdfk***gergerg***gerger***ergoeg***g"
},
{
"input": "86\ngdfggogogodfogogdofgoodfogodgodgfogoogdodhbofgobgbgfogoogdfogogogogogoogggsgfosogodgon",
"output": "gdfgg***df***gdofgoodf***dgodgf***ogdodhbofgobgbgf***ogdf***ogggsgfos***dgon"
},
{
"input": "83\nergegergoogogogerogoerorogroeorgefxcfgzdgawofqpwzxcvbnmsdfkfpweoriqgtogooggsdgogoog",
"output": "ergegergo***ger***erorogroeorgefxcfgzdgawofqpwzxcvbnmsdfkfpweoriqgt***oggsdg***og"
},
{
"input": "2\nog",
"output": "og"
},
{
"input": "4\nogog",
"output": "***g"
},
{
"input": "4\nogoo",
"output": "***o"
},
{
"input": "29\naaaogogogogogogogogogogogobbb",
"output": "aaa***bbb"
},
{
"input": "11\nogogogogogo",
"output": "***"
},
{
"input": "9\nogogogogo",
"output": "***"
},
{
"input": "3\nogo",
"output": "***"
},
{
"input": "15\nogogogogogogogo",
"output": "***"
},
{
"input": "25\nogogogogogogogogogogogogo",
"output": "***"
}
] | 1,479,633,374 | 674 | Python 3 | OK | TESTS | 48 | 62 | 307,200 | ans = ["ogo","ogogo","ogogogo","ogogogogo","ogogogogogo","ogogogogogogo","ogogogogogogogo","ogogogogogogogogo","ogogogogogogogogogo","ogogogogogogogogogogo","ogogogogogogogogogogogo","ogogogogogogogogogogogogo","ogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogog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n = int(input())
s = input()
for i in range(len(ans) - 1, -1, -1):
s = s.replace(ans[i], "***")
print(s) | Title: Interview with Oleg
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters.
Output Specification:
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Demo Input:
['7\naogogob\n', '13\nogogmgogogogo\n', '9\nogoogoogo\n']
Demo Output:
['a***b\n', '***gmg***\n', '*********\n']
Note:
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | ```python
ans = ["ogo","ogogo","ogogogo","ogogogogo","ogogogogogo","ogogogogogogo","ogogogogogogogo","ogogogogogogogogo","ogogogogogogogogogo","ogogogogogogogogogogo","ogogogogogogogogogogogo","ogogogogogogogogogogogogo","ogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo","ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo"]
n = int(input())
s = input()
for i in range(len(ans) - 1, -1, -1):
s = s.replace(ans[i], "***")
print(s)
``` | 3 | |
873 | B | Balanced Substring | PROGRAMMING | 1,500 | [
"dp",
"implementation"
] | null | null | You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of *s*. | The first line contains *n* (1<=≤<=*n*<=≤<=100000) — the number of characters in *s*.
The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*. | If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring. | [
"8\n11010111\n",
"3\n111\n"
] | [
"4\n",
"0\n"
] | In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | 0 | [
{
"input": "8\n11010111",
"output": "4"
},
{
"input": "3\n111",
"output": "0"
},
{
"input": "11\n00001000100",
"output": "2"
},
{
"input": "10\n0100000000",
"output": "2"
},
{
"input": "13\n0001000011010",
"output": "6"
},
{
"input": "14\n00000100101011",
"output": "10"
},
{
"input": "14\n01111101111111",
"output": "2"
},
{
"input": "18\n110010101101111111",
"output": "10"
},
{
"input": "11\n00010000011",
"output": "4"
},
{
"input": "10\n1000010110",
"output": "6"
},
{
"input": "15\n100000100000011",
"output": "4"
},
{
"input": "18\n011010101110111101",
"output": "8"
},
{
"input": "10\n0011011111",
"output": "6"
},
{
"input": "3\n011",
"output": "2"
},
{
"input": "14\n11111111111111",
"output": "0"
},
{
"input": "65\n11001000110001001011011110111100000000101001001010101111000100010",
"output": "48"
},
{
"input": "10\n0011111000",
"output": "10"
},
{
"input": "13\n1110000001110",
"output": "12"
},
{
"input": "3\n110",
"output": "2"
},
{
"input": "4\n1110",
"output": "2"
},
{
"input": "9\n001011001",
"output": "8"
},
{
"input": "2\n10",
"output": "2"
},
{
"input": "2\n01",
"output": "2"
},
{
"input": "12\n110000110000",
"output": "8"
},
{
"input": "3\n001",
"output": "2"
},
{
"input": "14\n11000011000000",
"output": "8"
},
{
"input": "19\n0000011111111100000",
"output": "18"
},
{
"input": "45\n011010001100001111110001011100000001101100111",
"output": "44"
},
{
"input": "18\n000001111111100000",
"output": "16"
},
{
"input": "4\n0101",
"output": "4"
},
{
"input": "12\n000101011001",
"output": "10"
}
] | 1,570,720,198 | 2,147,483,647 | PyPy 3 | OK | TESTS | 53 | 187 | 10,444,800 | n = int(input())
s = input()
from collections import defaultdict
hash = {0:-1}
b = 0
a = 0
for j,i in enumerate(s):
b+=2*int(i)-1
if b in hash:
a = max(a,j-hash[b])
else:
hash[b] = j
print(a)
| Title: Balanced Substring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of *s*.
Input Specification:
The first line contains *n* (1<=≤<=*n*<=≤<=100000) — the number of characters in *s*.
The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*.
Output Specification:
If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring.
Demo Input:
['8\n11010111\n', '3\n111\n']
Demo Output:
['4\n', '0\n']
Note:
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | ```python
n = int(input())
s = input()
from collections import defaultdict
hash = {0:-1}
b = 0
a = 0
for j,i in enumerate(s):
b+=2*int(i)-1
if b in hash:
a = max(a,j-hash[b])
else:
hash[b] = j
print(a)
``` | 3 | |
459 | B | Pashmak and Flowers | PROGRAMMING | 1,300 | [
"combinatorics",
"implementation",
"sortings"
] | null | null | Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way. | The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109). | The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively. | [
"2\n1 2\n",
"3\n1 4 5\n",
"5\n3 1 2 3 1\n"
] | [
"1 1",
"4 1",
"2 4"
] | In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers. | 500 | [
{
"input": "2\n1 2",
"output": "1 1"
},
{
"input": "3\n1 4 5",
"output": "4 1"
},
{
"input": "5\n3 1 2 3 1",
"output": "2 4"
},
{
"input": "2\n1 1",
"output": "0 1"
},
{
"input": "3\n1 1 1",
"output": "0 3"
},
{
"input": "4\n1 1 1 1",
"output": "0 6"
},
{
"input": "5\n1 1 1 1 1",
"output": "0 10"
},
{
"input": "5\n2 2 2 2 2",
"output": "0 10"
},
{
"input": "10\n2 2 2 2 2 2 2 2 2 2",
"output": "0 45"
},
{
"input": "3\n2 2 2",
"output": "0 3"
},
{
"input": "3\n3 3 3",
"output": "0 3"
},
{
"input": "2\n10000000 100000000",
"output": "90000000 1"
},
{
"input": "5\n5 5 5 5 5",
"output": "0 10"
},
{
"input": "5\n3 3 3 3 3",
"output": "0 10"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "0 15"
},
{
"input": "2\n5 6",
"output": "1 1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "0 45"
},
{
"input": "10\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "0 45"
},
{
"input": "4\n4 4 4 4",
"output": "0 6"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "0 21"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "0 55"
},
{
"input": "8\n8 8 8 8 8 8 8 8",
"output": "0 28"
},
{
"input": "2\n3 2",
"output": "1 1"
}
] | 1,656,167,612 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 108 | 14,745,600 | n = int(input())
list1 = list(map(int, input().split()))
maxi = max(list1)
maxiCount = list1.count(maxi)
mini = min(list1)
miniCount = list1.count(mini)
diff = maxi - mini
if maxi == mini:
a = n * (n-1) / 2
else:
a = maxiCount * miniCount
print(diff, a)
| Title: Pashmak and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109).
Output Specification:
The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.
Demo Input:
['2\n1 2\n', '3\n1 4 5\n', '5\n3 1 2 3 1\n']
Demo Output:
['1 1', '4 1', '2 4']
Note:
In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers. | ```python
n = int(input())
list1 = list(map(int, input().split()))
maxi = max(list1)
maxiCount = list1.count(maxi)
mini = min(list1)
miniCount = list1.count(mini)
diff = maxi - mini
if maxi == mini:
a = n * (n-1) / 2
else:
a = maxiCount * miniCount
print(diff, a)
``` | 0 | |
522 | A | Reposts | PROGRAMMING | 1,200 | [
"*special",
"dfs and similar",
"dp",
"graphs",
"trees"
] | null | null | One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user. | Print a single integer — the maximum length of a repost chain. | [
"5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n",
"6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n",
"1\nSoMeStRaNgEgUe reposted PoLyCaRp\n"
] | [
"6\n",
"2\n",
"2\n"
] | none | 500 | [
{
"input": "5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya",
"output": "6"
},
{
"input": "6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp",
"output": "2"
},
{
"input": "1\nSoMeStRaNgEgUe reposted PoLyCaRp",
"output": "2"
},
{
"input": "1\niuNtwVf reposted POlYcarP",
"output": "2"
},
{
"input": "10\ncs reposted poLYCaRp\nAFIkDrY7Of4V7Mq reposted CS\nsoBiwyN7KOvoFUfbhux reposted aFikDry7Of4v7MQ\nvb6LbwA reposted sObIWYN7KOvoFufBHUx\nDtWKIcVwIHgj4Rcv reposted vb6lbwa\nkt reposted DTwKicvwihgJ4rCV\n75K reposted kT\njKzyxx1 reposted 75K\nuoS reposted jkZyXX1\npZJskHTCIqE3YyZ5ME reposted uoS",
"output": "11"
},
{
"input": "10\nvxrUpCXvx8Isq reposted pOLYcaRP\nICb1 reposted vXRUpCxvX8ISq\nJFMt4b8jZE7iF2m8by7y2 reposted Icb1\nqkG6ZkMIf9QRrBFQU reposted ICb1\nnawsNfcR2palIMnmKZ reposted pOlYcaRP\nKksyH reposted jFMT4b8JzE7If2M8by7y2\nwJtWwQS5FvzN0h8CxrYyL reposted NawsNfcR2paLIMnmKz\nDpBcBPYAcTXEdhldI6tPl reposted NaWSnFCr2pALiMnmkZ\nlEnwTVnlwdQg2vaIRQry reposted kKSYh\nQUVFgwllaWO reposted Wjtwwqs5FVzN0H8cxRyyl",
"output": "6"
},
{
"input": "10\nkkuLGEiHv reposted POLYcArp\n3oX1AoUqyw1eR3nCADY9hLwd reposted kkuLGeIHV\nwf97dqq5bx1dPIchCoT reposted 3OX1AOuQYW1eR3ncAdY9hLwD\nWANr8h reposted Wf97dQQ5bx1dpIcHcoT\n3Fb736lkljZK2LtSbfL reposted wANR8h\n6nq9xLOn reposted 3fB736lKlJZk2LtSbFL\nWL reposted 3Fb736lKLjZk2LTSbfl\ndvxn4Xtc6SBcvKf1 reposted wF97DQq5bX1dPiChCOt\nMCcPLIMISqxDzrj reposted 6nQ9XLOn\nxsQL4Z2Iu reposted MCcpLiMiSqxdzrj",
"output": "9"
},
{
"input": "10\nsMA4 reposted pOLyCARP\nlq3 reposted pOlycARp\nEa16LSFTQxLJnE reposted polYcARp\nkvZVZhJwXcWsnC7NA1DV2WvS reposted polYCArp\nEYqqlrjRwddI reposted pOlyCArP\nsPqQCA67Y6PBBbcaV3EhooO reposted ea16LSFTqxLJne\njjPnneZdF6WLZ3v reposted Ea16LSFTqxLjNe\nWEoi6UpnfBUx79 reposted ea16LSFtqXljNe\nqi4yra reposted eYqqlRJrWDDI\ncw7E1UCSUD reposted eYqqLRJRwDdI",
"output": "3"
}
] | 1,564,584,776 | 2,147,483,647 | PyPy 3 | OK | TESTS | 36 | 140 | 0 | n = int(input())
mp = {
"polycarp" : 1
}
mx = 1
for i in range(n) :
d , r , s = input().split();
mp[d.lower()] = mp[s.lower()] + 1
mx = max(mx , mp[d.lower()])
print(mx) | Title: Reposts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user.
Output Specification:
Print a single integer — the maximum length of a repost chain.
Demo Input:
['5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n', '6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n', '1\nSoMeStRaNgEgUe reposted PoLyCaRp\n']
Demo Output:
['6\n', '2\n', '2\n']
Note:
none | ```python
n = int(input())
mp = {
"polycarp" : 1
}
mx = 1
for i in range(n) :
d , r , s = input().split();
mp[d.lower()] = mp[s.lower()] + 1
mx = max(mx , mp[d.lower()])
print(mx)
``` | 3 | |
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,613,521,666 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 154 | 0 | s = list(input())
t = list(input())
t_rev = t.copy()
t_rev.reverse()
#print(s)
#print(t_rev)
if str(s) == str(t_rev):
print('YES')
else:
print('NO')
| Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
s = list(input())
t = list(input())
t_rev = t.copy()
t_rev.reverse()
#print(s)
#print(t_rev)
if str(s) == str(t_rev):
print('YES')
else:
print('NO')
``` | 3.9615 |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
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{
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{
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{
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{
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{
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{
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},
{
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"output": "01000111100111001011110010100011111111110010101100001101"
},
{
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"output": "110001010001000011000101110101000100001011111001011001001"
},
{
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{
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},
{
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"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
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{
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{
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},
{
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},
{
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},
{
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},
{
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"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
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"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
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"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
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"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
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"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
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"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
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"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
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"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
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"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
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"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
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"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
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{
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},
{
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{
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{
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},
{
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{
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{
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},
{
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"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
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"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
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},
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},
{
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"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
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"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,673,268,362 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 46 | 0 | s=str(input())
s2=str(input())
s1=''
for i in range(len(s)):
if(s[i]==s2[i]):
s1=s1+'0'
else:
s1=s1+'1'
print(s1) | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
s=str(input())
s2=str(input())
s1=''
for i in range(len(s)):
if(s[i]==s2[i]):
s1=s1+'0'
else:
s1=s1+'1'
print(s1)
``` | 3.9885 |
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,650,273,263 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | мa=input()
a2=input()
b=len(a)
b1=len(a2)
b2=b
m=0
k=0
c=-1
n=0
for i in range(b):
while b>0:
if a[n]==a2[c]:
m=m+1
else:
k=k+1
b=b-1
n=n+1
c=c-1
if b2==m:
print('YES')
elif b1>b or b>b1 or k>0:
print('NO') | Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
мa=input()
a2=input()
b=len(a)
b1=len(a2)
b2=b
m=0
k=0
c=-1
n=0
for i in range(b):
while b>0:
if a[n]==a2[c]:
m=m+1
else:
k=k+1
b=b-1
n=n+1
c=c-1
if b2==m:
print('YES')
elif b1>b or b>b1 or k>0:
print('NO')
``` | -1 |
0 | none | none | none | 0 | [
"none"
] | null | null | ...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with *n* vertices and *m* edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops. | The first line contains two integers — the number of vertices *n* and the number of edges *m* of the graph (1<=≤<=*n*<=≤<=100, 0<=≤<=*m*<=≤<=).
Each of the following *m* lines contains a pair of integers *x* and *y*, that show that an edge exists between vertices *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*,<=*x*<=≠<=*y*). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself. | Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is. | [
"6 6\n6 3\n6 4\n5 1\n2 5\n1 4\n5 4\n",
"6 5\n5 6\n4 6\n3 1\n5 1\n1 2\n"
] | [
"FHTAGN!",
"NO"
] | Let us denote as a simple cycle a set of *v* vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., *v* - 1 and *v*, *v* and 1.
A tree is a connected undirected graph consisting of *n* vertices and *n* - 1 edges (*n* > 0).
A rooted tree is a tree where one vertex is selected to be the root. | 0 | [
{
"input": "6 6\n6 3\n6 4\n5 1\n2 5\n1 4\n5 4",
"output": "FHTAGN!"
},
{
"input": "6 5\n5 6\n4 6\n3 1\n5 1\n1 2",
"output": "NO"
},
{
"input": "10 10\n4 10\n8 5\n2 8\n4 9\n9 3\n2 7\n10 6\n10 2\n9 8\n1 8",
"output": "FHTAGN!"
},
{
"input": "5 4\n1 5\n1 3\n1 4\n3 2",
"output": "NO"
},
{
"input": "12 12\n4 12\n4 7\n4 9\n7 2\n5 12\n2 1\n5 9\n8 6\n10 12\n2 5\n10 9\n12 3",
"output": "NO"
},
{
"input": "12 15\n3 2\n11 12\n1 9\n2 1\n1 8\n9 6\n11 5\n9 5\n9 10\n11 3\n7 11\n5 6\n11 10\n4 6\n4 2",
"output": "NO"
},
{
"input": "12 10\n1 11\n3 6\n5 7\n4 7\n6 8\n11 7\n3 12\n11 12\n7 9\n12 2",
"output": "NO"
},
{
"input": "1 0",
"output": "NO"
},
{
"input": "2 1\n1 2",
"output": "NO"
},
{
"input": "3 1\n1 3",
"output": "NO"
},
{
"input": "3 2\n1 2\n2 3",
"output": "NO"
},
{
"input": "3 3\n1 2\n2 3\n3 1",
"output": "FHTAGN!"
},
{
"input": "4 4\n1 2\n3 4\n4 1\n2 4",
"output": "FHTAGN!"
},
{
"input": "6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4",
"output": "NO"
},
{
"input": "2 0",
"output": "NO"
},
{
"input": "3 0",
"output": "NO"
},
{
"input": "100 0",
"output": "NO"
},
{
"input": "100 1\n11 23",
"output": "NO"
},
{
"input": "10 10\n5 7\n8 1\n10 3\n6 4\n10 6\n5 3\n5 6\n2 6\n4 3\n2 10",
"output": "NO"
},
{
"input": "20 20\n9 10\n4 19\n9 20\n12 20\n1 15\n2 12\n19 10\n19 15\n4 10\n4 8\n8 9\n20 8\n6 2\n2 15\n7 19\n20 4\n3 16\n1 20\n9 1\n20 10",
"output": "NO"
},
{
"input": "30 30\n17 6\n16 29\n16 13\n16 20\n29 26\n17 5\n27 28\n24 16\n7 18\n24 10\n1 27\n12 17\n27 30\n6 1\n3 30\n5 19\n18 13\n16 2\n30 1\n5 8\n14 16\n26 18\n7 19\n5 6\n23 14\n6 8\n23 8\n18 8\n18 3\n5 21",
"output": "NO"
},
{
"input": "100 66\n41 14\n19 13\n70 43\n79 62\n9 62\n71 40\n53 86\n80 4\n34 33\n72 68\n40 96\n84 59\n36 77\n55 50\n40 3\n79 81\n3 43\n33 47\n22 98\n33 90\n56 49\n69 28\n73 30\n65 22\n98 20\n9 52\n54 20\n32 70\n51 80\n63 12\n21 48\n35 17\n48 87\n25 43\n65 80\n42 3\n86 35\n95 98\n43 59\n51 46\n66 37\n88 34\n32 47\n24 42\n21 44\n92 59\n81 6\n100 82\n85 6\n58 25\n66 6\n14 32\n59 85\n3 98\n44 4\n85 51\n69 41\n80 70\n81 24\n75 71\n93 9\n82 55\n70 46\n66 32\n77 58\n11 46",
"output": "NO"
},
{
"input": "4 4\n1 2\n4 3\n2 3\n3 1",
"output": "FHTAGN!"
},
{
"input": "5 5\n2 3\n2 4\n5 4\n4 1\n1 2",
"output": "FHTAGN!"
},
{
"input": "10 10\n1 10\n5 9\n6 2\n8 9\n9 1\n5 4\n2 8\n1 3\n6 3\n4 1",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4",
"output": "NO"
},
{
"input": "4 3\n1 2\n2 3\n3 1",
"output": "NO"
},
{
"input": "6 5\n1 2\n2 3\n3 1\n1 4\n1 5",
"output": "NO"
}
] | 1,544,706,345 | 2,147,483,647 | Python 3 | OK | TESTS | 47 | 248 | 0 | n,m=map(int,input().split())
t=[]
l=[]
l.append(1)
def hambandi(q):
for k in t[q]:
if(not(k in l)):
l.append(k)
hambandi(k)
if(n!=m):
print('NO')
else:
for i in range(n+1):
t.append([])
for kl in range(m):
a,b=map(int,input().split())
t[a].append(b)
t[b].append(a)
hambandi(1)
if(len(l)==n):
print('FHTAGN!')
else:
print('NO')
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with *n* vertices and *m* edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
Input Specification:
The first line contains two integers — the number of vertices *n* and the number of edges *m* of the graph (1<=≤<=*n*<=≤<=100, 0<=≤<=*m*<=≤<=).
Each of the following *m* lines contains a pair of integers *x* and *y*, that show that an edge exists between vertices *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*,<=*x*<=≠<=*y*). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output Specification:
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Demo Input:
['6 6\n6 3\n6 4\n5 1\n2 5\n1 4\n5 4\n', '6 5\n5 6\n4 6\n3 1\n5 1\n1 2\n']
Demo Output:
['FHTAGN!', 'NO']
Note:
Let us denote as a simple cycle a set of *v* vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., *v* - 1 and *v*, *v* and 1.
A tree is a connected undirected graph consisting of *n* vertices and *n* - 1 edges (*n* > 0).
A rooted tree is a tree where one vertex is selected to be the root. | ```python
n,m=map(int,input().split())
t=[]
l=[]
l.append(1)
def hambandi(q):
for k in t[q]:
if(not(k in l)):
l.append(k)
hambandi(k)
if(n!=m):
print('NO')
else:
for i in range(n+1):
t.append([])
for kl in range(m):
a,b=map(int,input().split())
t[a].append(b)
t[b].append(a)
hambandi(1)
if(len(l)==n):
print('FHTAGN!')
else:
print('NO')
``` | 3 | |
427 | A | Police Recruits | PROGRAMMING | 800 | [
"implementation"
] | null | null | The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated. | The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time. | Print a single integer, the number of crimes which will go untreated. | [
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] | [
"2\n",
"1\n",
"8\n"
] | Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated. | 500 | [
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 -1 2 2 1 6 5 -1 5",
"output": "5"
},
{
"input": "98\n-1 -1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -1 1 1 1 -1 1 1 -1 -1 -1 1 1 1 -1 -1 -1 1 -1 1 1",
"output": "13"
},
{
"input": "3\n-1 5 4",
"output": "1"
},
{
"input": "146\n4 -1 -1 -1 -1 -1 -1 -1 -1 -1 4 -1 3 -1 3 -1 -1 1 4 -1 2 -1 -1 3 -1 -1 -1 4 1 -1 -1 -1 4 -1 -1 -1 -1 -1 -1 3 2 -1 3 2 4 5 2 4 1 5 -1 -1 2 -1 -1 1 -1 5 3 -1 1 2 2 3 5 3 -1 -1 3 -1 -1 3 5 5 -1 -1 5 -1 4 4 1 -1 -1 -1 2 1 -1 -1 -1 2 5 3 -1 -1 -1 3 -1 5 4 -1 1 -1 -1 3 -1 -1 3 1 1 2 -1 -1 -1 1 3 1 -1 2 -1 -1 5 5 -1 -1 3 4 5 1 -1 2 -1 -1 -1 3 -1 5 3 2 -1 2 -1 -1 5 -1 3 -1",
"output": "5"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n1 -1",
"output": "0"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n-1 -1",
"output": "2"
},
{
"input": "4\n10 -1 -1 -1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "4\n-1 -1 1 1",
"output": "2"
}
] | 1,687,578,602 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 62 | 8,704,000 | n = int(input())
ev = map( int , input().split())
pm = 0
count = 0
for e in ev:
if e == -1 :
if pm > 0 :
pm -= 1
else :
count += 1
else :
pm += e
print(count) | Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated. | ```python
n = int(input())
ev = map( int , input().split())
pm = 0
count = 0
for e in ev:
if e == -1 :
if pm > 0 :
pm -= 1
else :
count += 1
else :
pm += e
print(count)
``` | 3 | |
895 | A | Pizza Separation | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360. | Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya. | [
"4\n90 90 90 90\n",
"3\n100 100 160\n",
"1\n360\n",
"4\n170 30 150 10\n"
] | [
"0\n",
"40\n",
"360\n",
"0\n"
] | In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | 500 | [
{
"input": "4\n90 90 90 90",
"output": "0"
},
{
"input": "3\n100 100 160",
"output": "40"
},
{
"input": "1\n360",
"output": "360"
},
{
"input": "4\n170 30 150 10",
"output": "0"
},
{
"input": "5\n10 10 10 10 320",
"output": "280"
},
{
"input": "8\n45 45 45 45 45 45 45 45",
"output": "0"
},
{
"input": "3\n120 120 120",
"output": "120"
},
{
"input": "5\n110 90 70 50 40",
"output": "40"
},
{
"input": "2\n170 190",
"output": "20"
},
{
"input": "15\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 10",
"output": "10"
},
{
"input": "5\n30 60 180 60 30",
"output": "0"
},
{
"input": "2\n359 1",
"output": "358"
},
{
"input": "5\n100 100 30 100 30",
"output": "40"
},
{
"input": "5\n36 34 35 11 244",
"output": "128"
},
{
"input": "5\n96 94 95 71 4",
"output": "18"
},
{
"input": "2\n85 275",
"output": "190"
},
{
"input": "3\n281 67 12",
"output": "202"
},
{
"input": "5\n211 113 25 9 2",
"output": "62"
},
{
"input": "13\n286 58 6 1 1 1 1 1 1 1 1 1 1",
"output": "212"
},
{
"input": "15\n172 69 41 67 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "20\n226 96 2 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "92"
},
{
"input": "50\n148 53 32 11 4 56 8 2 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "3\n1 1 358",
"output": "356"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 341",
"output": "322"
},
{
"input": "33\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 328",
"output": "296"
},
{
"input": "70\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 291",
"output": "222"
},
{
"input": "130\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 231",
"output": "102"
},
{
"input": "200\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 161",
"output": "0"
},
{
"input": "222\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 139",
"output": "0"
},
{
"input": "10\n8 3 11 4 1 10 10 1 8 304",
"output": "248"
},
{
"input": "12\n8 7 7 3 11 2 10 1 10 8 10 283",
"output": "206"
},
{
"input": "13\n10 8 9 10 5 9 4 1 10 11 1 7 275",
"output": "190"
},
{
"input": "14\n1 6 3 11 9 5 9 8 5 6 7 3 7 280",
"output": "200"
},
{
"input": "15\n10 11 5 4 11 5 4 1 5 4 5 5 9 6 275",
"output": "190"
},
{
"input": "30\n8 7 5 8 3 7 2 4 3 8 11 3 9 11 2 4 1 4 5 6 11 5 8 3 6 3 11 2 11 189",
"output": "18"
},
{
"input": "70\n5 3 6 8 9 2 8 9 11 5 2 8 9 11 7 6 6 9 7 11 7 6 3 8 2 4 4 8 4 3 2 2 3 5 6 5 11 2 7 7 5 8 10 5 2 1 10 9 4 10 7 1 8 10 9 1 5 1 1 1 2 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "29\n2 10 1 5 7 2 9 11 9 9 10 8 4 11 2 5 4 1 4 9 6 10 8 3 1 3 8 9 189",
"output": "18"
},
{
"input": "35\n3 4 11 4 4 2 3 4 3 9 7 10 2 7 8 3 11 3 6 4 6 7 11 10 8 7 6 7 2 8 5 3 2 2 168",
"output": "0"
},
{
"input": "60\n4 10 3 10 6 3 11 8 11 9 3 5 9 2 6 5 6 9 4 10 1 1 3 7 2 10 5 5 3 10 5 2 1 2 9 11 11 9 11 4 11 7 5 6 10 9 3 4 7 8 7 3 6 7 8 5 1 1 1 5",
"output": "0"
},
{
"input": "71\n3 11 8 1 10 1 7 9 6 4 11 10 11 2 4 1 11 7 9 10 11 4 8 7 11 3 8 4 1 8 4 2 9 9 7 10 10 9 5 7 9 7 2 1 7 6 5 11 5 9 4 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "63\n2 11 5 8 7 9 9 8 10 5 9 10 11 8 10 2 3 5 3 7 5 10 2 9 4 8 1 8 5 9 7 7 1 8 7 7 9 10 10 10 8 7 7 2 2 8 9 7 10 8 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "81\n5 8 7 11 2 7 1 1 5 8 7 2 3 11 4 9 7 6 4 4 2 1 1 7 9 4 1 8 3 1 4 10 7 9 9 8 11 3 4 3 10 8 6 4 7 2 4 3 6 11 11 10 7 10 2 10 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "47\n5 3 7 4 2 7 8 1 9 10 5 11 10 7 7 5 1 3 2 11 3 8 6 1 6 10 8 3 2 10 5 6 8 6 9 7 10 9 7 4 8 11 10 1 5 11 68",
"output": "0"
},
{
"input": "100\n5 8 9 3 2 3 9 8 11 10 4 8 1 1 1 1 6 5 10 9 5 3 7 7 2 11 10 2 3 2 2 8 7 3 5 5 10 9 2 5 10 6 7 7 4 7 7 8 2 8 9 9 2 4 1 1 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "120\n9 11 3 7 3 7 9 1 10 7 11 4 1 5 3 5 6 3 1 11 8 8 11 7 3 5 1 9 1 7 10 10 10 10 9 5 4 8 2 8 2 1 4 5 3 11 3 5 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "200\n7 7 9 8 2 8 5 8 3 9 7 10 2 9 11 8 11 7 5 2 6 3 11 9 5 1 10 2 1 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "220\n3 2 8 1 3 5 5 11 1 5 2 6 9 2 2 6 8 10 7 1 3 2 10 9 10 10 4 10 9 5 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "6\n27 15 28 34 41 215",
"output": "70"
},
{
"input": "7\n41 38 41 31 22 41 146",
"output": "14"
},
{
"input": "8\n24 27 34 23 29 23 30 170",
"output": "20"
},
{
"input": "9\n11 11 20 20 33 32 35 26 172",
"output": "6"
},
{
"input": "10\n36 13 28 13 33 34 23 25 34 121",
"output": "0"
},
{
"input": "11\n19 37 13 41 37 15 32 12 19 35 100",
"output": "10"
},
{
"input": "12\n37 25 34 38 21 24 34 38 11 29 28 41",
"output": "2"
},
{
"input": "13\n24 40 20 26 25 29 39 29 35 28 19 18 28",
"output": "2"
},
{
"input": "14\n11 21 40 19 28 34 13 16 23 30 34 22 25 44",
"output": "4"
},
{
"input": "3\n95 91 174",
"output": "12"
},
{
"input": "4\n82 75 78 125",
"output": "46"
},
{
"input": "6\n87 75 88 94 15 1",
"output": "4"
},
{
"input": "10\n27 52 58 64 45 64 1 19 2 28",
"output": "12"
},
{
"input": "50\n14 12 11 8 1 6 11 6 7 8 4 11 4 5 7 3 5 4 7 24 10 2 3 4 6 13 2 1 8 7 5 13 10 8 5 20 1 2 23 7 14 3 4 4 2 8 8 2 6 1",
"output": "0"
},
{
"input": "100\n3 3 4 3 3 6 3 2 8 2 13 3 1 1 2 1 3 4 1 7 1 2 2 6 3 2 10 3 1 2 5 6 2 3 3 2 3 11 8 3 2 6 1 3 3 4 7 7 2 2 1 2 6 3 3 2 3 1 3 8 2 6 4 2 1 12 2 2 2 1 4 1 4 1 3 1 3 1 5 2 6 6 7 1 2 3 2 4 4 2 5 9 8 2 4 6 5 1 1 3",
"output": "0"
},
{
"input": "150\n1 5 1 2 2 2 1 4 2 2 2 3 1 2 1 2 2 2 2 1 2 2 2 1 5 3 4 1 3 4 5 2 4 2 1 2 2 1 1 2 3 2 4 2 2 3 3 1 1 5 2 3 2 1 9 2 1 1 2 1 4 1 1 3 2 2 2 1 2 2 2 1 3 3 4 2 2 1 3 3 3 1 4 3 4 1 2 2 1 1 1 2 2 5 4 1 1 1 2 1 2 3 2 2 6 3 3 3 1 2 1 1 2 8 2 2 4 3 4 5 3 1 4 2 2 2 2 1 4 4 1 1 2 2 4 9 6 3 1 1 2 1 3 4 1 3 2 2 2 1",
"output": "0"
},
{
"input": "200\n1 2 1 3 1 3 1 2 1 4 6 1 2 2 2 2 1 1 1 1 3 2 1 2 2 2 1 2 2 2 2 1 1 1 3 2 3 1 1 2 1 1 2 1 1 1 1 1 1 2 1 2 2 4 1 3 1 2 1 2 2 1 2 1 3 1 1 2 2 1 1 1 1 2 4 1 2 1 1 1 2 1 3 1 1 3 1 2 2 4 1 1 2 1 2 1 2 2 2 2 1 1 2 1 2 1 3 3 1 1 1 2 1 3 3 1 2 1 3 1 3 3 1 2 2 1 4 1 2 2 1 2 2 4 2 5 1 2 2 1 2 1 2 1 5 2 1 2 2 1 2 4 1 2 2 4 2 3 2 3 1 2 1 1 2 2 2 1 1 2 1 4 1 2 1 1 2 1 2 3 1 1 1 2 2 3 1 3 2 2 3 1 2 1 2 1 1 2 1 2",
"output": "0"
},
{
"input": "5\n35 80 45 100 100",
"output": "40"
},
{
"input": "4\n90 179 90 1",
"output": "2"
},
{
"input": "5\n50 50 20 160 80",
"output": "0"
},
{
"input": "5\n30 175 30 5 120",
"output": "10"
},
{
"input": "4\n170 30 10 150",
"output": "20"
},
{
"input": "6\n90 30 90 30 90 30",
"output": "60"
},
{
"input": "4\n70 80 110 100",
"output": "20"
},
{
"input": "7\n35 45 70 100 10 10 90",
"output": "0"
},
{
"input": "6\n50 90 10 90 20 100",
"output": "20"
},
{
"input": "6\n10 155 162 1 26 6",
"output": "18"
},
{
"input": "7\n80 90 80 45 10 10 45",
"output": "20"
},
{
"input": "4\n18 36 162 144",
"output": "36"
},
{
"input": "5\n20 50 50 160 80",
"output": "40"
},
{
"input": "5\n10 30 140 20 160",
"output": "0"
},
{
"input": "6\n90 80 60 50 40 40",
"output": "20"
},
{
"input": "9\n40 20 20 20 20 20 20 40 160",
"output": "40"
},
{
"input": "4\n90 54 90 126",
"output": "72"
},
{
"input": "4\n150 170 30 10",
"output": "20"
},
{
"input": "8\n130 12 13 85 41 67 5 7",
"output": "26"
},
{
"input": "7\n70 170 20 10 30 30 30",
"output": "20"
},
{
"input": "8\n100 100 50 50 15 15 15 15",
"output": "40"
},
{
"input": "4\n100 70 80 110",
"output": "20"
},
{
"input": "5\n160 130 40 20 10",
"output": "20"
},
{
"input": "4\n20 149 151 40",
"output": "22"
},
{
"input": "4\n100 10 100 150",
"output": "60"
},
{
"input": "6\n19 64 105 168 1 3",
"output": "16"
},
{
"input": "8\n10 10 70 70 90 90 10 10",
"output": "0"
}
] | 1,511,714,201 | 1,901 | Python 3 | WRONG_ANSWER | PRETESTS | 7 | 186 | 11,161,600 | import sys
import itertools
sys.setrecursionlimit(100000000)
# def input():
# return sys.stdin.readline()[:-1]
N = int(input())
arr = [int(x) for x in input().split()]
result = 360
for i in range(1, N+1):
temp = list(itertools.permutations(arr, i))
for j in range(len(temp)):
vasya = sum(temp[j])
petya = sum(arr) - vasya
check = abs(vasya - petya)
result = min(result, check)
print(result)
| Title: Pizza Separation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360.
Output Specification:
Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya.
Demo Input:
['4\n90 90 90 90\n', '3\n100 100 160\n', '1\n360\n', '4\n170 30 150 10\n']
Demo Output:
['0\n', '40\n', '360\n', '0\n']
Note:
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | ```python
import sys
import itertools
sys.setrecursionlimit(100000000)
# def input():
# return sys.stdin.readline()[:-1]
N = int(input())
arr = [int(x) for x in input().split()]
result = 360
for i in range(1, N+1):
temp = list(itertools.permutations(arr, i))
for j in range(len(temp)):
vasya = sum(temp[j])
petya = sum(arr) - vasya
check = abs(vasya - petya)
result = min(result, check)
print(result)
``` | 0 | |
337 | A | Puzzles | PROGRAMMING | 900 | [
"greedy"
] | null | null | The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*. | The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop. | Print a single integer — the least possible difference the teacher can obtain. | [
"4 6\n10 12 10 7 5 22\n"
] | [
"5\n"
] | Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5. | 500 | [
{
"input": "4 6\n10 12 10 7 5 22",
"output": "5"
},
{
"input": "2 2\n4 4",
"output": "0"
},
{
"input": "2 10\n4 5 6 7 8 9 10 11 12 12",
"output": "0"
},
{
"input": "4 5\n818 136 713 59 946",
"output": "759"
},
{
"input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903",
"output": "13"
},
{
"input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94",
"output": "3"
},
{
"input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190",
"output": "31"
},
{
"input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997",
"output": "0"
},
{
"input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176",
"output": "412"
},
{
"input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886",
"output": "938"
},
{
"input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993",
"output": "937"
},
{
"input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7",
"output": "0"
},
{
"input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19",
"output": "12"
},
{
"input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17",
"output": "31"
},
{
"input": "2 2\n1000 4",
"output": "996"
},
{
"input": "2 3\n4 502 1000",
"output": "498"
},
{
"input": "3 3\n4 1000 4",
"output": "996"
}
] | 1,691,076,480 | 2,147,483,647 | Python 3 | OK | TESTS | 18 | 62 | 0 | def find_min_difference(n, m, puzzle_pieces):
puzzle_pieces.sort()
min_difference = float('inf')
for i in range(m - n + 1):
min_difference = min(min_difference, puzzle_pieces[i + n - 1] - puzzle_pieces[i])
return min_difference
def main():
n, m = map(int, input().split())
puzzle_pieces = list(map(int, input().split()))
result = find_min_difference(n, m, puzzle_pieces)
print(result)
if __name__ == "__main__":
main()
| Title: Puzzles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
Input Specification:
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Output Specification:
Print a single integer — the least possible difference the teacher can obtain.
Demo Input:
['4 6\n10 12 10 7 5 22\n']
Demo Output:
['5\n']
Note:
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5. | ```python
def find_min_difference(n, m, puzzle_pieces):
puzzle_pieces.sort()
min_difference = float('inf')
for i in range(m - n + 1):
min_difference = min(min_difference, puzzle_pieces[i + n - 1] - puzzle_pieces[i])
return min_difference
def main():
n, m = map(int, input().split())
puzzle_pieces = list(map(int, input().split()))
result = find_min_difference(n, m, puzzle_pieces)
print(result)
if __name__ == "__main__":
main()
``` | 3 | |
39 | D | Cubical Planet | PROGRAMMING | 1,100 | [
"math"
] | D. Cubical Planet | 2 | 64 | You can find anything whatsoever in our Galaxy! A cubical planet goes round an icosahedral star. Let us introduce a system of axes so that the edges of the cubical planet are parallel to the coordinate axes and two opposite vertices lay in the points (0,<=0,<=0) and (1,<=1,<=1). Two flies live on the planet. At the moment they are sitting on two different vertices of the cubical planet. Your task is to determine whether they see each other or not. The flies see each other when the vertices they occupy lie on the same face of the cube. | The first line contains three space-separated integers (0 or 1) — the coordinates of the first fly, the second line analogously contains the coordinates of the second fly. | Output "YES" (without quotes) if the flies see each other. Otherwise, output "NO". | [
"0 0 0\n0 1 0\n",
"1 1 0\n0 1 0\n",
"0 0 0\n1 1 1\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "0 0 0\n0 1 0",
"output": "YES"
},
{
"input": "1 1 0\n0 1 0",
"output": "YES"
},
{
"input": "0 0 0\n1 1 1",
"output": "NO"
},
{
"input": "0 0 0\n1 0 0",
"output": "YES"
},
{
"input": "0 0 0\n0 1 0",
"output": "YES"
},
{
"input": "0 0 0\n1 1 0",
"output": "YES"
},
{
"input": "0 0 0\n0 0 1",
"output": "YES"
},
{
"input": "0 0 0\n1 0 1",
"output": "YES"
},
{
"input": "0 0 0\n0 1 1",
"output": "YES"
},
{
"input": "0 0 0\n1 1 1",
"output": "NO"
},
{
"input": "1 0 0\n0 0 0",
"output": "YES"
},
{
"input": "1 0 0\n0 1 0",
"output": "YES"
},
{
"input": "1 0 0\n1 1 0",
"output": "YES"
},
{
"input": "1 0 0\n0 0 1",
"output": "YES"
},
{
"input": "1 0 0\n1 0 1",
"output": "YES"
},
{
"input": "1 0 0\n0 1 1",
"output": "NO"
},
{
"input": "1 0 0\n1 1 1",
"output": "YES"
},
{
"input": "0 1 0\n0 0 0",
"output": "YES"
},
{
"input": "0 1 0\n1 0 0",
"output": "YES"
},
{
"input": "0 1 0\n1 1 0",
"output": "YES"
},
{
"input": "0 1 0\n0 0 1",
"output": "YES"
},
{
"input": "0 1 0\n1 0 1",
"output": "NO"
},
{
"input": "0 1 0\n0 1 1",
"output": "YES"
},
{
"input": "0 1 0\n1 1 1",
"output": "YES"
},
{
"input": "1 1 0\n0 0 0",
"output": "YES"
},
{
"input": "1 1 0\n1 0 0",
"output": "YES"
},
{
"input": "1 1 0\n0 1 0",
"output": "YES"
},
{
"input": "1 1 0\n0 0 1",
"output": "NO"
},
{
"input": "1 1 0\n1 0 1",
"output": "YES"
},
{
"input": "1 1 0\n0 1 1",
"output": "YES"
},
{
"input": "1 1 0\n1 1 1",
"output": "YES"
},
{
"input": "0 0 1\n0 0 0",
"output": "YES"
},
{
"input": "0 0 1\n1 0 0",
"output": "YES"
},
{
"input": "0 0 1\n0 1 0",
"output": "YES"
},
{
"input": "0 0 1\n1 1 0",
"output": "NO"
},
{
"input": "0 0 1\n1 0 1",
"output": "YES"
},
{
"input": "0 0 1\n0 1 1",
"output": "YES"
},
{
"input": "0 0 1\n1 1 1",
"output": "YES"
},
{
"input": "1 0 1\n0 0 0",
"output": "YES"
},
{
"input": "1 0 1\n1 0 0",
"output": "YES"
},
{
"input": "1 0 1\n0 1 0",
"output": "NO"
},
{
"input": "1 0 1\n1 1 0",
"output": "YES"
},
{
"input": "1 0 1\n0 0 1",
"output": "YES"
},
{
"input": "1 0 1\n0 1 1",
"output": "YES"
},
{
"input": "1 0 1\n1 1 1",
"output": "YES"
},
{
"input": "0 1 1\n0 0 0",
"output": "YES"
},
{
"input": "0 1 1\n1 0 0",
"output": "NO"
},
{
"input": "0 1 1\n0 1 0",
"output": "YES"
},
{
"input": "0 1 1\n1 1 0",
"output": "YES"
},
{
"input": "0 1 1\n0 0 1",
"output": "YES"
},
{
"input": "0 1 1\n1 0 1",
"output": "YES"
},
{
"input": "0 1 1\n1 1 1",
"output": "YES"
},
{
"input": "1 1 1\n0 0 0",
"output": "NO"
},
{
"input": "1 1 1\n1 0 0",
"output": "YES"
},
{
"input": "1 1 1\n0 1 0",
"output": "YES"
},
{
"input": "1 1 1\n1 1 0",
"output": "YES"
},
{
"input": "1 1 1\n0 0 1",
"output": "YES"
},
{
"input": "1 1 1\n1 0 1",
"output": "YES"
},
{
"input": "1 1 1\n0 1 1",
"output": "YES"
}
] | 1,699,435,445 | 2,147,483,647 | Python 3 | OK | TESTS | 59 | 92 | 0 | # LUOGU_RID: 133989237
a,b,c=map(int,input().split())
w,e,r=map(int,input().split())
if a==w or b==e or c==r:
print("YES")
else:
print("NO") | Title: Cubical Planet
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
You can find anything whatsoever in our Galaxy! A cubical planet goes round an icosahedral star. Let us introduce a system of axes so that the edges of the cubical planet are parallel to the coordinate axes and two opposite vertices lay in the points (0,<=0,<=0) and (1,<=1,<=1). Two flies live on the planet. At the moment they are sitting on two different vertices of the cubical planet. Your task is to determine whether they see each other or not. The flies see each other when the vertices they occupy lie on the same face of the cube.
Input Specification:
The first line contains three space-separated integers (0 or 1) — the coordinates of the first fly, the second line analogously contains the coordinates of the second fly.
Output Specification:
Output "YES" (without quotes) if the flies see each other. Otherwise, output "NO".
Demo Input:
['0 0 0\n0 1 0\n', '1 1 0\n0 1 0\n', '0 0 0\n1 1 1\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none | ```python
# LUOGU_RID: 133989237
a,b,c=map(int,input().split())
w,e,r=map(int,input().split())
if a==w or b==e or c==r:
print("YES")
else:
print("NO")
``` | 3.977 |
897 | B | Chtholly's request | PROGRAMMING | 1,300 | [
"brute force"
] | null | null | — I experienced so many great things.
— You gave me memories like dreams... But I have to leave now...
— One last request, can you...
— Help me solve a Codeforces problem?
— ......
— What?
Chtholly has been thinking about a problem for days:
If a number is palindrome and length of its decimal representation without leading zeros is even, we call it a zcy number. A number is palindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example 12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and 12321 is not.
Given integers *k* and *p*, calculate the sum of the *k* smallest zcy numbers and output this sum modulo *p*.
Unfortunately, Willem isn't good at solving this kind of problems, so he asks you for help! | The first line contains two integers *k* and *p* (1<=≤<=*k*<=≤<=105,<=1<=≤<=*p*<=≤<=109). | Output single integer — answer to the problem. | [
"2 100\n",
"5 30\n"
] | [
"33\n",
"15\n"
] | In the first example, the smallest zcy number is 11, and the second smallest zcy number is 22.
In the second example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/68fffad54395f7d920ad0384e07c6215ddc64141.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 1,000 | [
{
"input": "2 100",
"output": "33"
},
{
"input": "5 30",
"output": "15"
},
{
"input": "42147 412393322",
"output": "251637727"
},
{
"input": "77809 868097296",
"output": "440411873"
},
{
"input": "5105 443422097",
"output": "363192634"
},
{
"input": "75615 376679484",
"output": "373089399"
},
{
"input": "22951 23793302",
"output": "1898631"
},
{
"input": "12785 993582106",
"output": "286204743"
},
{
"input": "60276 428978808",
"output": "376477293"
},
{
"input": "84776 104860385",
"output": "10209596"
},
{
"input": "41984 653766991",
"output": "17823101"
},
{
"input": "100000 1000000000",
"output": "495495496"
},
{
"input": "41163 472310076",
"output": "207304047"
},
{
"input": "6983 765352180",
"output": "586866999"
},
{
"input": "33493 967727004",
"output": "305705165"
},
{
"input": "90898 94010922",
"output": "65928728"
},
{
"input": "67298 349286579",
"output": "156435206"
},
{
"input": "92452 296773064",
"output": "229486976"
},
{
"input": "58832 563860457",
"output": "16775206"
},
{
"input": "90234 156145441",
"output": "44023160"
},
{
"input": "91454 977186148",
"output": "681779748"
},
{
"input": "11108 444095250",
"output": "188075844"
},
{
"input": "46304 584475527",
"output": "275627129"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 1000000000",
"output": "11"
},
{
"input": "100000 1",
"output": "0"
}
] | 1,629,642,312 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 155 | 6,758,400 | k,p = [int(i) for i in input().split()]
sum = 0
for i in range(1,k+1):
sum+= int(str(i)+str(i)[::-1])
print(sum%p)
| Title: Chtholly's request
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
— I experienced so many great things.
— You gave me memories like dreams... But I have to leave now...
— One last request, can you...
— Help me solve a Codeforces problem?
— ......
— What?
Chtholly has been thinking about a problem for days:
If a number is palindrome and length of its decimal representation without leading zeros is even, we call it a zcy number. A number is palindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example 12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and 12321 is not.
Given integers *k* and *p*, calculate the sum of the *k* smallest zcy numbers and output this sum modulo *p*.
Unfortunately, Willem isn't good at solving this kind of problems, so he asks you for help!
Input Specification:
The first line contains two integers *k* and *p* (1<=≤<=*k*<=≤<=105,<=1<=≤<=*p*<=≤<=109).
Output Specification:
Output single integer — answer to the problem.
Demo Input:
['2 100\n', '5 30\n']
Demo Output:
['33\n', '15\n']
Note:
In the first example, the smallest zcy number is 11, and the second smallest zcy number is 22.
In the second example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/68fffad54395f7d920ad0384e07c6215ddc64141.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
k,p = [int(i) for i in input().split()]
sum = 0
for i in range(1,k+1):
sum+= int(str(i)+str(i)[::-1])
print(sum%p)
``` | 3 | |
797 | B | Odd sum | PROGRAMMING | 1,400 | [
"dp",
"greedy",
"implementation"
] | null | null | You are given sequence *a*1,<=*a*2,<=...,<=*a**n* of integer numbers of length *n*. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.
Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
You should write a program which finds sum of the best subsequence. | The first line contains integer number *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=104<=≤<=*a**i*<=≤<=104). The sequence contains at least one subsequence with odd sum. | Print sum of resulting subseqeuence. | [
"4\n-2 2 -3 1\n",
"3\n2 -5 -3\n"
] | [
"3\n",
"-1\n"
] | In the first example sum of the second and the fourth elements is 3. | 0 | [
{
"input": "4\n-2 2 -3 1",
"output": "3"
},
{
"input": "3\n2 -5 -3",
"output": "-1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "15\n-6004 4882 9052 413 6056 4306 9946 -4616 -6135 906 -1718 5252 -2866 9061 4046",
"output": "53507"
},
{
"input": "2\n-5439 -6705",
"output": "-5439"
},
{
"input": "2\n2850 6843",
"output": "9693"
},
{
"input": "2\n144 9001",
"output": "9145"
},
{
"input": "10\n7535 -819 2389 4933 5495 4887 -5181 -9355 7955 5757",
"output": "38951"
},
{
"input": "10\n-9169 -1574 3580 -8579 -7177 -3216 7490 3470 3465 -1197",
"output": "18005"
},
{
"input": "10\n941 7724 2220 -4704 -8374 -8249 7606 9502 612 -9097",
"output": "28605"
},
{
"input": "10\n4836 -2331 -3456 2312 -1574 3134 -670 -204 512 -5504",
"output": "8463"
},
{
"input": "10\n1184 5136 1654 3254 6576 6900 6468 327 179 7114",
"output": "38613"
},
{
"input": "10\n-2152 -1776 -1810 -9046 -6090 -2324 -8716 -6103 -787 -812",
"output": "-787"
},
{
"input": "3\n1 1 1",
"output": "3"
},
{
"input": "5\n5 5 5 3 -1",
"output": "17"
},
{
"input": "5\n-1 -2 5 3 0",
"output": "7"
},
{
"input": "5\n-3 -2 5 -1 3",
"output": "7"
},
{
"input": "3\n-2 2 -1",
"output": "1"
},
{
"input": "5\n5 0 7 -2 3",
"output": "15"
},
{
"input": "2\n-2 -5",
"output": "-5"
},
{
"input": "3\n-1 -3 0",
"output": "-1"
},
{
"input": "5\n2 -1 0 -3 -2",
"output": "1"
},
{
"input": "4\n2 3 0 5",
"output": "7"
},
{
"input": "5\n-5 3 -2 2 5",
"output": "7"
},
{
"input": "59\n8593 5929 3016 -859 4366 -6842 8435 -3910 -2458 -8503 -3612 -9793 -5360 -9791 -362 -7180 727 -6245 -8869 -7316 8214 -7944 7098 3788 -5436 -6626 -1131 -2410 -5647 -7981 263 -5879 8786 709 6489 5316 -4039 4909 -4340 7979 -89 9844 -906 172 -7674 -3371 -6828 9505 3284 5895 3646 6680 -1255 3635 -9547 -5104 -1435 -7222 2244",
"output": "129433"
},
{
"input": "17\n-6170 2363 6202 -9142 7889 779 2843 -5089 2313 -3952 1843 5171 462 -3673 5098 -2519 9565",
"output": "43749"
},
{
"input": "26\n-8668 9705 1798 -1766 9644 3688 8654 -3077 -5462 2274 6739 2732 3635 -4745 -9144 -9175 -7488 -2010 1637 1118 8987 1597 -2873 -5153 -8062 146",
"output": "60757"
},
{
"input": "51\n8237 -7239 -3545 -6059 -5110 4066 -4148 -7641 -5797 -994 963 1144 -2785 -8765 -1216 5410 1508 -6312 -6313 -680 -7657 4579 -6898 7379 2015 -5087 -5417 -6092 3819 -9101 989 -8380 9161 -7519 -9314 -3838 7160 5180 567 -1606 -3842 -9665 -2266 1296 -8417 -3976 7436 -2075 -441 -4565 3313",
"output": "73781"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n-2 1",
"output": "1"
},
{
"input": "2\n3 2",
"output": "5"
},
{
"input": "2\n1 2",
"output": "3"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n0 -1",
"output": "-1"
},
{
"input": "2\n2 1",
"output": "3"
},
{
"input": "2\n3 0",
"output": "3"
},
{
"input": "2\n0 -1",
"output": "-1"
},
{
"input": "3\n-3 1 -1",
"output": "1"
},
{
"input": "3\n3 -1 1",
"output": "3"
},
{
"input": "3\n1 3 1",
"output": "5"
},
{
"input": "3\n-1 0 1",
"output": "1"
},
{
"input": "3\n-3 -3 -2",
"output": "-3"
},
{
"input": "3\n3 -1 1",
"output": "3"
},
{
"input": "3\n3 -1 1",
"output": "3"
},
{
"input": "3\n-2 -2 1",
"output": "1"
},
{
"input": "4\n0 -1 -3 -4",
"output": "-1"
},
{
"input": "4\n5 3 2 1",
"output": "11"
},
{
"input": "4\n-1 -2 4 -2",
"output": "3"
},
{
"input": "4\n-1 -3 0 -3",
"output": "-1"
},
{
"input": "4\n1 -4 -3 -4",
"output": "1"
},
{
"input": "4\n5 3 3 4",
"output": "15"
},
{
"input": "4\n-1 -3 -1 2",
"output": "1"
},
{
"input": "4\n3 2 -1 -4",
"output": "5"
},
{
"input": "5\n-5 -4 -3 -5 2",
"output": "-1"
},
{
"input": "5\n5 5 1 2 -2",
"output": "13"
},
{
"input": "5\n-2 -1 -5 -1 4",
"output": "3"
},
{
"input": "5\n-5 -5 -4 4 0",
"output": "-1"
},
{
"input": "5\n2 -3 -1 -4 -5",
"output": "1"
},
{
"input": "5\n4 3 4 2 3",
"output": "13"
},
{
"input": "5\n0 -2 -5 3 3",
"output": "3"
},
{
"input": "5\n4 -2 -2 -3 0",
"output": "1"
},
{
"input": "6\n6 7 -1 1 5 -1",
"output": "19"
},
{
"input": "6\n-1 7 2 -3 -4 -5",
"output": "9"
},
{
"input": "6\n0 -1 -3 -5 2 -6",
"output": "1"
},
{
"input": "6\n4 -1 0 3 6 1",
"output": "13"
},
{
"input": "6\n5 3 3 4 4 -3",
"output": "19"
},
{
"input": "6\n0 -3 5 -4 5 -4",
"output": "7"
},
{
"input": "6\n-5 -3 1 -1 -5 -3",
"output": "1"
},
{
"input": "6\n-2 1 3 -2 7 4",
"output": "15"
},
{
"input": "7\n0 7 6 2 7 0 6",
"output": "21"
},
{
"input": "7\n6 -6 -1 -5 7 1 7",
"output": "21"
},
{
"input": "7\n2 3 -5 0 -4 0 -4",
"output": "5"
},
{
"input": "7\n-6 3 -3 -1 -6 -6 -5",
"output": "3"
},
{
"input": "7\n7 6 3 2 4 2 0",
"output": "21"
},
{
"input": "7\n-2 3 -3 4 4 0 -1",
"output": "11"
},
{
"input": "7\n-5 -7 4 0 5 -3 -5",
"output": "9"
},
{
"input": "7\n-3 -5 -4 1 3 -4 -7",
"output": "3"
},
{
"input": "8\n5 2 4 5 7 -2 7 3",
"output": "33"
},
{
"input": "8\n-8 -3 -1 3 -8 -4 -4 4",
"output": "7"
},
{
"input": "8\n-6 -7 -7 -5 -4 -9 -2 -7",
"output": "-5"
},
{
"input": "8\n8 7 6 8 3 4 8 -2",
"output": "41"
},
{
"input": "8\n6 7 0 -6 6 5 4 7",
"output": "35"
},
{
"input": "8\n0 -7 -5 -5 5 -1 -8 -7",
"output": "5"
},
{
"input": "8\n1 -6 -5 7 -3 -4 2 -2",
"output": "9"
},
{
"input": "8\n1 -8 -6 -6 -6 -7 -5 -1",
"output": "1"
},
{
"input": "9\n-3 -1 4 4 8 -8 -5 9 -2",
"output": "25"
},
{
"input": "9\n-9 -1 3 -2 -7 2 -9 -1 -4",
"output": "5"
},
{
"input": "9\n-6 -9 -3 -8 -5 2 -6 0 -5",
"output": "-1"
},
{
"input": "9\n5 4 3 3 6 7 8 5 9",
"output": "47"
},
{
"input": "9\n5 3 9 1 5 2 -3 7 0",
"output": "31"
},
{
"input": "9\n-3 -9 -1 -7 5 6 -4 -6 -6",
"output": "11"
},
{
"input": "9\n-6 -5 6 -5 -2 0 1 2 -9",
"output": "9"
},
{
"input": "9\n8 3 6 1 -3 5 2 9 1",
"output": "35"
},
{
"input": "10\n-6 -4 -7 -1 -9 -10 -10 1 0 -3",
"output": "1"
},
{
"input": "10\n-2 -10 -5 -6 -10 -3 -6 -3 -8 -8",
"output": "-3"
},
{
"input": "10\n8 5 9 2 3 3 -6 1 -1 8",
"output": "39"
},
{
"input": "10\n2 10 -7 6 -1 -1 7 -9 -4 -6",
"output": "25"
},
{
"input": "10\n-10 -2 -2 -1 -10 -7 1 0 -4 -5",
"output": "1"
},
{
"input": "10\n4 3 10 -2 -1 0 10 6 7 0",
"output": "39"
},
{
"input": "10\n-2 6 6 5 0 10 6 7 -1 1",
"output": "41"
},
{
"input": "10\n-10 2 8 -6 -1 -5 1 -10 -10 -1",
"output": "11"
}
] | 1,659,171,569 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 3 | 77 | 2,867,200 | import sys
input = sys.stdin.readline
import bisect
n = int(input())
w = sorted(map(int, input().split()))
c = bisect.bisect_right(w, 0)
s = sum(w[c:])
if s % 2:
print(s)
exit(0)
a, b = 0, 0
for i in range(c, n):
if w[i] % 2:
a = w[i]
break
for i in range(c, -1, -1):
if w[i] % 2:
b = w[i]
break
if a == 0 and b == 0:
print(-1)
elif a == 0 or a > -b:
print(s+b)
else:
print(s-a)
| Title: Odd sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given sequence *a*1,<=*a*2,<=...,<=*a**n* of integer numbers of length *n*. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.
Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
You should write a program which finds sum of the best subsequence.
Input Specification:
The first line contains integer number *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=104<=≤<=*a**i*<=≤<=104). The sequence contains at least one subsequence with odd sum.
Output Specification:
Print sum of resulting subseqeuence.
Demo Input:
['4\n-2 2 -3 1\n', '3\n2 -5 -3\n']
Demo Output:
['3\n', '-1\n']
Note:
In the first example sum of the second and the fourth elements is 3. | ```python
import sys
input = sys.stdin.readline
import bisect
n = int(input())
w = sorted(map(int, input().split()))
c = bisect.bisect_right(w, 0)
s = sum(w[c:])
if s % 2:
print(s)
exit(0)
a, b = 0, 0
for i in range(c, n):
if w[i] % 2:
a = w[i]
break
for i in range(c, -1, -1):
if w[i] % 2:
b = w[i]
break
if a == 0 and b == 0:
print(-1)
elif a == 0 or a > -b:
print(s+b)
else:
print(s-a)
``` | -1 | |
707 | A | Brain's Photos | PROGRAMMING | 800 | [
"implementation"
] | null | null | Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'. | Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line. | [
"2 2\nC M\nY Y\n",
"3 2\nW W\nW W\nB B\n",
"1 1\nW\n"
] | [
"#Color",
"#Black&White",
"#Black&White"
] | none | 500 | [
{
"input": "2 2\nC M\nY Y",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G Y",
"output": "#Color"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "5 5\nW G B Y M\nG B Y M C\nB Y M C W\nY M C W G\nM C W G B",
"output": "#Color"
},
{
"input": "1 6\nC M Y W G B",
"output": "#Color"
},
{
"input": "1 3\nW G B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "5 5\nW G B W G\nG B W G B\nB W G B W\nW G B W G\nG B W G B",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G C",
"output": "#Color"
},
{
"input": "2 3\nW W W\nB G M",
"output": "#Color"
},
{
"input": "3 3\nC B W\nB Y M\nB B W",
"output": "#Color"
},
{
"input": "1 3\nW C W",
"output": "#Color"
},
{
"input": "3 3\nB W B\nB C W\nB W W",
"output": "#Color"
},
{
"input": "1 2\nW Y",
"output": "#Color"
},
{
"input": "1 1\nG",
"output": "#Black&White"
},
{
"input": "1 3\nB G W",
"output": "#Black&White"
},
{
"input": "2 2\nW W\nB C",
"output": "#Color"
},
{
"input": "1 1\nM",
"output": "#Color"
},
{
"input": "1 2\nW C",
"output": "#Color"
},
{
"input": "2 3\nW W M\nW W M",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB C",
"output": "#Color"
},
{
"input": "2 3\nW W C\nW W W",
"output": "#Color"
},
{
"input": "1 3\nG G G",
"output": "#Black&White"
},
{
"input": "1 1\nC",
"output": "#Color"
},
{
"input": "1 2\nC W",
"output": "#Color"
},
{
"input": "1 3\nW W C",
"output": "#Color"
},
{
"input": "2 2\nW B\nB G",
"output": "#Black&White"
},
{
"input": "2 2\nB B\nY Y",
"output": "#Color"
},
{
"input": "2 2\nG G\nC C",
"output": "#Color"
},
{
"input": "1 2\nB C",
"output": "#Color"
},
{
"input": "1 4\nG G G C",
"output": "#Color"
},
{
"input": "1 1\nB",
"output": "#Black&White"
},
{
"input": "2 1\nY\nB",
"output": "#Color"
}
] | 1,656,679,102 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 46 | 0 | n,m = map(int,input().split())
a="".join(input() for i in range(n))
if "C" in a or "M" in a or "Y" in a: print("#Color")
else: print("#Black&White")
| Title: Brain's Photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
Output Specification:
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
Demo Input:
['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n']
Demo Output:
['#Color', '#Black&White', '#Black&White']
Note:
none | ```python
n,m = map(int,input().split())
a="".join(input() for i in range(n))
if "C" in a or "M" in a or "Y" in a: print("#Color")
else: print("#Black&White")
``` | 3 | |
17 | D | Notepad | PROGRAMMING | 2,400 | [
"number theory"
] | D. Notepad | 2 | 64 | Nick is attracted by everything unconventional. He doesn't like decimal number system any more, and he decided to study other number systems. A number system with base *b* caught his attention. Before he starts studying it, he wants to write in his notepad all the numbers of length *n* without leading zeros in this number system. Each page in Nick's notepad has enough space for *c* numbers exactly. Nick writes every suitable number only once, starting with the first clean page and leaving no clean spaces. Nick never writes number 0 as he has unpleasant memories about zero divide.
Would you help Nick find out how many numbers will be written on the last page. | The only input line contains three space-separated integers *b*, *n* and *c* (2<=≤<=*b*<=<<=10106, 1<=≤<=*n*<=<<=10106, 1<=≤<=*c*<=≤<=109). You may consider that Nick has infinite patience, endless amount of paper and representations of digits as characters. The numbers doesn't contain leading zeros. | In the only line output the amount of numbers written on the same page as the last number. | [
"2 3 3\n",
"2 3 4\n"
] | [
"1",
"4"
] | In both samples there are exactly 4 numbers of length 3 in binary number system. In the first sample Nick writes 3 numbers on the first page and 1 on the second page. In the second sample all the 4 numbers can be written on the first page. | 0 | [
{
"input": "2 3 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "4"
},
{
"input": "9 1 79",
"output": "8"
},
{
"input": "9 1 345",
"output": "8"
},
{
"input": "9 9 999982045",
"output": "344373768"
},
{
"input": "4 42 44",
"output": "12"
},
{
"input": "6 43 659",
"output": "365"
},
{
"input": "8 54 999992388",
"output": "741886148"
},
{
"input": "861 11 17",
"output": "14"
},
{
"input": "89 34 119",
"output": "80"
},
{
"input": "84 67 999993310",
"output": "829809148"
},
{
"input": "9219 537 98",
"output": "98"
},
{
"input": "763 582 510",
"output": "96"
},
{
"input": "6355 60160 999982994",
"output": "904671182"
},
{
"input": "396882961 9936448 752",
"output": "528"
},
{
"input": "394292559875270 34297300532732 28",
"output": "28"
},
{
"input": "8523703220091 953421047275844 163",
"output": "30"
},
{
"input": "713030357739784847 61197710123555584 999992531",
"output": "207016405"
},
{
"input": "75903940600326238527 492179977057716178 954",
"output": "450"
},
{
"input": "8085477143815539692925721 57241684823084777591460 968",
"output": "304"
},
{
"input": "67609394386924890416446 78162115935271414671181267 999987217",
"output": "926946271"
},
{
"input": "3351262437484130462277638791970372162118802730187825044167229944871677684706592699530322737272222086076517455404652584348 147310576952932829345029460612849431175622785231399764423717734155248977073541821053441627535488066058597900989095431439 999998948",
"output": "930694076"
},
{
"input": "61063034544457239668509642598956869508193198481915116469015956878854905975766584002919896320353661294612971855029955483257741525207429239630069409321331850413146512850720681578339422084340720535114848966742045420860633093949996367883 965415513080902927493169838825380834798188421277961155726185690857844534367611949025561401481462737822765050755128163519122172969767981851117402342816829930821131453945898813517587656899608854645391515043085723743408226445117376493281975889755859761322184701256801 999998603",
"output": "60342257"
},
{
"input": "9 1000000000000000000000000000000000000000000000000000000 345",
"output": "192"
},
{
"input": "8053063680000000000000000000000000002 268435456000000000000005 805306368",
"output": "268435456"
},
{
"input": "8053063680000000000000000000000000002 2684354560000000000000025 805306368",
"output": "268435456"
},
{
"input": "8053063680000000000000000000000000002 2684354560000000000000026 805306368",
"output": "536870912"
},
{
"input": "8053063680000000000000000000000000002 2684354560000000000000027 805306368",
"output": "268435456"
},
{
"input": "8053063680000000000000000000000000002 2684354560000000000000028 805306368",
"output": "536870912"
},
{
"input": "8053063680000000000000000000000000002 2684354560000000000000029 805306368",
"output": "268435456"
},
{
"input": "8053063680000000000000000000000000002 2684354560000000000000030 805306368",
"output": "536870912"
},
{
"input": "8053063680000000000000000000000000002 2684354560000000000000031 805306368",
"output": "268435456"
},
{
"input": "2271048430505293080737093330373572593316324321603522463486966273671353266974713306925326907468317965879775893196923719457524955744 8990615363653447573832140957083458603886706189959668013719622351914533208654357508127820477597609318856255372184258450991108060161 53727872",
"output": "26470400"
},
{
"input": "244741007655429712 1 297825872",
"output": "297825871"
}
] | 1,688,725,974 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 27 | 2,000 | 1,228,800 | def bin_exp(b, e, m):
p = 1
while e:
if e & 1:
p = p * b % m
e >>= 1
b = b * b % m
return p
b, n, c = list(map(int, input().split()))
b = b % c
s = (b + c - 1) % c * bin_exp(b, n - 1, c) % c
if s == 0:
s = c
print(s)
| Title: Notepad
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Nick is attracted by everything unconventional. He doesn't like decimal number system any more, and he decided to study other number systems. A number system with base *b* caught his attention. Before he starts studying it, he wants to write in his notepad all the numbers of length *n* without leading zeros in this number system. Each page in Nick's notepad has enough space for *c* numbers exactly. Nick writes every suitable number only once, starting with the first clean page and leaving no clean spaces. Nick never writes number 0 as he has unpleasant memories about zero divide.
Would you help Nick find out how many numbers will be written on the last page.
Input Specification:
The only input line contains three space-separated integers *b*, *n* and *c* (2<=≤<=*b*<=<<=10106, 1<=≤<=*n*<=<<=10106, 1<=≤<=*c*<=≤<=109). You may consider that Nick has infinite patience, endless amount of paper and representations of digits as characters. The numbers doesn't contain leading zeros.
Output Specification:
In the only line output the amount of numbers written on the same page as the last number.
Demo Input:
['2 3 3\n', '2 3 4\n']
Demo Output:
['1', '4']
Note:
In both samples there are exactly 4 numbers of length 3 in binary number system. In the first sample Nick writes 3 numbers on the first page and 1 on the second page. In the second sample all the 4 numbers can be written on the first page. | ```python
def bin_exp(b, e, m):
p = 1
while e:
if e & 1:
p = p * b % m
e >>= 1
b = b * b % m
return p
b, n, c = list(map(int, input().split()))
b = b % c
s = (b + c - 1) % c * bin_exp(b, n - 1, c) % c
if s == 0:
s = c
print(s)
``` | 0 |
353 | A | Domino | PROGRAMMING | 1,200 | [
"implementation",
"math"
] | null | null | Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even.
To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half. | Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1. | [
"2\n4 2\n6 4\n",
"1\n2 3\n",
"3\n1 4\n2 3\n4 4\n"
] | [
"0\n",
"-1\n",
"1\n"
] | In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything.
In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd.
In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8. | 500 | [
{
"input": "2\n4 2\n6 4",
"output": "0"
},
{
"input": "1\n2 3",
"output": "-1"
},
{
"input": "3\n1 4\n2 3\n4 4",
"output": "1"
},
{
"input": "5\n5 4\n5 4\n1 5\n5 5\n3 3",
"output": "1"
},
{
"input": "20\n1 3\n5 2\n5 2\n2 6\n2 4\n1 1\n1 3\n1 4\n2 6\n4 2\n5 6\n2 2\n6 2\n4 3\n2 1\n6 2\n6 5\n4 5\n2 4\n1 4",
"output": "-1"
},
{
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},
{
"input": "90\n1 4\n3 5\n4 2\n2 5\n4 3\n2 6\n2 6\n3 2\n4 4\n6 1\n4 3\n2 3\n5 3\n6 6\n2 2\n6 3\n4 1\n4 4\n5 6\n6 4\n4 2\n5 6\n4 6\n4 4\n6 4\n4 1\n5 3\n3 2\n4 4\n5 2\n5 4\n6 4\n1 2\n3 3\n3 4\n6 4\n1 6\n4 2\n3 2\n1 1\n2 2\n5 1\n6 6\n4 1\n5 2\n3 6\n2 1\n2 2\n4 6\n6 5\n4 4\n5 5\n5 6\n1 6\n1 4\n5 6\n3 6\n6 3\n5 6\n6 5\n5 1\n6 1\n6 6\n6 3\n1 5\n4 5\n3 1\n6 6\n3 4\n6 2\n1 4\n2 2\n3 2\n5 6\n2 4\n1 4\n6 3\n4 6\n1 4\n5 2\n1 2\n6 5\n1 5\n1 4\n4 2\n2 5\n3 2\n5 1\n5 4\n5 3",
"output": "-1"
},
{
"input": "95\n4 3\n3 2\n5 5\n5 3\n1 6\n4 4\n5 5\n6 5\n3 5\n1 5\n4 2\n5 1\n1 2\n2 3\n6 4\n2 3\n6 3\n6 5\n5 6\n1 4\n2 6\n2 6\n2 5\n2 1\n3 1\n3 5\n2 2\n6 1\n2 4\n4 6\n6 6\n6 4\n3 2\n5 1\n4 3\n6 5\n2 3\n4 1\n2 5\n6 5\n6 5\n6 5\n5 1\n5 4\n4 6\n3 2\n2 5\n2 6\n4 6\n6 3\n6 4\n5 6\n4 6\n2 4\n3 4\n1 4\n2 4\n2 3\n5 6\n6 4\n3 1\n5 1\n3 6\n3 5\n2 6\n6 3\n4 3\n3 1\n6 1\n2 2\n6 3\n2 2\n2 2\n6 4\n6 1\n2 1\n5 6\n5 4\n5 2\n3 4\n3 6\n2 1\n1 6\n5 5\n2 6\n2 3\n3 6\n1 3\n1 5\n5 1\n1 2\n2 2\n5 3\n6 4\n4 5",
"output": "0"
},
{
"input": "95\n4 5\n5 6\n3 2\n5 1\n4 3\n4 1\n6 1\n5 2\n2 4\n5 3\n2 3\n6 4\n4 1\n1 6\n2 6\n2 3\n4 6\n2 4\n3 4\n4 2\n5 5\n1 1\n1 5\n4 3\n4 5\n6 2\n6 1\n6 3\n5 5\n4 1\n5 1\n2 3\n5 1\n3 6\n6 6\n4 5\n4 4\n4 3\n1 6\n6 6\n4 6\n6 4\n1 2\n6 2\n4 6\n6 6\n5 5\n6 1\n5 2\n4 5\n6 6\n6 5\n4 4\n1 5\n4 6\n4 1\n3 6\n5 1\n3 1\n4 6\n4 5\n1 3\n5 4\n4 5\n2 2\n6 1\n5 2\n6 5\n2 2\n1 1\n6 3\n6 1\n2 6\n3 3\n2 1\n4 6\n2 4\n5 5\n5 2\n3 2\n1 2\n6 6\n6 2\n5 1\n2 6\n5 2\n2 2\n5 5\n3 5\n3 3\n2 6\n5 3\n4 3\n1 6\n5 4",
"output": "-1"
},
{
"input": "100\n1 1\n3 5\n2 1\n1 2\n3 4\n5 6\n5 6\n6 1\n5 5\n2 4\n5 5\n5 6\n6 2\n6 6\n2 6\n1 4\n2 2\n3 2\n1 3\n5 5\n6 3\n5 6\n1 1\n1 2\n1 2\n2 1\n2 3\n1 6\n4 3\n1 1\n2 5\n2 4\n4 4\n1 5\n3 3\n6 1\n3 5\n1 1\n3 6\n3 1\n4 2\n4 3\n3 6\n6 6\n1 6\n6 2\n2 5\n5 4\n6 3\n1 4\n2 6\n6 2\n3 4\n6 1\n6 5\n4 6\n6 5\n4 4\n3 1\n6 3\n5 1\n2 4\n5 1\n1 2\n2 4\n2 1\n6 6\n5 3\n4 6\n6 3\n5 5\n3 3\n1 1\n6 5\n4 3\n2 6\n1 5\n3 5\n2 4\n4 5\n1 6\n2 3\n6 3\n5 5\n2 6\n2 6\n3 4\n3 2\n6 1\n3 4\n6 4\n3 3\n2 3\n5 1\n3 1\n6 2\n2 3\n6 4\n1 4\n1 2",
"output": "-1"
},
{
"input": "100\n1 1\n5 5\n1 2\n5 3\n5 5\n2 2\n1 5\n3 4\n3 2\n1 3\n5 6\n4 5\n2 1\n5 5\n2 2\n1 6\n6 1\n5 1\n4 1\n4 6\n3 5\n6 1\n2 3\n5 6\n3 6\n2 3\n5 6\n1 6\n3 2\n2 2\n3 3\n6 5\n5 5\n1 4\n5 6\n6 4\n1 4\n1 2\n2 6\n3 2\n6 4\n5 3\n3 3\n6 4\n4 6\n2 2\n5 6\n5 1\n1 2\n3 4\n4 5\n1 1\n3 4\n5 2\n4 5\n3 3\n1 1\n3 4\n1 6\n2 4\n1 3\n3 2\n6 5\n1 6\n3 6\n2 3\n2 6\n5 1\n5 5\n5 6\n4 1\n6 2\n3 6\n5 3\n2 2\n2 4\n6 6\n3 6\n4 6\n2 5\n5 3\n1 2\n3 4\n3 4\n6 2\n2 4\n2 2\n4 6\n3 5\n4 2\n5 6\n4 2\n2 3\n6 2\n5 6\n2 1\n3 3\n6 6\n4 3\n4 2",
"output": "1"
},
{
"input": "1\n2 2",
"output": "0"
},
{
"input": "3\n2 4\n6 6\n3 3",
"output": "-1"
},
{
"input": "2\n3 6\n4 1",
"output": "1"
},
{
"input": "3\n1 1\n1 1\n3 3",
"output": "-1"
},
{
"input": "3\n2 3\n1 1\n2 3",
"output": "1"
},
{
"input": "3\n2 2\n2 1\n1 2",
"output": "1"
},
{
"input": "3\n1 1\n1 1\n1 1",
"output": "-1"
}
] | 1,629,834,284 | 2,147,483,647 | Python 3 | OK | TESTS | 59 | 154 | 6,963,200 | a = 0
b = 0
check = 0
for _ in range(int(input())):
p, q = map(int, input().split())
a += p
b += q
if (a % 2 == 1 and b % 2 == 0) or (a % 2 == 0 and b % 2 == 1):
check += 1
if a % 2 == b % 2 == 0:
print(0)
elif a % 2 == b % 2 == 1 and check > 0:
print(1)
else:
print(-1) | Title: Domino
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even.
To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half.
Output Specification:
Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1.
Demo Input:
['2\n4 2\n6 4\n', '1\n2 3\n', '3\n1 4\n2 3\n4 4\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything.
In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd.
In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8. | ```python
a = 0
b = 0
check = 0
for _ in range(int(input())):
p, q = map(int, input().split())
a += p
b += q
if (a % 2 == 1 and b % 2 == 0) or (a % 2 == 0 and b % 2 == 1):
check += 1
if a % 2 == b % 2 == 0:
print(0)
elif a % 2 == b % 2 == 1 and check > 0:
print(1)
else:
print(-1)
``` | 3 | |
864 | A | Fair Game | PROGRAMMING | 1,000 | [
"implementation",
"sortings"
] | null | null | Petya and Vasya decided to play a game. They have *n* cards (*n* is an even number). A single integer is written on each card.
Before the game Petya will choose an integer and after that Vasya will choose another integer (different from the number that Petya chose). During the game each player takes all the cards with number he chose. For example, if Petya chose number 5 before the game he will take all cards on which 5 is written and if Vasya chose number 10 before the game he will take all cards on which 10 is written.
The game is considered fair if Petya and Vasya can take all *n* cards, and the number of cards each player gets is the same.
Determine whether Petya and Vasya can choose integer numbers before the game so that the game is fair. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — number of cards. It is guaranteed that *n* is an even number.
The following *n* lines contain a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (one integer per line, 1<=≤<=*a**i*<=≤<=100) — numbers written on the *n* cards. | If it is impossible for Petya and Vasya to choose numbers in such a way that the game will be fair, print "NO" (without quotes) in the first line. In this case you should not print anything more.
In the other case print "YES" (without quotes) in the first line. In the second line print two distinct integers — number that Petya should choose and the number that Vasya should choose to make the game fair. If there are several solutions, print any of them. | [
"4\n11\n27\n27\n11\n",
"2\n6\n6\n",
"6\n10\n20\n30\n20\n10\n20\n",
"6\n1\n1\n2\n2\n3\n3\n"
] | [
"YES\n11 27\n",
"NO\n",
"NO\n",
"NO\n"
] | In the first example the game will be fair if, for example, Petya chooses number 11, and Vasya chooses number 27. Then the will take all cards — Petya will take cards 1 and 4, and Vasya will take cards 2 and 3. Thus, each of them will take exactly two cards.
In the second example fair game is impossible because the numbers written on the cards are equal, but the numbers that Petya and Vasya should choose should be distinct.
In the third example it is impossible to take all cards. Petya and Vasya can take at most five cards — for example, Petya can choose number 10 and Vasya can choose number 20. But for the game to be fair it is necessary to take 6 cards. | 500 | [
{
"input": "4\n11\n27\n27\n11",
"output": "YES\n11 27"
},
{
"input": "2\n6\n6",
"output": "NO"
},
{
"input": "6\n10\n20\n30\n20\n10\n20",
"output": "NO"
},
{
"input": "6\n1\n1\n2\n2\n3\n3",
"output": "NO"
},
{
"input": "2\n1\n100",
"output": "YES\n1 100"
},
{
"input": "2\n1\n1",
"output": "NO"
},
{
"input": "2\n100\n100",
"output": "NO"
},
{
"input": "14\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43",
"output": "NO"
},
{
"input": "100\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32",
"output": "YES\n14 32"
},
{
"input": "2\n50\n100",
"output": "YES\n50 100"
},
{
"input": "2\n99\n100",
"output": "YES\n99 100"
},
{
"input": "4\n4\n4\n5\n5",
"output": "YES\n4 5"
},
{
"input": "10\n10\n10\n10\n10\n10\n23\n23\n23\n23\n23",
"output": "YES\n10 23"
},
{
"input": "20\n34\n34\n34\n34\n34\n34\n34\n34\n34\n34\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11",
"output": "YES\n11 34"
},
{
"input": "40\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30",
"output": "YES\n20 30"
},
{
"input": "58\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "YES\n1 100"
},
{
"input": "98\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99",
"output": "YES\n2 99"
},
{
"input": "100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "YES\n1 100"
},
{
"input": "100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2",
"output": "YES\n1 2"
},
{
"input": "100\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12",
"output": "YES\n12 49"
},
{
"input": "100\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94",
"output": "YES\n15 94"
},
{
"input": "100\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42",
"output": "YES\n33 42"
},
{
"input": "100\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35",
"output": "YES\n16 35"
},
{
"input": "100\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44",
"output": "YES\n33 44"
},
{
"input": "100\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98",
"output": "YES\n54 98"
},
{
"input": "100\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12",
"output": "YES\n12 81"
},
{
"input": "100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "NO"
},
{
"input": "100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "NO"
},
{
"input": "40\n20\n20\n30\n30\n20\n20\n20\n30\n30\n20\n20\n30\n30\n30\n30\n20\n30\n30\n30\n30\n20\n20\n30\n30\n30\n20\n30\n20\n30\n20\n30\n20\n20\n20\n30\n20\n20\n20\n30\n30",
"output": "NO"
},
{
"input": "58\n100\n100\n100\n100\n100\n1\n1\n1\n1\n1\n1\n100\n100\n1\n100\n1\n100\n100\n1\n1\n100\n100\n1\n100\n1\n100\n100\n1\n1\n100\n1\n1\n1\n100\n1\n1\n1\n1\n100\n1\n100\n100\n100\n100\n100\n1\n1\n100\n100\n100\n100\n1\n100\n1\n1\n1\n1\n1",
"output": "NO"
},
{
"input": "98\n2\n99\n99\n99\n99\n2\n99\n99\n99\n2\n2\n99\n2\n2\n2\n2\n99\n99\n2\n99\n2\n2\n99\n99\n99\n99\n2\n2\n99\n2\n99\n99\n2\n2\n99\n2\n99\n2\n99\n2\n2\n2\n99\n2\n2\n2\n2\n99\n99\n99\n99\n2\n2\n2\n2\n2\n2\n2\n2\n99\n2\n99\n99\n2\n2\n99\n99\n99\n99\n99\n99\n99\n99\n2\n99\n2\n99\n2\n2\n2\n99\n99\n99\n99\n99\n99\n2\n99\n99\n2\n2\n2\n2\n2\n99\n99\n99\n2",
"output": "NO"
},
{
"input": "100\n100\n1\n100\n1\n1\n100\n1\n1\n1\n100\n100\n1\n100\n1\n100\n100\n1\n1\n1\n100\n1\n100\n1\n100\n100\n1\n100\n1\n100\n1\n1\n1\n1\n1\n100\n1\n100\n100\n100\n1\n100\n100\n1\n100\n1\n1\n100\n100\n100\n1\n100\n100\n1\n1\n100\n100\n1\n100\n1\n100\n1\n1\n100\n100\n100\n100\n100\n100\n1\n100\n100\n1\n100\n100\n1\n100\n1\n1\n1\n100\n100\n1\n100\n1\n100\n1\n1\n1\n1\n100\n1\n1\n100\n1\n100\n100\n1\n100\n1\n100",
"output": "NO"
},
{
"input": "100\n100\n100\n100\n1\n100\n1\n1\n1\n100\n1\n1\n1\n1\n100\n1\n100\n1\n100\n1\n100\n100\n100\n1\n100\n1\n1\n1\n100\n1\n1\n1\n1\n1\n100\n100\n1\n100\n1\n1\n100\n1\n1\n100\n1\n100\n100\n100\n1\n100\n100\n100\n1\n100\n1\n100\n100\n100\n1\n1\n100\n100\n100\n100\n1\n100\n36\n100\n1\n100\n1\n100\n100\n100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n100\n1\n1\n100\n100\n100\n100\n100\n1\n100\n1\n100\n1\n1\n100\n100\n1\n100",
"output": "NO"
},
{
"input": "100\n2\n1\n1\n2\n2\n1\n1\n1\n1\n2\n1\n1\n1\n2\n2\n2\n1\n1\n1\n2\n1\n2\n2\n2\n2\n1\n1\n2\n1\n1\n2\n1\n27\n1\n1\n1\n2\n2\n2\n1\n2\n1\n2\n1\n1\n2\n2\n2\n2\n2\n2\n2\n2\n1\n2\n2\n2\n2\n1\n2\n1\n1\n1\n1\n1\n2\n1\n1\n1\n2\n2\n2\n2\n2\n2\n1\n1\n1\n1\n2\n2\n1\n2\n2\n1\n1\n1\n2\n1\n2\n2\n1\n1\n2\n1\n1\n1\n2\n2\n1",
"output": "NO"
},
{
"input": "100\n99\n99\n100\n99\n99\n100\n100\n100\n99\n100\n99\n99\n100\n99\n99\n99\n99\n99\n99\n100\n100\n100\n99\n100\n100\n99\n100\n99\n100\n100\n99\n100\n99\n99\n99\n100\n99\n10\n99\n100\n100\n100\n99\n100\n100\n100\n100\n100\n100\n100\n99\n100\n100\n100\n99\n99\n100\n99\n100\n99\n100\n100\n99\n99\n99\n99\n100\n99\n100\n100\n100\n100\n100\n100\n99\n99\n100\n100\n99\n99\n99\n99\n99\n99\n100\n99\n99\n100\n100\n99\n100\n99\n99\n100\n99\n99\n99\n99\n100\n100",
"output": "NO"
},
{
"input": "100\n29\n43\n43\n29\n43\n29\n29\n29\n43\n29\n29\n29\n29\n43\n29\n29\n29\n29\n43\n29\n29\n29\n43\n29\n29\n29\n43\n43\n43\n43\n43\n43\n29\n29\n43\n43\n43\n29\n43\n43\n43\n29\n29\n29\n43\n29\n29\n29\n43\n43\n43\n43\n29\n29\n29\n29\n43\n29\n43\n43\n29\n29\n43\n43\n29\n29\n95\n29\n29\n29\n43\n43\n29\n29\n29\n29\n29\n43\n43\n43\n43\n29\n29\n43\n43\n43\n43\n43\n43\n29\n43\n43\n43\n43\n43\n43\n29\n43\n29\n43",
"output": "NO"
},
{
"input": "100\n98\n98\n98\n88\n88\n88\n88\n98\n98\n88\n98\n88\n98\n88\n88\n88\n88\n88\n98\n98\n88\n98\n98\n98\n88\n88\n88\n98\n98\n88\n88\n88\n98\n88\n98\n88\n98\n88\n88\n98\n98\n98\n88\n88\n98\n98\n88\n88\n88\n88\n88\n98\n98\n98\n88\n98\n88\n88\n98\n98\n88\n98\n88\n88\n98\n88\n88\n98\n27\n88\n88\n88\n98\n98\n88\n88\n98\n98\n98\n98\n98\n88\n98\n88\n98\n98\n98\n98\n88\n88\n98\n88\n98\n88\n98\n98\n88\n98\n98\n88",
"output": "NO"
},
{
"input": "100\n50\n1\n1\n50\n50\n50\n50\n1\n50\n100\n50\n50\n50\n100\n1\n100\n1\n100\n50\n50\n50\n50\n50\n1\n50\n1\n100\n1\n1\n50\n100\n50\n50\n100\n50\n50\n100\n1\n50\n50\n100\n1\n1\n50\n1\n100\n50\n50\n100\n100\n1\n100\n1\n50\n100\n50\n50\n1\n1\n50\n100\n50\n100\n100\n100\n50\n50\n1\n1\n50\n100\n1\n50\n100\n100\n1\n50\n50\n50\n100\n50\n50\n100\n1\n50\n50\n50\n50\n1\n50\n50\n50\n50\n1\n50\n50\n100\n1\n50\n100",
"output": "NO"
},
{
"input": "100\n45\n45\n45\n45\n45\n45\n44\n44\n44\n43\n45\n44\n44\n45\n44\n44\n45\n44\n43\n44\n43\n43\n43\n45\n43\n45\n44\n45\n43\n44\n45\n45\n45\n45\n45\n45\n45\n45\n43\n45\n43\n43\n45\n44\n45\n45\n45\n44\n45\n45\n45\n45\n45\n45\n44\n43\n45\n45\n43\n44\n45\n45\n45\n45\n44\n45\n45\n45\n43\n43\n44\n44\n43\n45\n43\n45\n45\n45\n44\n44\n43\n43\n44\n44\n44\n43\n45\n43\n44\n43\n45\n43\n43\n45\n45\n44\n45\n43\n43\n45",
"output": "NO"
},
{
"input": "100\n12\n12\n97\n15\n97\n12\n15\n97\n12\n97\n12\n12\n97\n12\n15\n12\n12\n15\n12\n12\n97\n12\n12\n15\n15\n12\n97\n15\n12\n97\n15\n12\n12\n15\n15\n15\n97\n15\n97\n12\n12\n12\n12\n12\n97\n12\n97\n12\n15\n15\n12\n15\n12\n15\n12\n12\n12\n12\n12\n12\n12\n12\n97\n97\n12\n12\n97\n12\n97\n97\n15\n97\n12\n97\n97\n12\n12\n12\n97\n97\n15\n12\n12\n15\n12\n15\n97\n97\n12\n15\n12\n12\n97\n12\n15\n15\n15\n15\n12\n12",
"output": "NO"
},
{
"input": "12\n2\n3\n1\n3\n3\n1\n2\n1\n2\n1\n3\n2",
"output": "NO"
},
{
"input": "48\n99\n98\n100\n100\n99\n100\n99\n100\n100\n98\n99\n98\n98\n99\n98\n99\n98\n100\n100\n98\n100\n98\n99\n100\n98\n99\n98\n99\n99\n100\n98\n99\n99\n98\n100\n99\n98\n99\n98\n100\n100\n100\n99\n98\n99\n98\n100\n100",
"output": "NO"
},
{
"input": "4\n1\n3\n3\n3",
"output": "NO"
},
{
"input": "6\n1\n1\n1\n1\n2\n2",
"output": "NO"
},
{
"input": "4\n1\n1\n1\n2",
"output": "NO"
},
{
"input": "4\n1\n2\n2\n2",
"output": "NO"
},
{
"input": "4\n1\n2\n3\n4",
"output": "NO"
},
{
"input": "8\n1\n1\n2\n2\n3\n3\n4\n4",
"output": "NO"
},
{
"input": "4\n1\n3\n2\n4",
"output": "NO"
},
{
"input": "4\n10\n10\n10\n20",
"output": "NO"
},
{
"input": "4\n11\n12\n13\n13",
"output": "NO"
},
{
"input": "4\n1\n1\n1\n3",
"output": "NO"
},
{
"input": "6\n1\n1\n2\n2\n2\n2",
"output": "NO"
},
{
"input": "10\n1\n1\n2\n2\n2\n3\n3\n4\n4\n4",
"output": "NO"
}
] | 1,591,266,563 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 28 | 140 | 0 | #for i in range(int(input())):
#arr=list(map(int,input().split()))
#ls=sorted(arr)
#a,b,f,k=map(int,input().split())
n=int(input())
d={}
for i in range(n):
s=int(input())
if s not in d:
d[s]=1
else:
d[s]+=1
if len(d.keys())==2:
print("YES")
for i in d.keys():
print(i,end=" ")
else:
print("NO")
| Title: Fair Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya and Vasya decided to play a game. They have *n* cards (*n* is an even number). A single integer is written on each card.
Before the game Petya will choose an integer and after that Vasya will choose another integer (different from the number that Petya chose). During the game each player takes all the cards with number he chose. For example, if Petya chose number 5 before the game he will take all cards on which 5 is written and if Vasya chose number 10 before the game he will take all cards on which 10 is written.
The game is considered fair if Petya and Vasya can take all *n* cards, and the number of cards each player gets is the same.
Determine whether Petya and Vasya can choose integer numbers before the game so that the game is fair.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — number of cards. It is guaranteed that *n* is an even number.
The following *n* lines contain a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (one integer per line, 1<=≤<=*a**i*<=≤<=100) — numbers written on the *n* cards.
Output Specification:
If it is impossible for Petya and Vasya to choose numbers in such a way that the game will be fair, print "NO" (without quotes) in the first line. In this case you should not print anything more.
In the other case print "YES" (without quotes) in the first line. In the second line print two distinct integers — number that Petya should choose and the number that Vasya should choose to make the game fair. If there are several solutions, print any of them.
Demo Input:
['4\n11\n27\n27\n11\n', '2\n6\n6\n', '6\n10\n20\n30\n20\n10\n20\n', '6\n1\n1\n2\n2\n3\n3\n']
Demo Output:
['YES\n11 27\n', 'NO\n', 'NO\n', 'NO\n']
Note:
In the first example the game will be fair if, for example, Petya chooses number 11, and Vasya chooses number 27. Then the will take all cards — Petya will take cards 1 and 4, and Vasya will take cards 2 and 3. Thus, each of them will take exactly two cards.
In the second example fair game is impossible because the numbers written on the cards are equal, but the numbers that Petya and Vasya should choose should be distinct.
In the third example it is impossible to take all cards. Petya and Vasya can take at most five cards — for example, Petya can choose number 10 and Vasya can choose number 20. But for the game to be fair it is necessary to take 6 cards. | ```python
#for i in range(int(input())):
#arr=list(map(int,input().split()))
#ls=sorted(arr)
#a,b,f,k=map(int,input().split())
n=int(input())
d={}
for i in range(n):
s=int(input())
if s not in d:
d[s]=1
else:
d[s]+=1
if len(d.keys())==2:
print("YES")
for i in d.keys():
print(i,end=" ")
else:
print("NO")
``` | 0 | |
327 | A | Flipping Game | PROGRAMMING | 1,200 | [
"brute force",
"dp",
"implementation"
] | null | null | Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1. | Print an integer — the maximal number of 1s that can be obtained after exactly one move. | [
"5\n1 0 0 1 0\n",
"4\n1 0 0 1\n"
] | [
"4\n",
"4\n"
] | In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1. | 500 | [
{
"input": "5\n1 0 0 1 0",
"output": "4"
},
{
"input": "4\n1 0 0 1",
"output": "4"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "8\n1 0 0 0 1 0 0 0",
"output": "7"
},
{
"input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "18"
},
{
"input": "23\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "22"
},
{
"input": "100\n0 1 0 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 0 1 1 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1",
"output": "70"
},
{
"input": "100\n0 1 1 0 1 0 0 1 0 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 0 0 1 0 0 0 0 1 1 1 1",
"output": "60"
},
{
"input": "18\n0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 0",
"output": "11"
},
{
"input": "25\n0 1 0 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 1 0 0 1 1 0 1",
"output": "18"
},
{
"input": "55\n0 0 1 1 0 0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 1 1 1 0 1 1 0 0 0 0 1 1 0 1 1 1 1",
"output": "36"
},
{
"input": "75\n1 1 0 1 0 1 1 0 0 0 0 0 1 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 1 0 1 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 0 1 1 0 0 0 1 0 0 0",
"output": "44"
},
{
"input": "100\n0 0 1 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 0 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 1 0 0 0 0 1 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 0 1 0 1 0 0 1 0 1 1 1 0 0 0 0 1 0 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1",
"output": "61"
},
{
"input": "100\n0 0 0 1 0 0 0 1 0 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 1 1 0 0 0 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 1 1 0 1 1 0",
"output": "61"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "100"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "99"
},
{
"input": "100\n0 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 1 1 1 1 1 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 0 1 1 1 0 1 1 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 1 0 0 0 0 1 0",
"output": "61"
},
{
"input": "100\n0 1 1 1 1 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 0 1 0 0 0 0 1 0 1 0 1 0 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 1",
"output": "59"
},
{
"input": "99\n1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 1 1 0 0 1 0 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 1 1",
"output": "61"
},
{
"input": "2\n1 1",
"output": "1"
}
] | 1,683,776,063 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 6 | 92 | 1,638,400 | n = int(input())
a = list(map(int, input().split()))
ans = 0
if n == 1 and a[0] == 0:
print(1)
else:
for i in range(n):
for j in range(i):
s = sum(a)
for k in range(j, i + 1):
if a[k] == 1:
s -= 1
if a[k] == 0:
s += 1
ans = max(ans, s)
print(ans)
# 4
# 1 0 0 1
# i 0 1 2 3 4
# j 0 1
# a 1 0 0 1
# ans 0
| Title: Flipping Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1.
Output Specification:
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Demo Input:
['5\n1 0 0 1 0\n', '4\n1 0 0 1\n']
Demo Output:
['4\n', '4\n']
Note:
In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1. | ```python
n = int(input())
a = list(map(int, input().split()))
ans = 0
if n == 1 and a[0] == 0:
print(1)
else:
for i in range(n):
for j in range(i):
s = sum(a)
for k in range(j, i + 1):
if a[k] == 1:
s -= 1
if a[k] == 0:
s += 1
ans = max(ans, s)
print(ans)
# 4
# 1 0 0 1
# i 0 1 2 3 4
# j 0 1
# a 1 0 0 1
# ans 0
``` | 0 | |
17 | A | Noldbach problem | PROGRAMMING | 1,000 | [
"brute force",
"math",
"number theory"
] | A. Noldbach problem | 2 | 64 | Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong. | The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000). | Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO. | [
"27 2\n",
"45 7\n"
] | [
"YES",
"NO"
] | In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form. | 0 | [
{
"input": "27 2",
"output": "YES"
},
{
"input": "45 7",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
},
{
"input": "15 1",
"output": "YES"
},
{
"input": "17 1",
"output": "YES"
},
{
"input": "34 5",
"output": "NO"
},
{
"input": "37 4",
"output": "YES"
},
{
"input": "43 5",
"output": "YES"
},
{
"input": "47 7",
"output": "NO"
},
{
"input": "50 5",
"output": "YES"
},
{
"input": "57 6",
"output": "YES"
},
{
"input": "60 8",
"output": "NO"
},
{
"input": "62 7",
"output": "YES"
},
{
"input": "76 9",
"output": "NO"
},
{
"input": "69 7",
"output": "YES"
},
{
"input": "113 10",
"output": "YES"
},
{
"input": "141 11",
"output": "YES"
},
{
"input": "207 16",
"output": "NO"
},
{
"input": "231 18",
"output": "NO"
},
{
"input": "296 19",
"output": "YES"
},
{
"input": "332 20",
"output": "YES"
},
{
"input": "378 24",
"output": "NO"
},
{
"input": "428 23",
"output": "YES"
},
{
"input": "497 27",
"output": "NO"
},
{
"input": "640 32",
"output": "YES"
},
{
"input": "798 35",
"output": "YES"
},
{
"input": "802 35",
"output": "YES"
},
{
"input": "864 40",
"output": "NO"
},
{
"input": "953 44",
"output": "NO"
},
{
"input": "995 44",
"output": "YES"
},
{
"input": "1000 44",
"output": "YES"
},
{
"input": "1000 0",
"output": "YES"
},
{
"input": "1000 1000",
"output": "NO"
},
{
"input": "2 1000",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
}
] | 1,444,987,733 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 31 | 92 | 0 | def sieve(n):
np1 = n + 1
s = list(range(np1))
s[1] = 0
sqrtn = int(round(n**0.5))
for i in range(2, sqrtn + 1):
if s[i]:
s[i*i: np1: i] = [0] * len(range(i*i, np1, i))
return filter(None, s)
n, k = [int(i) for i in input().split()]
numbers = list(sieve(n))
ans = 0
for i in range(5, len(numbers)):
current = numbers[i] - 1
for idx2 in range(0, i):
if idx2 == i - 1:
break
if numbers[idx2] + numbers[idx2 +1] == current:
ans += 1
if ans == k:
print('YES')
else: print('NO') | Title: Noldbach problem
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong.
Input Specification:
The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000).
Output Specification:
Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO.
Demo Input:
['27 2\n', '45 7\n']
Demo Output:
['YES', 'NO']
Note:
In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form. | ```python
def sieve(n):
np1 = n + 1
s = list(range(np1))
s[1] = 0
sqrtn = int(round(n**0.5))
for i in range(2, sqrtn + 1):
if s[i]:
s[i*i: np1: i] = [0] * len(range(i*i, np1, i))
return filter(None, s)
n, k = [int(i) for i in input().split()]
numbers = list(sieve(n))
ans = 0
for i in range(5, len(numbers)):
current = numbers[i] - 1
for idx2 in range(0, i):
if idx2 == i - 1:
break
if numbers[idx2] + numbers[idx2 +1] == current:
ans += 1
if ans == k:
print('YES')
else: print('NO')
``` | 0 |
771 | A | Bear and Friendship Condition | PROGRAMMING | 1,500 | [
"dfs and similar",
"dsu",
"graphs"
] | null | null | Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes. | The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input. | If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes). | [
"4 3\n1 3\n3 4\n1 4\n",
"4 4\n3 1\n2 3\n3 4\n1 2\n",
"10 4\n4 3\n5 10\n8 9\n1 2\n",
"3 2\n1 2\n2 3\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"NO\n"
] | The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not. | 250 | [
{
"input": "4 3\n1 3\n3 4\n1 4",
"output": "YES"
},
{
"input": "4 4\n3 1\n2 3\n3 4\n1 2",
"output": "NO"
},
{
"input": "10 4\n4 3\n5 10\n8 9\n1 2",
"output": "YES"
},
{
"input": "3 2\n1 2\n2 3",
"output": "NO"
},
{
"input": "3 0",
"output": "YES"
},
{
"input": "15 42\n8 1\n3 14\n7 14\n12 3\n7 9\n6 7\n6 12\n14 12\n3 10\n10 14\n6 3\n3 13\n13 10\n7 12\n7 2\n6 10\n11 4\n9 3\n8 4\n7 3\n2 3\n2 10\n9 13\n2 14\n6 14\n13 2\n1 4\n13 6\n7 10\n13 14\n12 10\n13 7\n12 2\n9 10\n13 12\n2 6\n9 14\n6 9\n12 9\n11 1\n2 9\n11 8",
"output": "YES"
},
{
"input": "20 80\n17 4\n10 1\n11 10\n17 7\n15 10\n14 15\n13 1\n18 13\n3 13\n12 7\n9 13\n10 12\n14 12\n18 11\n4 7\n10 13\n11 3\n19 8\n14 7\n10 17\n14 3\n7 11\n11 14\n19 5\n10 14\n15 17\n3 1\n9 10\n11 1\n4 1\n11 4\n9 1\n12 3\n13 7\n1 14\n11 12\n7 1\n9 12\n18 15\n17 3\n7 15\n4 10\n7 18\n7 9\n12 17\n14 18\n3 18\n18 17\n9 15\n14 4\n14 9\n9 18\n12 4\n7 10\n15 4\n4 18\n15 13\n1 12\n7 3\n13 11\n4 13\n5 8\n12 18\n12 15\n17 9\n11 15\n3 10\n18 10\n4 3\n15 3\n13 12\n9 4\n9 11\n14 17\n13 17\n3 9\n13 14\n1 17\n15 1\n17 11",
"output": "NO"
},
{
"input": "99 26\n64 17\n48 70\n71 50\n3 50\n9 60\n61 64\n53 50\n25 12\n3 71\n71 53\n3 53\n65 70\n9 25\n9 12\n59 56\n39 60\n64 69\n65 94\n70 94\n25 60\n60 12\n94 48\n17 69\n61 17\n65 48\n61 69",
"output": "NO"
},
{
"input": "3 1\n1 2",
"output": "YES"
},
{
"input": "3 2\n3 2\n1 3",
"output": "NO"
},
{
"input": "3 3\n2 3\n1 2\n1 3",
"output": "YES"
},
{
"input": "4 2\n4 1\n2 1",
"output": "NO"
},
{
"input": "4 3\n3 1\n2 1\n3 2",
"output": "YES"
},
{
"input": "5 9\n1 2\n5 1\n3 1\n1 4\n2 4\n5 3\n5 4\n2 3\n5 2",
"output": "NO"
},
{
"input": "10 5\n9 5\n1 2\n6 8\n6 3\n10 6",
"output": "NO"
},
{
"input": "10 8\n10 7\n9 7\n5 7\n6 8\n3 5\n8 10\n3 4\n7 8",
"output": "NO"
},
{
"input": "10 20\n8 2\n8 3\n1 8\n9 5\n2 4\n10 1\n10 5\n7 5\n7 8\n10 7\n6 5\n3 7\n1 9\n9 8\n7 2\n2 10\n2 1\n6 4\n9 7\n4 3",
"output": "NO"
},
{
"input": "150000 10\n62562 50190\n48849 60549\n139470 18456\n21436 25159\n66845 120884\n99972 114453\n11631 99153\n62951 134848\n78114 146050\n136760 131762",
"output": "YES"
},
{
"input": "150000 0",
"output": "YES"
},
{
"input": "4 4\n1 2\n2 3\n3 4\n1 4",
"output": "NO"
},
{
"input": "30 73\n25 2\n2 16\n20 12\n16 20\n7 18\n11 15\n13 11\n30 29\n16 12\n12 25\n2 1\n18 14\n9 8\n28 16\n2 9\n22 21\n1 25\n12 28\n14 7\n4 9\n26 7\n14 27\n12 2\n29 22\n1 9\n13 15\n3 10\n1 12\n8 20\n30 24\n25 20\n4 1\n4 12\n20 1\n8 4\n2 28\n25 16\n16 8\n20 4\n9 12\n21 30\n23 11\n19 6\n28 4\n29 21\n9 28\n30 10\n22 24\n25 8\n27 26\n25 4\n28 20\n9 25\n24 29\n20 9\n18 26\n1 28\n30 22\n23 15\n28 27\n8 2\n23 13\n12 8\n14 26\n16 4\n28 25\n8 1\n4 2\n9 16\n20 2\n18 27\n28 8\n27 7",
"output": "NO"
},
{
"input": "5 4\n1 2\n2 5\n3 4\n4 5",
"output": "NO"
},
{
"input": "4 4\n1 2\n2 3\n3 4\n4 1",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 4\n4 3\n1 5\n5 6\n6 3",
"output": "NO"
},
{
"input": "3 2\n1 2\n1 3",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 4\n4 5\n5 6\n1 6",
"output": "NO"
},
{
"input": "4 4\n1 2\n1 3\n2 4\n3 4",
"output": "NO"
},
{
"input": "6 9\n1 4\n1 5\n1 6\n2 4\n2 5\n2 6\n3 4\n3 5\n3 6",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n3 4",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n2 4",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 4\n4 5\n5 6\n6 1",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n1 4",
"output": "NO"
},
{
"input": "150000 3\n150000 149999\n149998 149999\n149998 150000",
"output": "YES"
},
{
"input": "10 15\n1 2\n1 3\n2 4\n2 5\n3 4\n3 5\n4 5\n1 6\n6 7\n6 8\n7 9\n7 10\n8 9\n8 10\n9 10",
"output": "NO"
},
{
"input": "5 5\n1 2\n2 3\n3 4\n4 5\n5 1",
"output": "NO"
},
{
"input": "5 4\n1 2\n1 3\n1 4\n1 5",
"output": "NO"
},
{
"input": "9 9\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 1",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n1 4\n2 3\n3 4",
"output": "NO"
},
{
"input": "4 4\n1 2\n1 3\n4 2\n4 3",
"output": "NO"
},
{
"input": "5 5\n1 4\n5 3\n5 1\n2 3\n4 5",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n2 3\n1 4\n2 4",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n1 4\n2 3\n2 4",
"output": "NO"
},
{
"input": "6 12\n1 2\n1 4\n1 5\n1 6\n2 3\n2 5\n2 6\n3 4\n3 5\n3 6\n4 5\n4 6",
"output": "NO"
},
{
"input": "1000 4\n100 132\n100 164\n196 132\n196 164",
"output": "NO"
}
] | 1,587,957,385 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 1,000 | 30,003,200 | from collections import defaultdict as dd
def solve():
n, m = map(int, input().split())
edges = dd(set)
for i in range(m):
u, v = map(int, input().split())
edges[u].add(v)
edges[v].add(u)
for u in range(1, n + 1):
for v in edges[u]:
for w in edges[v]:
if w not in edges[u] and w != u:
print ("NO")
return
print ("YES")
return
solve()
| Title: Bear and Friendship Condition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.
Input Specification:
The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input.
Output Specification:
If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).
Demo Input:
['4 3\n1 3\n3 4\n1 4\n', '4 4\n3 1\n2 3\n3 4\n1 2\n', '10 4\n4 3\n5 10\n8 9\n1 2\n', '3 2\n1 2\n2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n']
Note:
The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not. | ```python
from collections import defaultdict as dd
def solve():
n, m = map(int, input().split())
edges = dd(set)
for i in range(m):
u, v = map(int, input().split())
edges[u].add(v)
edges[v].add(u)
for u in range(1, n + 1):
for v in edges[u]:
for w in edges[v]:
if w not in edges[u] and w != u:
print ("NO")
return
print ("YES")
return
solve()
``` | 0 | |
379 | A | New Year Candles | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number. | The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000). | Print a single integer — the number of hours Vasily can light up the room for. | [
"4 2\n",
"6 3\n"
] | [
"7\n",
"8\n"
] | Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours. | 500 | [
{
"input": "4 2",
"output": "7"
},
{
"input": "6 3",
"output": "8"
},
{
"input": "1000 1000",
"output": "1001"
},
{
"input": "123 5",
"output": "153"
},
{
"input": "1000 2",
"output": "1999"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "3 2",
"output": "5"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "999 2",
"output": "1997"
},
{
"input": "1000 3",
"output": "1499"
},
{
"input": "1000 4",
"output": "1333"
},
{
"input": "1 1000",
"output": "1"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "10 4",
"output": "13"
},
{
"input": "4 3",
"output": "5"
},
{
"input": "91 5",
"output": "113"
},
{
"input": "777 17",
"output": "825"
},
{
"input": "100 4",
"output": "133"
},
{
"input": "5 3",
"output": "7"
},
{
"input": "6 4",
"output": "7"
},
{
"input": "26 8",
"output": "29"
},
{
"input": "9 4",
"output": "11"
},
{
"input": "20 3",
"output": "29"
},
{
"input": "17 3",
"output": "25"
}
] | 1,651,378,235 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 46 | 0 | a, b = [int(i) for i in input().split()]
t = a
while a > 1:
t += a//b
a /= b
print(int(t)) | Title: New Year Candles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input Specification:
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
Output Specification:
Print a single integer — the number of hours Vasily can light up the room for.
Demo Input:
['4 2\n', '6 3\n']
Demo Output:
['7\n', '8\n']
Note:
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours. | ```python
a, b = [int(i) for i in input().split()]
t = a
while a > 1:
t += a//b
a /= b
print(int(t))
``` | 0 | |
731 | A | Night at the Museum | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it. | The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters. | Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input. | [
"zeus\n",
"map\n",
"ares\n"
] | [
"18\n",
"35\n",
"34\n"
] | To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations). | 500 | [
{
"input": "zeus",
"output": "18"
},
{
"input": "map",
"output": "35"
},
{
"input": "ares",
"output": "34"
},
{
"input": "l",
"output": "11"
},
{
"input": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv",
"output": "99"
},
{
"input": "gngvi",
"output": "44"
},
{
"input": "aaaaa",
"output": "0"
},
{
"input": "a",
"output": "0"
},
{
"input": "z",
"output": "1"
},
{
"input": "vyadeehhikklnoqrs",
"output": "28"
},
{
"input": "jjiihhhhgggfedcccbazyxx",
"output": "21"
},
{
"input": "fyyptqqxuciqvwdewyppjdzur",
"output": "117"
},
{
"input": "fqcnzmzmbobmancqcoalzmanaobpdse",
"output": "368"
},
{
"input": "zzzzzaaaaaaazzzzzzaaaaaaazzzzzzaaaazzzza",
"output": "8"
},
{
"input": "aucnwhfixuruefkypvrvnvznwtjgwlghoqtisbkhuwxmgzuljvqhmnwzisnsgjhivnjmbknptxatdkelhzkhsuxzrmlcpeoyukiy",
"output": "644"
},
{
"input": "sssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss",
"output": "8"
},
{
"input": "nypjygrdtpzpigzyrisqeqfriwgwlengnezppgttgtndbrryjdl",
"output": "421"
},
{
"input": "pnllnnmmmmoqqqqqrrtssssuuvtsrpopqoonllmonnnpppopnonoopooqpnopppqppqstuuuwwwwvxzxzzaa",
"output": "84"
},
{
"input": "btaoahqgxnfsdmzsjxgvdwjukcvereqeskrdufqfqgzqfsftdqcthtkcnaipftcnco",
"output": "666"
},
{
"input": "eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeerrrrrrrrrrrrrrrrwwwwwwwwww",
"output": "22"
},
{
"input": "uyknzcrwjyzmscqucclvacmorepdgmnyhmakmmnygqwglrxkxhkpansbmruwxdeoprxzmpsvwackopujxbbkpwyeggsvjykpxh",
"output": "643"
},
{
"input": "gzwpooohffcxwtpjgfzwtooiccxsrrokezutoojdzwsrmmhecaxwrojcbyrqlfdwwrliiib",
"output": "245"
},
{
"input": "dbvnkktasjdwqsrzfwwtmjgbcxggdxsoeilecihduypktkkbwfbruxzzhlttrssicgdwqruddwrlbtxgmhdbatzvdxbbro",
"output": "468"
},
{
"input": "mdtvowlktxzzbuaeiuebfeorgbdczauxsovbucactkvyvemsknsjfhifqgycqredzchipmkvzbxdjkcbyukomjlzvxzoswumned",
"output": "523"
},
{
"input": "kkkkkkkaaaaxxaaaaaaaxxxxxxxxaaaaaaxaaaaaaaaaakkkkkkkkkaaaaaaannnnnxxxxkkkkkkkkaannnnnnna",
"output": "130"
},
{
"input": "dffiknqqrsvwzcdgjkmpqtuwxadfhkkkmpqrtwxyadfggjmpppsuuwyyzcdgghhknnpsvvvwwwyabccffiloqruwwyyzabeeehh",
"output": "163"
},
{
"input": "qpppmmkjihgecbyvvsppnnnkjiffeebaaywutrrqpmkjhgddbzzzywtssssqnmmljheddbbaxvusrqonmlifedbbzyywwtqnkheb",
"output": "155"
},
{
"input": "wvvwwwvvwxxxyyyxxwwvwwvuttttttuvvwxxwxxyxxwwwwwvvuttssrssstsssssrqpqqppqrssrsrrssrssssrrsrqqrrqpppqp",
"output": "57"
},
{
"input": "dqcpcobpcobnznamznamzlykxkxlxlylzmaobnaobpbnanbpcoaobnboaoboanzlymzmykylymylzlylymanboanaocqdqesfrfs",
"output": "1236"
},
{
"input": "nnnnnnnnnnnnnnnnnnnnaaaaaaaaaaaaaaaaaaaakkkkkkkkkkkkkkkkkkkkkkaaaaaaaaaaaaaaaaaaaaxxxxxxxxxxxxxxxxxx",
"output": "49"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "cgilqsuwzaffilptwwbgmnttyyejkorxzflqvzbddhmnrvxchijpuwaeiimosxyycejlpquuwbfkpvbgijkqvxybdjjjptxcfkqt",
"output": "331"
},
{
"input": "ufsepwgtzgtgjssxaitgpailuvgqweoppszjwhoxdhhhpwwdorwfrdjwcdekxiktwziqwbkvbknrtvajpyeqbjvhiikxxaejjpte",
"output": "692"
},
{
"input": "uhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuh",
"output": "1293"
},
{
"input": "vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvgggggggggggggggggggggggggggggggggggggggggggggggggg",
"output": "16"
},
{
"input": "lyidmjyzbszgiwkxhhpnnthfwcvvstueionspfrvqgkvngmwyhezlosrpdnbvtcjjxxsykixwnepbumaacdzadlqhnjlcejovple",
"output": "616"
},
{
"input": "etzqqbaveffalkdguunfmyyrzkccnxmlluxeasqmopxzfvlkbhipqdwjgrttoemruohgwukfisdhznqyvhswbbypoxgtxyappcrl",
"output": "605"
},
{
"input": "lizussgedcbdjhrbeskhgatyozvwwekanlggcstijrniivupmcoofbaxfqrxddyzzptwxcftlhajsmmkkriarrqtkoauhcqefyud",
"output": "549"
},
{
"input": "dvjuvgfdogpknmbowlsfjzcimnygbtjiucyeeroqwhmzwpjqxlbjkqawrdtmvxbiqufllfuqibxvmtdrwaqkjblxqjpwzmhwqore",
"output": "688"
},
{
"input": "eeycuijtbgynmiczjfslwobmnkpgodfgvujvduyfeqchuaoktqrrairkkmmsjahltfcxwtpzzyddxrqfxabfoocmpuviinrjitsc",
"output": "604"
},
{
"input": "cgglnakewwvzoytaghksebrhjdbcdegssuzilrcppayxtgxopybbwshvyqnzhdsifkuwghourmeottrgjwdqpihbklvfzxpomqsa",
"output": "572"
},
{
"input": "aexullmxncckzryymfnuugdklaffevabqqztelpvojecljnhqldazdcaamubpenwxikysxxjjctvbndprsolzehywmgnvkgqvrfp",
"output": "609"
},
{
"input": "psnoieutsvvcwfhtnnphhxkwigzsbzyjmdiyl",
"output": "223"
},
{
"input": "aa",
"output": "0"
}
] | 1,645,777,465 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 31 | 0 | __author__ = 'X230874'
str = input()
len = len(str)
ans = int(0)
ans = min( abs(ord(str[0])-ord('a')) , 26-abs(ord(str[0])-ord('a')) )
#print(ans)
for i in range(len-1):
# print( abs(ord(str[i])-ord(str[i+1]) ) , (26-abs(ord(str[i])-ord(str[i+1])) ) )
ans += min( abs(ord(str[i])-ord(str[i+1]) ) , (26-abs(ord(str[i])-ord(str[i+1])) ) )
print(ans) | Title: Night at the Museum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
Input Specification:
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
Output Specification:
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
Demo Input:
['zeus\n', 'map\n', 'ares\n']
Demo Output:
['18\n', '35\n', '34\n']
Note:
To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations). | ```python
__author__ = 'X230874'
str = input()
len = len(str)
ans = int(0)
ans = min( abs(ord(str[0])-ord('a')) , 26-abs(ord(str[0])-ord('a')) )
#print(ans)
for i in range(len-1):
# print( abs(ord(str[i])-ord(str[i+1]) ) , (26-abs(ord(str[i])-ord(str[i+1])) ) )
ans += min( abs(ord(str[i])-ord(str[i+1]) ) , (26-abs(ord(str[i])-ord(str[i+1])) ) )
print(ans)
``` | 3 | |
584 | A | Olesya and Rodion | PROGRAMMING | 1,000 | [
"math"
] | null | null | Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1. | The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by. | Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them. | [
"3 2\n"
] | [
"712"
] | none | 500 | [
{
"input": "3 2",
"output": "222"
},
{
"input": "2 2",
"output": "22"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 3",
"output": "33333"
},
{
"input": "10 7",
"output": "7777777777"
},
{
"input": "2 9",
"output": "99"
},
{
"input": "18 8",
"output": "888888888888888888"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 10",
"output": "-1"
},
{
"input": "100 5",
"output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "10 2",
"output": "2222222222"
},
{
"input": "18 10",
"output": "111111111111111110"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "7 6",
"output": "6666666"
},
{
"input": "4 4",
"output": "4444"
},
{
"input": "14 7",
"output": "77777777777777"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "2 8",
"output": "88"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 9",
"output": "99999"
},
{
"input": "4 8",
"output": "8888"
},
{
"input": "3 4",
"output": "444"
},
{
"input": "9 4",
"output": "444444444"
},
{
"input": "8 10",
"output": "11111110"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "20 3",
"output": "33333333333333333333"
},
{
"input": "15 10",
"output": "111111111111110"
},
{
"input": "31 4",
"output": "4444444444444444444444444444444"
},
{
"input": "18 9",
"output": "999999999999999999"
},
{
"input": "72 4",
"output": "444444444444444444444444444444444444444444444444444444444444444444444444"
},
{
"input": "76 8",
"output": "8888888888888888888888888888888888888888888888888888888888888888888888888888"
},
{
"input": "12 5",
"output": "555555555555"
},
{
"input": "54 5",
"output": "555555555555555555555555555555555555555555555555555555"
},
{
"input": "96 10",
"output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "15 9",
"output": "999999999999999"
},
{
"input": "100 2",
"output": "2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222"
},
{
"input": "99 3",
"output": "333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333"
},
{
"input": "98 4",
"output": "44444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444"
},
{
"input": "97 5",
"output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "100 6",
"output": "6666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666"
},
{
"input": "99 7",
"output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "98 8",
"output": "88888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888"
},
{
"input": "97 9",
"output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "100 10",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "3 10",
"output": "110"
},
{
"input": "2 4",
"output": "44"
},
{
"input": "2 10",
"output": "10"
},
{
"input": "10 10",
"output": "1111111110"
},
{
"input": "4 10",
"output": "1110"
},
{
"input": "25 10",
"output": "1111111111111111111111110"
},
{
"input": "50 10",
"output": "11111111111111111111111111111111111111111111111110"
},
{
"input": "5 10",
"output": "11110"
},
{
"input": "99 10",
"output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "100 10",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "20 10",
"output": "11111111111111111110"
},
{
"input": "99 7",
"output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "89 5",
"output": "55555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "6 10",
"output": "111110"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "100 7",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
}
] | 1,689,845,321 | 2,147,483,647 | Python 3 | OK | TESTS | 65 | 46 | 0 | l = input().split()
n = int(l[0])
t = int(l[1])
ans = 10**(n-1) -((10**(n-1))%t) + t
if len(str(ans))!=n:
print(-1)
else:
print(ans) | Title: Olesya and Rodion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
Input Specification:
The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.
Output Specification:
Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Demo Input:
['3 2\n']
Demo Output:
['712']
Note:
none | ```python
l = input().split()
n = int(l[0])
t = int(l[1])
ans = 10**(n-1) -((10**(n-1))%t) + t
if len(str(ans))!=n:
print(-1)
else:
print(ans)
``` | 3 | |
39 | J | Spelling Check | PROGRAMMING | 1,500 | [
"hashing",
"implementation",
"strings"
] | J. Spelling Check | 2 | 256 | Petya has noticed that when he types using a keyboard, he often presses extra buttons and adds extra letters to the words. Of course, the spell-checking system underlines the words for him and he has to click every word and choose the right variant. Petya got fed up with correcting his mistakes himself, that’s why he decided to invent the function that will correct the words itself. Petya started from analyzing the case that happens to him most of the time, when all one needs is to delete one letter for the word to match a word from the dictionary. Thus, Petya faces one mini-task: he has a printed word and a word from the dictionary, and he should delete one letter from the first word to get the second one. And now the very non-trivial question that Petya faces is: which letter should he delete? | The input data contains two strings, consisting of lower-case Latin letters. The length of each string is from 1 to 106 symbols inclusive, the first string contains exactly 1 symbol more than the second one. | In the first line output the number of positions of the symbols in the first string, after the deleting of which the first string becomes identical to the second one. In the second line output space-separated positions of these symbols in increasing order. The positions are numbered starting from 1. If it is impossible to make the first string identical to the second string by deleting one symbol, output one number 0. | [
"abdrakadabra\nabrakadabra\n",
"aa\na\n",
"competition\ncodeforces\n"
] | [
"1\n3\n",
"2\n1 2\n",
"0\n"
] | none | 0 | [
{
"input": "abdrakadabra\nabrakadabra",
"output": "1\n3 "
},
{
"input": "aa\na",
"output": "2\n1 2 "
},
{
"input": "competition\ncodeforces",
"output": "0"
},
{
"input": "ab\na",
"output": "1\n2 "
},
{
"input": "bb\nb",
"output": "2\n1 2 "
},
{
"input": "aab\nab",
"output": "2\n1 2 "
},
{
"input": "aabb\nabb",
"output": "2\n1 2 "
},
{
"input": "babaacaacaa\nbbaacaacaa",
"output": "1\n2 "
},
{
"input": "bccaabbcccc\nbccaabcccc",
"output": "2\n6 7 "
},
{
"input": "ababcaabaaa\nabacaabaaa",
"output": "1\n4 "
},
{
"input": "cccacaccacb\ncccacaccac",
"output": "1\n11 "
},
{
"input": "aaaaaaaaaaa\naaaaaaaaaa",
"output": "11\n1 2 3 4 5 6 7 8 9 10 11 "
},
{
"input": "lcaaxcbcjca\nccaaacccca",
"output": "0"
},
{
"input": "babbbtaamba\nbabbbaabba",
"output": "0"
},
{
"input": "xdfxmcnzpch\nazvotghvtk",
"output": "0"
},
{
"input": "ki\nb",
"output": "0"
},
{
"input": "vct\nie",
"output": "0"
},
{
"input": "feee\nsnl",
"output": "0"
},
{
"input": "cbxxxxzvks\ncbxxxzvks",
"output": "4\n3 4 5 6 "
},
{
"input": "qybldcgfhdhhhhhhhhhhopqkhuczzytzluiahwbqjltgafvvoecititchjwdoljiehubngmtjckqymldhoncgtqhxnqvoagnrmur\nqybldcgfhdhhhhhhhhhopqkhuczzytzluiahwbqjltgafvvoecititchjwdoljiehubngmtjckqymldhoncgtqhxnqvoagnrmur",
"output": "10\n11 12 13 14 15 16 17 18 19 20 "
}
] | 1,641,882,858 | 2,147,483,647 | PyPy 3-64 | MEMORY_LIMIT_EXCEEDED | TESTS | 15 | 996 | 268,390,400 | def det_pref(words):
cnt = dict(); pfx_len = [len(w) for w in words]
while any(pfx_len):
for i,w_pfx in [(i,words[i][:l]) for i,l in enumerate(pfx_len)]:
pfx_len[i] -= pfx_len[i] and 1 or 0; n = 1 + cnt.get(w_pfx,0)
if n/len(words)> 0.7: return len(w_pfx)
cnt[w_pfx] = n
return 0
k = input(); s = input(); l = det_pref([k, s]); r = det_pref([k[::-1], s[::-1]]); print(0 if l+1 < len(k)-r else min(l+1, len(k))+1 - max(len(k)-r, 1), '\n' + ' '.join([str(i) for i in range(max(len(k)-r, 1), min(l+1, len(k))+1)])) | Title: Spelling Check
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya has noticed that when he types using a keyboard, he often presses extra buttons and adds extra letters to the words. Of course, the spell-checking system underlines the words for him and he has to click every word and choose the right variant. Petya got fed up with correcting his mistakes himself, that’s why he decided to invent the function that will correct the words itself. Petya started from analyzing the case that happens to him most of the time, when all one needs is to delete one letter for the word to match a word from the dictionary. Thus, Petya faces one mini-task: he has a printed word and a word from the dictionary, and he should delete one letter from the first word to get the second one. And now the very non-trivial question that Petya faces is: which letter should he delete?
Input Specification:
The input data contains two strings, consisting of lower-case Latin letters. The length of each string is from 1 to 106 symbols inclusive, the first string contains exactly 1 symbol more than the second one.
Output Specification:
In the first line output the number of positions of the symbols in the first string, after the deleting of which the first string becomes identical to the second one. In the second line output space-separated positions of these symbols in increasing order. The positions are numbered starting from 1. If it is impossible to make the first string identical to the second string by deleting one symbol, output one number 0.
Demo Input:
['abdrakadabra\nabrakadabra\n', 'aa\na\n', 'competition\ncodeforces\n']
Demo Output:
['1\n3\n', '2\n1 2\n', '0\n']
Note:
none | ```python
def det_pref(words):
cnt = dict(); pfx_len = [len(w) for w in words]
while any(pfx_len):
for i,w_pfx in [(i,words[i][:l]) for i,l in enumerate(pfx_len)]:
pfx_len[i] -= pfx_len[i] and 1 or 0; n = 1 + cnt.get(w_pfx,0)
if n/len(words)> 0.7: return len(w_pfx)
cnt[w_pfx] = n
return 0
k = input(); s = input(); l = det_pref([k, s]); r = det_pref([k[::-1], s[::-1]]); print(0 if l+1 < len(k)-r else min(l+1, len(k))+1 - max(len(k)-r, 1), '\n' + ' '.join([str(i) for i in range(max(len(k)-r, 1), min(l+1, len(k))+1)]))
``` | 0 |
33 | A | What is for dinner? | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | A. What is for dinner? | 2 | 256 | In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one. | The first line contains three integers *n*, *m*, *k* (1<=≤<=*m*<=≤<=*n*<=≤<=1000,<=0<=≤<=*k*<=≤<=106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow *n* lines, each containing two integers: *r* (1<=≤<=*r*<=≤<=*m*) — index of the row, where belongs the corresponding tooth, and *c* (0<=≤<=*c*<=≤<=106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth. | In the first line output the maximum amount of crucians that Valerie can consume for dinner. | [
"4 3 18\n2 3\n1 2\n3 6\n2 3\n",
"2 2 13\n1 13\n2 12\n"
] | [
"11\n",
"13\n"
] | none | 500 | [
{
"input": "4 3 18\n2 3\n1 2\n3 6\n2 3",
"output": "11"
},
{
"input": "2 2 13\n1 13\n2 12",
"output": "13"
},
{
"input": "5 4 8\n4 6\n4 5\n1 3\n2 0\n3 3",
"output": "8"
},
{
"input": "1 1 0\n1 3",
"output": "0"
},
{
"input": "7 1 30\n1 8\n1 15\n1 5\n1 17\n1 9\n1 16\n1 16",
"output": "5"
},
{
"input": "4 2 8\n1 9\n1 10\n1 4\n2 6",
"output": "8"
},
{
"input": "10 4 14\n2 6\n1 5\n2 8\n2 6\n2 5\n4 1\n4 0\n2 4\n3 4\n1 0",
"output": "8"
},
{
"input": "54 22 1009\n15 7\n17 7\n11 9\n5 11\n12 9\n13 8\n13 12\n22 11\n20 9\n20 7\n16 11\n19 12\n3 12\n15 9\n1 12\n2 10\n16 10\n16 10\n14 10\n9 11\n9 9\n14 8\n10 10\n16 12\n1 8\n3 8\n21 11\n18 12\n2 6\n9 11\n11 7\n15 9\n20 11\n6 8\n8 8\n19 11\n15 7\n9 9\n22 7\n12 9\n17 9\n17 11\n12 7\n15 9\n5 11\n11 6\n18 8\n7 10\n4 10\n13 12\n4 8\n20 6\n15 12\n9 7",
"output": "180"
},
{
"input": "1 1 1000000\n1 1000000",
"output": "1000000"
},
{
"input": "4 3 181818\n3 1299\n1 1694\n3 1164\n2 1278",
"output": "4136"
},
{
"input": "50 50 0\n33 0\n26 0\n30 0\n41 0\n38 0\n36 0\n16 0\n7 0\n23 0\n5 0\n39 0\n29 0\n40 0\n47 0\n24 0\n43 0\n14 0\n10 0\n20 0\n3 0\n6 0\n34 0\n13 0\n18 0\n21 0\n9 0\n8 0\n22 0\n25 0\n4 0\n11 0\n17 0\n32 0\n44 0\n49 0\n50 0\n27 0\n28 0\n45 0\n12 0\n35 0\n15 0\n42 0\n31 0\n2 0\n1 0\n48 0\n19 0\n37 0\n46 0",
"output": "0"
},
{
"input": "19 12 199\n7 1\n8 6\n6 14\n1 7\n4 1\n6 6\n3 4\n1 5\n9 2\n5 3\n11 3\n9 4\n1 12\n4 7\n7 3\n12 14\n2 1\n10 8\n6 12",
"output": "54"
}
] | 1,616,073,352 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 60 | 0 | a=int(input())
b=input().split()
s=0
d=0
for i in range(1,a):
for n in range(0,i):
if int(b[i])>int(b[n]):
s+=1
if s==i:
d+=1
s=0
print(d)
| Title: What is for dinner?
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one.
Input Specification:
The first line contains three integers *n*, *m*, *k* (1<=≤<=*m*<=≤<=*n*<=≤<=1000,<=0<=≤<=*k*<=≤<=106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow *n* lines, each containing two integers: *r* (1<=≤<=*r*<=≤<=*m*) — index of the row, where belongs the corresponding tooth, and *c* (0<=≤<=*c*<=≤<=106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth.
Output Specification:
In the first line output the maximum amount of crucians that Valerie can consume for dinner.
Demo Input:
['4 3 18\n2 3\n1 2\n3 6\n2 3\n', '2 2 13\n1 13\n2 12\n']
Demo Output:
['11\n', '13\n']
Note:
none | ```python
a=int(input())
b=input().split()
s=0
d=0
for i in range(1,a):
for n in range(0,i):
if int(b[i])>int(b[n]):
s+=1
if s==i:
d+=1
s=0
print(d)
``` | -1 |
984 | A | Game | PROGRAMMING | 800 | [
"sortings"
] | null | null | Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves. | The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$). | Print one number that will be left on the board. | [
"3\n2 1 3\n",
"3\n2 2 2\n"
] | [
"2",
"2"
] | In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players. | 500 | [
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 2 2",
"output": "2"
},
{
"input": "9\n44 53 51 80 5 27 74 79 94",
"output": "53"
},
{
"input": "10\n38 82 23 37 96 4 81 60 67 86",
"output": "60"
},
{
"input": "10\n58 26 77 15 53 81 68 48 22 65",
"output": "53"
},
{
"input": "1\n124",
"output": "124"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 1 1000",
"output": "1"
},
{
"input": "2\n322 322",
"output": "322"
},
{
"input": "3\n9 92 12",
"output": "12"
},
{
"input": "3\n1 2 2",
"output": "2"
}
] | 1,612,634,224 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 62 | 307,200 | n=int(input())
a=list(map(int, input().split()))
a.sort()
if int(n%2)==0:
print(a[int(n/2)-1])
else:print(a[int(n/2)]) | Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
Input Specification:
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
Output Specification:
Print one number that will be left on the board.
Demo Input:
['3\n2 1 3\n', '3\n2 2 2\n']
Demo Output:
['2', '2']
Note:
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players. | ```python
n=int(input())
a=list(map(int, input().split()))
a.sort()
if int(n%2)==0:
print(a[int(n/2)-1])
else:print(a[int(n/2)])
``` | 3 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,694,511,778 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 30 | 0 | s = input()
if s.count('h') >= 1 and s.count('e') >= 1 and s.count('l') >= 2 and s.count('o') >= 1:
if s.index('h') < s.index('e') < s.index('l') < s.rindex('l') < s.index('o'):
print('YES')
else:
print('NO')
else:
print('NO') | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
s = input()
if s.count('h') >= 1 and s.count('e') >= 1 and s.count('l') >= 2 and s.count('o') >= 1:
if s.index('h') < s.index('e') < s.index('l') < s.rindex('l') < s.index('o'):
print('YES')
else:
print('NO')
else:
print('NO')
``` | 0 |
16 | A | Flag | PROGRAMMING | 800 | [
"implementation"
] | A. Flag | 2 | 64 | According to a new ISO standard, a flag of every country should have a chequered field *n*<=×<=*m*, each square should be of one of 10 colours, and the flag should be «striped»: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets the new ISO standard. | The first line of the input contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), *n* — the amount of rows, *m* — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following *n* lines contain *m* characters. Each character is a digit between 0 and 9, and stands for the colour of the corresponding square. | Output YES, if the flag meets the new ISO standard, and NO otherwise. | [
"3 3\n000\n111\n222\n",
"3 3\n000\n000\n111\n",
"3 3\n000\n111\n002\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 0 | [
{
"input": "3 3\n000\n111\n222",
"output": "YES"
},
{
"input": "3 3\n000\n000\n111",
"output": "NO"
},
{
"input": "3 3\n000\n111\n002",
"output": "NO"
},
{
"input": "10 10\n2222222222\n5555555555\n0000000000\n4444444444\n1111111111\n3333333393\n3333333333\n5555555555\n0000000000\n8888888888",
"output": "NO"
},
{
"input": "10 13\n4442444444444\n8888888888888\n6666666666666\n0000000000000\n3333333333333\n4444444444444\n7777777777777\n8388888888888\n1111111111111\n5555555555555",
"output": "NO"
},
{
"input": "10 8\n33333333\n44444444\n11111115\n81888888\n44444444\n11111111\n66666666\n33330333\n33333333\n33333333",
"output": "NO"
},
{
"input": "5 5\n88888\n44444\n66666\n55555\n88888",
"output": "YES"
},
{
"input": "20 19\n1111111111111111111\n5555555555555555555\n0000000000000000000\n3333333333333333333\n1111111111111111111\n2222222222222222222\n4444444444444444444\n5555555555555555555\n0000000000000000000\n4444444444444444444\n0000000000000000000\n5555555555555555555\n7777777777777777777\n9999999999999999999\n2222222222222222222\n4444444444444444444\n1111111111111111111\n6666666666666666666\n7777777777777777777\n2222222222222222222",
"output": "YES"
},
{
"input": "1 100\n8888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888",
"output": "YES"
},
{
"input": "100 1\n5\n7\n9\n4\n7\n2\n5\n1\n6\n7\n2\n7\n6\n8\n7\n4\n0\n2\n9\n8\n9\n1\n6\n4\n3\n4\n7\n1\n9\n3\n0\n8\n3\n1\n7\n5\n3\n9\n5\n1\n3\n5\n8\n1\n9\n3\n9\n0\n6\n0\n7\n6\n5\n2\n8\n3\n7\n6\n5\n1\n8\n3\n6\n9\n6\n0\n5\n8\n5\n2\n9\n1\n0\n1\n8\n3\n2\n1\n0\n3\n9\n0\n5\n1\n0\n4\n9\n3\n0\n4\n8\n4\n8\n6\n3\n0\n4\n6\n8\n4",
"output": "YES"
},
{
"input": "1 1\n2",
"output": "YES"
},
{
"input": "1 100\n1111111111111111111111111111111111111111111111111111111111111111111111111111181111111111111111111111",
"output": "NO"
},
{
"input": "100 1\n3\n6\n4\n3\n0\n2\n8\n7\n3\n2\n1\n7\n1\n3\n2\n3\n6\n9\n0\n8\n5\n9\n7\n9\n2\n1\n4\n5\n1\n9\n2\n5\n1\n4\n6\n4\n9\n1\n0\n2\n1\n4\n7\n1\n4\n8\n0\n9\n2\n1\n6\n2\n8\n6\n9\n5\n8\n6\n4\n5\n9\n2\n7\n4\n1\n5\n8\n0\n9\n5\n4\n6\n5\n0\n6\n3\n6\n9\n7\n2\n0\n9\n7\n3\n2\n4\n9\n4\n7\n1\n2\n3\n1\n7\n9\n1\n9\n0\n4\n0",
"output": "YES"
}
] | 1,666,155,023 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 92 | 0 | """ According to a new ISO standard, a flag of every country should have a chequered field n x m,
each square should be of one of 10 colours, and the flag should be «striped»:
each horizontal row of the flag should contain squares of the same colour,
and the colours of adjacent horizontal rows should be different.
Berland's government asked you to find out whether their flag meets the new ISO standard.
Input
The first line of the input contains numbers n and m (1 <= n, m <= 100),
n - the amount of rows,
m - the amount of columns on the flag of Berland.
Then there follows the description of the flag: each of the following n lines contain m characters.
Each character is a digit between 0 and 9, and stands for the colour of the corresponding square.
Output
Output YES, if the flag meets the new ISO standard, and NO otherwise. """
result = "YES"
n, m = map(int, input().split())
# print("n = " + str(n))
# print("m = " + str(m))
before = ""
for i in range(n):
s = input()
c = s[0:1]
if m != s.count(c):
result = "No"
break
if i == 0:
before = c
continue
if c == before:
result = "No"
break
before = c
print(result) | Title: Flag
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
According to a new ISO standard, a flag of every country should have a chequered field *n*<=×<=*m*, each square should be of one of 10 colours, and the flag should be «striped»: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets the new ISO standard.
Input Specification:
The first line of the input contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), *n* — the amount of rows, *m* — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following *n* lines contain *m* characters. Each character is a digit between 0 and 9, and stands for the colour of the corresponding square.
Output Specification:
Output YES, if the flag meets the new ISO standard, and NO otherwise.
Demo Input:
['3 3\n000\n111\n222\n', '3 3\n000\n000\n111\n', '3 3\n000\n111\n002\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
""" According to a new ISO standard, a flag of every country should have a chequered field n x m,
each square should be of one of 10 colours, and the flag should be «striped»:
each horizontal row of the flag should contain squares of the same colour,
and the colours of adjacent horizontal rows should be different.
Berland's government asked you to find out whether their flag meets the new ISO standard.
Input
The first line of the input contains numbers n and m (1 <= n, m <= 100),
n - the amount of rows,
m - the amount of columns on the flag of Berland.
Then there follows the description of the flag: each of the following n lines contain m characters.
Each character is a digit between 0 and 9, and stands for the colour of the corresponding square.
Output
Output YES, if the flag meets the new ISO standard, and NO otherwise. """
result = "YES"
n, m = map(int, input().split())
# print("n = " + str(n))
# print("m = " + str(m))
before = ""
for i in range(n):
s = input()
c = s[0:1]
if m != s.count(c):
result = "No"
break
if i == 0:
before = c
continue
if c == before:
result = "No"
break
before = c
print(result)
``` | 3.977 |
893 | C | Rumor | PROGRAMMING | 1,300 | [
"dfs and similar",
"graphs",
"greedy"
] | null | null | Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely. | The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once. | Print one number — the minimum amount of gold Vova has to spend in order to finish the quest. | [
"5 2\n2 5 3 4 8\n1 4\n4 5\n",
"10 0\n1 2 3 4 5 6 7 8 9 10\n",
"10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n"
] | [
"10\n",
"55\n",
"15\n"
] | In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters. | 0 | [
{
"input": "5 2\n2 5 3 4 8\n1 4\n4 5",
"output": "10"
},
{
"input": "10 0\n1 2 3 4 5 6 7 8 9 10",
"output": "55"
},
{
"input": "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10",
"output": "15"
},
{
"input": "1 0\n0",
"output": "0"
},
{
"input": "1 0\n1000000000",
"output": "1000000000"
},
{
"input": "2 0\n0 0",
"output": "0"
},
{
"input": "2 0\n1000000000 0",
"output": "1000000000"
},
{
"input": "2 0\n0 1000000000",
"output": "1000000000"
},
{
"input": "2 0\n1000000000 1000000000",
"output": "2000000000"
},
{
"input": "2 1\n0 0\n1 2",
"output": "0"
}
] | 1,689,004,353 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 3 | 233 | 23,040,000 | import sys
sys.setrecursionlimit(100000)
def dfs(v, k):
global d
d.append(value[v])
used[v] = k
for to in a[v]:
if used[to] == 0:
dfs(to, k)
n, m = list(map(int, input().split()))
value = list(map(int, input().split()))
a = []
for i in range(n):
a.append([])
for i in range(m):
b = list(map(int, input().split()))
a[b[0]-1].append(b[1]-1)
a[b[1]-1].append(b[0]-1)
used = [0]*n
k = 0
d = []
ans = 0
for i in range(n):
if used[i] == 0:
k+=1
dfs(i, k)
ans += min(d)
d = []
print(ans)
# Mon Jul 10 2023 18:52:32 GMT+0300 (Moscow Standard Time)
| Title: Rumor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely.
Input Specification:
The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once.
Output Specification:
Print one number — the minimum amount of gold Vova has to spend in order to finish the quest.
Demo Input:
['5 2\n2 5 3 4 8\n1 4\n4 5\n', '10 0\n1 2 3 4 5 6 7 8 9 10\n', '10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n']
Demo Output:
['10\n', '55\n', '15\n']
Note:
In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters. | ```python
import sys
sys.setrecursionlimit(100000)
def dfs(v, k):
global d
d.append(value[v])
used[v] = k
for to in a[v]:
if used[to] == 0:
dfs(to, k)
n, m = list(map(int, input().split()))
value = list(map(int, input().split()))
a = []
for i in range(n):
a.append([])
for i in range(m):
b = list(map(int, input().split()))
a[b[0]-1].append(b[1]-1)
a[b[1]-1].append(b[0]-1)
used = [0]*n
k = 0
d = []
ans = 0
for i in range(n):
if used[i] == 0:
k+=1
dfs(i, k)
ans += min(d)
d = []
print(ans)
# Mon Jul 10 2023 18:52:32 GMT+0300 (Moscow Standard Time)
``` | -1 | |
981 | G | Magic multisets | PROGRAMMING | 2,500 | [
"data structures"
] | null | null | In the School of Magic in Dirtpolis a lot of interesting objects are studied on Computer Science lessons.
Consider, for example, the magic multiset. If you try to add an integer to it that is already presented in the multiset, each element in the multiset duplicates. For example, if you try to add the integer $2$ to the multiset $\{1, 2, 3, 3\}$, you will get $\{1, 1, 2, 2, 3, 3, 3, 3\}$.
If you try to add an integer that is not presented in the multiset, it is simply added to it. For example, if you try to add the integer $4$ to the multiset $\{1, 2, 3, 3\}$, you will get $\{1, 2, 3, 3, 4\}$.
Also consider an array of $n$ initially empty magic multisets, enumerated from $1$ to $n$.
You are to answer $q$ queries of the form "add an integer $x$ to all multisets with indices $l, l + 1, \ldots, r$" and "compute the sum of sizes of multisets with indices $l, l + 1, \ldots, r$". The answers for the second type queries can be large, so print the answers modulo $998244353$. | The first line contains two integers $n$ and $q$ ($1 \leq n, q \leq 2 \cdot 10^{5}$) — the number of magic multisets in the array and the number of queries, respectively.
The next $q$ lines describe queries, one per line. Each line starts with an integer $t$ ($1 \leq t \leq 2$) — the type of the query. If $t$ equals $1$, it is followed by three integers $l$, $r$, $x$ ($1 \leq l \leq r \leq n$, $1 \leq x \leq n$) meaning that you should add $x$ to all multisets with indices from $l$ to $r$ inclusive. If $t$ equals $2$, it is followed by two integers $l$, $r$ ($1 \leq l \leq r \leq n$) meaning that you should compute the sum of sizes of all multisets with indices from $l$ to $r$ inclusive. | For each query of the second type print the sum of sizes of multisets on the given segment.
The answers can be large, so print them modulo $998244353$. | [
"4 4\n1 1 2 1\n1 1 2 2\n1 1 4 1\n2 1 4\n",
"3 7\n1 1 1 3\n1 1 1 3\n1 1 1 2\n1 1 1 1\n2 1 1\n1 1 1 2\n2 1 1\n"
] | [
"10\n",
"4\n8\n"
] | In the first example after the first two queries the multisets are equal to $[\{1, 2\},\{1, 2\},\{\},\{\}]$, after the third query they are equal to $[\{1, 1, 2, 2\},\{1, 1, 2, 2\},\{1\},\{1\}]$.
In the second example the first multiset evolves as follows:
$\{\} \to \{3\} \to \{3, 3\} \to \{2, 3, 3\} \to \{1, 2, 3, 3\} \to \{1, 1, 2, 2, 3, 3, 3, 3\}$. | 3,250 | [] | 1,527,436,206 | 3,606 | Python 3 | WRONG_ANSWER | PRETESTS | 2 | 2,605 | 0 | n, q = map(int,input().split())
m = 998244353
a = [[set(),0] for i in range(n)]
for i in range(q):
line = list(map(int,input().split()))
if line[0] == 1:
for j in range(line[1]-1,line[2]):
if line[3] in a[j][0]:
a[j][1] *= 2
else:
a[j][0].add(line[3])
a[j][1] += 1
a[j][1] = a[j][1] % m
else:
print(sum(a[j][1] for j in range(n)) % m) | Title: Magic multisets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the School of Magic in Dirtpolis a lot of interesting objects are studied on Computer Science lessons.
Consider, for example, the magic multiset. If you try to add an integer to it that is already presented in the multiset, each element in the multiset duplicates. For example, if you try to add the integer $2$ to the multiset $\{1, 2, 3, 3\}$, you will get $\{1, 1, 2, 2, 3, 3, 3, 3\}$.
If you try to add an integer that is not presented in the multiset, it is simply added to it. For example, if you try to add the integer $4$ to the multiset $\{1, 2, 3, 3\}$, you will get $\{1, 2, 3, 3, 4\}$.
Also consider an array of $n$ initially empty magic multisets, enumerated from $1$ to $n$.
You are to answer $q$ queries of the form "add an integer $x$ to all multisets with indices $l, l + 1, \ldots, r$" and "compute the sum of sizes of multisets with indices $l, l + 1, \ldots, r$". The answers for the second type queries can be large, so print the answers modulo $998244353$.
Input Specification:
The first line contains two integers $n$ and $q$ ($1 \leq n, q \leq 2 \cdot 10^{5}$) — the number of magic multisets in the array and the number of queries, respectively.
The next $q$ lines describe queries, one per line. Each line starts with an integer $t$ ($1 \leq t \leq 2$) — the type of the query. If $t$ equals $1$, it is followed by three integers $l$, $r$, $x$ ($1 \leq l \leq r \leq n$, $1 \leq x \leq n$) meaning that you should add $x$ to all multisets with indices from $l$ to $r$ inclusive. If $t$ equals $2$, it is followed by two integers $l$, $r$ ($1 \leq l \leq r \leq n$) meaning that you should compute the sum of sizes of all multisets with indices from $l$ to $r$ inclusive.
Output Specification:
For each query of the second type print the sum of sizes of multisets on the given segment.
The answers can be large, so print them modulo $998244353$.
Demo Input:
['4 4\n1 1 2 1\n1 1 2 2\n1 1 4 1\n2 1 4\n', '3 7\n1 1 1 3\n1 1 1 3\n1 1 1 2\n1 1 1 1\n2 1 1\n1 1 1 2\n2 1 1\n']
Demo Output:
['10\n', '4\n8\n']
Note:
In the first example after the first two queries the multisets are equal to $[\{1, 2\},\{1, 2\},\{\},\{\}]$, after the third query they are equal to $[\{1, 1, 2, 2\},\{1, 1, 2, 2\},\{1\},\{1\}]$.
In the second example the first multiset evolves as follows:
$\{\} \to \{3\} \to \{3, 3\} \to \{2, 3, 3\} \to \{1, 2, 3, 3\} \to \{1, 1, 2, 2, 3, 3, 3, 3\}$. | ```python
n, q = map(int,input().split())
m = 998244353
a = [[set(),0] for i in range(n)]
for i in range(q):
line = list(map(int,input().split()))
if line[0] == 1:
for j in range(line[1]-1,line[2]):
if line[3] in a[j][0]:
a[j][1] *= 2
else:
a[j][0].add(line[3])
a[j][1] += 1
a[j][1] = a[j][1] % m
else:
print(sum(a[j][1] for j in range(n)) % m)
``` | 0 | |
985 | D | Sand Fortress | PROGRAMMING | 2,100 | [
"binary search",
"constructive algorithms",
"math"
] | null | null | You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought *n* packs of sand with you. Let height *h**i* of the sand pillar on some spot *i* be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with *H* sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
- *h*1<=≤<=*H*: no sand from the leftmost spot should go over the fence; - For any |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen; - : you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold. | The only line contains two integer numbers *n* and *H* (1<=≤<=*n*,<=*H*<=≤<=1018) — the number of sand packs you have and the height of the fence, respectively. | Print the minimum number of spots you can occupy so the all the castle building conditions hold. | [
"5 2\n",
"6 8\n"
] | [
"3\n",
"3\n"
] | Here are the heights of some valid castles:
- *n* = 5, *H* = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...] - *n* = 6, *H* = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
- *n* = 5, *H* = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...] - *n* = 6, *H* = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | 0 | [
{
"input": "5 2",
"output": "3"
},
{
"input": "6 8",
"output": "3"
},
{
"input": "20 4",
"output": "7"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1414213562"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 1000000000000000000",
"output": "1"
},
{
"input": "1000000000000000000 1",
"output": "1999999999"
},
{
"input": "1036191544337895 45523434",
"output": "45523434"
},
{
"input": "1036191544337896 45523434",
"output": "45523435"
},
{
"input": "1036191544337895 45523433",
"output": "45523435"
},
{
"input": "1036191544337895 1",
"output": "64379858"
},
{
"input": "1036191544337895 1000000000000000000",
"output": "45523434"
},
{
"input": "30 3",
"output": "9"
},
{
"input": "30 4",
"output": "8"
},
{
"input": "6 100",
"output": "3"
},
{
"input": "7 100",
"output": "4"
},
{
"input": "1000000000000000000 99999999999",
"output": "1414213562"
},
{
"input": "999999997351043581 1000000000000000000",
"output": "1414213561"
},
{
"input": "911343366122896086 1416605974",
"output": "1350069158"
},
{
"input": "828974163639871882 2010864527",
"output": "1287613423"
},
{
"input": "696616491401388220 958775125",
"output": "1191798158"
},
{
"input": "999999999000000000 1",
"output": "1999999998"
},
{
"input": "961245465290770608 1687994843",
"output": "1386539192"
},
{
"input": "692106376966414549 974053139",
"output": "1186035874"
},
{
"input": "12 1",
"output": "6"
},
{
"input": "806680349368385877 1068656310",
"output": "1278847474"
}
] | 1,526,934,996 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 3 | 2,000 | 0 | import sys
def s(x):
return x*(x+1)//2
def f(h2,r2):
return s(h2+r2-1)-s(h2-1)+s(h2+r2)
n,h=map(int,input().split())
if n<=s(h):
for i in range(h+1):
if s(i)>=n:
print(i)
sys.exit(0)
l=-1
r=n
while r-l>1:
m=(l+r)//2
if f(h,m)<n:
l=m
else:
r=m
ans1=h+2*r
l=-1
r=n
while r-l>1:
m=(l+r)//2
if h+f(h,m)<n:
l=m
else:
r=m
ans2=h+2*r+1
print(min(ans1,ans2)) | Title: Sand Fortress
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought *n* packs of sand with you. Let height *h**i* of the sand pillar on some spot *i* be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with *H* sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
- *h*1<=≤<=*H*: no sand from the leftmost spot should go over the fence; - For any |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen; - : you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input Specification:
The only line contains two integer numbers *n* and *H* (1<=≤<=*n*,<=*H*<=≤<=1018) — the number of sand packs you have and the height of the fence, respectively.
Output Specification:
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Demo Input:
['5 2\n', '6 8\n']
Demo Output:
['3\n', '3\n']
Note:
Here are the heights of some valid castles:
- *n* = 5, *H* = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...] - *n* = 6, *H* = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
- *n* = 5, *H* = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...] - *n* = 6, *H* = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | ```python
import sys
def s(x):
return x*(x+1)//2
def f(h2,r2):
return s(h2+r2-1)-s(h2-1)+s(h2+r2)
n,h=map(int,input().split())
if n<=s(h):
for i in range(h+1):
if s(i)>=n:
print(i)
sys.exit(0)
l=-1
r=n
while r-l>1:
m=(l+r)//2
if f(h,m)<n:
l=m
else:
r=m
ans1=h+2*r
l=-1
r=n
while r-l>1:
m=(l+r)//2
if h+f(h,m)<n:
l=m
else:
r=m
ans2=h+2*r+1
print(min(ans1,ans2))
``` | 0 | |
205 | A | Little Elephant and Rozdil | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities. | Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes). | [
"2\n7 4\n",
"7\n7 4 47 100 4 9 12\n"
] | [
"2\n",
"Still Rozdil\n"
] | In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil". | 500 | [
{
"input": "2\n7 4",
"output": "2"
},
{
"input": "7\n7 4 47 100 4 9 12",
"output": "Still Rozdil"
},
{
"input": "1\n47",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "Still Rozdil"
},
{
"input": "7\n7 6 5 4 3 2 1",
"output": "7"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "Still Rozdil"
},
{
"input": "4\n1000000000 100000000 1000000 1000000",
"output": "Still Rozdil"
},
{
"input": "20\n7 1 1 2 1 1 8 7 7 8 4 3 7 10 5 3 10 5 10 6",
"output": "Still Rozdil"
},
{
"input": "20\n3 3 6 9 8 2 4 1 7 3 2 9 7 7 9 7 2 6 2 7",
"output": "8"
},
{
"input": "47\n35 79 84 56 67 95 80 34 77 68 14 55 95 32 40 89 58 79 96 66 50 79 35 86 31 74 91 35 22 72 84 38 11 59 73 51 65 11 11 62 30 12 32 71 69 15 11",
"output": "Still Rozdil"
},
{
"input": "47\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "32"
},
{
"input": "25\n73 56 65 70 57 57 49 50 65 68 54 48 66 55 55 66 60 70 47 62 63 70 62 70 63",
"output": "19"
},
{
"input": "100\n922 938 103 689 526 118 314 496 373 47 224 544 495 656 804 754 269 356 354 536 817 597 787 178 516 659 790 993 786 169 866 779 997 350 674 437 219 546 564 893 402 449 751 628 238 686 392 951 716 439 332 563 215 146 746 311 866 773 548 675 731 183 890 802 488 944 473 25 662 559 732 534 777 144 502 593 839 692 84 337 679 718 1005 663 709 940 86 469 310 960 902 233 606 761 116 231 662 903 68 622",
"output": "68"
},
{
"input": "5\n5 5 2 3 1",
"output": "5"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "3\n7 7 4",
"output": "3"
},
{
"input": "3\n2 2 1",
"output": "3"
},
{
"input": "3\n4 4 3",
"output": "3"
},
{
"input": "2\n10000001 10000002",
"output": "1"
},
{
"input": "4\n3 3 2 1",
"output": "4"
},
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "2\n1000000000 100000000",
"output": "2"
},
{
"input": "1\n99999999",
"output": "1"
},
{
"input": "3\n5 5 3",
"output": "3"
},
{
"input": "3\n7 7 5",
"output": "3"
},
{
"input": "3\n3 3 1",
"output": "3"
},
{
"input": "4\n10 10 10 1",
"output": "4"
}
] | 1,588,589,012 | 2,147,483,647 | PyPy 3 | OK | TESTS | 45 | 436 | 10,752,000 | n = int(input())
f_line = input()
f_arr = [int(y) for y in f_line.split()]
min_ele = min(f_arr)
if f_arr.count(min_ele) > 1:
print('Still Rozdil')
else:
print(f_arr.index(min_ele) +1) | Title: Little Elephant and Rozdil
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities.
Output Specification:
Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes).
Demo Input:
['2\n7 4\n', '7\n7 4 47 100 4 9 12\n']
Demo Output:
['2\n', 'Still Rozdil\n']
Note:
In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil". | ```python
n = int(input())
f_line = input()
f_arr = [int(y) for y in f_line.split()]
min_ele = min(f_arr)
if f_arr.count(min_ele) > 1:
print('Still Rozdil')
else:
print(f_arr.index(min_ele) +1)
``` | 3 | |
519 | B | A and B and Compilation Errors | PROGRAMMING | 1,100 | [
"data structures",
"implementation",
"sortings"
] | null | null | A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected? | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one. | Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively. | [
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] | [
"8\n123\n",
"1\n3\n"
] | In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | 1,000 | [
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 656422756 9301599 25720377\n277246428 388684428 661369649 460626451 656422756 802090732 9301599 784303821 376287098\n376287098 802090732 388684428 9301599 656422756 784303821 460626451 277246428",
"output": "25720377\n661369649"
},
{
"input": "3\n796067435 964699482 819602309\n964699482 796067435\n964699482",
"output": "819602309\n796067435"
},
{
"input": "3\n374054998 726316780 902899520\n902899520 726316780\n726316780",
"output": "374054998\n902899520"
},
{
"input": "3\n168638990 939116221 323703261\n168638990 323703261\n168638990",
"output": "939116221\n323703261"
},
{
"input": "3\n77 77 77\n77 77\n77",
"output": "77\n77"
},
{
"input": "3\n84 30 9\n9 84\n9",
"output": "30\n84"
},
{
"input": "6\n5 4 3 3 5 5\n3 5 5 4 3\n3 5 4 3",
"output": "5\n5"
},
{
"input": "4\n1 5 7 8\n1 5 7\n1 5",
"output": "8\n7"
}
] | 1,636,944,777 | 2,147,483,647 | PyPy 3 | OK | TESTS | 33 | 389 | 16,896,000 | from sys import stdin
n = int(stdin.readline().strip())
arr1 = list(map(int, stdin.readline().strip().split()))
arr2 = list(map(int, stdin.readline().strip().split()))
arr3 = list(map(int, stdin.readline().strip().split()))
arr1.sort()
arr2.sort()
arr3.sort()
first_num = -1
second_num = -1
for i in range(n-1):
if arr1[i]!=arr2[i]:
first_num = arr1[i]
break
if first_num == -1:
first_num = arr1[n-1]
for i in range(n-2):
if arr2[i]!=arr3[i]:
second_num = arr2[i]
break
if second_num == -1:
second_num = arr2[n-2]
print(first_num)
print(second_num)
| Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | ```python
from sys import stdin
n = int(stdin.readline().strip())
arr1 = list(map(int, stdin.readline().strip().split()))
arr2 = list(map(int, stdin.readline().strip().split()))
arr3 = list(map(int, stdin.readline().strip().split()))
arr1.sort()
arr2.sort()
arr3.sort()
first_num = -1
second_num = -1
for i in range(n-1):
if arr1[i]!=arr2[i]:
first_num = arr1[i]
break
if first_num == -1:
first_num = arr1[n-1]
for i in range(n-2):
if arr2[i]!=arr3[i]:
second_num = arr2[i]
break
if second_num == -1:
second_num = arr2[n-2]
print(first_num)
print(second_num)
``` | 3 | |
949 | B | A Leapfrog in the Array | PROGRAMMING | 1,700 | [
"constructive algorithms",
"math"
] | null | null | Dima is a beginner programmer. During his working process, he regularly has to repeat the following operation again and again: to remove every second element from the array. One day he has been bored with easy solutions of this problem, and he has come up with the following extravagant algorithm.
Let's consider that initially array contains *n* numbers from 1 to *n* and the number *i* is located in the cell with the index 2*i*<=-<=1 (Indices are numbered starting from one) and other cells of the array are empty. Each step Dima selects a non-empty array cell with the maximum index and moves the number written in it to the nearest empty cell to the left of the selected one. The process continues until all *n* numbers will appear in the first *n* cells of the array. For example if *n*<==<=4, the array is changing as follows:
You have to write a program that allows you to determine what number will be in the cell with index *x* (1<=≤<=*x*<=≤<=*n*) after Dima's algorithm finishes. | The first line contains two integers *n* and *q* (1<=≤<=*n*<=≤<=1018, 1<=≤<=*q*<=≤<=200<=000), the number of elements in the array and the number of queries for which it is needed to find the answer.
Next *q* lines contain integers *x**i* (1<=≤<=*x**i*<=≤<=*n*), the indices of cells for which it is necessary to output their content after Dima's algorithm finishes. | For each of *q* queries output one integer number, the value that will appear in the corresponding array cell after Dima's algorithm finishes. | [
"4 3\n2\n3\n4\n",
"13 4\n10\n5\n4\n8\n"
] | [
"3\n2\n4\n",
"13\n3\n8\n9\n"
] | The first example is shown in the picture.
In the second example the final array is [1, 12, 2, 8, 3, 11, 4, 9, 5, 13, 6, 10, 7]. | 1,000 | [
{
"input": "4 3\n2\n3\n4",
"output": "3\n2\n4"
},
{
"input": "13 4\n10\n5\n4\n8",
"output": "13\n3\n8\n9"
},
{
"input": "2 2\n1\n2",
"output": "1\n2"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "3 3\n3\n2\n1",
"output": "2\n3\n1"
},
{
"input": "12 12\n9\n11\n5\n3\n7\n2\n8\n6\n4\n10\n12\n1",
"output": "5\n6\n3\n2\n4\n7\n12\n8\n10\n9\n11\n1"
}
] | 1,522,984,626 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 1,669 | 28,262,400 | import sys
def query(n, a):
while not (a & 1):
a += (n - a//2)
return a+1 >> 1
n, q = map(int, sys.stdin.readline().split())
arr = [int(sys.stdin.readline()) for _ in range(q)]
sys.stdout.write("\n".join(str(query(n, a)) for a in arr)) | Title: A Leapfrog in the Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima is a beginner programmer. During his working process, he regularly has to repeat the following operation again and again: to remove every second element from the array. One day he has been bored with easy solutions of this problem, and he has come up with the following extravagant algorithm.
Let's consider that initially array contains *n* numbers from 1 to *n* and the number *i* is located in the cell with the index 2*i*<=-<=1 (Indices are numbered starting from one) and other cells of the array are empty. Each step Dima selects a non-empty array cell with the maximum index and moves the number written in it to the nearest empty cell to the left of the selected one. The process continues until all *n* numbers will appear in the first *n* cells of the array. For example if *n*<==<=4, the array is changing as follows:
You have to write a program that allows you to determine what number will be in the cell with index *x* (1<=≤<=*x*<=≤<=*n*) after Dima's algorithm finishes.
Input Specification:
The first line contains two integers *n* and *q* (1<=≤<=*n*<=≤<=1018, 1<=≤<=*q*<=≤<=200<=000), the number of elements in the array and the number of queries for which it is needed to find the answer.
Next *q* lines contain integers *x**i* (1<=≤<=*x**i*<=≤<=*n*), the indices of cells for which it is necessary to output their content after Dima's algorithm finishes.
Output Specification:
For each of *q* queries output one integer number, the value that will appear in the corresponding array cell after Dima's algorithm finishes.
Demo Input:
['4 3\n2\n3\n4\n', '13 4\n10\n5\n4\n8\n']
Demo Output:
['3\n2\n4\n', '13\n3\n8\n9\n']
Note:
The first example is shown in the picture.
In the second example the final array is [1, 12, 2, 8, 3, 11, 4, 9, 5, 13, 6, 10, 7]. | ```python
import sys
def query(n, a):
while not (a & 1):
a += (n - a//2)
return a+1 >> 1
n, q = map(int, sys.stdin.readline().split())
arr = [int(sys.stdin.readline()) for _ in range(q)]
sys.stdout.write("\n".join(str(query(n, a)) for a in arr))
``` | 3 | |
989 | A | A Blend of Springtime | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | "What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order. | The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively. | Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower). | [
".BAC.\n",
"AA..CB\n"
] | [
"Yes\n",
"No\n"
] | In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | 500 | [
{
"input": ".BAC.",
"output": "Yes"
},
{
"input": "AA..CB",
"output": "No"
},
{
"input": ".",
"output": "No"
},
{
"input": "ACB.AAAAAA",
"output": "Yes"
},
{
"input": "B.BC.BBBCA",
"output": "Yes"
},
{
"input": "BA..CAB..B",
"output": "Yes"
},
{
"input": "CACCBAA.BC",
"output": "Yes"
},
{
"input": ".CAACCBBA.CBB.AC..BABCCBCCB..B.BC..CBC.CA.CC.C.CC.B.A.CC.BBCCBB..ACAACAC.CBCCB.AABAAC.CBCC.BA..CCBC.",
"output": "Yes"
},
{
"input": "A",
"output": "No"
},
{
"input": "..",
"output": "No"
},
{
"input": "BC",
"output": "No"
},
{
"input": "CAB",
"output": "Yes"
},
{
"input": "A.CB",
"output": "No"
},
{
"input": "B.ACAA.CA..CBCBBAA.B.CCBCB.CAC.ABC...BC.BCCC.BC.CB",
"output": "Yes"
},
{
"input": "B.B...CC.B..CCCB.CB..CBCB..CBCC.CCBC.B.CB..CA.C.C.",
"output": "No"
},
{
"input": "AA.CBAABABCCC..B..B.ABBABAB.B.B.CCA..CB.B...A..CBC",
"output": "Yes"
},
{
"input": "CA.ABB.CC.B.C.BBBABAAB.BBBAACACAAA.C.AACA.AAC.C.BCCB.CCBC.C..CCACA.CBCCB.CCAABAAB.AACAA..A.AAA.",
"output": "No"
},
{
"input": "CBC...AC.BBBB.BBABABA.CAAACC.AAABB..A.BA..BC.CBBBC.BBBBCCCAA.ACCBB.AB.C.BA..CC..AAAC...AB.A.AAABBA.A",
"output": "No"
},
{
"input": "CC.AAAC.BA.BBB.AABABBCCAA.A.CBCCB.B.BC.ABCBCBBAA.CACA.CCCA.CB.CCB.A.BCCCB...C.A.BCCBC..B.ABABB.C.BCB",
"output": "Yes"
},
{
"input": "CCC..A..CACACCA.CA.ABAAB.BBA..C.AAA...ACB.ACA.CA.B.AB.A..C.BC.BC.A.C....ABBCCACCCBCC.BBBAA.ACCACB.BB",
"output": "Yes"
},
{
"input": "BC.ABACAACC..AC.A..CCCAABBCCACAC.AA.CC.BAABABABBCBB.BA..C.C.C.A.BBA.C..BC.ACACCC.AAAACCCCC.AAC.AC.AB",
"output": "Yes"
},
{
"input": "ACAC.BAA.C..CAAC..ABBAACC..BAA...CC...ACCBBCA.BAABABAACCAC.A.BBCACCC..BCB.BABAAAACCBCB.BCAABBC.C.BBB",
"output": "Yes"
},
{
"input": "CCAC.BCBC.A.ABBAB.C.C.BC.CCABBCBCCBC..B.AA.C.BC...B..BAA.ACCCCBBB.AAAACA.CAACCB.CCB.CC.BCCAB.BBBBABB",
"output": "Yes"
},
{
"input": ".AACAA.AAAAC.BBBB.BC...CCACCACAAA.A..CCA..BCC.AB.ABAAB..AABA...B.C.CBAB.BAAB.A.C.AAC.BBBA.ACAAA.BB.C",
"output": "Yes"
},
{
"input": "CC.ACCC.BCCCCAA.BBAACB.ABABAAAA.A.CBAB.CBACBBC..C.CA.AAA..AA..ABBB.A.C..CBBCAAACC.B..CC.AC..CAABACB.",
"output": "Yes"
},
{
"input": ".BAB.",
"output": "No"
},
{
"input": "BBBBBBBBB",
"output": "No"
},
{
"input": "..AAC..",
"output": "No"
},
{
"input": ".AAABBBCCC.",
"output": "No"
},
{
"input": "AAABC",
"output": "Yes"
},
{
"input": "BBB",
"output": "No"
},
{
"input": "AAAAABABAAAAA",
"output": "No"
},
{
"input": "AABBCC",
"output": "No"
},
{
"input": ".BA",
"output": "No"
},
{
"input": "CAAAAB",
"output": "No"
}
] | 1,528,725,058 | 958 | Python 3 | CHALLENGED | CHALLENGES | 8 | 77 | 0 | s=input()
l=len(s)
counter=0
for i in range(l-2):
if(s[i]!='.')and (s[i+1]!='.')and(s[i+2]!='.'):
counter+=1
if(counter>=1):
print ("Yes")
else:
print("No") | Title: A Blend of Springtime
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input Specification:
The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output Specification:
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['.BAC.\n', 'AA..CB\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | ```python
s=input()
l=len(s)
counter=0
for i in range(l-2):
if(s[i]!='.')and (s[i+1]!='.')and(s[i+2]!='.'):
counter+=1
if(counter>=1):
print ("Yes")
else:
print("No")
``` | -1 | |
975 | A | Aramic script | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | In Aramic language words can only represent objects.
Words in Aramic have special properties:
- A word is a root if it does not contain the same letter more than once. - A root and all its permutations represent the same object. - The root $x$ of a word $y$ is the word that contains all letters that appear in $y$ in a way that each letter appears once. For example, the root of "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab". - Any word in Aramic represents the same object as its root.
You have an ancient script in Aramic. What is the number of different objects mentioned in the script? | The first line contains one integer $n$ ($1 \leq n \leq 10^3$) — the number of words in the script.
The second line contains $n$ words $s_1, s_2, \ldots, s_n$ — the script itself. The length of each string does not exceed $10^3$.
It is guaranteed that all characters of the strings are small latin letters. | Output one integer — the number of different objects mentioned in the given ancient Aramic script. | [
"5\na aa aaa ab abb\n",
"3\namer arem mrea\n"
] | [
"2",
"1"
] | In the first test, there are two objects mentioned. The roots that represent them are "a","ab".
In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer". | 500 | [
{
"input": "5\na aa aaa ab abb",
"output": "2"
},
{
"input": "3\namer arem mrea",
"output": "1"
},
{
"input": "10\nbda bbb cda dca dda dcb bcd dcb ada ddd",
"output": "6"
},
{
"input": "2\nfhjlqs aceginpr",
"output": "2"
},
{
"input": "2\nbcdfghimn efghijlmo",
"output": "2"
}
] | 1,563,579,823 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 155 | 3,072,000 | import sys
n = sys.stdin.readline()
words = sys.stdin.readline().split()
known = set()
def to_root(s):
return "".join(sorted(set([c for c in s])))
for word in words:
known.add(to_root(word))
print(len(known)) | Title: Aramic script
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Aramic language words can only represent objects.
Words in Aramic have special properties:
- A word is a root if it does not contain the same letter more than once. - A root and all its permutations represent the same object. - The root $x$ of a word $y$ is the word that contains all letters that appear in $y$ in a way that each letter appears once. For example, the root of "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab". - Any word in Aramic represents the same object as its root.
You have an ancient script in Aramic. What is the number of different objects mentioned in the script?
Input Specification:
The first line contains one integer $n$ ($1 \leq n \leq 10^3$) — the number of words in the script.
The second line contains $n$ words $s_1, s_2, \ldots, s_n$ — the script itself. The length of each string does not exceed $10^3$.
It is guaranteed that all characters of the strings are small latin letters.
Output Specification:
Output one integer — the number of different objects mentioned in the given ancient Aramic script.
Demo Input:
['5\na aa aaa ab abb\n', '3\namer arem mrea\n']
Demo Output:
['2', '1']
Note:
In the first test, there are two objects mentioned. The roots that represent them are "a","ab".
In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer". | ```python
import sys
n = sys.stdin.readline()
words = sys.stdin.readline().split()
known = set()
def to_root(s):
return "".join(sorted(set([c for c in s])))
for word in words:
known.add(to_root(word))
print(len(known))
``` | 3 | |
902 | A | Visiting a Friend | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | Pig is visiting a friend.
Pig's house is located at point 0, and his friend's house is located at point *m* on an axis.
Pig can use teleports to move along the axis.
To use a teleport, Pig should come to a certain point (where the teleport is located) and choose where to move: for each teleport there is the rightmost point it can move Pig to, this point is known as the limit of the teleport.
Formally, a teleport located at point *x* with limit *y* can move Pig from point *x* to any point within the segment [*x*;<=*y*], including the bounds.
Determine if Pig can visit the friend using teleports only, or he should use his car. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100) — the number of teleports and the location of the friend's house.
The next *n* lines contain information about teleports.
The *i*-th of these lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=*b**i*<=≤<=*m*), where *a**i* is the location of the *i*-th teleport, and *b**i* is its limit.
It is guaranteed that *a**i*<=≥<=*a**i*<=-<=1 for every *i* (2<=≤<=*i*<=≤<=*n*). | Print "YES" if there is a path from Pig's house to his friend's house that uses only teleports, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower). | [
"3 5\n0 2\n2 4\n3 5\n",
"3 7\n0 4\n2 5\n6 7\n"
] | [
"YES\n",
"NO\n"
] | The first example is shown on the picture below:
Pig can use the first teleport from his house (point 0) to reach point 2, then using the second teleport go from point 2 to point 3, then using the third teleport go from point 3 to point 5, where his friend lives.
The second example is shown on the picture below:
You can see that there is no path from Pig's house to his friend's house that uses only teleports. | 500 | [
{
"input": "3 5\n0 2\n2 4\n3 5",
"output": "YES"
},
{
"input": "3 7\n0 4\n2 5\n6 7",
"output": "NO"
},
{
"input": "1 1\n0 0",
"output": "NO"
},
{
"input": "30 10\n0 7\n1 2\n1 2\n1 4\n1 4\n1 3\n2 2\n2 4\n2 6\n2 9\n2 2\n3 5\n3 8\n4 8\n4 5\n4 6\n5 6\n5 7\n6 6\n6 9\n6 7\n6 9\n7 7\n7 7\n8 8\n8 8\n9 9\n9 9\n10 10\n10 10",
"output": "NO"
},
{
"input": "30 100\n0 27\n4 82\n11 81\n14 32\n33 97\n33 34\n37 97\n38 52\n45 91\n49 56\n50 97\n57 70\n59 94\n59 65\n62 76\n64 65\n65 95\n67 77\n68 100\n71 73\n80 94\n81 92\n84 85\n85 100\n88 91\n91 95\n92 98\n92 98\n99 100\n100 100",
"output": "YES"
},
{
"input": "70 10\n0 4\n0 4\n0 8\n0 9\n0 1\n0 5\n0 7\n1 3\n1 8\n1 8\n1 6\n1 6\n1 2\n1 3\n1 2\n1 3\n2 5\n2 4\n2 3\n2 4\n2 6\n2 2\n2 5\n2 7\n3 7\n3 4\n3 7\n3 4\n3 8\n3 4\n3 9\n3 3\n3 7\n3 9\n3 3\n3 9\n4 6\n4 7\n4 5\n4 7\n5 8\n5 5\n5 9\n5 7\n5 5\n6 6\n6 9\n6 7\n6 8\n6 9\n6 8\n7 7\n7 8\n7 7\n7 8\n8 9\n8 8\n8 9\n8 8\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "NO"
},
{
"input": "30 10\n0 7\n1 2\n1 2\n1 4\n1 4\n1 3\n2 2\n2 4\n2 6\n2 9\n2 2\n3 5\n3 8\n4 8\n4 5\n4 6\n5 6\n5 7\n6 6\n6 9\n6 7\n6 9\n7 7\n7 7\n8 10\n8 10\n9 9\n9 9\n10 10\n10 10",
"output": "YES"
},
{
"input": "50 100\n0 95\n1 100\n1 38\n2 82\n5 35\n7 71\n8 53\n11 49\n15 27\n17 84\n17 75\n18 99\n18 43\n18 69\n21 89\n27 60\n27 29\n38 62\n38 77\n39 83\n40 66\n48 80\n48 100\n50 51\n50 61\n53 77\n53 63\n55 58\n56 68\n60 82\n62 95\n66 74\n67 83\n69 88\n69 81\n69 88\n69 98\n70 91\n70 76\n71 90\n72 99\n81 99\n85 87\n88 97\n88 93\n90 97\n90 97\n92 98\n98 99\n100 100",
"output": "YES"
},
{
"input": "70 10\n0 4\n0 4\n0 8\n0 9\n0 1\n0 5\n0 7\n1 3\n1 8\n1 8\n1 10\n1 9\n1 6\n1 2\n1 3\n1 2\n2 6\n2 5\n2 4\n2 3\n2 10\n2 2\n2 6\n2 2\n3 10\n3 7\n3 7\n3 4\n3 7\n3 4\n3 8\n3 4\n3 10\n3 5\n3 3\n3 7\n4 8\n4 8\n4 9\n4 6\n5 7\n5 10\n5 7\n5 8\n5 5\n6 8\n6 9\n6 10\n6 6\n6 9\n6 7\n7 8\n7 9\n7 10\n7 10\n8 8\n8 8\n8 9\n8 10\n9 10\n9 9\n9 10\n9 10\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "YES"
},
{
"input": "85 10\n0 9\n0 4\n0 2\n0 5\n0 1\n0 8\n0 7\n1 2\n1 4\n1 5\n1 9\n1 1\n1 6\n1 6\n2 5\n2 7\n2 7\n2 7\n2 7\n3 4\n3 7\n3 9\n3 5\n3 3\n4 4\n4 6\n4 5\n5 6\n5 6\n5 6\n5 6\n5 7\n5 8\n5 5\n5 7\n5 8\n5 9\n5 8\n6 8\n6 7\n6 8\n6 9\n6 9\n6 6\n6 9\n6 7\n7 7\n7 7\n7 7\n7 8\n7 7\n7 8\n7 8\n7 9\n8 8\n8 8\n8 8\n8 8\n8 8\n8 9\n8 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "NO"
},
{
"input": "30 40\n0 0\n4 8\n5 17\n7 32\n7 16\n8 16\n10 19\n12 22\n12 27\n13 21\n13 28\n13 36\n14 28\n14 18\n18 21\n21 26\n21 36\n22 38\n23 32\n24 30\n26 35\n29 32\n29 32\n31 34\n31 31\n33 33\n33 35\n35 40\n38 38\n40 40",
"output": "NO"
},
{
"input": "70 100\n0 99\n1 87\n1 94\n1 4\n2 72\n3 39\n3 69\n4 78\n5 85\n7 14\n8 59\n12 69\n14 15\n14 76\n17 17\n19 53\n19 57\n19 21\n21 35\n21 83\n24 52\n24 33\n27 66\n27 97\n30 62\n30 74\n30 64\n32 63\n35 49\n37 60\n40 99\n40 71\n41 83\n42 66\n42 46\n45 83\n51 76\n53 69\n54 82\n54 96\n54 88\n55 91\n56 88\n58 62\n62 87\n64 80\n67 90\n67 69\n68 92\n72 93\n74 93\n77 79\n77 91\n78 97\n78 98\n81 85\n81 83\n81 83\n84 85\n86 88\n89 94\n89 92\n92 97\n96 99\n97 98\n97 99\n99 99\n100 100\n100 100\n100 100",
"output": "NO"
},
{
"input": "1 10\n0 10",
"output": "YES"
},
{
"input": "70 40\n0 34\n1 16\n3 33\n4 36\n4 22\n5 9\n5 9\n7 16\n8 26\n9 29\n9 25\n10 15\n10 22\n10 29\n10 20\n11 27\n11 26\n11 12\n12 19\n13 21\n14 31\n14 36\n15 34\n15 37\n16 21\n17 31\n18 22\n20 27\n20 32\n20 20\n20 29\n21 29\n21 34\n21 30\n22 40\n23 23\n23 28\n24 29\n25 38\n26 35\n27 37\n28 39\n28 33\n28 40\n28 33\n29 31\n29 33\n30 38\n30 36\n30 30\n30 38\n31 37\n31 35\n31 32\n31 36\n33 39\n33 40\n35 38\n36 38\n37 38\n37 40\n38 39\n38 40\n38 39\n39 39\n39 40\n40 40\n40 40\n40 40\n40 40",
"output": "YES"
},
{
"input": "50 40\n0 9\n1 26\n1 27\n2 33\n2 5\n3 30\n4 28\n5 31\n5 27\n5 29\n7 36\n8 32\n8 13\n9 24\n10 10\n10 30\n11 26\n11 22\n11 40\n11 31\n12 26\n13 25\n14 32\n17 19\n21 29\n22 36\n24 27\n25 39\n25 27\n27 32\n27 29\n27 39\n27 29\n28 38\n30 38\n32 40\n32 38\n33 33\n33 40\n34 35\n34 34\n34 38\n34 38\n35 37\n36 39\n36 39\n37 37\n38 40\n39 39\n40 40",
"output": "YES"
},
{
"input": "70 40\n0 34\n1 16\n3 33\n4 36\n4 22\n5 9\n5 9\n7 16\n8 26\n9 29\n9 25\n10 15\n10 22\n10 29\n10 20\n11 27\n11 26\n11 12\n12 19\n13 21\n14 31\n14 36\n15 34\n15 37\n16 21\n17 31\n18 22\n20 27\n20 32\n20 20\n20 29\n21 29\n21 34\n21 30\n22 22\n23 28\n23 39\n24 24\n25 27\n26 38\n27 39\n28 33\n28 39\n28 34\n28 33\n29 30\n29 35\n30 30\n30 38\n30 34\n30 31\n31 36\n31 31\n31 32\n31 38\n33 34\n33 34\n35 36\n36 38\n37 38\n37 39\n38 38\n38 38\n38 38\n39 39\n39 39\n40 40\n40 40\n40 40\n40 40",
"output": "NO"
},
{
"input": "10 100\n0 34\n8 56\n17 79\n24 88\n28 79\n45 79\n48 93\n55 87\n68 93\n88 99",
"output": "NO"
},
{
"input": "10 10\n0 2\n3 8\n3 5\n3 3\n3 9\n3 8\n5 7\n6 10\n7 10\n9 10",
"output": "NO"
},
{
"input": "50 10\n0 2\n0 2\n0 6\n1 9\n1 3\n1 2\n1 6\n1 1\n1 1\n2 7\n2 6\n2 4\n3 9\n3 8\n3 8\n3 8\n3 6\n3 4\n3 7\n3 4\n3 6\n3 5\n4 8\n5 5\n5 7\n6 7\n6 6\n7 7\n7 7\n7 7\n7 8\n7 8\n8 8\n8 8\n8 9\n8 8\n8 9\n9 9\n9 9\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "NO"
},
{
"input": "10 40\n0 21\n1 19\n4 33\n6 26\n8 39\n15 15\n20 24\n27 27\n29 39\n32 37",
"output": "NO"
},
{
"input": "50 10\n0 2\n0 2\n0 6\n1 9\n1 3\n1 2\n1 6\n1 1\n1 1\n2 7\n2 6\n2 4\n3 9\n3 8\n3 8\n3 8\n3 6\n3 4\n3 7\n3 4\n3 6\n3 10\n4 6\n5 9\n5 5\n6 7\n6 10\n7 8\n7 7\n7 7\n7 7\n7 10\n8 8\n8 8\n8 10\n8 8\n8 8\n9 10\n9 10\n9 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "YES"
},
{
"input": "1 1\n0 1",
"output": "YES"
},
{
"input": "30 40\n0 0\n4 8\n5 17\n7 32\n7 16\n8 16\n10 19\n12 22\n12 27\n13 21\n13 28\n13 36\n14 28\n14 18\n18 21\n21 26\n21 36\n22 38\n23 32\n24 30\n26 35\n29 32\n29 32\n31 34\n31 31\n33 33\n33 35\n35 36\n38 38\n40 40",
"output": "NO"
},
{
"input": "30 100\n0 27\n4 82\n11 81\n14 32\n33 97\n33 34\n37 97\n38 52\n45 91\n49 56\n50 97\n57 70\n59 94\n59 65\n62 76\n64 65\n65 95\n67 77\n68 82\n71 94\n80 90\n81 88\n84 93\n85 89\n88 92\n91 97\n92 99\n92 97\n99 99\n100 100",
"output": "NO"
},
{
"input": "10 100\n0 34\n8 56\n17 79\n24 88\n28 79\n45 79\n48 93\n55 87\n68 93\n79 100",
"output": "YES"
},
{
"input": "10 40\n0 21\n1 19\n4 33\n6 26\n8 39\n15 15\n20 24\n27 27\n29 39\n37 40",
"output": "YES"
},
{
"input": "85 10\n0 9\n0 4\n0 2\n0 5\n0 1\n0 8\n0 7\n1 2\n1 10\n1 2\n1 5\n1 10\n1 8\n1 1\n2 8\n2 7\n2 5\n2 5\n2 7\n3 5\n3 7\n3 5\n3 4\n3 7\n4 7\n4 8\n4 6\n5 7\n5 10\n5 5\n5 6\n5 6\n5 6\n5 6\n5 7\n5 8\n5 5\n5 7\n6 10\n6 9\n6 7\n6 10\n6 8\n6 7\n6 10\n6 10\n7 8\n7 9\n7 8\n7 8\n7 8\n7 8\n7 7\n7 7\n8 8\n8 8\n8 10\n8 9\n8 9\n8 9\n8 9\n9 9\n9 10\n9 9\n9 9\n9 9\n9 9\n9 10\n9 10\n9 9\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10",
"output": "YES"
},
{
"input": "50 100\n0 95\n1 7\n1 69\n2 83\n5 67\n7 82\n8 31\n11 25\n15 44\n17 75\n17 27\n18 43\n18 69\n18 40\n21 66\n27 29\n27 64\n38 77\n38 90\n39 52\n40 60\n48 91\n48 98\n50 89\n50 63\n53 54\n53 95\n55 76\n56 59\n60 96\n62 86\n66 70\n67 77\n69 88\n69 98\n69 80\n69 95\n70 74\n70 77\n71 99\n72 73\n81 87\n85 99\n88 96\n88 91\n90 97\n90 99\n92 92\n98 99\n100 100",
"output": "NO"
},
{
"input": "50 40\n0 9\n1 26\n1 27\n2 33\n2 5\n3 30\n4 28\n5 31\n5 27\n5 29\n7 36\n8 32\n8 13\n9 24\n10 10\n10 30\n11 26\n11 22\n11 35\n11 23\n12 36\n13 31\n14 31\n17 17\n21 25\n22 33\n24 26\n25 32\n25 25\n27 39\n27 29\n27 34\n27 32\n28 34\n30 36\n32 37\n32 33\n33 35\n33 33\n34 38\n34 38\n34 36\n34 36\n35 36\n36 36\n36 39\n37 37\n38 39\n39 39\n40 40",
"output": "NO"
},
{
"input": "10 10\n0 2\n3 8\n3 5\n3 3\n3 9\n3 8\n5 7\n6 9\n7 7\n9 9",
"output": "NO"
},
{
"input": "70 100\n0 99\n1 87\n1 94\n1 4\n2 72\n3 39\n3 69\n4 78\n5 85\n7 14\n8 59\n12 69\n14 15\n14 76\n17 17\n19 53\n19 57\n19 21\n21 35\n21 83\n24 52\n24 33\n27 66\n27 97\n30 62\n30 74\n30 64\n32 63\n35 49\n37 60\n40 99\n40 71\n41 83\n42 66\n42 46\n45 83\n51 76\n53 69\n54 82\n54 96\n54 88\n55 91\n56 88\n58 62\n62 87\n64 80\n67 90\n67 69\n68 92\n72 93\n74 93\n77 79\n77 91\n78 97\n78 98\n81 85\n81 83\n81 83\n84 85\n86 88\n89 94\n89 100\n92 97\n96 96\n97 98\n97 100\n99 100\n100 100\n100 100\n100 100",
"output": "YES"
},
{
"input": "2 2\n0 2\n0 1",
"output": "YES"
},
{
"input": "2 100\n0 49\n50 100",
"output": "NO"
},
{
"input": "2 100\n0 50\n50 100",
"output": "YES"
},
{
"input": "10 100\n0 10\n10 20\n20 30\n30 40\n40 50\n50 60\n60 70\n70 80\n80 90\n90 100",
"output": "YES"
},
{
"input": "100 10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 10",
"output": "YES"
},
{
"input": "1 4\n1 4",
"output": "NO"
},
{
"input": "3 5\n0 2\n2 5\n3 4",
"output": "YES"
},
{
"input": "2 5\n0 5\n1 2",
"output": "YES"
},
{
"input": "5 5\n0 2\n0 2\n3 5\n3 5\n3 5",
"output": "NO"
},
{
"input": "4 10\n0 3\n2 10\n4 10\n5 7",
"output": "YES"
},
{
"input": "3 8\n0 0\n1 8\n2 8",
"output": "NO"
},
{
"input": "6 8\n0 3\n1 5\n2 3\n5 6\n6 7\n7 8",
"output": "YES"
}
] | 1,645,946,398 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 31 | 0 | n,m=map(int,input().split())
p=q=f=0
for _ in range(n):
a,b=map(int,input().split())
if p<=a and a<=q:
f+=1
p=a
q=b
if f==n:
if p<=m and q>=m:
print("YES")
else:print("NO")
else:
print("NO") | Title: Visiting a Friend
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pig is visiting a friend.
Pig's house is located at point 0, and his friend's house is located at point *m* on an axis.
Pig can use teleports to move along the axis.
To use a teleport, Pig should come to a certain point (where the teleport is located) and choose where to move: for each teleport there is the rightmost point it can move Pig to, this point is known as the limit of the teleport.
Formally, a teleport located at point *x* with limit *y* can move Pig from point *x* to any point within the segment [*x*;<=*y*], including the bounds.
Determine if Pig can visit the friend using teleports only, or he should use his car.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100) — the number of teleports and the location of the friend's house.
The next *n* lines contain information about teleports.
The *i*-th of these lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=*b**i*<=≤<=*m*), where *a**i* is the location of the *i*-th teleport, and *b**i* is its limit.
It is guaranteed that *a**i*<=≥<=*a**i*<=-<=1 for every *i* (2<=≤<=*i*<=≤<=*n*).
Output Specification:
Print "YES" if there is a path from Pig's house to his friend's house that uses only teleports, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['3 5\n0 2\n2 4\n3 5\n', '3 7\n0 4\n2 5\n6 7\n']
Demo Output:
['YES\n', 'NO\n']
Note:
The first example is shown on the picture below:
Pig can use the first teleport from his house (point 0) to reach point 2, then using the second teleport go from point 2 to point 3, then using the third teleport go from point 3 to point 5, where his friend lives.
The second example is shown on the picture below:
You can see that there is no path from Pig's house to his friend's house that uses only teleports. | ```python
n,m=map(int,input().split())
p=q=f=0
for _ in range(n):
a,b=map(int,input().split())
if p<=a and a<=q:
f+=1
p=a
q=b
if f==n:
if p<=m and q>=m:
print("YES")
else:print("NO")
else:
print("NO")
``` | 0 | |
270 | A | Fancy Fence | PROGRAMMING | 1,100 | [
"geometry",
"implementation",
"math"
] | null | null | Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*? | The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees. | For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible. | [
"3\n30\n60\n90\n"
] | [
"NO\nYES\nYES\n"
] | In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | 500 | [
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,645,885,251 | 2,147,483,647 | PyPy 3 | OK | TESTS | 3 | 154 | 0 | """
$$$$$ $$$$$ $$$$$$ $ $
$ $ $ $ $ $$ $$
$ $ $ $ $$$$$ $ $$ $
$$$$$ $$$$$ $ $ $
$ $ $ $ $ $
$ $ $ $$$$$$ $ $
"""
for _ in range(int(input())):
Q = int(input())
n = 360/(180-Q)
N=360//(180-Q)
if n==N:
print('YES')
else :
print('NO')
| Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | ```python
"""
$$$$$ $$$$$ $$$$$$ $ $
$ $ $ $ $ $$ $$
$ $ $ $ $$$$$ $ $$ $
$$$$$ $$$$$ $ $ $
$ $ $ $ $ $
$ $ $ $$$$$$ $ $
"""
for _ in range(int(input())):
Q = int(input())
n = 360/(180-Q)
N=360//(180-Q)
if n==N:
print('YES')
else :
print('NO')
``` | 3 | |
381 | A | Sereja and Dima | PROGRAMMING | 800 | [
"greedy",
"implementation",
"two pointers"
] | null | null | Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her. | The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000. | On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game. | [
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] | [
"12 5\n",
"16 12\n"
] | In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | 500 | [
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24",
"output": "644 500"
},
{
"input": "1\n3",
"output": "3 0"
},
{
"input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646",
"output": "6848 6568"
},
{
"input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727",
"output": "9562 9561"
},
{
"input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12",
"output": "315 315"
},
{
"input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304",
"output": "3238 2222"
},
{
"input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325",
"output": "5246 4864"
},
{
"input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71",
"output": "8147 7807"
},
{
"input": "1\n1",
"output": "1 0"
}
] | 1,668,457,184 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 46 | 0 | input()
k = map(int, (input().split()))
lst = list(k)
sreja = 0
dima = 0
k = 0
while len(lst) != 0:
big_element = max(lst[0], lst[-1])
if k % 2 == 0:
sreja += big_element
else:
dima += big_element
lst.remove(big_element)
k += 1
print(f"{sreja} {dima}")
| Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | ```python
input()
k = map(int, (input().split()))
lst = list(k)
sreja = 0
dima = 0
k = 0
while len(lst) != 0:
big_element = max(lst[0], lst[-1])
if k % 2 == 0:
sreja += big_element
else:
dima += big_element
lst.remove(big_element)
k += 1
print(f"{sreja} {dima}")
``` | 3 | |
912 | B | New Year's Eve | PROGRAMMING | 1,300 | [
"bitmasks",
"constructive algorithms",
"number theory"
] | null | null | Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains *n* sweet candies from the good ol' bakery, each labeled from 1 to *n* corresponding to its tastiness. No two candies have the same tastiness.
The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!
A xor-sum of a sequence of integers *a*1,<=*a*2,<=...,<=*a**m* is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than *k* candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain. | The sole string contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1018). | Output one number — the largest possible xor-sum. | [
"4 3\n",
"6 6\n"
] | [
"7\n",
"7\n"
] | In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.
In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7. | 1,000 | [
{
"input": "4 3",
"output": "7"
},
{
"input": "6 6",
"output": "7"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "1022 10",
"output": "1023"
},
{
"input": "415853337373441 52",
"output": "562949953421311"
},
{
"input": "75 12",
"output": "127"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1152921504606846975"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000000000000 2",
"output": "1152921504606846975"
},
{
"input": "49194939 22",
"output": "67108863"
},
{
"input": "228104606 17",
"output": "268435455"
},
{
"input": "817034381 7",
"output": "1073741823"
},
{
"input": "700976748 4",
"output": "1073741823"
},
{
"input": "879886415 9",
"output": "1073741823"
},
{
"input": "18007336 10353515",
"output": "33554431"
},
{
"input": "196917003 154783328",
"output": "268435455"
},
{
"input": "785846777 496205300",
"output": "1073741823"
},
{
"input": "964756444 503568330",
"output": "1073741823"
},
{
"input": "848698811 317703059",
"output": "1073741823"
},
{
"input": "676400020444788 1",
"output": "676400020444788"
},
{
"input": "502643198528213 1",
"output": "502643198528213"
},
{
"input": "815936580997298686 684083143940282566",
"output": "1152921504606846975"
},
{
"input": "816762824175382110 752185261508428780",
"output": "1152921504606846975"
},
{
"input": "327942415253132295 222598158321260499",
"output": "576460752303423487"
},
{
"input": "328768654136248423 284493129147496637",
"output": "576460752303423487"
},
{
"input": "329594893019364551 25055600080496801",
"output": "576460752303423487"
},
{
"input": "921874985256864012 297786684518764536",
"output": "1152921504606846975"
},
{
"input": "922701224139980141 573634416190460758",
"output": "1152921504606846975"
},
{
"input": "433880815217730325 45629641110945892",
"output": "576460752303423487"
},
{
"input": "434707058395813749 215729375494216481",
"output": "576460752303423487"
},
{
"input": "435533301573897173 34078453236225189",
"output": "576460752303423487"
},
{
"input": "436359544751980597 199220719961060641",
"output": "576460752303423487"
},
{
"input": "437185783635096725 370972992240105630",
"output": "576460752303423487"
},
{
"input": "438012026813180149 111323110116193830",
"output": "576460752303423487"
},
{
"input": "438838269991263573 295468957052046146",
"output": "576460752303423487"
},
{
"input": "439664513169346997 46560240538186155",
"output": "576460752303423487"
},
{
"input": "440490752052463125 216165966013438147",
"output": "576460752303423487"
},
{
"input": "441316995230546549 401964286420555423",
"output": "576460752303423487"
},
{
"input": "952496582013329437 673506882352402278",
"output": "1152921504606846975"
},
{
"input": "1000000000000000000 1",
"output": "1000000000000000000"
},
{
"input": "2147483647 1",
"output": "2147483647"
},
{
"input": "2147483647 2",
"output": "2147483647"
},
{
"input": "2147483647 31",
"output": "2147483647"
},
{
"input": "8 2",
"output": "15"
},
{
"input": "3 3",
"output": "3"
},
{
"input": "4 1",
"output": "4"
},
{
"input": "10 2",
"output": "15"
},
{
"input": "288230376151711743 2",
"output": "288230376151711743"
},
{
"input": "5 2",
"output": "7"
},
{
"input": "576460752303423487 2",
"output": "576460752303423487"
},
{
"input": "36028797018963967 123",
"output": "36028797018963967"
},
{
"input": "1125899906842623 2",
"output": "1125899906842623"
},
{
"input": "576460752303423489 5",
"output": "1152921504606846975"
},
{
"input": "288230376151711743 3",
"output": "288230376151711743"
},
{
"input": "36028797018963967 345",
"output": "36028797018963967"
},
{
"input": "18014398509481984 30",
"output": "36028797018963967"
},
{
"input": "8 8",
"output": "15"
},
{
"input": "8 1",
"output": "8"
}
] | 1,522,422,804 | 2,147,483,647 | Python 3 | OK | TESTS | 58 | 93 | 7,065,600 | n, k = map(int, input().split())
if k==1:
print(n)
else:
max_sum = ''
for i in range(n.bit_length()):
max_sum += '1'
print(int(max_sum, 2)) | Title: New Year's Eve
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains *n* sweet candies from the good ol' bakery, each labeled from 1 to *n* corresponding to its tastiness. No two candies have the same tastiness.
The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!
A xor-sum of a sequence of integers *a*1,<=*a*2,<=...,<=*a**m* is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than *k* candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.
Input Specification:
The sole string contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1018).
Output Specification:
Output one number — the largest possible xor-sum.
Demo Input:
['4 3\n', '6 6\n']
Demo Output:
['7\n', '7\n']
Note:
In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.
In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7. | ```python
n, k = map(int, input().split())
if k==1:
print(n)
else:
max_sum = ''
for i in range(n.bit_length()):
max_sum += '1'
print(int(max_sum, 2))
``` | 3 | |
245 | A | System Administrator | PROGRAMMING | 800 | [
"implementation"
] | null | null | Polycarpus is a system administrator. There are two servers under his strict guidance — *a* and *b*. To stay informed about the servers' performance, Polycarpus executes commands "ping a" and "ping b". Each ping command sends exactly ten packets to the server specified in the argument of the command. Executing a program results in two integers *x* and *y* (*x*<=+<=*y*<==<=10; *x*,<=*y*<=≥<=0). These numbers mean that *x* packets successfully reached the corresponding server through the network and *y* packets were lost.
Today Polycarpus has performed overall *n* ping commands during his workday. Now for each server Polycarpus wants to know whether the server is "alive" or not. Polycarpus thinks that the server is "alive", if at least half of the packets that we send to this server reached it successfully along the network.
Help Polycarpus, determine for each server, whether it is "alive" or not by the given commands and their results. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of commands Polycarpus has fulfilled. Each of the following *n* lines contains three integers — the description of the commands. The *i*-th of these lines contains three space-separated integers *t**i*, *x**i*, *y**i* (1<=≤<=*t**i*<=≤<=2; *x**i*,<=*y**i*<=≥<=0; *x**i*<=+<=*y**i*<==<=10). If *t**i*<==<=1, then the *i*-th command is "ping a", otherwise the *i*-th command is "ping b". Numbers *x**i*, *y**i* represent the result of executing this command, that is, *x**i* packets reached the corresponding server successfully and *y**i* packets were lost.
It is guaranteed that the input has at least one "ping a" command and at least one "ping b" command. | In the first line print string "LIVE" (without the quotes) if server *a* is "alive", otherwise print "DEAD" (without the quotes).
In the second line print the state of server *b* in the similar format. | [
"2\n1 5 5\n2 6 4\n",
"3\n1 0 10\n2 0 10\n1 10 0\n"
] | [
"LIVE\nLIVE\n",
"LIVE\nDEAD\n"
] | Consider the first test case. There 10 packets were sent to server *a*, 5 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall there were 10 packets sent to server *b*, 6 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network.
Consider the second test case. There were overall 20 packages sent to server *a*, 10 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall 10 packets were sent to server *b*, 0 of them reached it. Therefore, less than half of all packets sent to this server successfully reached it through the network. | 0 | [
{
"input": "2\n1 5 5\n2 6 4",
"output": "LIVE\nLIVE"
},
{
"input": "3\n1 0 10\n2 0 10\n1 10 0",
"output": "LIVE\nDEAD"
},
{
"input": "10\n1 3 7\n2 4 6\n1 2 8\n2 5 5\n2 10 0\n2 10 0\n1 8 2\n2 2 8\n2 10 0\n1 1 9",
"output": "DEAD\nLIVE"
},
{
"input": "11\n1 8 2\n1 6 4\n1 9 1\n1 7 3\n2 0 10\n2 0 10\n1 8 2\n2 2 8\n2 6 4\n2 7 3\n2 9 1",
"output": "LIVE\nDEAD"
},
{
"input": "12\n1 5 5\n1 0 10\n1 4 6\n1 2 8\n1 2 8\n1 5 5\n1 9 1\n2 9 1\n1 5 5\n1 1 9\n2 9 1\n2 7 3",
"output": "DEAD\nLIVE"
},
{
"input": "13\n1 8 2\n1 4 6\n1 5 5\n1 5 5\n2 10 0\n2 9 1\n1 3 7\n2 6 4\n2 6 4\n2 5 5\n1 7 3\n2 3 7\n2 9 1",
"output": "LIVE\nLIVE"
},
{
"input": "14\n1 7 3\n1 0 10\n1 7 3\n1 1 9\n2 2 8\n2 0 10\n1 1 9\n2 8 2\n2 6 4\n1 3 7\n1 3 7\n2 6 4\n2 1 9\n2 7 3",
"output": "DEAD\nDEAD"
}
] | 1,661,438,725 | 2,147,483,647 | Python 3 | OK | TESTS | 13 | 92 | 0 | n=int(input())
ax=0;bx=0
ay=0;by=0
for i in range(n):
t,x,y=map(int,input().split())
if t==1:
ax+=x
ay+=y
else:
bx+=x
by+=y
if ax>=(ax+ay)//2:
print('LIVE')
else:
print('DEAD')
if bx>=(bx+by)//2:
print('LIVE')
else:
print('DEAD') | Title: System Administrator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus is a system administrator. There are two servers under his strict guidance — *a* and *b*. To stay informed about the servers' performance, Polycarpus executes commands "ping a" and "ping b". Each ping command sends exactly ten packets to the server specified in the argument of the command. Executing a program results in two integers *x* and *y* (*x*<=+<=*y*<==<=10; *x*,<=*y*<=≥<=0). These numbers mean that *x* packets successfully reached the corresponding server through the network and *y* packets were lost.
Today Polycarpus has performed overall *n* ping commands during his workday. Now for each server Polycarpus wants to know whether the server is "alive" or not. Polycarpus thinks that the server is "alive", if at least half of the packets that we send to this server reached it successfully along the network.
Help Polycarpus, determine for each server, whether it is "alive" or not by the given commands and their results.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of commands Polycarpus has fulfilled. Each of the following *n* lines contains three integers — the description of the commands. The *i*-th of these lines contains three space-separated integers *t**i*, *x**i*, *y**i* (1<=≤<=*t**i*<=≤<=2; *x**i*,<=*y**i*<=≥<=0; *x**i*<=+<=*y**i*<==<=10). If *t**i*<==<=1, then the *i*-th command is "ping a", otherwise the *i*-th command is "ping b". Numbers *x**i*, *y**i* represent the result of executing this command, that is, *x**i* packets reached the corresponding server successfully and *y**i* packets were lost.
It is guaranteed that the input has at least one "ping a" command and at least one "ping b" command.
Output Specification:
In the first line print string "LIVE" (without the quotes) if server *a* is "alive", otherwise print "DEAD" (without the quotes).
In the second line print the state of server *b* in the similar format.
Demo Input:
['2\n1 5 5\n2 6 4\n', '3\n1 0 10\n2 0 10\n1 10 0\n']
Demo Output:
['LIVE\nLIVE\n', 'LIVE\nDEAD\n']
Note:
Consider the first test case. There 10 packets were sent to server *a*, 5 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall there were 10 packets sent to server *b*, 6 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network.
Consider the second test case. There were overall 20 packages sent to server *a*, 10 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall 10 packets were sent to server *b*, 0 of them reached it. Therefore, less than half of all packets sent to this server successfully reached it through the network. | ```python
n=int(input())
ax=0;bx=0
ay=0;by=0
for i in range(n):
t,x,y=map(int,input().split())
if t==1:
ax+=x
ay+=y
else:
bx+=x
by+=y
if ax>=(ax+ay)//2:
print('LIVE')
else:
print('DEAD')
if bx>=(bx+by)//2:
print('LIVE')
else:
print('DEAD')
``` | 3 | |
30 | A | Accounting | PROGRAMMING | 1,400 | [
"brute force",
"math"
] | A. Accounting | 2 | 256 | A long time ago in some far country lived king Copa. After the recent king's reform, he got so large powers that started to keep the books by himself.
The total income *A* of his kingdom during 0-th year is known, as well as the total income *B* during *n*-th year (these numbers can be negative — it means that there was a loss in the correspondent year).
King wants to show financial stability. To do this, he needs to find common coefficient *X* — the coefficient of income growth during one year. This coefficient should satisfy the equation:
Surely, the king is not going to do this job by himself, and demands you to find such number *X*.
It is necessary to point out that the fractional numbers are not used in kingdom's economy. That's why all input numbers as well as coefficient *X* must be integers. The number *X* may be zero or negative. | The input contains three integers *A*, *B*, *n* (|*A*|,<=|*B*|<=≤<=1000, 1<=≤<=*n*<=≤<=10). | Output the required integer coefficient *X*, or «No solution», if such a coefficient does not exist or it is fractional. If there are several possible solutions, output any of them. | [
"2 18 2\n",
"-1 8 3\n",
"0 0 10\n",
"1 16 5\n"
] | [
"3",
"-2",
"5",
"No solution"
] | none | 500 | [
{
"input": "2 18 2",
"output": "3"
},
{
"input": "-1 8 3",
"output": "-2"
},
{
"input": "0 0 10",
"output": "5"
},
{
"input": "1 16 5",
"output": "No solution"
},
{
"input": "0 1 2",
"output": "No solution"
},
{
"input": "3 0 4",
"output": "0"
},
{
"input": "1 1000 1",
"output": "1000"
},
{
"input": "7 896 7",
"output": "2"
},
{
"input": "4 972 1",
"output": "243"
},
{
"input": "-1 -1 5",
"output": "1"
},
{
"input": "-1 0 4",
"output": "0"
},
{
"input": "-7 0 1",
"output": "0"
},
{
"input": "-5 -5 3",
"output": "1"
},
{
"input": "-5 -5 9",
"output": "1"
},
{
"input": "-5 -5 6",
"output": "1"
},
{
"input": "-4 0 1",
"output": "0"
},
{
"input": "-5 0 3",
"output": "0"
},
{
"input": "-4 4 9",
"output": "-1"
},
{
"input": "10 0 6",
"output": "0"
},
{
"input": "-5 3 4",
"output": "No solution"
},
{
"input": "0 3 6",
"output": "No solution"
},
{
"input": "3 6 10",
"output": "No solution"
},
{
"input": "-3 7 5",
"output": "No solution"
},
{
"input": "-526 526 1",
"output": "-1"
},
{
"input": "-373 373 3",
"output": "-1"
},
{
"input": "-141 0 8",
"output": "0"
},
{
"input": "7 175 1",
"output": "25"
},
{
"input": "-5 -560 1",
"output": "112"
},
{
"input": "-1 -512 10",
"output": "No solution"
},
{
"input": "-3 -768 8",
"output": "2"
},
{
"input": "-3 -768 7",
"output": "No solution"
},
{
"input": "-3 -768 9",
"output": "No solution"
},
{
"input": "-3 -768 4",
"output": "4"
},
{
"input": "4 972 4",
"output": "No solution"
},
{
"input": "4 972 5",
"output": "3"
},
{
"input": "4 972 6",
"output": "No solution"
},
{
"input": "4 972 1",
"output": "243"
},
{
"input": "4 972 2",
"output": "No solution"
},
{
"input": "1 1000 1",
"output": "1000"
},
{
"input": "1 961 2",
"output": "31"
},
{
"input": "1 1000 3",
"output": "10"
},
{
"input": "1 625 4",
"output": "5"
},
{
"input": "4 972 5",
"output": "3"
},
{
"input": "1 729 6",
"output": "3"
},
{
"input": "7 896 7",
"output": "2"
},
{
"input": "3 768 8",
"output": "2"
},
{
"input": "1 512 9",
"output": "2"
},
{
"input": "1 1000 1",
"output": "1000"
},
{
"input": "1 961 2",
"output": "31"
},
{
"input": "1 1000 3",
"output": "10"
},
{
"input": "1 625 4",
"output": "5"
},
{
"input": "4 972 5",
"output": "3"
},
{
"input": "1 729 6",
"output": "3"
},
{
"input": "7 896 7",
"output": "2"
},
{
"input": "3 768 8",
"output": "2"
},
{
"input": "1 512 9",
"output": "2"
},
{
"input": "1 1000 1",
"output": "1000"
},
{
"input": "1 961 2",
"output": "31"
},
{
"input": "1 1000 3",
"output": "10"
},
{
"input": "1 625 4",
"output": "5"
},
{
"input": "4 972 5",
"output": "3"
},
{
"input": "1 729 6",
"output": "3"
},
{
"input": "7 896 7",
"output": "2"
},
{
"input": "3 768 8",
"output": "2"
},
{
"input": "1 512 9",
"output": "2"
},
{
"input": "1 1 5",
"output": "1"
},
{
"input": "1 1 4",
"output": "1"
},
{
"input": "1 -1 1",
"output": "-1"
}
] | 1,625,266,291 | 2,147,483,647 | PyPy 3 | OK | TESTS | 68 | 248 | 1,536,000 | a, b, n = map(int, input().split())
for i in range(-1000, 1001):
if a*(i**n) == b:
print(i)
exit()
print("No solution") | Title: Accounting
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A long time ago in some far country lived king Copa. After the recent king's reform, he got so large powers that started to keep the books by himself.
The total income *A* of his kingdom during 0-th year is known, as well as the total income *B* during *n*-th year (these numbers can be negative — it means that there was a loss in the correspondent year).
King wants to show financial stability. To do this, he needs to find common coefficient *X* — the coefficient of income growth during one year. This coefficient should satisfy the equation:
Surely, the king is not going to do this job by himself, and demands you to find such number *X*.
It is necessary to point out that the fractional numbers are not used in kingdom's economy. That's why all input numbers as well as coefficient *X* must be integers. The number *X* may be zero or negative.
Input Specification:
The input contains three integers *A*, *B*, *n* (|*A*|,<=|*B*|<=≤<=1000, 1<=≤<=*n*<=≤<=10).
Output Specification:
Output the required integer coefficient *X*, or «No solution», if such a coefficient does not exist or it is fractional. If there are several possible solutions, output any of them.
Demo Input:
['2 18 2\n', '-1 8 3\n', '0 0 10\n', '1 16 5\n']
Demo Output:
['3', '-2', '5', 'No solution']
Note:
none | ```python
a, b, n = map(int, input().split())
for i in range(-1000, 1001):
if a*(i**n) == b:
print(i)
exit()
print("No solution")
``` | 3.935139 |
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,670,667,317 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 92 | 0 | s = input()
t = input()
def reverse(string):
string = string[::-1]
return string
if reverse(s) == t:
print("YES")
else:
print("NO")
quit()
| Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
s = input()
t = input()
def reverse(string):
string = string[::-1]
return string
if reverse(s) == t:
print("YES")
else:
print("NO")
quit()
``` | 3.977 |
730 | A | Toda 2 | PROGRAMMING | 1,800 | [
"greedy",
"implementation"
] | null | null | A group of *n* friends enjoys playing popular video game Toda 2. There is a rating system describing skill level of each player, initially the rating of the *i*-th friend is *r**i*.
The friends decided to take part in the championship as a team. But they should have equal ratings to be allowed to compose a single team consisting of all *n* friends. So the friends are faced with the problem: how to make all their ratings equal.
One way to change ratings is to willingly lose in some matches. Friends can form a party consisting of two to five (but not more than *n*) friends and play a match in the game. When the party loses, the rating of each of its members decreases by 1. A rating can't become negative, so *r**i*<==<=0 doesn't change after losing.
The friends can take part in multiple matches, each time making a party from any subset of friends (but remember about constraints on party size: from 2 to 5 members).
The friends want to make their ratings equal but as high as possible.
Help the friends develop a strategy of losing the matches so that all their ratings become equal and the resulting rating is maximum possible. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the number of friends.
The second line contains *n* non-negative integers *r*1,<=*r*2,<=...,<=*r**n* (0<=≤<=*r**i*<=≤<=100), where *r**i* is the initial rating of the *i*-th friend. | In the first line, print a single integer *R* — the final rating of each of the friends.
In the second line, print integer *t* — the number of matches the friends have to play. Each of the following *t* lines should contain *n* characters '0' or '1', where the *j*-th character of the *i*-th line is equal to:
- '0', if friend *j* should not play in match *i*, - '1', if friend *j* should play in match *i*.
Each line should contain between two and five characters '1', inclusive.
The value *t* should not exceed 104, it is guaranteed that such solution exists.
Remember that you shouldn't minimize the value *t*, but you should maximize *R*. If there are multiple solutions, print any of them. | [
"5\n4 5 1 7 4\n",
"2\n1 2\n",
"3\n1 1 1\n"
] | [
"1\n8\n01010\n00011\n01010\n10010\n00011\n11000\n00011\n11000\n",
"0\n2\n11\n11\n",
"1\n0\n"
] | none | 0 | [
{
"input": "5\n4 5 1 7 4",
"output": "1\n8\n01010\n00011\n01010\n10010\n00011\n11000\n00011\n11000"
},
{
"input": "2\n1 2",
"output": "0\n2\n11\n11"
},
{
"input": "3\n1 1 1",
"output": "1\n0"
},
{
"input": "10\n6 8 7 6 8 7 6 7 8 7",
"output": "6\n4\n0100100010\n0000000111\n0000110000\n0110000000"
},
{
"input": "5\n4 4 4 7 3",
"output": "2\n6\n00110\n01010\n10010\n00011\n00110\n11000"
},
{
"input": "5\n4 7 5 2 2",
"output": "2\n5\n01100\n01100\n11000\n01100\n11000"
},
{
"input": "6\n5 4 2 4 3 2",
"output": "2\n4\n100100\n110000\n000110\n110000"
},
{
"input": "7\n7 8 2 7 10 11 5",
"output": "2\n17\n0000110\n0000110\n0000110\n0100010\n0001110\n1100000\n0001110\n1100000\n0000011\n0001100\n1100000\n0000011\n0001100\n1100000\n0000011\n0001100\n1100000"
},
{
"input": "10\n2 3 3 3 2 6 2 5 3 5",
"output": "2\n6\n0000010001\n0000010100\n0000010101\n0000000111\n0001010000\n0110000000"
},
{
"input": "90\n45 69 0 10 8 58 25 66 22 2 4 62 64 90 82 83 67 32 56 80 64 51 78 21 2 90 65 55 11 51 1 43 6 32 25 46 22 46 26 6 43 14 50 40 74 52 44 60 76 35 21 10 3 49 87 23 89 17 65 75 7 3 42 12 39 73 9 88 60 91 3 49 9 29 35 2 37 63 48 31 60 62 50 4 15 71 8 49 66 31",
"output": "0\n1847\n000000000000000000000000010000000000000000000000000000000000000000000100000000000000000000\n000000000000010000000000000000000000000000000000000000000000000000000100000000000000000000\n000000000000000000000000000000000000000000000000000000001000000000000100000000000000000000\n000000000000010000000000010000000000000000000000000000000000000000000000000000000000000000\n000000000000000000000000000000000000000000000000000000001000000000010100000000000000000000\n000000000000010000000000010000000000000000..."
},
{
"input": "100\n20 11 15 10 20 19 10 11 16 12 13 20 11 18 16 16 14 16 13 13 11 12 10 18 14 13 16 15 15 11 12 19 19 11 17 16 11 12 15 12 15 14 20 13 17 13 10 15 18 12 13 16 12 13 18 17 18 18 17 13 15 10 16 15 14 14 12 15 15 15 19 14 15 20 10 15 19 10 15 10 13 19 19 18 18 14 14 14 13 14 19 15 11 11 100 17 11 13 20 12",
"output": "10\n272\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000100010\n0000000000000000000000000000000000000000000000000000000000000000000000000100000000000000000000100000\n0000000000000000000000000000000000000000001000000000000000000000000000000000000000000000000000100000\n0000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000100000\n0000100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "100\n29 26 22 21 28 29 28 24 27 29 25 29 25 25 27 20 25 27 20 26 27 26 21 26 20 30 23 21 22 30 28 25 21 28 29 28 28 24 23 22 25 23 23 30 25 24 27 25 27 25 23 28 25 21 24 25 30 24 29 21 26 28 23 30 30 27 28 20 24 30 28 26 27 23 30 30 26 28 21 23 27 28 23 29 24 90 26 25 21 21 25 22 22 28 23 26 24 29 24 21",
"output": "20\n300\n0000000000000000000000000000000000000000000000000000000000000000000000000001000000000100000000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000010000000000100000000000000\n0000000000000000000000000000000000000000000000000000000000000000000001000000000000000100000000000000\n0000000000000000000000000000000000000000000000000000000000000000100000000000000000000100000000000000\n0000000000000000000000000000000000000000000000000000000000000001000000000000000000000100000000..."
},
{
"input": "100\n34 31 31 50 24 48 43 48 22 32 23 22 32 22 32 23 42 20 28 40 32 31 21 52 44 36 29 25 46 20 37 41 36 20 20 46 25 45 26 35 34 25 37 29 38 47 42 25 26 27 48 44 42 45 32 20 25 20 34 50 37 37 20 50 23 27 23 47 39 45 38 23 20 44 48 34 22 49 30 42 24 45 48 28 46 46 42 27 34 23 37 50 39 39 27 44 22 23 34 37",
"output": "20\n701\n0000000000000000000000010000000000000000000000000000000000000000000000000000000000000000000100000000\n0000000000000000000000010000000000000000000000000000000000000001000000000000000000000000000000000000\n0001000000000000000000010000000000000000000000000000000000010000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000100000000\n0000000000000000000000000000000000000000000000000000000000010001000000000000000000000000000000..."
},
{
"input": "100\n84 53 98 75 42 42 37 74 22 94 47 8 53 15 28 80 95 74 60 71 71 21 96 31 97 54 31 55 20 76 13 57 10 30 28 66 68 76 24 100 16 51 2 35 56 37 37 84 22 18 15 97 66 25 48 17 95 47 41 56 45 30 29 99 98 8 3 33 41 22 32 11 67 0 50 73 90 55 50 37 24 81 6 4 66 50 51 19 49 10 36 29 62 51 25 10 89 13 46 98",
"output": "0\n2342\n0000000000000000000000000000000000000001000000000000000000000001000000000000000000000000000000000000\n0000000000000000000000000000000000000001000000000000000000000000000000000000000000000000000000000001\n0000000000000000000000000000000000000000000000000000000000000001100000000000000000000000000000000000\n0010000000000000000000000000000000000001000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000000000000000001100000000000000000000000000000..."
},
{
"input": "100\n4 3 2 5 5 4 4 5 3 5 5 5 4 2 4 3 5 3 3 3 3 100 3 3 4 4 3 5 5 5 5 3 4 2 2 5 5 3 5 4 3 3 3 4 2 2 2 5 4 2 2 2 4 4 3 4 2 2 2 3 2 5 4 2 5 5 2 5 3 4 3 2 4 5 5 3 5 5 4 3 2 4 2 5 4 3 3 3 4 2 3 3 5 3 5 3 5 5 3 5",
"output": "2\n128\n0000000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000000001\n0000000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000000100\n0000000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000001000\n0000000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000100000\n00000000000000000000010000000000000000000000000000000000000000000000000000000000000000000000100..."
},
{
"input": "100\n4 2 4 1 1 4 1 1 2 4 1 1 4 4 1 1 1 3 2 4 2 3 2 4 3 4 3 4 3 2 1 4 4 4 3 2 4 3 3 2 2 3 1 4 3 4 1 1 4 3 3 1 1 3 2 4 1 4 2 2 2 2 2 2 1 4 3 1 4 4 100 4 4 3 1 2 4 3 3 2 1 2 1 2 2 3 3 2 2 1 2 4 2 4 4 1 4 1 4 3",
"output": "1\n127\n0000000000000000000000000000000000000000000000000000000000000000000000100000000000000000000000000010\n0000000000000000000000000000000000000000000000000000000000000000000000100000000000000000000000001000\n0000000000000000000000000000000000000000000000000000000000000000000000100000000000000000000000100000\n0000000000000000000000000000000000000000000000000000000000000000000000100000000000000000000001000000\n00000000000000000000000000000000000000000000000000000000000000000000001000000000000000000001000..."
},
{
"input": "100\n34 37 20 32 28 21 26 35 38 27 23 39 32 21 27 26 36 23 32 33 29 24 22 38 22 29 38 21 32 38 21 28 36 36 27 40 33 33 32 21 20 33 21 33 30 45 33 21 39 32 28 22 34 29 38 39 38 36 30 40 29 25 22 28 35 40 22 28 38 39 40 40 34 40 38 33 38 36 40 25 35 28 39 36 21 21 34 35 21 28 35 37 22 24 36 30 37 34 29 40",
"output": "20\n559\n0000000000000000000000000000000000000000000001000000000000000000000000000000000000000000000000000001\n0000000000000000000000000000000000000000000001000000000000000000000000000000001000000000000000000000\n0000000000000000000000000000000000000000000001000000000000000000000000000100000000000000000000000000\n0000000000000000000000000000000000000000000001000000000000000000000000010000000000000000000000000000\n0000000000000000000000000000000000000000000001000000000000000000000000100000000000000000000000..."
},
{
"input": "99\n77 79 94 77 79 73 85 56 60 94 99 53 79 100 88 87 86 3 95 79 63 79 76 74 67 88 94 57 93 83 84 75 97 78 80 90 65 91 80 66 100 93 73 90 90 57 99 95 76 91 62 50 93 71 50 84 65 79 55 64 52 96 91 77 96 78 88 64 74 70 66 58 84 68 52 85 85 90 55 71 87 69 97 95 67 100 81 78 62 57 98 74 83 85 89 57 77 71 78",
"output": "3\n3670\n000000000000010000000000000000000000000010000000000000000000000000000000000000000000010000000000000\n000000000000000000000000000000000000000010000010000000000000000000000000000000000000010000000000000\n000000000010010000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n000000000000000000000000000000000000000000000000000000000000000000000000000000000000010000100000000\n00000000000000000000000000000000000000001000001000000000000000000000000000000000000000000000000000..."
},
{
"input": "2\n62 64",
"output": "0\n64\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11"
},
{
"input": "2\n71 70",
"output": "0\n71\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11"
},
{
"input": "5\n4 5 4 3 3",
"output": "3\n2\n01100\n11000"
},
{
"input": "6\n4 2 3 2 2 2",
"output": "1\n4\n101000\n100001\n001110\n110000"
},
{
"input": "7\n4 6 5 8 8 2 6",
"output": "2\n12\n0001100\n0001100\n0000101\n0101000\n0001101\n0110000\n0000101\n0011000\n1100000\n0000101\n0011000\n1100000"
},
{
"input": "10\n4 5 3 3 3 4 2 3 2 2",
"output": "2\n5\n0100010000\n1100000000\n0000110100\n0011000000\n1100000000"
},
{
"input": "90\n39 52 38 12 50 81 18 43 47 28 36 50 7 1 57 71 7 6 14 34 22 30 47 8 19 46 66 36 25 31 87 75 10 63 38 18 54 40 61 8 10 59 37 63 77 8 44 51 81 60 43 4 65 66 5 2 58 43 75 52 18 19 81 7 22 60 68 23 64 28 20 77 49 41 12 64 4 1 24 67 90 77 5 33 70 47 35 25 89 48",
"output": "1\n1754\n000000000000000000000000000000000000000000000000000000000000000000000000000000001000000010\n000000000000000000000000000000000000000000000000000000000000000000000000000000001000000010\n000000000000000000000000000000000000000000000000000000000000000000000000000000001000000010\n000000000000000000000000000000100000000000000000000000000000000000000000000000001000000000\n000000000000000000000000000000100000000000000000000000000000000000000000000000001000000010\n000000000000000000000000000000100000000000..."
},
{
"input": "100\n10 19 10 19 17 11 15 20 15 16 20 10 20 18 20 15 18 17 11 19 10 16 17 18 20 18 17 15 16 11 11 12 19 14 16 20 16 16 20 10 11 15 20 11 16 10 14 12 15 17 20 18 18 10 19 11 13 17 19 15 13 18 10 16 17 11 16 14 10 12 17 12 11 20 11 14 13 13 17 14 12 14 19 10 20 19 11 14 19 11 18 20 17 18 16 20 12 19 20 14",
"output": "10\n266\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010010\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000100000000\n0000000000000000000000000000000000000000000000000010000000000000000000000100000000000000000000000000\n0000000000000000000000000000000000000010001000000000000000000000000000000000000000000000000000000000\n0000000000000000000000001000000000010000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "100\n25 25 22 30 24 21 23 30 28 29 20 22 28 23 20 29 24 26 29 28 23 23 24 27 22 25 26 22 20 22 27 20 24 29 22 21 20 27 23 25 23 20 22 22 21 26 21 27 25 26 25 24 21 28 21 28 25 27 20 20 28 27 23 20 24 20 20 27 30 28 21 29 20 27 26 20 23 22 28 20 23 25 26 25 26 30 26 26 22 27 23 23 26 28 25 24 26 20 24 29",
"output": "20\n216\n0000000000000000000000000000000000000000000000000000000000000000000010000000000000000100000000000000\n0001000100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000001\n0000000000000000000000000000000000000000000000000000000000000000000010010000000000000000000000000000\n0000000000000000001000000000000001000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "100\n50 21 46 32 43 47 44 41 31 20 35 49 33 35 29 34 48 40 32 44 20 24 37 50 26 39 42 42 20 33 41 27 49 47 35 27 23 40 23 38 37 48 41 46 38 24 46 20 42 47 43 47 35 50 46 43 27 35 20 26 46 32 50 26 35 22 29 42 28 47 38 45 46 28 45 24 28 46 46 49 41 37 30 35 25 45 36 30 35 45 24 38 24 45 35 46 23 30 33 26",
"output": "20\n809\n0000000000000000000000000000000000000000000000000000010000000010000000000000000000000000000000000000\n1000000000000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000000010000000010000000000000000100000000000000000000\n0000000000000000000000010000000010000000000000000000000000000000000000000000000000000000000000000000\n1000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "100\n71 48 79 53 91 33 40 14 35 23 77 32 80 63 63 24 50 71 58 3 43 42 17 67 47 28 29 53 2 55 51 93 90 95 97 29 24 23 9 62 15 6 98 9 58 87 93 59 9 62 69 33 87 21 15 17 35 79 51 3 86 28 89 6 75 62 90 25 60 38 4 17 86 78 65 65 9 4 53 10 77 1 91 79 22 91 16 27 65 2 83 38 29 29 2 19 51 66 64 9",
"output": "1\n2270\n0000000000000000000000000000000000100000001000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000100000001000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000100000001000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000001000000001000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000001100000001000000000000000000000000000000000000000000000000000..."
},
{
"input": "100\n4 2 5 2 5 2 5 5 4 2 5 4 2 2 2 5 4 3 5 2 3 5 3 5 5 3 2 5 2 2 2 3 4 3 5 5 3 4 4 3 3 3 4 5 5 3 2 2 5 5 3 4 2 4 5 2 4 5 4 5 5 4 5 3 5 5 2 3 3 5 4 2 5 2 4 4 4 2 3 5 5 5 2 5 5 2 3 5 3 2 2 3 4 4 4 5 2 2 3 2",
"output": "2\n78\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000001001000000010000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000001010000000000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000110000000000000000000\n0000000000000000000000000000000000000000000000000000000000000000000001001000000000000000000000000000\n000000000000000000000000000000000000000000000000000000000000000011000000000000000000000000000000..."
},
{
"input": "100\n4 3 4 3 1 3 2 1 2 4 1 3 3 3 4 1 4 4 4 2 2 2 1 1 4 4 3 1 2 1 2 2 1 1 2 1 1 1 3 4 2 3 3 4 4 3 3 4 3 4 2 1 3 1 2 4 1 1 4 2 4 3 4 1 1 2 3 4 3 3 3 3 4 2 3 4 2 4 1 3 3 4 2 3 2 1 1 3 4 2 4 3 4 2 1 4 2 4 2 3",
"output": "1\n79\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001010000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000001000000100000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000001010000000000000000000000\n000000000000000000000000000000000000000000000000000000000000000000010000100000000000000000000000..."
},
{
"input": "100\n29 30 36 24 21 22 21 35 35 36 20 20 37 37 26 27 33 30 38 22 28 31 30 22 25 32 32 21 36 30 22 34 38 27 32 23 32 27 35 31 23 36 40 20 27 36 20 27 27 39 22 27 29 36 26 34 20 39 30 28 34 24 28 38 24 33 29 33 30 35 37 31 30 34 29 20 27 23 30 33 33 21 25 39 27 40 35 38 36 28 33 26 38 34 38 37 30 21 29 33",
"output": "20\n493\n0000000000000000000000000000000000000000001000000000000000000000000000000000000000000100000000000000\n0000000000000000000000000000000000000000000000000000000001000000000000000000000000010100000000000000\n0000000000000000000000000000000000000000001000000100000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000001000010100000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000010100000000..."
},
{
"input": "99\n77 93 81 86 86 84 62 55 93 58 50 53 61 91 62 96 83 58 93 87 68 100 53 76 80 60 82 54 88 53 84 85 83 98 55 73 62 64 88 89 72 69 68 85 83 89 72 100 97 83 58 63 80 60 98 72 64 55 62 71 73 99 89 98 96 65 52 68 66 53 62 78 75 81 95 90 97 58 94 87 100 60 59 94 88 79 88 64 89 86 59 67 59 65 81 78 91 95 62",
"output": "50\n1285\n000000000000000000000100000000000000000000000001000000000000000000000000000000001000000000000000000\n000000000000000000000000000000000000000000000000000000000000010000000000000000001000000000000000000\n000000000000000000000100000000000000000000000001000000000000000000000000000000000000000000000000000\n000000000000000000000000000000000000000000000000000000000000010100000000000000001000000000000000000\n0000000000000000000000000000000000000000000000010000001000000000000000000000000000000000000000000..."
},
{
"input": "2\n61 60",
"output": "0\n61\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11"
},
{
"input": "2\n70 70",
"output": "70\n0"
},
{
"input": "100\n6 9 34 78 60 53 6 51 73 96 93 93 23 47 75 85 40 2 84 74 63 94 98 5 47 36 40 97 16 5 21 68 64 24 42 20 32 94 8 99 36 39 7 8 98 85 2 3 68 76 68 93 37 45 88 61 52 70 62 21 65 34 12 95 58 82 90 60 36 36 2 80 75 44 13 44 97 94 48 11 16 25 76 62 47 31 83 80 5 42 96 10 75 53 30 4 60 81 10 2",
"output": "2\n2398\n0000000000000000000000000000000000000001000010000000000000000000000000000000000000000000000000000000\n0000000000000000000000100000000000000001000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000001000010000000000000000000000000000000100000000000000000000000\n0000000000000000000000100001000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000010000000000000000000000000000000100000000000001000..."
},
{
"input": "100\n52 52 44 69 57 29 44 98 94 98 89 77 67 51 21 91 7 76 91 11 30 49 58 30 34 29 64 53 99 47 99 82 29 28 27 11 66 75 94 37 1 68 75 34 1 78 87 60 79 21 84 84 12 44 85 91 76 8 75 88 56 31 87 37 25 82 85 78 15 23 7 9 91 44 4 56 9 94 17 33 99 75 11 49 24 69 47 30 94 64 1 77 36 48 78 69 85 60 94 86",
"output": "1\n2672\n0000000000000000000000000000101000000000000000000000000000000000000000000000000010000000000000000000\n0000000000000000000000000000101000000000000000000000000000000000000000000000000010000000000000000000\n0000000101000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000101000000000000000000000000000000000000000000000000010000000000000000000\n0000000101000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99\n32 83 75 47 92 14 21 45 67 49 52 47 1 4 57 95 38 49 23 17 14 29 88 92 2 73 2 95 29 94 74 6 32 94 84 59 63 57 90 81 38 8 85 48 96 69 71 75 39 73 47 36 22 1 15 40 82 54 73 52 6 42 67 21 82 56 4 38 20 79 56 97 22 88 64 21 62 35 43 14 18 38 42 84 76 22 13 68 79 12 63 58 64 1 19 74 39 75 13",
"output": "1\n2360\n000000000000000000000000000000000000000000001000000000000000000000000001000000000000000000000000000\n000000000000000000000000000000000000000000001000000000000000000000000001000000000000000000000000000\n000000000000000100000000000100000000000000000000000000000000000000000001000000000000000000000000000\n000000000000000000000000000000000000000000001000000000000000000000000001000000000000000000000000000\n00000000000000000000000000000100010000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "100\n2 1 2 2 1 1 1 2 1 1 1 1 2 1 1 2 2 2 1 2 1 1 2 1 2 1 2 2 1 2 2 2 1 2 1 2 1 2 1 1 2 2 2 2 2 2 2 1 1 2 1 1 2 1 1 1 2 2 1 2 1 1 1 2 2 2 1 2 2 2 1 1 1 1 2 2 2 1 1 2 2 1 1 1 2 2 1 2 1 2 2 2 2 1 1 1 2 2 1 1",
"output": "1\n25\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010001100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001100000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000001010000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000001100000000000000\n000000000000000000000000000000000000000000000000000000000000000000000000000000011000000000000000..."
},
{
"input": "4\n1 1 6 2",
"output": "0\n6\n0011\n0011\n0110\n1010\n0011\n0011"
},
{
"input": "3\n1 2 6",
"output": "0\n6\n011\n011\n101\n011\n011\n011"
},
{
"input": "7\n2 5 4 1 6 3 4",
"output": "1\n8\n0100100\n0000101\n0110100\n0000111\n0110000\n0000011\n0010100\n1100000"
},
{
"input": "99\n91 3 12 82 69 36 95 18 97 43 13 18 80 73 93 21 87 71 33 45 38 93 75 80 13 86 74 84 18 35 97 12 41 68 1 33 48 93 99 74 23 66 79 52 81 55 95 27 65 97 60 71 46 63 16 47 7 90 59 94 32 90 38 65 91 94 19 46 67 52 98 60 57 87 42 21 86 27 46 82 39 1 36 23 29 4 30 32 33 44 79 61 93 7 75 12 65 35 68",
"output": "1\n2619\n000000000000000000000000000000000000001000000000000000000000000000000010000000000000000000000000000\n000000000000000000000000000000000000001000000000000000000000000000000010000000000000000000000000000\n000000000000000000000000000000000000001000000000010000000000000000000000000000000000000000000000000\n000000001000000000000000000000100000000000000000000000000000000000000000000000000000000000000000000\n00000000000000000000000000000000000000100000000001000000000000000000001000000000000000000000000000..."
},
{
"input": "100\n1 2 1 1 1 1 1 2 1 2 1 2 2 2 2 2 2 1 2 2 2 1 1 2 1 1 2 2 2 1 1 2 2 1 2 2 1 1 2 1 1 1 2 2 2 1 2 2 2 2 1 1 2 2 2 2 1 2 1 2 2 1 2 2 2 1 1 2 2 2 2 2 1 2 1 1 2 2 1 1 2 1 1 1 2 2 2 2 1 2 1 2 1 2 1 2 1 1 1 2",
"output": "1\n28\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010001\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000101000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000001010000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000110000000000000\n000000000000000000000000000000000000000000000000000000000000000000000000000000001000100000000000..."
},
{
"input": "95\n15 20 18 19 13 10 14 17 11 14 20 17 17 22 15 7 12 17 12 7 21 14 21 8 10 14 21 15 21 16 19 14 8 15 11 22 14 9 7 7 8 7 7 9 15 16 10 16 9 10 11 18 17 9 17 8 16 8 16 21 9 22 14 16 14 20 22 15 7 15 80 13 10 10 15 8 8 11 9 8 12 17 19 9 11 22 15 10 21 18 16 8 12 16 15",
"output": "7\n356\n00000000000000000000000000000000000000000000000000000000000000000000001000000000000001000000000\n00000000000000000000000000000000000000000000000000000000000000000010001000000000000000000000000\n00000000000000000000000000000000000000000000000000000000000001000000001000000000000000000000000\n00000000000000000000000000000000000100000000000000000000000000000000001000000000000000000000000\n00000000000001000000000000000000000000000000000000000000000000000000001000000000000000000000000\n000000000000000000..."
},
{
"input": "100\n10 22 21 7 3 5 14 23 17 6 4 2 8 10 7 15 8 23 16 16 18 11 20 17 19 12 13 12 21 10 4 20 19 18 2 6 4 15 6 14 8 19 15 22 16 20 15 23 18 21 5 23 4 8 2 9 15 3 92 6 7 4 16 4 3 3 16 2 3 20 14 22 17 23 11 19 4 2 15 22 19 8 7 13 5 13 11 2 19 15 8 12 17 7 18 4 17 12 15 15",
"output": "2\n549\n0000000000000000000000000000000000000000000000000000000000100000000000000100000000000000000000000000\n0000000000000000000000000000000000000000000000000001000000100000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000010000000000100000000000000000000000000000000000000000\n0000000000000000010000000000000000000000000000000000000000100000000000000000000000000000000000000000\n00000001000000000000000000000000000000000000000000000000001000000000000000000000000000000000000..."
},
{
"input": "92\n12 10 17 16 16 10 10 19 17 13 10 10 12 16 20 20 16 14 12 19 14 11 20 15 11 16 10 80 14 20 11 19 16 10 10 19 20 11 13 15 15 18 19 20 10 13 12 20 14 17 10 13 10 20 17 19 12 15 10 17 17 18 10 17 20 14 17 11 12 11 11 10 14 14 11 10 18 14 12 20 16 12 16 13 17 20 14 20 12 16 15 17",
"output": "10\n245\n00000000000000000000000000010000000000000000000000000000000000000000000000000000000000010000\n00000000000000000000000000010000000000000000000000000000000000000000000000000000000001000000\n00000000000000000000000000010000000000000000000000000000000000000000000000000001000000000000\n00000000000000000000000000010000000000000000000000000000000000001000000000000000000000000000\n00000000000000000000000000010000000000000000000000000100000000000000000000000000000000000000\n00000000000000000000000000010000..."
},
{
"input": "91\n14 23 8 8 12 9 28 25 20 9 16 14 22 26 17 9 16 9 26 26 9 19 10 28 19 27 20 13 19 18 21 15 20 13 27 7 14 16 10 78 8 20 13 20 27 20 13 20 15 9 7 7 7 14 23 7 7 25 7 26 28 8 9 11 18 19 8 11 8 26 21 19 12 20 26 18 22 22 28 9 21 27 13 11 9 11 9 27 27 25 28",
"output": "7\n471\n0000000000000000000000000000000000000001000000000000000000000000000000000000000000000000001\n0000000000000000000000000000000000000001000000000000000000000000000000000000001000000000000\n0000000000000000000000000000000000000001000000000000000000001000000000000000000000000000000\n0000000000000000000000010000000000000001000000000000000000000000000000000000000000000000000\n0000001000000000000000000000000000000001000000000000000000000000000000000000000000000000000\n00000000000000000000000000000000000000..."
},
{
"input": "97\n16 5 5 8 19 11 6 10 7 16 5 10 10 18 15 6 9 8 6 16 13 12 17 14 14 5 7 14 19 19 18 11 12 18 19 15 18 12 8 16 11 13 6 7 17 18 8 8 19 5 15 11 14 15 8 18 19 16 6 77 12 8 12 12 20 9 19 10 9 16 12 17 20 10 16 20 20 11 5 15 20 6 7 15 18 14 18 6 17 16 10 16 12 13 16 19 6",
"output": "5\n404\n0000000000000000000000000000000000000000000000000000000000010000000000000000000010000000000000000\n0000000000000000000000000000000000000000000000000000000000010000000000000000100000000000000000000\n0000000000000000000000000000000000000000000000000000000000010000000000000001000000000000000000000\n0000000000000000000000000000000000000000000000000000000000010000000000001000000000000000000000000\n0000000000000000000000000000000000000000000000000000000000010000100000000000000000000000000000000\n00000000..."
},
{
"input": "93\n12 8 17 8 17 16 8 16 98 19 13 14 20 12 14 19 13 13 19 14 18 13 20 17 8 14 15 16 19 13 20 12 17 19 11 18 19 19 20 11 13 8 13 8 19 18 11 19 15 20 20 17 8 20 16 17 9 11 8 8 16 13 20 9 11 18 15 9 18 11 17 11 18 15 17 18 13 12 9 15 17 11 17 17 20 18 20 17 19 16 16 17 18",
"output": "8\n364\n000000001000000000000000000000000000000000000000000000000000000000000000000000000000001000000\n000000001000000000000000000000000000000000000000000000000000000000000000000000000000100000000\n000000001000000000000000000000000000000000000000000000000000001000000000000000000000000000000\n000000001000000000000000000000000000000000000000000001000000000000000000000000000000000000000\n000000001000000000000000000000000000000000000000001000000000000000000000000000000000000000000\n0000000010000000000000000000..."
},
{
"input": "10\n67 46 46 67 46 46 67 67 46 67",
"output": "46\n42\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n0000001101\n1001000000\n..."
},
{
"input": "100\n86 59 68 74 86 35 86 44 86 86 86 86 86 25 86 86 86 77 54 86 54 30 51 55 86 38 61 68 86 49 86 22 52 86 38 86 86 56 86 37 86 71 48 86 61 57 86 86 61 72 86 44 56 39 19 86 86 86 86 86 56 41 86 77 86 46 45 23 86 69 86 86 35 86 86 66 86 86 86 48 41 86 86 43 86 77 86 86 86 86 28 27 19 86 86 52 86 51 86 51",
"output": "19\n2422\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001010\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001100000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000110000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000011000000000000\n000000000000000000000000000000000000000000000000000000000000000000000000000000000010100000000..."
},
{
"input": "100\n64 76 71 82 74 87 51 80 70 53 86 53 57 76 52 98 86 69 64 73 82 44 59 77 78 53 59 91 66 75 99 48 67 94 51 46 61 98 46 54 60 89 52 85 64 69 71 55 94 66 88 75 65 51 58 79 95 50 94 50 94 89 58 99 75 87 51 49 51 98 64 63 61 70 94 82 85 81 45 56 48 73 55 64 47 82 88 50 50 78 72 54 74 44 91 47 85 72 47 88",
"output": "44\n1256\n0000000000000000000000000000001000000000000000000000000000000001000000000000000000000000000000000000\n0000000000000000000000000000000000000100000000000000000000000001000001000000000000000000000000000000\n0000000000000001000000000000001000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000100000000000000000000000001000001000000000000000000000000000000\n000000000000000100000000000000100000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "100\n80 82 72 75 73 79 72 80 79 82 81 81 78 75 79 77 82 78 79 78 77 79 73 81 72 79 77 78 77 77 75 74 78 77 72 76 79 77 81 82 82 78 74 82 72 74 73 81 76 75 95 74 76 77 81 95 82 73 81 78 77 82 77 73 95 75 77 82 78 79 79 73 77 75 77 77 75 78 74 77 81 76 79 77 81 77 74 73 75 76 82 81 73 73 81 75 77 81 72 79",
"output": "72\n281\n0000000000000000000000000000000000000000000000000010000100000000100000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000010000100000000100000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000010000100000000100000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000010000100000000100000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000010000100000000100000000000000000000000000000..."
},
{
"input": "100\n61 41 11 61 12 23 7 22 35 20 3 20 2 47 38 3 8 7 7 40 1 34 31 8 18 24 30 41 28 26 17 26 20 27 12 36 30 45 46 47 8 15 10 27 30 38 10 23 47 10 42 37 42 3 16 39 16 9 61 7 2 1 5 43 34 25 11 6 22 19 42 0 43 17 28 17 2 16 36 21 20 37 9 5 44 2 17 46 43 26 24 61 32 4 18 11 40 14 22 18",
"output": "0\n1182\n0000000000000000000000000000000000000000000000000000000000100000000000000000000000000000000100000000\n1001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000000000000100000000000000000000000000000000100000000\n1001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000000000000100000000000000000000000000000000100..."
},
{
"input": "100\n40 55 53 29 45 24 35 37 46 56 51 51 58 44 54 54 25 33 38 55 45 40 57 62 66 57 30 43 88 47 51 43 30 50 65 51 26 66 26 54 65 36 59 24 52 56 37 29 53 39 60 53 28 46 49 41 50 54 62 38 29 67 50 49 41 88 27 42 27 44 45 44 42 39 62 44 37 48 42 88 88 44 67 26 59 60 44 40 67 28 29 38 41 59 26 38 44 57 88 59",
"output": "24\n1161\n0000000000000000000000000000000000000000000000000000000000000000000000000000000110000000000000000010\n0000000000000000000000000000100000000000000000000000000000000000010000000000000000000000000000000000\n0000000000000000000000000000000000000000000000000000000000000000000000000000000110000000000000000010\n0000000000000000000000000000100000000000000000000000000000000000010000000000000000000000000000000000\n000000000000000000000000000000000000000000000000000000000000000000000000000000011000000000000..."
},
{
"input": "100\n94 81 57 66 82 53 92 96 75 78 94 82 96 90 84 83 58 96 49 85 88 65 94 96 85 92 60 86 50 57 90 60 73 70 69 70 78 63 56 84 84 87 85 80 88 86 95 73 78 76 91 85 85 62 75 90 85 68 62 91 49 51 54 77 79 86 81 83 74 64 56 79 53 55 65 86 49 52 68 56 93 96 54 91 70 74 82 55 84 54 81 73 89 65 96 85 52 85 78 81",
"output": "49\n1317\n0000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000100000\n0000000000000000010000010000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000100001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000000000000000100000000000000000000000000000000001000000000000100000\n000000000000000001000001000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "100\n83 91 87 78 85 76 82 80 91 88 77 75 73 81 74 91 75 75 84 82 80 91 87 72 80 82 91 75 85 82 82 75 77 79 76 84 87 80 81 78 78 84 75 74 80 87 73 86 79 82 75 84 77 77 87 80 74 76 77 86 83 73 80 76 91 83 85 73 83 87 86 78 74 83 77 80 83 82 85 72 83 81 77 82 83 87 87 77 85 82 85 87 76 74 72 82 84 75 75 91",
"output": "72\n436\n0000000000000000000000000010000000000000000000000000000000000000100000000000000000000000000000000001\n0000000000000001000001000000000000000000000000000000000000000000000000000000000000000000000000000000\n0100000010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000000000000000000000010000000000000000000000000000000000000100000000000000000000000000000000001\n0000000000000001000001000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "100\n7 22 26 17 16 24 27 2 45 16 9 2 45 20 6 24 18 27 19 19 21 25 23 6 15 13 15 24 1 27 2 23 6 27 15 22 5 3 17 5 1 22 27 22 4 26 0 24 5 2 45 8 16 15 24 9 4 6 2 27 23 45 23 22 5 2 1 0 27 13 24 45 15 17 8 5 12 0 20 4 3 7 15 24 45 28 11 45 28 0 17 45 5 0 19 17 28 10 3 27",
"output": "0\n824\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000001000100000000\n0000000000000000000000000000000000000000000000000000000000000000000000010000000000001000000000000000\n0000000000000000000000000000000000000000000000000010000000000100000000000000000000000000000000000000\n0000000010001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n00000000000000000000000000000000000000000000000000000000000000000000000000000000000000010001000..."
},
{
"input": "100\n60 65 37 37 41 64 72 71 64 63 59 56 43 32 45 68 28 78 63 67 63 44 28 27 54 78 60 70 44 58 64 66 40 44 26 69 35 47 49 78 78 30 68 58 25 28 68 78 26 63 71 58 53 32 72 25 64 52 51 67 28 67 66 41 33 69 34 57 62 52 55 59 68 42 70 73 72 78 78 32 37 71 36 27 49 60 34 66 33 68 62 78 78 33 24 35 56 58 70 51",
"output": "24\n1478\n0000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000110000000\n0000000000000000000000000000000000000000000000010000000000000000000000000000010000000000000000000000\n0000000000000000000000000000000000000001100000000000000000000000000000000000000000000000000000000000\n0000000000000000010000000100000000000000000000000000000000000000000000000000000000000000000000000000\n000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000011..."
},
{
"input": "100\n95 60 75 63 59 57 95 72 66 57 61 68 66 71 72 68 65 59 60 71 75 70 56 58 69 68 67 95 70 60 65 70 75 95 73 95 66 74 61 68 65 75 69 71 64 95 62 95 59 95 62 75 63 70 57 69 72 67 73 69 69 95 60 64 60 57 71 70 75 59 70 72 61 74 60 95 61 72 71 70 60 59 61 56 56 65 73 59 71 71 75 74 59 75 75 69 58 57 74 68",
"output": "56\n653\n0000000000000000000000000000000000000000000000000000000000000100000000000001000000000000000000000000\n0000000000000000000000000000000000000000000000010100000000000000000000000000000000000000000000000000\n0000000000000000000000000000000000010000000001000000000000000000000000000000000000000000000000000000\n0000000000000000000000000001000001000000000000000000000000000000000000000000000000000000000000000000\n1000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "100\n84 84 86 84 87 85 87 84 87 92 87 85 85 85 85 86 85 84 84 84 92 86 92 86 92 84 86 84 84 85 85 87 86 84 84 85 92 86 85 84 85 87 86 87 86 92 87 86 87 86 85 85 92 87 85 85 86 85 85 84 84 92 86 85 87 87 87 84 84 87 85 85 85 85 85 84 92 86 86 87 87 86 92 86 86 86 86 92 84 87 86 85 87 87 86 87 85 84 84 84",
"output": "84\n105\n0000000000000000000000000000000000000000000000000000000000000000000000000000100000100001000000000000\n0000000000000000000000000000000000000000000000000000100000000100000000000000000000000000000000000000\n0000000000000000000000000000000000001000000001000000000000000000000000000000000000000000000000000000\n0000000000000000000000101000000000000000000000000000000000000000000000000000000000000000000000000000\n0000000001000000000010000000000000000000000000000000000000000000000000000000000000000000000000..."
}
] | 1,478,665,878 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 12 | 140 | 307,200 | n = int(input())
d = list(map(int, input().split()))
answer = []
while not all(i == d[0] for i in d):
i = max(range(n), key = lambda x: d[x])
max1 = d[i]
d[i] = 0
j = max(range(n), key = lambda x: d[x])
d[i] = max1
d[i] = max(d[i] - 1, 0)
d[j] = max(d[j] - 1, 0)
answer.append(''.join('1' if k == i or k == j else '0' for k in range(n)))
print(d[0])
print(len(answer))
for i in answer:
print(i)
| Title: Toda 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A group of *n* friends enjoys playing popular video game Toda 2. There is a rating system describing skill level of each player, initially the rating of the *i*-th friend is *r**i*.
The friends decided to take part in the championship as a team. But they should have equal ratings to be allowed to compose a single team consisting of all *n* friends. So the friends are faced with the problem: how to make all their ratings equal.
One way to change ratings is to willingly lose in some matches. Friends can form a party consisting of two to five (but not more than *n*) friends and play a match in the game. When the party loses, the rating of each of its members decreases by 1. A rating can't become negative, so *r**i*<==<=0 doesn't change after losing.
The friends can take part in multiple matches, each time making a party from any subset of friends (but remember about constraints on party size: from 2 to 5 members).
The friends want to make their ratings equal but as high as possible.
Help the friends develop a strategy of losing the matches so that all their ratings become equal and the resulting rating is maximum possible.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the number of friends.
The second line contains *n* non-negative integers *r*1,<=*r*2,<=...,<=*r**n* (0<=≤<=*r**i*<=≤<=100), where *r**i* is the initial rating of the *i*-th friend.
Output Specification:
In the first line, print a single integer *R* — the final rating of each of the friends.
In the second line, print integer *t* — the number of matches the friends have to play. Each of the following *t* lines should contain *n* characters '0' or '1', where the *j*-th character of the *i*-th line is equal to:
- '0', if friend *j* should not play in match *i*, - '1', if friend *j* should play in match *i*.
Each line should contain between two and five characters '1', inclusive.
The value *t* should not exceed 104, it is guaranteed that such solution exists.
Remember that you shouldn't minimize the value *t*, but you should maximize *R*. If there are multiple solutions, print any of them.
Demo Input:
['5\n4 5 1 7 4\n', '2\n1 2\n', '3\n1 1 1\n']
Demo Output:
['1\n8\n01010\n00011\n01010\n10010\n00011\n11000\n00011\n11000\n', '0\n2\n11\n11\n', '1\n0\n']
Note:
none | ```python
n = int(input())
d = list(map(int, input().split()))
answer = []
while not all(i == d[0] for i in d):
i = max(range(n), key = lambda x: d[x])
max1 = d[i]
d[i] = 0
j = max(range(n), key = lambda x: d[x])
d[i] = max1
d[i] = max(d[i] - 1, 0)
d[j] = max(d[j] - 1, 0)
answer.append(''.join('1' if k == i or k == j else '0' for k in range(n)))
print(d[0])
print(len(answer))
for i in answer:
print(i)
``` | 0 | |
764 | A | Taymyr is calling you | PROGRAMMING | 800 | [
"brute force",
"implementation",
"math"
] | null | null | Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute. | The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104). | Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls. | [
"1 1 10\n",
"1 2 5\n",
"2 3 9\n"
] | [
"10\n",
"2\n",
"1\n"
] | Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | 500 | [
{
"input": "1 1 10",
"output": "10"
},
{
"input": "1 2 5",
"output": "2"
},
{
"input": "2 3 9",
"output": "1"
},
{
"input": "4 8 9",
"output": "1"
},
{
"input": "7 9 2",
"output": "0"
},
{
"input": "10000 10000 10000",
"output": "1"
},
{
"input": "24 22 9235",
"output": "34"
},
{
"input": "74 8 417",
"output": "1"
},
{
"input": "972 1 203",
"output": "0"
},
{
"input": "550 1 754",
"output": "1"
},
{
"input": "860 1 884",
"output": "1"
},
{
"input": "358 2 809",
"output": "2"
},
{
"input": "33 27 216",
"output": "0"
},
{
"input": "2940 1 9311",
"output": "3"
},
{
"input": "4624 1 1953",
"output": "0"
},
{
"input": "2696 2 7345",
"output": "2"
},
{
"input": "3443 2 6701",
"output": "0"
},
{
"input": "3 613 2275",
"output": "1"
},
{
"input": "1 10000 10000",
"output": "1"
},
{
"input": "10000 1 10000",
"output": "1"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "1 1 10000",
"output": "10000"
},
{
"input": "34 27 10000",
"output": "10"
},
{
"input": "2 2 9999",
"output": "4999"
},
{
"input": "2 2 1",
"output": "0"
},
{
"input": "6 4 36",
"output": "3"
},
{
"input": "33 6 3005",
"output": "45"
},
{
"input": "5 1 20",
"output": "4"
},
{
"input": "1 2 10",
"output": "5"
},
{
"input": "2 1 100",
"output": "50"
},
{
"input": "10 20 10000",
"output": "500"
},
{
"input": "8 12 12",
"output": "0"
}
] | 1,549,208,044 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 109 | 0 | import math
f=lambda:map(int,input().split())
n,m,z=f()
c=0
lcm=n*m//math.gcd(n,m)
while lcm*(c+1)<=z:
c+=1
print(c) | Title: Taymyr is calling you
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input Specification:
The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).
Output Specification:
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Demo Input:
['1 1 10\n', '1 2 5\n', '2 3 9\n']
Demo Output:
['10\n', '2\n', '1\n']
Note:
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | ```python
import math
f=lambda:map(int,input().split())
n,m,z=f()
c=0
lcm=n*m//math.gcd(n,m)
while lcm*(c+1)<=z:
c+=1
print(c)
``` | 3 | |
938 | A | Word Correction | PROGRAMMING | 800 | [
"implementation"
] | null | null | Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.
Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.
You are given a word *s*. Can you predict what will it become after correction?
In this problem letters a, e, i, o, u and y are considered to be vowels. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of letters in word *s* before the correction.
The second line contains a string *s* consisting of exactly *n* lowercase Latin letters — the word before the correction. | Output the word *s* after the correction. | [
"5\nweird\n",
"4\nword\n",
"5\naaeaa\n"
] | [
"werd\n",
"word\n",
"a\n"
] | Explanations of the examples:
1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a. | 0 | [
{
"input": "5\nweird",
"output": "werd"
},
{
"input": "4\nword",
"output": "word"
},
{
"input": "5\naaeaa",
"output": "a"
},
{
"input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyw",
"output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyw"
},
{
"input": "69\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "12\nmmmmmmmmmmmm",
"output": "mmmmmmmmmmmm"
},
{
"input": "18\nyaywptqwuyiqypwoyw",
"output": "ywptqwuqypwow"
},
{
"input": "85\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "13\nmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmm"
},
{
"input": "10\nmmmmmmmmmm",
"output": "mmmmmmmmmm"
},
{
"input": "11\nmmmmmmmmmmm",
"output": "mmmmmmmmmmm"
},
{
"input": "15\nmmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmmm"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "14\nmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmm"
},
{
"input": "33\nmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm"
},
{
"input": "79\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "90\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "2\naa",
"output": "a"
},
{
"input": "18\niuiuqpyyaoaetiwliu",
"output": "iqpytiwli"
},
{
"input": "5\nxxxxx",
"output": "xxxxx"
},
{
"input": "6\nxxxahg",
"output": "xxxahg"
},
{
"input": "3\nzcv",
"output": "zcv"
},
{
"input": "4\naepo",
"output": "apo"
},
{
"input": "5\nqqqqq",
"output": "qqqqq"
},
{
"input": "6\naaaaaa",
"output": "a"
},
{
"input": "4\naeta",
"output": "ata"
},
{
"input": "20\nttyttlwaoieulyiluuri",
"output": "ttyttlwalyluri"
},
{
"input": "1\nb",
"output": "b"
},
{
"input": "3\nanc",
"output": "anc"
},
{
"input": "1\ne",
"output": "e"
},
{
"input": "3\naie",
"output": "a"
},
{
"input": "3\nvio",
"output": "vi"
},
{
"input": "2\nea",
"output": "e"
},
{
"input": "3\nuas",
"output": "us"
},
{
"input": "2\nba",
"output": "ba"
},
{
"input": "2\naq",
"output": "aq"
},
{
"input": "2\nya",
"output": "y"
},
{
"input": "2\nou",
"output": "o"
},
{
"input": "2\nbb",
"output": "bb"
},
{
"input": "7\nayylmao",
"output": "alma"
},
{
"input": "2\nab",
"output": "ab"
},
{
"input": "19\nyuouiyaoiiweqrryqqp",
"output": "yweqrryqqp"
},
{
"input": "25\niqypwqpriiioetiuqqqttouei",
"output": "iqypwqpritiqqqtto"
},
{
"input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyz",
"output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyz"
},
{
"input": "17\naccccccccccccccca",
"output": "accccccccccccccca"
},
{
"input": "5\nababa",
"output": "ababa"
},
{
"input": "10\naaaaaaaaaa",
"output": "a"
},
{
"input": "22\naaaaabbbbboyoyoyoyoyac",
"output": "abbbbboc"
},
{
"input": "7\nmahmoud",
"output": "mahmod"
}
] | 1,683,702,341 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 46 | 0 | n = int(input())
s = input()
vowels = set('aeiouy')
result = []
for c in s:
if result and result[-1] in vowels and c in vowels:
continue
result.append(c)
result = ''.join(result)
if not result:
print()
else:
print(result)
| Title: Word Correction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.
Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.
You are given a word *s*. Can you predict what will it become after correction?
In this problem letters a, e, i, o, u and y are considered to be vowels.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of letters in word *s* before the correction.
The second line contains a string *s* consisting of exactly *n* lowercase Latin letters — the word before the correction.
Output Specification:
Output the word *s* after the correction.
Demo Input:
['5\nweird\n', '4\nword\n', '5\naaeaa\n']
Demo Output:
['werd\n', 'word\n', 'a\n']
Note:
Explanations of the examples:
1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a. | ```python
n = int(input())
s = input()
vowels = set('aeiouy')
result = []
for c in s:
if result and result[-1] in vowels and c in vowels:
continue
result.append(c)
result = ''.join(result)
if not result:
print()
else:
print(result)
``` | 3 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,660,621,420 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | s = input()
it = iter(s)
print(all(c in it for c in 'hello')) | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
s = input()
it = iter(s)
print(all(c in it for c in 'hello'))
``` | 0 |
222 | C | Reducing Fractions | PROGRAMMING | 1,800 | [
"implementation",
"math",
"number theory",
"sortings"
] | null | null | To confuse the opponents, the Galactic Empire represents fractions in an unusual format. The fractions are represented as two sets of integers. The product of numbers from the first set gives the fraction numerator, the product of numbers from the second set gives the fraction denominator. However, it turned out that the programs that work with fractions in this representations aren't complete, they lack supporting the operation of reducing fractions. Implement this operation and the Empire won't forget you. | The first input line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) that show how many numbers the first set (the numerator) and the second set (the denominator) contain, correspondingly.
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=107) — the numbers that are multiplied to produce the numerator.
The third line contains *m* space-separated integers: *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=107) — the numbers that are multiplied to produce the denominator. | Print the answer to the problem in the form, similar to the form of the input data. The number of values in the sets you print *n**out*,<=*m**out* must satisfy the inequality 1<=≤<=*n**out*,<=*m**out*<=≤<=105, and the actual values in the sets *a**out*,<=*i* and *b**out*,<=*i* must satisfy the inequality 1<=≤<=*a**out*,<=*i*,<=*b**out*,<=*i*<=≤<=107.
Separate the values in the lines by spaces. The printed fraction must be reduced, that is, there mustn't be such integer *x* (*x*<=><=1), that the numerator and the denominator of the printed fraction are divisible by *x*. If there are several matching answers, print any of them. | [
"3 2\n100 5 2\n50 10\n",
"4 3\n2 5 10 20\n100 1 3\n"
] | [
"2 3\n2 1\n1 1 1\n",
"1 1\n20\n3\n"
] | In the first test sample the numerator equals 1000, the denominator equals 500. If we reduce fraction 1000/500 by the greatest common divisor of the numerator and the denominator (by 500), we obtain fraction 2/1.
In the second test sample the numerator equals 2000, the denominator equals 300. If we reduce fraction 2000/300 by the greatest common divisor of the numerator and the denominator (by 100), we obtain fraction 20/3. | 1,500 | [
{
"input": "3 2\n100 5 2\n50 10",
"output": "2 3\n2 1\n1 1 1"
},
{
"input": "4 3\n2 5 10 20\n100 1 3",
"output": "1 1\n20\n3"
},
{
"input": "2 3\n50 10\n100 5 2",
"output": "2 3\n1 1 \n2 1 1 "
},
{
"input": "1 1\n1\n1",
"output": "1 1\n1 \n1 "
},
{
"input": "3 2\n100 5 2\n10 100",
"output": "3 2\n1 1 1 \n1 1 "
},
{
"input": "5 3\n16 24 36 54 81\n4 6 9",
"output": "5 3\n16 24 9 27 3 \n1 1 1 "
},
{
"input": "10 10\n2 5 11 17 23 31 41 47 59 67\n3 7 13 19 29 37 43 53 61 71",
"output": "10 10\n2 5 11 17 23 31 41 47 59 67 \n3 7 13 19 29 37 43 53 61 71 "
},
{
"input": "5 5\n2 9 8 3 5\n5 7 8 1 1",
"output": "5 5\n2 9 1 3 1 \n1 7 1 1 1 "
},
{
"input": "10 11\n7 43 39 13 25 23 33 40 5 2\n42 22 39 1 23 37 12 48 46 2 2",
"output": "10 11\n1 43 13 1 25 1 1 5 5 1 \n6 2 3 1 23 37 4 8 1 1 1 "
},
{
"input": "1 5\n99\n3 55 18 1 19",
"output": "1 5\n1 \n3 5 2 1 19 "
}
] | 1,675,716,726 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 61 | 1,840 | 201,216,000 | import sys
def read(T):
return [T(i) for i in sys.stdin.readline().split()]
lsp,pr=[],[]
def sieve(n=int(1e7)):
global lsp
lsp=[i for i in range(n+1)]
for i in range(2,n+1):
if lsp[i]!=i:
continue
for j in range(i*i,n+1,i):
lsp[j]=i
for i in range(2,n+1):
if lsp[i]==i:
pr.append(i)
def solve():
[[n,m],a,b]=read(int),read(int),read(int)
nax=int(1e7+5)
d=[0 for i in range(nax)]
for i in range(n):
cur=a[i]
while lsp[cur]!=1:
p,c=lsp[cur],0
while cur%p==0:
cur//=p
c+=1
d[p]+=c
for i in range(m):
cur=b[i]
while lsp[cur]!=1:
p,c=lsp[cur],0
while cur%p==0:
cur//=p
c+=1
d[p]-=c
num,den=[],[]
print(n,m)
for i in range(n):
cur=a[i]
res=1
while lsp[cur]!=1:
p,c=lsp[cur],0
while cur%p==0:
cur//=p
c+=1
if d[p]<=0:
continue
c=min(c,d[p])
d[p]-=c
res*=p**c
print(res,end=" ")
print()
for i in range(m):
cur=b[i]
res=1
while lsp[cur]!=1:
p,c=lsp[cur],0
while cur%p==0:
cur//=p
c+=1
if d[p]>=0:
continue
c=min(c,abs(d[p]))
d[p]+=c
res*=p**c
print(res,end=" ")
def main():
sieve()
solve()
main() | Title: Reducing Fractions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To confuse the opponents, the Galactic Empire represents fractions in an unusual format. The fractions are represented as two sets of integers. The product of numbers from the first set gives the fraction numerator, the product of numbers from the second set gives the fraction denominator. However, it turned out that the programs that work with fractions in this representations aren't complete, they lack supporting the operation of reducing fractions. Implement this operation and the Empire won't forget you.
Input Specification:
The first input line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) that show how many numbers the first set (the numerator) and the second set (the denominator) contain, correspondingly.
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=107) — the numbers that are multiplied to produce the numerator.
The third line contains *m* space-separated integers: *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=107) — the numbers that are multiplied to produce the denominator.
Output Specification:
Print the answer to the problem in the form, similar to the form of the input data. The number of values in the sets you print *n**out*,<=*m**out* must satisfy the inequality 1<=≤<=*n**out*,<=*m**out*<=≤<=105, and the actual values in the sets *a**out*,<=*i* and *b**out*,<=*i* must satisfy the inequality 1<=≤<=*a**out*,<=*i*,<=*b**out*,<=*i*<=≤<=107.
Separate the values in the lines by spaces. The printed fraction must be reduced, that is, there mustn't be such integer *x* (*x*<=><=1), that the numerator and the denominator of the printed fraction are divisible by *x*. If there are several matching answers, print any of them.
Demo Input:
['3 2\n100 5 2\n50 10\n', '4 3\n2 5 10 20\n100 1 3\n']
Demo Output:
['2 3\n2 1\n1 1 1\n', '1 1\n20\n3\n']
Note:
In the first test sample the numerator equals 1000, the denominator equals 500. If we reduce fraction 1000/500 by the greatest common divisor of the numerator and the denominator (by 500), we obtain fraction 2/1.
In the second test sample the numerator equals 2000, the denominator equals 300. If we reduce fraction 2000/300 by the greatest common divisor of the numerator and the denominator (by 100), we obtain fraction 20/3. | ```python
import sys
def read(T):
return [T(i) for i in sys.stdin.readline().split()]
lsp,pr=[],[]
def sieve(n=int(1e7)):
global lsp
lsp=[i for i in range(n+1)]
for i in range(2,n+1):
if lsp[i]!=i:
continue
for j in range(i*i,n+1,i):
lsp[j]=i
for i in range(2,n+1):
if lsp[i]==i:
pr.append(i)
def solve():
[[n,m],a,b]=read(int),read(int),read(int)
nax=int(1e7+5)
d=[0 for i in range(nax)]
for i in range(n):
cur=a[i]
while lsp[cur]!=1:
p,c=lsp[cur],0
while cur%p==0:
cur//=p
c+=1
d[p]+=c
for i in range(m):
cur=b[i]
while lsp[cur]!=1:
p,c=lsp[cur],0
while cur%p==0:
cur//=p
c+=1
d[p]-=c
num,den=[],[]
print(n,m)
for i in range(n):
cur=a[i]
res=1
while lsp[cur]!=1:
p,c=lsp[cur],0
while cur%p==0:
cur//=p
c+=1
if d[p]<=0:
continue
c=min(c,d[p])
d[p]-=c
res*=p**c
print(res,end=" ")
print()
for i in range(m):
cur=b[i]
res=1
while lsp[cur]!=1:
p,c=lsp[cur],0
while cur%p==0:
cur//=p
c+=1
if d[p]>=0:
continue
c=min(c,abs(d[p]))
d[p]+=c
res*=p**c
print(res,end=" ")
def main():
sieve()
solve()
main()
``` | 3 | |
522 | B | Photo to Remember | PROGRAMMING | 1,100 | [
"*special",
"data structures",
"dp",
"implementation"
] | null | null | One day *n* friends met at a party, they hadn't seen each other for a long time and so they decided to make a group photo together.
Simply speaking, the process of taking photos can be described as follows. On the photo, each photographed friend occupies a rectangle of pixels: the *i*-th of them occupies the rectangle of width *w**i* pixels and height *h**i* pixels. On the group photo everybody stands in a line, thus the minimum pixel size of the photo including all the photographed friends, is *W*<=×<=*H*, where *W* is the total sum of all widths and *H* is the maximum height of all the photographed friends.
As is usually the case, the friends made *n* photos — the *j*-th (1<=≤<=*j*<=≤<=*n*) photo had everybody except for the *j*-th friend as he was the photographer.
Print the minimum size of each made photo in pixels. | The first line contains integer *n* (2<=≤<=*n*<=≤<=200<=000) — the number of friends.
Then *n* lines follow: the *i*-th line contains information about the *i*-th friend. The line contains a pair of integers *w**i*,<=*h**i* (1<=≤<=*w**i*<=≤<=10,<=1<=≤<=*h**i*<=≤<=1000) — the width and height in pixels of the corresponding rectangle. | Print *n* space-separated numbers *b*1,<=*b*2,<=...,<=*b**n*, where *b**i* — the total number of pixels on the minimum photo containing all friends expect for the *i*-th one. | [
"3\n1 10\n5 5\n10 1\n",
"3\n2 1\n1 2\n2 1\n"
] | [
"75 110 60 ",
"6 4 6 "
] | none | 1,000 | [
{
"input": "3\n1 10\n5 5\n10 1",
"output": "75 110 60 "
},
{
"input": "3\n2 1\n1 2\n2 1",
"output": "6 4 6 "
},
{
"input": "2\n1 5\n2 3",
"output": "6 5 "
},
{
"input": "2\n2 3\n1 1",
"output": "1 6 "
},
{
"input": "3\n1 10\n2 10\n3 10",
"output": "50 40 30 "
},
{
"input": "3\n2 10\n1 9\n3 7",
"output": "36 50 30 "
},
{
"input": "3\n1 1\n3 2\n2 3",
"output": "15 9 8 "
},
{
"input": "3\n3 123\n1 456\n2 789",
"output": "2367 3945 1824 "
},
{
"input": "3\n2 987\n3 654\n1 321",
"output": "2616 2961 4935 "
},
{
"input": "3\n3 143\n2 543\n1 893",
"output": "2679 3572 2715 "
},
{
"input": "2\n1 1\n1 2",
"output": "2 1 "
},
{
"input": "3\n2 22\n1 11\n2 22",
"output": "66 88 66 "
},
{
"input": "3\n1 11\n1 12\n1 13",
"output": "26 26 24 "
},
{
"input": "3\n1 11\n1 12\n2 10",
"output": "36 33 24 "
},
{
"input": "10\n6 20\n1 175\n1 758\n1 169\n2 490\n2 600\n4 463\n7 377\n9 40\n4 961",
"output": "29791 34596 34596 34596 33635 33635 31713 28830 26908 25014 "
},
{
"input": "10\n8 158\n1 709\n6 766\n4 335\n5 356\n2 972\n1 108\n4 235\n3 631\n1 414",
"output": "26244 33048 28188 30132 29160 25278 33048 30132 31104 33048 "
},
{
"input": "10\n7 549\n9 115\n8 141\n3 650\n5 730\n3 841\n7 18\n9 170\n2 217\n1 155",
"output": "39527 37845 38686 42891 41209 37230 39527 37845 43732 44573 "
},
{
"input": "10\n6 386\n9 816\n9 268\n9 481\n8 284\n10 715\n9 351\n7 580\n4 327\n7 392",
"output": "58752 49335 56304 56304 57120 55488 56304 57936 60384 57936 "
},
{
"input": "10\n9 292\n4 6\n6 638\n8 461\n10 970\n10 488\n9 769\n10 644\n8 280\n5 334",
"output": "67900 72750 70810 68870 53061 66930 67900 66930 68870 71780 "
},
{
"input": "10\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000",
"output": "90000 90000 90000 90000 90000 90000 90000 90000 90000 90000 "
}
] | 1,639,408,904 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 842 | 12,185,600 | n = int(input())
w = []
h = []
for i in range(n):
wi,hi = list(map(int,input().split()))
w.append(wi)
h.append(hi)
W = 0
H = -1
H2 = -1
for i in range(n):
W += w[i]
if h[i] > H:
H = h[i]
if h.count(H) > 1:
H2 = H
else:
for i in range(n):
if h[i] > H2 and not h[i] == H:
H2 = h[i]
sum = []
for i in range(n):
x = W - w[i]
y = 0
if h[i] == H:
y = H2
else:
y = H
sum.append(x*y)
print(*sum)
| Title: Photo to Remember
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day *n* friends met at a party, they hadn't seen each other for a long time and so they decided to make a group photo together.
Simply speaking, the process of taking photos can be described as follows. On the photo, each photographed friend occupies a rectangle of pixels: the *i*-th of them occupies the rectangle of width *w**i* pixels and height *h**i* pixels. On the group photo everybody stands in a line, thus the minimum pixel size of the photo including all the photographed friends, is *W*<=×<=*H*, where *W* is the total sum of all widths and *H* is the maximum height of all the photographed friends.
As is usually the case, the friends made *n* photos — the *j*-th (1<=≤<=*j*<=≤<=*n*) photo had everybody except for the *j*-th friend as he was the photographer.
Print the minimum size of each made photo in pixels.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=200<=000) — the number of friends.
Then *n* lines follow: the *i*-th line contains information about the *i*-th friend. The line contains a pair of integers *w**i*,<=*h**i* (1<=≤<=*w**i*<=≤<=10,<=1<=≤<=*h**i*<=≤<=1000) — the width and height in pixels of the corresponding rectangle.
Output Specification:
Print *n* space-separated numbers *b*1,<=*b*2,<=...,<=*b**n*, where *b**i* — the total number of pixels on the minimum photo containing all friends expect for the *i*-th one.
Demo Input:
['3\n1 10\n5 5\n10 1\n', '3\n2 1\n1 2\n2 1\n']
Demo Output:
['75 110 60 ', '6 4 6 ']
Note:
none | ```python
n = int(input())
w = []
h = []
for i in range(n):
wi,hi = list(map(int,input().split()))
w.append(wi)
h.append(hi)
W = 0
H = -1
H2 = -1
for i in range(n):
W += w[i]
if h[i] > H:
H = h[i]
if h.count(H) > 1:
H2 = H
else:
for i in range(n):
if h[i] > H2 and not h[i] == H:
H2 = h[i]
sum = []
for i in range(n):
x = W - w[i]
y = 0
if h[i] == H:
y = H2
else:
y = H
sum.append(x*y)
print(*sum)
``` | 3 | |
669 | A | Little Artem and Presents | PROGRAMMING | 800 | [
"math"
] | null | null | Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha? | The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday. | Print the maximum possible number of times Artem can give presents to Masha. | [
"1\n",
"2\n",
"3\n",
"4\n"
] | [
"1\n",
"1\n",
"2\n",
"3\n"
] | In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again. | 500 | [
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "3"
},
{
"input": "100",
"output": "67"
},
{
"input": "101",
"output": "67"
},
{
"input": "102",
"output": "68"
},
{
"input": "1000000000",
"output": "666666667"
},
{
"input": "5",
"output": "3"
},
{
"input": "6",
"output": "4"
},
{
"input": "999999999",
"output": "666666666"
},
{
"input": "999999998",
"output": "666666665"
},
{
"input": "999999997",
"output": "666666665"
},
{
"input": "999999996",
"output": "666666664"
},
{
"input": "999999995",
"output": "666666663"
},
{
"input": "999999994",
"output": "666666663"
},
{
"input": "999999993",
"output": "666666662"
},
{
"input": "999999992",
"output": "666666661"
},
{
"input": "999999991",
"output": "666666661"
},
{
"input": "1000",
"output": "667"
},
{
"input": "10000",
"output": "6667"
},
{
"input": "100000",
"output": "66667"
},
{
"input": "1000000",
"output": "666667"
},
{
"input": "10000000",
"output": "6666667"
},
{
"input": "100000000",
"output": "66666667"
},
{
"input": "7",
"output": "5"
}
] | 1,579,433,387 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 2,000 | 0 | n=int(input())
t=0
for j in range(1,n+1):
if j%2!=0:
n-=1
t+=1
else:
if n<2:
print(t)
break
n-=2
t+=1
if n==0:
print(t)
break
| Title: Little Artem and Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.
Output Specification:
Print the maximum possible number of times Artem can give presents to Masha.
Demo Input:
['1\n', '2\n', '3\n', '4\n']
Demo Output:
['1\n', '1\n', '2\n', '3\n']
Note:
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again. | ```python
n=int(input())
t=0
for j in range(1,n+1):
if j%2!=0:
n-=1
t+=1
else:
if n<2:
print(t)
break
n-=2
t+=1
if n==0:
print(t)
break
``` | 0 | |
149 | A | Business trip | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters. | The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100). | Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1. | [
"5\n1 1 1 1 2 2 3 2 2 1 1 1\n",
"0\n0 0 0 0 0 0 0 1 1 2 3 0\n",
"11\n1 1 4 1 1 5 1 1 4 1 1 1\n"
] | [
"2\n",
"0\n",
"3\n"
] | Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all. | 500 | [
{
"input": "5\n1 1 1 1 2 2 3 2 2 1 1 1",
"output": "2"
},
{
"input": "0\n0 0 0 0 0 0 0 1 1 2 3 0",
"output": "0"
},
{
"input": "11\n1 1 4 1 1 5 1 1 4 1 1 1",
"output": "3"
},
{
"input": "15\n20 1 1 1 1 2 2 1 2 2 1 1",
"output": "1"
},
{
"input": "7\n8 9 100 12 14 17 21 10 11 100 23 10",
"output": "1"
},
{
"input": "52\n1 12 3 11 4 5 10 6 9 7 8 2",
"output": "6"
},
{
"input": "50\n2 2 3 4 5 4 4 5 7 3 2 7",
"output": "-1"
},
{
"input": "0\n55 81 28 48 99 20 67 95 6 19 10 93",
"output": "0"
},
{
"input": "93\n85 40 93 66 92 43 61 3 64 51 90 21",
"output": "1"
},
{
"input": "99\n36 34 22 0 0 0 52 12 0 0 33 47",
"output": "2"
},
{
"input": "99\n28 32 31 0 10 35 11 18 0 0 32 28",
"output": "3"
},
{
"input": "99\n19 17 0 1 18 11 29 9 29 22 0 8",
"output": "4"
},
{
"input": "76\n2 16 11 10 12 0 20 4 4 14 11 14",
"output": "5"
},
{
"input": "41\n2 1 7 7 4 2 4 4 9 3 10 0",
"output": "6"
},
{
"input": "47\n8 2 2 4 3 1 9 4 2 7 7 8",
"output": "7"
},
{
"input": "58\n6 11 7 0 5 6 3 9 4 9 5 1",
"output": "8"
},
{
"input": "32\n5 2 4 1 5 0 5 1 4 3 0 3",
"output": "9"
},
{
"input": "31\n6 1 0 4 4 5 1 0 5 3 2 0",
"output": "9"
},
{
"input": "35\n2 3 0 0 6 3 3 4 3 5 0 6",
"output": "9"
},
{
"input": "41\n3 1 3 4 3 6 6 1 4 4 0 6",
"output": "11"
},
{
"input": "97\n0 5 3 12 10 16 22 8 21 17 21 10",
"output": "5"
},
{
"input": "100\n21 21 0 0 4 13 0 26 0 0 0 15",
"output": "6"
},
{
"input": "100\n0 0 16 5 22 0 5 0 25 0 14 13",
"output": "7"
},
{
"input": "97\n17 0 10 0 0 0 18 0 14 23 15 0",
"output": "6"
},
{
"input": "100\n0 9 0 18 7 0 0 14 33 3 0 16",
"output": "7"
},
{
"input": "95\n5 2 13 0 15 18 17 0 6 11 0 8",
"output": "9"
},
{
"input": "94\n11 13 0 9 15 8 8 16 3 7 1 3",
"output": "11"
},
{
"input": "96\n8 4 12 15 8 0 4 10 6 6 12 11",
"output": "11"
},
{
"input": "100\n5 5 3 8 6 5 0 3 3 8 1 3",
"output": "-1"
},
{
"input": "100\n1 0 0 1 1 0 1 1 1 1 2 1",
"output": "-1"
},
{
"input": "100\n6 3 2 0 4 1 2 2 2 2 1 1",
"output": "-1"
},
{
"input": "0\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "-1"
},
{
"input": "0\n100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100",
"output": "1"
},
{
"input": "12\n1 1 1 1 1 1 1 1 1 1 1 1",
"output": "12"
},
{
"input": "13\n1 1 1 1 1 1 1 1 1 1 1 2",
"output": "12"
},
{
"input": "15\n10 1 1 1 1 1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "-1"
}
] | 1,648,988,135 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 6 | 92 | 0 | n = int(input())
a = [int(x) for x in input().split()]
a = sorted(a)
a = a[::-1]
s = 0
i = 0
while(s < n and i < 12):
s += a[i]
i+=1
print(i)
| Title: Business trip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters.
Input Specification:
The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100).
Output Specification:
Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1.
Demo Input:
['5\n1 1 1 1 2 2 3 2 2 1 1 1\n', '0\n0 0 0 0 0 0 0 1 1 2 3 0\n', '11\n1 1 4 1 1 5 1 1 4 1 1 1\n']
Demo Output:
['2\n', '0\n', '3\n']
Note:
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all. | ```python
n = int(input())
a = [int(x) for x in input().split()]
a = sorted(a)
a = a[::-1]
s = 0
i = 0
while(s < n and i < 12):
s += a[i]
i+=1
print(i)
``` | 0 | |
768 | A | Oath of the Night's Watch | PROGRAMMING | 900 | [
"constructive algorithms",
"sortings"
] | null | null | "Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support? | First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow.
Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards. | Output a single integer representing the number of stewards which Jon will feed. | [
"2\n1 5\n",
"3\n1 2 5\n"
] | [
"0",
"1"
] | In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | 500 | [
{
"input": "2\n1 5",
"output": "0"
},
{
"input": "3\n1 2 5",
"output": "1"
},
{
"input": "4\n1 2 3 4",
"output": "2"
},
{
"input": "8\n7 8 9 4 5 6 1 2",
"output": "6"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n100",
"output": "0"
},
{
"input": "205\n5 5 3 3 6 2 9 3 8 9 6 6 10 8 1 5 3 3 1 2 9 9 9 3 9 10 3 9 8 3 5 6 6 4 6 9 2 9 10 9 5 6 6 7 4 2 6 3 4 1 10 1 7 2 7 7 3 2 6 5 5 2 9 3 8 8 7 6 6 4 2 2 6 2 3 5 7 2 2 10 1 4 6 9 2 3 7 2 2 7 4 4 9 10 7 5 8 6 5 3 6 10 2 7 5 6 6 8 3 3 9 4 3 5 7 9 3 2 1 1 3 2 1 9 3 1 4 4 10 2 5 5 8 1 4 8 5 3 1 10 8 6 5 8 3 5 4 5 4 4 6 7 2 8 10 8 7 6 6 9 6 7 1 10 3 2 5 10 4 4 5 4 3 4 8 5 3 8 10 3 10 9 7 2 1 8 6 4 6 5 8 10 2 6 7 4 9 4 5 1 8 7 10 3 1",
"output": "174"
},
{
"input": "4\n1000000000 99999999 1000000000 1000000000",
"output": "0"
},
{
"input": "3\n2 2 2",
"output": "0"
},
{
"input": "5\n1 1 1 1 1",
"output": "0"
},
{
"input": "3\n1 1 1",
"output": "0"
},
{
"input": "6\n1 1 3 3 2 2",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "4\n1 1 2 5",
"output": "1"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n1 1 1 1 5",
"output": "0"
},
{
"input": "5\n1 1 2 3 3",
"output": "1"
},
{
"input": "3\n1 1 3",
"output": "0"
},
{
"input": "3\n2 2 3",
"output": "0"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "5\n1 5 3 5 1",
"output": "1"
},
{
"input": "7\n1 2 2 2 2 2 3",
"output": "5"
},
{
"input": "4\n2 2 2 2",
"output": "0"
},
{
"input": "9\n2 2 2 3 4 5 6 6 6",
"output": "3"
},
{
"input": "10\n1 1 1 2 3 3 3 3 3 3",
"output": "1"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "0"
},
{
"input": "3\n0 0 1",
"output": "0"
},
{
"input": "9\n1 1 1 2 2 2 3 3 3",
"output": "3"
},
{
"input": "3\n1 2 2",
"output": "0"
},
{
"input": "6\n2 2 2 2 2 2",
"output": "0"
},
{
"input": "5\n2 2 2 2 2",
"output": "0"
},
{
"input": "5\n5 5 5 5 5",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "6\n1 2 5 5 5 5",
"output": "1"
},
{
"input": "5\n1 2 3 3 3",
"output": "1"
},
{
"input": "3\n1 1 2",
"output": "0"
},
{
"input": "6\n1 1 1 1 1 2",
"output": "0"
},
{
"input": "5\n1 1 2 4 4",
"output": "1"
},
{
"input": "3\n999999 5999999 9999999",
"output": "1"
},
{
"input": "4\n1 1 5 5",
"output": "0"
},
{
"input": "9\n1 1 1 2 2 2 4 4 4",
"output": "3"
},
{
"input": "5\n1 3 4 5 1",
"output": "2"
},
{
"input": "5\n3 3 3 3 3",
"output": "0"
},
{
"input": "5\n1 1 2 2 2",
"output": "0"
},
{
"input": "5\n2 1 1 1 3",
"output": "1"
},
{
"input": "5\n0 0 0 1 2",
"output": "1"
},
{
"input": "4\n2 2 2 3",
"output": "0"
},
{
"input": "7\n1 1 1 1 5 5 5",
"output": "0"
},
{
"input": "5\n1 2 3 4 4",
"output": "2"
},
{
"input": "2\n5 4",
"output": "0"
},
{
"input": "4\n5 5 5 5",
"output": "0"
},
{
"input": "5\n1 1 1 5 5",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "1\n3",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "0"
},
{
"input": "4\n1 2 2 2",
"output": "0"
},
{
"input": "8\n1000000000 1000000000 1000000000 999999999 999999999 999999999 999999998 999999998",
"output": "3"
},
{
"input": "5\n1 1 3 4 4",
"output": "1"
},
{
"input": "6\n1 1 2 2 3 3",
"output": "2"
},
{
"input": "4\n1 1 1 1",
"output": "0"
},
{
"input": "9\n1 2 3 4 1 5 6 7 8",
"output": "6"
},
{
"input": "8\n5 4 4 6 6 4 4 3",
"output": "5"
},
{
"input": "8\n4 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "7\n4 3 3 3 3 3 3",
"output": "0"
},
{
"input": "6\n4 3 3 3 3 3",
"output": "0"
},
{
"input": "5\n4 3 3 3 3",
"output": "0"
}
] | 1,672,686,434 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 2,000 | 7,065,600 | n=input()
S=input()
S_list=S.split(" ")
i=0
for item in S_list:
S_list[i]=int(item)
i=i+1
i=0
S_NoDup=[]
for x in S_list:
if x not in S_NoDup:
S_NoDup.append(x)
x=len(S_NoDup)
if x==2 or x==1:
print(0)
else:
print(x-2) | Title: Oath of the Night's Watch
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input Specification:
First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow.
Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards.
Output Specification:
Output a single integer representing the number of stewards which Jon will feed.
Demo Input:
['2\n1 5\n', '3\n1 2 5\n']
Demo Output:
['0', '1']
Note:
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | ```python
n=input()
S=input()
S_list=S.split(" ")
i=0
for item in S_list:
S_list[i]=int(item)
i=i+1
i=0
S_NoDup=[]
for x in S_list:
if x not in S_NoDup:
S_NoDup.append(x)
x=len(S_NoDup)
if x==2 or x==1:
print(0)
else:
print(x-2)
``` | 0 | |
78 | B | Easter Eggs | PROGRAMMING | 1,200 | [
"constructive algorithms",
"implementation"
] | B. Easter Eggs | 2 | 256 | The Easter Rabbit laid *n* eggs in a circle and is about to paint them.
Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied:
- Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors.
Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible. | The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100). | Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet.
If there are several answers, print any of them. | [
"8\n",
"13\n"
] | [
"ROYGRBIV\n",
"ROYGBIVGBIVYG\n"
] | The way the eggs will be painted in the first sample is shown on the picture: | 1,000 | [
{
"input": "8",
"output": "ROYGBIVG"
},
{
"input": "13",
"output": "ROYGBIVOYGBIV"
},
{
"input": "7",
"output": "ROYGBIV"
},
{
"input": "10",
"output": "ROYGBIVYGB"
},
{
"input": "14",
"output": "ROYGBIVROYGBIV"
},
{
"input": "50",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "9",
"output": "ROYGBIVGB"
},
{
"input": "11",
"output": "ROYGBIVYGBI"
},
{
"input": "12",
"output": "ROYGBIVOYGBI"
},
{
"input": "15",
"output": "ROYGBIVROYGBIVG"
},
{
"input": "16",
"output": "ROYGBIVROYGBIVGB"
},
{
"input": "17",
"output": "ROYGBIVROYGBIVYGB"
},
{
"input": "18",
"output": "ROYGBIVROYGBIVYGBI"
},
{
"input": "19",
"output": "ROYGBIVROYGBIVOYGBI"
},
{
"input": "20",
"output": "ROYGBIVROYGBIVOYGBIV"
},
{
"input": "21",
"output": "ROYGBIVROYGBIVROYGBIV"
},
{
"input": "22",
"output": "ROYGBIVROYGBIVROYGBIVG"
},
{
"input": "23",
"output": "ROYGBIVROYGBIVROYGBIVGB"
},
{
"input": "24",
"output": "ROYGBIVROYGBIVROYGBIVYGB"
},
{
"input": "25",
"output": "ROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "28",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIV"
},
{
"input": "29",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "34",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVOYGBIV"
},
{
"input": "43",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "61",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBI"
},
{
"input": "79",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVGB"
},
{
"input": "81",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "92",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "95",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "96",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBI"
},
{
"input": "97",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBIV"
},
{
"input": "98",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIV"
},
{
"input": "99",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "100",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVGB"
}
] | 1,599,522,541 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 218 | 0 | n = int(input())
x = n-3
ans = []
adder = ['R', "O", 'Y', 'G']
for i in range(x//4):
ans.append('ROYG')
if x%4 != 0:
for i in range (x%4):
ans.append(adder[i])
ans.append('BIV')
fans = ''.join(ans)
print(fans) | Title: Easter Eggs
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Easter Rabbit laid *n* eggs in a circle and is about to paint them.
Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied:
- Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors.
Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible.
Input Specification:
The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100).
Output Specification:
Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet.
If there are several answers, print any of them.
Demo Input:
['8\n', '13\n']
Demo Output:
['ROYGRBIV\n', 'ROYGBIVGBIVYG\n']
Note:
The way the eggs will be painted in the first sample is shown on the picture: | ```python
n = int(input())
x = n-3
ans = []
adder = ['R', "O", 'Y', 'G']
for i in range(x//4):
ans.append('ROYG')
if x%4 != 0:
for i in range (x%4):
ans.append(adder[i])
ans.append('BIV')
fans = ''.join(ans)
print(fans)
``` | 3.9455 |
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,637,511,124 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 154 | 0 | n = int(input())
c_1, c_2 = 0, 0
rez_mass = [[], []]
for x in [*map(int, input().split())]:
if c_1 >= 2 and c_2 != 0:
print(rez_mass[0][0])
break
if c_2 >= 2 and c_1 != 0:
print(rez_mass[1][0])
break
if x % 2:
c_1 += 1
rez_mass[1].append(c_1+c_2)
else:
c_2 += 1
rez_mass[0].append(c_1+c_2)
| Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
n = int(input())
c_1, c_2 = 0, 0
rez_mass = [[], []]
for x in [*map(int, input().split())]:
if c_1 >= 2 and c_2 != 0:
print(rez_mass[0][0])
break
if c_2 >= 2 and c_1 != 0:
print(rez_mass[1][0])
break
if x % 2:
c_1 += 1
rez_mass[1].append(c_1+c_2)
else:
c_2 += 1
rez_mass[0].append(c_1+c_2)
``` | 0 |
686 | A | Free Ice Cream | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids. | The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place. | Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress. | [
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] | [
"22 1\n",
"3 2\n"
] | Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | 500 | [
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
"input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46",
"output": "46 0"
},
{
"input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100",
"output": "700 0"
},
{
"input": "1 0\n- 526403222",
"output": "0 1"
},
{
"input": "1 897986543\n- 371188251",
"output": "526798292 0"
},
{
"input": "1 0\n+ 1",
"output": "1 0"
},
{
"input": "1 0\n- 1",
"output": "0 1"
},
{
"input": "1 10\n+ 10",
"output": "20 0"
},
{
"input": "1 3\n- 5",
"output": "3 1"
},
{
"input": "1 0\n- 5",
"output": "0 1"
},
{
"input": "1 0\n+ 5",
"output": "5 0"
}
] | 1,631,927,756 | 2,147,483,647 | PyPy 3 | OK | TESTS | 34 | 124 | 21,606,400 | a,b=map(int,input().split())
count,sad=b,0
for _ in range(a):
string,n=input().split(maxsplit=1)
if string=="+":
count+=int(n)
elif count<int(n):
sad+=1
else:
count-=int(n)
print(count,sad) | Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | ```python
a,b=map(int,input().split())
count,sad=b,0
for _ in range(a):
string,n=input().split(maxsplit=1)
if string=="+":
count+=int(n)
elif count<int(n):
sad+=1
else:
count-=int(n)
print(count,sad)
``` | 3 | |
266 | A | Stones on the Table | PROGRAMMING | 800 | [
"implementation"
] | null | null | There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them. | The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue. | Print a single integer — the answer to the problem. | [
"3\nRRG\n",
"5\nRRRRR\n",
"4\nBRBG\n"
] | [
"1\n",
"4\n",
"0\n"
] | none | 500 | [
{
"input": "3\nRRG",
"output": "1"
},
{
"input": "5\nRRRRR",
"output": "4"
},
{
"input": "4\nBRBG",
"output": "0"
},
{
"input": "1\nB",
"output": "0"
},
{
"input": "2\nBG",
"output": "0"
},
{
"input": "3\nBGB",
"output": "0"
},
{
"input": "4\nRBBR",
"output": "1"
},
{
"input": "5\nRGGBG",
"output": "1"
},
{
"input": "10\nGGBRBRGGRB",
"output": "2"
},
{
"input": "50\nGRBGGRBRGRBGGBBBBBGGGBBBBRBRGBRRBRGBBBRBBRRGBGGGRB",
"output": "18"
},
{
"input": "15\nBRRBRGGBBRRRRGR",
"output": "6"
},
{
"input": "20\nRRGBBRBRGRGBBGGRGRRR",
"output": "6"
},
{
"input": "25\nBBGBGRBGGBRRBGRRBGGBBRBRB",
"output": "6"
},
{
"input": "30\nGRGGGBGGRGBGGRGRBGBGBRRRRRRGRB",
"output": "9"
},
{
"input": "35\nGBBGBRGBBGGRBBGBRRGGRRRRRRRBRBBRRGB",
"output": "14"
},
{
"input": "40\nGBBRRGBGGGRGGGRRRRBRBGGBBGGGBGBBBBBRGGGG",
"output": "20"
},
{
"input": "45\nGGGBBRBBRRGRBBGGBGRBRGGBRBRGBRRGBGRRBGRGRBRRG",
"output": "11"
},
{
"input": "50\nRBGGBGGRBGRBBBGBBGRBBBGGGRBBBGBBBGRGGBGGBRBGBGRRGG",
"output": "17"
},
{
"input": "50\nGGGBBRGGGGGRRGGRBGGRGBBRBRRBGRGBBBGBRBGRGBBGRGGBRB",
"output": "16"
},
{
"input": "50\nGBGRGRRBRRRRRGGBBGBRRRBBBRBBBRRGRBBRGBRBGGRGRBBGGG",
"output": "19"
},
{
"input": "10\nGRRBRBRBGR",
"output": "1"
},
{
"input": "10\nBRBGBGRRBR",
"output": "1"
},
{
"input": "20\nGBGBGGRRRRGRBBGRGRGR",
"output": "5"
},
{
"input": "20\nRRGGRBBGBBRBGRRBRRBG",
"output": "6"
},
{
"input": "30\nBGBRGBBBGRGBBRGBGRBBBRGGRRGRRB",
"output": "8"
},
{
"input": "30\nBBBBGGBRBGBBGBGBGBGGGRGRRGGBBB",
"output": "11"
},
{
"input": "40\nGBRRGRBGBRRGBRGGGBRGBGBRGBBRRGRGGBBGBGBB",
"output": "9"
},
{
"input": "40\nBRGRGGRGGRBBRRRBRBBGGGRRGBGBBGRBBRGBRRGG",
"output": "13"
},
{
"input": "50\nRBGBGGRRGGRGGBGBGRRBGGBGBRRBBGBBGBBBGBBRBBRBRBRGRG",
"output": "13"
},
{
"input": "50\nRBRRGBGRRRBGRRBGRRGRBBRBBRRBRGGBRBRRBGGRBGGBRBRGRB",
"output": "12"
},
{
"input": "2\nBB",
"output": "1"
},
{
"input": "50\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "49"
},
{
"input": "50\nRRRRRRRRGRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "47"
},
{
"input": "50\nRRRRRRRRRRRRGGRRRRRRRRRBRRRRRRRRRRRRRRBBRRRRRRRRRR",
"output": "43"
}
] | 1,697,010,899 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 62 | 0 | n = int(input())
stones = [i for i in input()]
i = 1
count = 0
last_stone = stones[0]
while i < n:
if stones[i] == last_stone:
count += 1
i += 1
else:
i += 1
print(count) | Title: Stones on the Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['3\nRRG\n', '5\nRRRRR\n', '4\nBRBG\n']
Demo Output:
['1\n', '4\n', '0\n']
Note:
none | ```python
n = int(input())
stones = [i for i in input()]
i = 1
count = 0
last_stone = stones[0]
while i < n:
if stones[i] == last_stone:
count += 1
i += 1
else:
i += 1
print(count)
``` | 0 | |
522 | A | Reposts | PROGRAMMING | 1,200 | [
"*special",
"dfs and similar",
"dp",
"graphs",
"trees"
] | null | null | One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user. | Print a single integer — the maximum length of a repost chain. | [
"5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n",
"6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n",
"1\nSoMeStRaNgEgUe reposted PoLyCaRp\n"
] | [
"6\n",
"2\n",
"2\n"
] | none | 500 | [
{
"input": "5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya",
"output": "6"
},
{
"input": "6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp",
"output": "2"
},
{
"input": "1\nSoMeStRaNgEgUe reposted PoLyCaRp",
"output": "2"
},
{
"input": "1\niuNtwVf reposted POlYcarP",
"output": "2"
},
{
"input": "10\ncs reposted poLYCaRp\nAFIkDrY7Of4V7Mq reposted CS\nsoBiwyN7KOvoFUfbhux reposted aFikDry7Of4v7MQ\nvb6LbwA reposted sObIWYN7KOvoFufBHUx\nDtWKIcVwIHgj4Rcv reposted vb6lbwa\nkt reposted DTwKicvwihgJ4rCV\n75K reposted kT\njKzyxx1 reposted 75K\nuoS reposted jkZyXX1\npZJskHTCIqE3YyZ5ME reposted uoS",
"output": "11"
},
{
"input": "10\nvxrUpCXvx8Isq reposted pOLYcaRP\nICb1 reposted vXRUpCxvX8ISq\nJFMt4b8jZE7iF2m8by7y2 reposted Icb1\nqkG6ZkMIf9QRrBFQU reposted ICb1\nnawsNfcR2palIMnmKZ reposted pOlYcaRP\nKksyH reposted jFMT4b8JzE7If2M8by7y2\nwJtWwQS5FvzN0h8CxrYyL reposted NawsNfcR2paLIMnmKz\nDpBcBPYAcTXEdhldI6tPl reposted NaWSnFCr2pALiMnmkZ\nlEnwTVnlwdQg2vaIRQry reposted kKSYh\nQUVFgwllaWO reposted Wjtwwqs5FVzN0H8cxRyyl",
"output": "6"
},
{
"input": "10\nkkuLGEiHv reposted POLYcArp\n3oX1AoUqyw1eR3nCADY9hLwd reposted kkuLGeIHV\nwf97dqq5bx1dPIchCoT reposted 3OX1AOuQYW1eR3ncAdY9hLwD\nWANr8h reposted Wf97dQQ5bx1dpIcHcoT\n3Fb736lkljZK2LtSbfL reposted wANR8h\n6nq9xLOn reposted 3fB736lKlJZk2LtSbFL\nWL reposted 3Fb736lKLjZk2LTSbfl\ndvxn4Xtc6SBcvKf1 reposted wF97DQq5bX1dPiChCOt\nMCcPLIMISqxDzrj reposted 6nQ9XLOn\nxsQL4Z2Iu reposted MCcpLiMiSqxdzrj",
"output": "9"
},
{
"input": "10\nsMA4 reposted pOLyCARP\nlq3 reposted pOlycARp\nEa16LSFTQxLJnE reposted polYcARp\nkvZVZhJwXcWsnC7NA1DV2WvS reposted polYCArp\nEYqqlrjRwddI reposted pOlyCArP\nsPqQCA67Y6PBBbcaV3EhooO reposted ea16LSFTqxLJne\njjPnneZdF6WLZ3v reposted Ea16LSFTqxLjNe\nWEoi6UpnfBUx79 reposted ea16LSFtqXljNe\nqi4yra reposted eYqqlRJrWDDI\ncw7E1UCSUD reposted eYqqLRJRwDdI",
"output": "3"
}
] | 1,425,818,603 | 78,203 | Python 3 | OK | TESTS | 36 | 62 | 0 | n = int(input())
i=0
s=[]
k=[]
x=[]
while i<n:
s.append(input().lower().split(' reposted '))
i=i+1
k.append(["polycarp", 1])
for l in k:
for z in s:
if z[1]==l[0]:
k.append([z[0],l[1]+1])
for l in k:
x.append(l[1])
print(max(x)) | Title: Reposts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user.
Output Specification:
Print a single integer — the maximum length of a repost chain.
Demo Input:
['5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n', '6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n', '1\nSoMeStRaNgEgUe reposted PoLyCaRp\n']
Demo Output:
['6\n', '2\n', '2\n']
Note:
none | ```python
n = int(input())
i=0
s=[]
k=[]
x=[]
while i<n:
s.append(input().lower().split(' reposted '))
i=i+1
k.append(["polycarp", 1])
for l in k:
for z in s:
if z[1]==l[0]:
k.append([z[0],l[1]+1])
for l in k:
x.append(l[1])
print(max(x))
``` | 3 | |
231 | A | Team | PROGRAMMING | 800 | [
"brute force",
"greedy"
] | null | null | One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. | Print a single integer — the number of problems the friends will implement on the contest. | [
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] | [
"2\n",
"1\n"
] | In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | 500 | [
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,698,518,180 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 122 | 0 | import random
n = int(input("n:"))
s = 0
for i in range(n):
arr = [random.randint(0, 1) for i in range(3)]
g = 0
for j in arr:
if j == 1:
g += 1
if g > 1:
s += 1
print(s) | Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | ```python
import random
n = int(input("n:"))
s = 0
for i in range(n):
arr = [random.randint(0, 1) for i in range(3)]
g = 0
for j in arr:
if j == 1:
g += 1
if g > 1:
s += 1
print(s)
``` | 0 | |
912 | A | Tricky Alchemy | PROGRAMMING | 800 | [
"implementation"
] | null | null | During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls. | The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained. | Print a single integer — the minimum number of crystals that Grisha should acquire in addition. | [
"4 3\n2 1 1\n",
"3 9\n1 1 3\n",
"12345678 87654321\n43043751 1000000000 53798715\n"
] | [
"2\n",
"1\n",
"2147483648\n"
] | In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue. | 500 | [
{
"input": "4 3\n2 1 1",
"output": "2"
},
{
"input": "3 9\n1 1 3",
"output": "1"
},
{
"input": "12345678 87654321\n43043751 1000000000 53798715",
"output": "2147483648"
},
{
"input": "12 12\n3 5 2",
"output": "0"
},
{
"input": "770 1390\n170 442 311",
"output": "12"
},
{
"input": "3555165 6693472\n1499112 556941 3075290",
"output": "3089339"
},
{
"input": "0 0\n1000000000 1000000000 1000000000",
"output": "7000000000"
},
{
"input": "1 1\n0 1 0",
"output": "0"
},
{
"input": "117708228 562858833\n118004008 360437130 154015822",
"output": "738362681"
},
{
"input": "999998118 700178721\n822106746 82987112 547955384",
"output": "1753877029"
},
{
"input": "566568710 765371101\n60614022 80126928 809950465",
"output": "1744607222"
},
{
"input": "448858599 829062060\n764716760 97644201 203890025",
"output": "1178219122"
},
{
"input": "626115781 966381948\n395190569 820194184 229233367",
"output": "1525971878"
},
{
"input": "803372962 103701834\n394260597 837711458 623172928",
"output": "3426388098"
},
{
"input": "980630143 241021722\n24734406 928857659 312079781",
"output": "1624075280"
},
{
"input": "862920032 378341609\n360240924 241342224 337423122",
"output": "974174021"
},
{
"input": "40177212 515661496\n64343660 963892207 731362684",
"output": "3694721078"
},
{
"input": "217434393 579352456\n694817470 981409480 756706026",
"output": "4825785129"
},
{
"input": "394691574 716672343\n398920207 72555681 150645586",
"output": "475704521"
},
{
"input": "276981463 853992230\n29394015 90072954 839552440",
"output": "1754738044"
},
{
"input": "843552056 919184611\n341530221 423649259 101547519",
"output": "263157645"
},
{
"input": "20809236 56504497\n972004030 441166533 495487081",
"output": "4235488636"
},
{
"input": "198066417 825228166\n602477839 532312735 520830423",
"output": "2808777834"
},
{
"input": "80356306 962548053\n601547868 549830008 914769984",
"output": "4004161345"
},
{
"input": "257613487 394835231\n642087093 567347282 308709545",
"output": "2692548667"
},
{
"input": "139903376 532155119\n641157122 289897263 629020178",
"output": "3077110809"
},
{
"input": "612127849 669475006\n271630930 676010757 22959739",
"output": "682559736"
},
{
"input": "0 0\n0 0 0",
"output": "0"
},
{
"input": "1000000000 1000000000\n499999998 4 333333332",
"output": "0"
},
{
"input": "1000000000 1000000000\n1000000000 1000000000 1000000000",
"output": "5000000000"
},
{
"input": "4 3\n1 0 1",
"output": "0"
},
{
"input": "4 12\n1 2 3",
"output": "0"
},
{
"input": "4 20\n1 2 1",
"output": "0"
},
{
"input": "100 10\n2 3 4",
"output": "5"
},
{
"input": "6 0\n1 1 1",
"output": "4"
},
{
"input": "25 5\n3 3 3",
"output": "7"
},
{
"input": "48 27\n22 39 20",
"output": "107"
},
{
"input": "4 0\n1 1 1",
"output": "4"
}
] | 1,572,789,086 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 124 | 0 | a,b=[int(i) for i in input().split()]
x,y,z=[int(i) for i in input().split()]
a_needs=x*2+y-a
b_needs=z*3+y-b
if a_needs+b_needs<=0:
print("0")
exit()
print(a_needs+b_needs)
| Title: Tricky Alchemy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input Specification:
The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained.
Output Specification:
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Demo Input:
['4 3\n2 1 1\n', '3 9\n1 1 3\n', '12345678 87654321\n43043751 1000000000 53798715\n']
Demo Output:
['2\n', '1\n', '2147483648\n']
Note:
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue. | ```python
a,b=[int(i) for i in input().split()]
x,y,z=[int(i) for i in input().split()]
a_needs=x*2+y-a
b_needs=z*3+y-b
if a_needs+b_needs<=0:
print("0")
exit()
print(a_needs+b_needs)
``` | 0 | |
554 | B | Ohana Cleans Up | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"strings"
] | null | null | Ohana Matsumae is trying to clean a room, which is divided up into an *n* by *n* grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column.
Return the maximum number of rows that she can make completely clean. | The first line of input will be a single integer *n* (1<=≤<=*n*<=≤<=100).
The next *n* lines will describe the state of the room. The *i*-th line will contain a binary string with *n* characters denoting the state of the *i*-th row of the room. The *j*-th character on this line is '1' if the *j*-th square in the *i*-th row is clean, and '0' if it is dirty. | The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean. | [
"4\n0101\n1000\n1111\n0101\n",
"3\n111\n111\n111\n"
] | [
"2\n",
"3\n"
] | In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean.
In the second sample, everything is already clean, so Ohana doesn't need to do anything. | 500 | [
{
"input": "4\n0101\n1000\n1111\n0101",
"output": "2"
},
{
"input": "3\n111\n111\n111",
"output": "3"
},
{
"input": "10\n0100000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000",
"output": "9"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n0111010011\n0111010011\n1010010001\n0111010011\n0000110000\n0111010011\n0111010011\n0000110000\n1010010001\n0111010011",
"output": "6"
},
{
"input": "20\n10101011101000011010\n11111010001100110101\n01011100010000001111\n10110100000101010011\n11010001110111101101\n00100110011011101010\n01000110101011001100\n01101100111101101101\n10111010010100111100\n00010010110001101110\n10111110010000101010\n10010111110100100100\n11010111001111110100\n11110111101100000001\n00011010100111011000\n11110001011000011010\n10001101010000011011\n01010101110010000111\n11100110111101101111\n11011111110010001111",
"output": "1"
},
{
"input": "10\n1001000000\n0111101111\n1111001011\n1000010100\n0111101111\n0101100110\n1001000000\n1000010100\n0111101111\n1001000000",
"output": "3"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000",
"output": "10"
},
{
"input": "10\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111\n1111111111",
"output": "10"
},
{
"input": "10\n1000000000\n0100000000\n0010000000\n0001000000\n0000100000\n0000010000\n0000001000\n0000000100\n0000000010\n0000000001",
"output": "1"
},
{
"input": "2\n10\n01",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "4\n0000\n0000\n1111\n1111",
"output": "2"
},
{
"input": "11\n10000000001\n10000000001\n10000000001\n10000000001\n10001000001\n10001000000\n10001000001\n10001000001\n10001000000\n10001000000\n10001000100",
"output": "4"
}
] | 1,435,167,124 | 3,724 | Python 3 | WRONG_ANSWER | PRETESTS | 6 | 46 | 0 | n = int(input())
l = [input() for i in range(n)]
s = set(l)
print(len(l)-len(s)+1) | Title: Ohana Cleans Up
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ohana Matsumae is trying to clean a room, which is divided up into an *n* by *n* grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column.
Return the maximum number of rows that she can make completely clean.
Input Specification:
The first line of input will be a single integer *n* (1<=≤<=*n*<=≤<=100).
The next *n* lines will describe the state of the room. The *i*-th line will contain a binary string with *n* characters denoting the state of the *i*-th row of the room. The *j*-th character on this line is '1' if the *j*-th square in the *i*-th row is clean, and '0' if it is dirty.
Output Specification:
The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.
Demo Input:
['4\n0101\n1000\n1111\n0101\n', '3\n111\n111\n111\n']
Demo Output:
['2\n', '3\n']
Note:
In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean.
In the second sample, everything is already clean, so Ohana doesn't need to do anything. | ```python
n = int(input())
l = [input() for i in range(n)]
s = set(l)
print(len(l)-len(s)+1)
``` | 0 | |
991 | E | Bus Number | PROGRAMMING | 1,800 | [
"brute force",
"combinatorics",
"math"
] | null | null | This night wasn't easy on Vasya. His favorite team lost, and he didn't find himself victorious either — although he played perfectly, his teammates let him down every time. He had to win at least one more time, but the losestreak only grew longer and longer... It's no wonder he didn't get any sleep this night at all.
In the morning, Vasya was waiting the bus to the university on the bus stop. Vasya's thoughts were hazy and so he couldn't remember the right bus' number quite right and got onto the bus with the number $n$.
In the bus, Vasya thought that he could get the order of the digits in the number of the bus wrong. Futhermore, he could "see" some digits several times, but the digits he saw were definitely in the real number of the bus. For example, if Vasya saw the number 2028, it could mean that the real bus number could be 2028, 8022, 2820 or just 820. However, numbers 80, 22208, 52 definitely couldn't be the number of the bus. Also, real bus number couldn't start with the digit 0, this meaning that, for example, number 082 couldn't be the real bus number too.
Given $n$, determine the total number of possible bus number variants. | The first line contains one integer $n$ ($1 \leq n \leq 10^{18}$) — the number of the bus that was seen by Vasya. It is guaranteed that this number does not start with $0$. | Output a single integer — the amount of possible variants of the real bus number. | [
"97\n",
"2028\n"
] | [
"2\n",
"13\n"
] | In the first sample, only variants $97$ and $79$ are possible.
In the second sample, the variants (in the increasing order) are the following: $208$, $280$, $802$, $820$, $2028$, $2082$, $2208$, $2280$, $2802$, $2820$, $8022$, $8202$, $8220$. | 2,000 | [
{
"input": "97",
"output": "2"
},
{
"input": "2028",
"output": "13"
},
{
"input": "1",
"output": "1"
},
{
"input": "10",
"output": "1"
},
{
"input": "168",
"output": "6"
},
{
"input": "999999",
"output": "6"
},
{
"input": "987654320023456789",
"output": "29340299842560"
},
{
"input": "1000000000000000000",
"output": "18"
},
{
"input": "74774",
"output": "28"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "1"
},
{
"input": "5",
"output": "1"
},
{
"input": "6",
"output": "1"
},
{
"input": "7",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "101010101",
"output": "246"
},
{
"input": "1010101010",
"output": "456"
},
{
"input": "707070707070707070",
"output": "92368"
},
{
"input": "19293",
"output": "84"
},
{
"input": "987650",
"output": "600"
},
{
"input": "123456",
"output": "720"
},
{
"input": "900008",
"output": "28"
},
{
"input": "1000000",
"output": "6"
},
{
"input": "9900111",
"output": "404"
},
{
"input": "11112222",
"output": "242"
},
{
"input": "88888880",
"output": "28"
},
{
"input": "100000009",
"output": "70"
},
{
"input": "203456799",
"output": "196560"
},
{
"input": "890009800",
"output": "1120"
},
{
"input": "900000000",
"output": "8"
},
{
"input": "987654321",
"output": "362880"
},
{
"input": "999999999",
"output": "9"
},
{
"input": "1000000000",
"output": "9"
},
{
"input": "999999999999999999",
"output": "18"
},
{
"input": "987654321123456789",
"output": "33007837322880"
},
{
"input": "987654321123456780",
"output": "55657759288320"
},
{
"input": "888888888888888888",
"output": "18"
},
{
"input": "888884444444448888",
"output": "184736"
},
{
"input": "880000000008888888",
"output": "92368"
},
{
"input": "122661170586643693",
"output": "4205605773600"
},
{
"input": "166187867387753706",
"output": "224244425700"
},
{
"input": "54405428089931205",
"output": "417074011200"
},
{
"input": "96517150587709082",
"output": "417074011200"
},
{
"input": "234906817379759421",
"output": "22773236965920"
},
{
"input": "470038695054731020",
"output": "5099960335680"
},
{
"input": "888413836884649324",
"output": "76835760120"
},
{
"input": "978691308972024154",
"output": "33638772575520"
},
{
"input": "484211136976275613",
"output": "6471643862880"
},
{
"input": "824250067279351651",
"output": "21519859273920"
},
{
"input": "269041787841325833",
"output": "22773236965920"
},
{
"input": "462534182594129378",
"output": "13498126800480"
},
{
"input": "79318880250640214",
"output": "2075276790720"
},
{
"input": "58577142509378476",
"output": "1126629393120"
},
{
"input": "973088698775609061",
"output": "1646603038080"
},
{
"input": "529916324588161451",
"output": "3614537707200"
},
{
"input": "406105326393716536",
"output": "2760291011520"
},
{
"input": "490977896148785607",
"output": "2054415328560"
},
{
"input": "547694365350162078",
"output": "21519859273920"
},
{
"input": "868572419889505545",
"output": "1124978369760"
}
] | 1,611,407,908 | 2,147,483,647 | PyPy 3 | OK | TESTS | 61 | 187 | 7,577,600 | from math import factorial as f
from itertools import permutations
s = input()
c = [s.count(str(i)) for i in range(10)]
ans = 0
for d0 in range(c[0] != 0, c[0] + 1):
for d1 in range(c[1] != 0, c[1] + 1):
for d2 in range(c[2] != 0, c[2] + 1):
for d3 in range(c[3] != 0, c[3] + 1):
for d4 in range(c[4] != 0, c[4] + 1):
for d5 in range(c[5] != 0, c[5] + 1):
for d6 in range(c[6] != 0, c[6] + 1):
for d7 in range(c[7] != 0, c[7] + 1):
for d8 in range(c[8] != 0, c[8] + 1):
for d9 in range(c[9] != 0, c[9] + 1):
scnt = [d0, d1, d2, d3,
d4, d5, d6, d7, d8, d9]
li = []
for i in range(10):
li += [i] * scnt[i]
for i in range(1, 10):
if not scnt[i]:
continue
k = scnt[:]
k[i] -= 1
x = f(sum(k))
for j in range(10):
x //= f(k[j])
ans += x
print(ans)
| Title: Bus Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This night wasn't easy on Vasya. His favorite team lost, and he didn't find himself victorious either — although he played perfectly, his teammates let him down every time. He had to win at least one more time, but the losestreak only grew longer and longer... It's no wonder he didn't get any sleep this night at all.
In the morning, Vasya was waiting the bus to the university on the bus stop. Vasya's thoughts were hazy and so he couldn't remember the right bus' number quite right and got onto the bus with the number $n$.
In the bus, Vasya thought that he could get the order of the digits in the number of the bus wrong. Futhermore, he could "see" some digits several times, but the digits he saw were definitely in the real number of the bus. For example, if Vasya saw the number 2028, it could mean that the real bus number could be 2028, 8022, 2820 or just 820. However, numbers 80, 22208, 52 definitely couldn't be the number of the bus. Also, real bus number couldn't start with the digit 0, this meaning that, for example, number 082 couldn't be the real bus number too.
Given $n$, determine the total number of possible bus number variants.
Input Specification:
The first line contains one integer $n$ ($1 \leq n \leq 10^{18}$) — the number of the bus that was seen by Vasya. It is guaranteed that this number does not start with $0$.
Output Specification:
Output a single integer — the amount of possible variants of the real bus number.
Demo Input:
['97\n', '2028\n']
Demo Output:
['2\n', '13\n']
Note:
In the first sample, only variants $97$ and $79$ are possible.
In the second sample, the variants (in the increasing order) are the following: $208$, $280$, $802$, $820$, $2028$, $2082$, $2208$, $2280$, $2802$, $2820$, $8022$, $8202$, $8220$. | ```python
from math import factorial as f
from itertools import permutations
s = input()
c = [s.count(str(i)) for i in range(10)]
ans = 0
for d0 in range(c[0] != 0, c[0] + 1):
for d1 in range(c[1] != 0, c[1] + 1):
for d2 in range(c[2] != 0, c[2] + 1):
for d3 in range(c[3] != 0, c[3] + 1):
for d4 in range(c[4] != 0, c[4] + 1):
for d5 in range(c[5] != 0, c[5] + 1):
for d6 in range(c[6] != 0, c[6] + 1):
for d7 in range(c[7] != 0, c[7] + 1):
for d8 in range(c[8] != 0, c[8] + 1):
for d9 in range(c[9] != 0, c[9] + 1):
scnt = [d0, d1, d2, d3,
d4, d5, d6, d7, d8, d9]
li = []
for i in range(10):
li += [i] * scnt[i]
for i in range(1, 10):
if not scnt[i]:
continue
k = scnt[:]
k[i] -= 1
x = f(sum(k))
for j in range(10):
x //= f(k[j])
ans += x
print(ans)
``` | 3 | |
110 | A | Nearly Lucky Number | PROGRAMMING | 800 | [
"implementation"
] | A. Nearly Lucky Number | 2 | 256 | Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number. | The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator. | Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes). | [
"40047\n",
"7747774\n",
"1000000000000000000\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO". | 500 | [
{
"input": "40047",
"output": "NO"
},
{
"input": "7747774",
"output": "YES"
},
{
"input": "1000000000000000000",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "4",
"output": "NO"
},
{
"input": "474404774",
"output": "NO"
},
{
"input": "4744000695826",
"output": "YES"
},
{
"input": "10000000004744744",
"output": "YES"
},
{
"input": "446486416781684178",
"output": "YES"
},
{
"input": "999999999",
"output": "NO"
},
{
"input": "7777",
"output": "YES"
},
{
"input": "87414417444",
"output": "NO"
},
{
"input": "111222333444555667",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "4700",
"output": "NO"
},
{
"input": "3794555488744477",
"output": "NO"
},
{
"input": "444444444444444444",
"output": "NO"
},
{
"input": "474447447774444774",
"output": "NO"
},
{
"input": "777777777777777",
"output": "NO"
},
{
"input": "34777745021000000",
"output": "NO"
},
{
"input": "963",
"output": "NO"
},
{
"input": "855474448854788540",
"output": "NO"
},
{
"input": "999999999999994744",
"output": "YES"
},
{
"input": "400000000474",
"output": "YES"
},
{
"input": "123456789123456789",
"output": "YES"
},
{
"input": "740577777584945874",
"output": "NO"
},
{
"input": "7777777",
"output": "YES"
},
{
"input": "4444000111222333",
"output": "YES"
},
{
"input": "9847745885202111",
"output": "YES"
},
{
"input": "123456000000",
"output": "NO"
},
{
"input": "4744447444444",
"output": "NO"
},
{
"input": "7477",
"output": "YES"
},
{
"input": "4747477",
"output": "YES"
},
{
"input": "777777777444444444",
"output": "NO"
}
] | 1,694,514,038 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 92 | 0 | a=input()
l=list(a)
c=0
for i in l:
if i!='4' and i!='7':
c=c+1
else:
continue
if c==0:
print("YES")
else:
print("NO") | Title: Nearly Lucky Number
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number.
Input Specification:
The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output Specification:
Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes).
Demo Input:
['40047\n', '7747774\n', '1000000000000000000\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO". | ```python
a=input()
l=list(a)
c=0
for i in l:
if i!='4' and i!='7':
c=c+1
else:
continue
if c==0:
print("YES")
else:
print("NO")
``` | 0 |
429 | A | Xor-tree | PROGRAMMING | 1,300 | [
"dfs and similar",
"trees"
] | null | null | Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having *n* nodes, numbered from 1 to *n*. Each node *i* has an initial value *init**i*, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node *x*. Right after someone has picked node *x*, the value of node *x* flips, the values of sons of *x* remain the same, the values of sons of sons of *x* flips, the values of sons of sons of sons of *x* remain the same and so on.
The goal of the game is to get each node *i* to have value *goal**i*, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). Each of the next *n*<=-<=1 lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u**i*<=≠<=*v**i*) meaning there is an edge between nodes *u**i* and *v**i*.
The next line contains *n* integer numbers, the *i*-th of them corresponds to *init**i* (*init**i* is either 0 or 1). The following line also contains *n* integer numbers, the *i*-th number corresponds to *goal**i* (*goal**i* is either 0 or 1). | In the first line output an integer number *cnt*, representing the minimal number of operations you perform. Each of the next *cnt* lines should contain an integer *x**i*, representing that you pick a node *x**i*. | [
"10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1\n"
] | [
"2\n4\n7\n"
] | none | 500 | [
{
"input": "10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1",
"output": "2\n4\n7"
},
{
"input": "15\n2 1\n3 2\n4 3\n5 4\n6 5\n7 6\n8 7\n9 8\n10 9\n11 10\n12 11\n13 12\n14 13\n15 14\n0 1 0 0 1 1 1 1 1 1 0 0 0 1 1\n1 1 1 1 0 0 1 1 0 1 0 0 1 1 0",
"output": "7\n1\n4\n7\n8\n9\n11\n13"
},
{
"input": "20\n2 1\n3 2\n4 3\n5 4\n6 4\n7 1\n8 2\n9 4\n10 2\n11 6\n12 9\n13 2\n14 12\n15 14\n16 8\n17 9\n18 13\n19 2\n20 17\n1 0 0 1 0 0 0 1 0 1 0 0 0 1 1 0 1 0 1 0\n1 0 0 1 0 0 1 1 0 0 1 0 0 1 0 0 0 1 0 1",
"output": "8\n11\n15\n17\n20\n10\n18\n19\n7"
},
{
"input": "30\n2 1\n3 2\n4 3\n5 3\n6 5\n7 3\n8 3\n9 2\n10 3\n11 2\n12 11\n13 6\n14 4\n15 5\n16 11\n17 9\n18 14\n19 6\n20 2\n21 19\n22 9\n23 19\n24 20\n25 14\n26 22\n27 1\n28 6\n29 13\n30 27\n1 0 1 1 1 1 0 1 0 0 1 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 1 1 0\n0 1 0 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 1 0 1 0 0 1 0 1 1 0 0 0",
"output": "15\n1\n2\n4\n5\n6\n13\n29\n19\n21\n23\n28\n7\n22\n26\n30"
},
{
"input": "15\n2 1\n3 1\n4 1\n5 1\n6 3\n7 1\n8 1\n9 1\n10 5\n11 9\n12 3\n13 5\n14 5\n15 4\n1 1 0 0 0 0 1 1 1 0 1 1 1 0 0\n1 0 1 1 0 1 1 1 1 1 1 1 1 1 0",
"output": "6\n2\n3\n6\n4\n10\n14"
},
{
"input": "20\n2 1\n3 1\n4 2\n5 2\n6 3\n7 1\n8 6\n9 2\n10 3\n11 6\n12 2\n13 3\n14 2\n15 1\n16 8\n17 15\n18 2\n19 14\n20 14\n0 0 0 0 1 1 1 1 1 1 0 1 1 1 0 0 0 1 1 1\n0 1 0 1 1 1 0 0 1 0 1 1 0 1 0 1 0 0 1 0",
"output": "10\n2\n4\n19\n18\n8\n16\n11\n10\n13\n7"
},
{
"input": "23\n2 1\n3 2\n4 1\n5 1\n6 5\n7 3\n8 2\n9 8\n10 5\n11 6\n12 9\n13 3\n14 11\n15 5\n16 2\n17 3\n18 10\n19 16\n20 14\n21 19\n22 17\n23 7\n0 1 0 1 1 1 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0 0 0\n0 0 0 0 1 1 1 1 1 0 1 0 1 0 0 1 1 0 0 0 0 0 1",
"output": "8\n2\n23\n13\n17\n9\n4\n11\n20"
},
{
"input": "1\n0\n0",
"output": "0"
},
{
"input": "10\n1 10\n1 9\n10 2\n10 3\n3 7\n3 8\n9 4\n9 5\n5 6\n1 0 1 1 0 1 0 1 0 1\n0 0 0 0 0 0 0 0 0 0",
"output": "6\n1\n10\n2\n7\n5\n6"
}
] | 1,614,338,319 | 2,147,483,647 | PyPy 3 | OK | TESTS | 35 | 498 | 21,401,600 | from sys import stdin,stdout
from collections import defaultdict,deque
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
def bfs():
global ans
q=deque()
p[1]=-1
q.append(1)
while q:
src=q.popleft()
if p[src] != -1 and p[p[src]] != -1:
dp[src] = dp[p[p[src]]]
if dp[src]^init[src-1]!=goal[src-1]:
dp[src]^=1
ans+=[src]
for neigh in g[src]:
if neigh==p[src]:continue
p[neigh]=src
q.append(neigh)
for _ in range(1):#nmbr():
n=nmbr()
g=defaultdict(list)
for i in range(n-1):
u,v=lst()
g[u]+=[v]
g[v]+=[u]
init=lst()
goal=lst()
p=[0]*(1+n)
dp=[0]*(1+n)
ans=[]
bfs()
print(len(ans))
print(*sorted(ans),sep='\n')
| Title: Xor-tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having *n* nodes, numbered from 1 to *n*. Each node *i* has an initial value *init**i*, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node *x*. Right after someone has picked node *x*, the value of node *x* flips, the values of sons of *x* remain the same, the values of sons of sons of *x* flips, the values of sons of sons of sons of *x* remain the same and so on.
The goal of the game is to get each node *i* to have value *goal**i*, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). Each of the next *n*<=-<=1 lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u**i*<=≠<=*v**i*) meaning there is an edge between nodes *u**i* and *v**i*.
The next line contains *n* integer numbers, the *i*-th of them corresponds to *init**i* (*init**i* is either 0 or 1). The following line also contains *n* integer numbers, the *i*-th number corresponds to *goal**i* (*goal**i* is either 0 or 1).
Output Specification:
In the first line output an integer number *cnt*, representing the minimal number of operations you perform. Each of the next *cnt* lines should contain an integer *x**i*, representing that you pick a node *x**i*.
Demo Input:
['10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1\n']
Demo Output:
['2\n4\n7\n']
Note:
none | ```python
from sys import stdin,stdout
from collections import defaultdict,deque
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
def bfs():
global ans
q=deque()
p[1]=-1
q.append(1)
while q:
src=q.popleft()
if p[src] != -1 and p[p[src]] != -1:
dp[src] = dp[p[p[src]]]
if dp[src]^init[src-1]!=goal[src-1]:
dp[src]^=1
ans+=[src]
for neigh in g[src]:
if neigh==p[src]:continue
p[neigh]=src
q.append(neigh)
for _ in range(1):#nmbr():
n=nmbr()
g=defaultdict(list)
for i in range(n-1):
u,v=lst()
g[u]+=[v]
g[v]+=[u]
init=lst()
goal=lst()
p=[0]*(1+n)
dp=[0]*(1+n)
ans=[]
bfs()
print(len(ans))
print(*sorted(ans),sep='\n')
``` | 3 | |
131 | A | cAPS lOCK | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged. | The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive. | Print the result of the given word's processing. | [
"cAPS\n",
"Lock\n"
] | [
"Caps",
"Lock\n"
] | none | 500 | [
{
"input": "cAPS",
"output": "Caps"
},
{
"input": "Lock",
"output": "Lock"
},
{
"input": "cAPSlOCK",
"output": "cAPSlOCK"
},
{
"input": "CAPs",
"output": "CAPs"
},
{
"input": "LoCK",
"output": "LoCK"
},
{
"input": "OOPS",
"output": "oops"
},
{
"input": "oops",
"output": "oops"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "a"
},
{
"input": "aA",
"output": "Aa"
},
{
"input": "Zz",
"output": "Zz"
},
{
"input": "Az",
"output": "Az"
},
{
"input": "zA",
"output": "Za"
},
{
"input": "AAA",
"output": "aaa"
},
{
"input": "AAa",
"output": "AAa"
},
{
"input": "AaR",
"output": "AaR"
},
{
"input": "Tdr",
"output": "Tdr"
},
{
"input": "aTF",
"output": "Atf"
},
{
"input": "fYd",
"output": "fYd"
},
{
"input": "dsA",
"output": "dsA"
},
{
"input": "fru",
"output": "fru"
},
{
"input": "hYBKF",
"output": "Hybkf"
},
{
"input": "XweAR",
"output": "XweAR"
},
{
"input": "mogqx",
"output": "mogqx"
},
{
"input": "eOhEi",
"output": "eOhEi"
},
{
"input": "nkdku",
"output": "nkdku"
},
{
"input": "zcnko",
"output": "zcnko"
},
{
"input": "lcccd",
"output": "lcccd"
},
{
"input": "vwmvg",
"output": "vwmvg"
},
{
"input": "lvchf",
"output": "lvchf"
},
{
"input": "IUNVZCCHEWENCHQQXQYPUJCRDZLUXCLJHXPHBXEUUGNXOOOPBMOBRIBHHMIRILYJGYYGFMTMFSVURGYHUWDRLQVIBRLPEVAMJQYO",
"output": "iunvzcchewenchqqxqypujcrdzluxcljhxphbxeuugnxooopbmobribhhmirilyjgyygfmtmfsvurgyhuwdrlqvibrlpevamjqyo"
},
{
"input": "OBHSZCAMDXEJWOZLKXQKIVXUUQJKJLMMFNBPXAEFXGVNSKQLJGXHUXHGCOTESIVKSFMVVXFVMTEKACRIWALAGGMCGFEXQKNYMRTG",
"output": "obhszcamdxejwozlkxqkivxuuqjkjlmmfnbpxaefxgvnskqljgxhuxhgcotesivksfmvvxfvmtekacriwalaggmcgfexqknymrtg"
},
{
"input": "IKJYZIKROIYUUCTHSVSKZTETNNOCMAUBLFJCEVANCADASMZRCNLBZPQRXESHEEMOMEPCHROSRTNBIDXYMEPJSIXSZQEBTEKKUHFS",
"output": "ikjyzikroiyuucthsvskztetnnocmaublfjcevancadasmzrcnlbzpqrxesheemomepchrosrtnbidxymepjsixszqebtekkuhfs"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "uCKJZRGZJCPPLEEYJTUNKOQSWGBMTBQEVPYFPIPEKRVYQNTDPANOIXKMPINNFUSZWCURGBDPYTEKBEKCPMVZPMWAOSHJYMGKOMBQ",
"output": "Uckjzrgzjcppleeyjtunkoqswgbmtbqevpyfpipekrvyqntdpanoixkmpinnfuszwcurgbdpytekbekcpmvzpmwaoshjymgkombq"
},
{
"input": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR",
"output": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR"
},
{
"input": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE",
"output": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE"
},
{
"input": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ",
"output": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ"
},
{
"input": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm",
"output": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm"
},
{
"input": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm",
"output": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm"
},
{
"input": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg",
"output": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg"
},
{
"input": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc",
"output": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc"
},
{
"input": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv",
"output": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv"
},
{
"input": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect",
"output": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu",
"output": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu"
},
{
"input": "aBACABa",
"output": "aBACABa"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAaa",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAaa"
},
{
"input": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA"
},
{
"input": "dDDDDDDDDDDDDD",
"output": "Dddddddddddddd"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "z",
"output": "Z"
},
{
"input": "AZ",
"output": "az"
},
{
"input": "Z",
"output": "z"
},
{
"input": "aAAAA",
"output": "Aaaaa"
},
{
"input": "F",
"output": "f"
}
] | 1,691,166,670 | 2,147,483,647 | Python 3 | OK | TESTS | 56 | 46 | 0 | s = input()
if len(s) == 1:
s = s.swapcase()
else:
if s[0].islower() and s[1:].isupper():
s = s.swapcase()
if s.isupper():
s = s.lower()
print(s)
| Title: cAPS lOCK
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input Specification:
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output Specification:
Print the result of the given word's processing.
Demo Input:
['cAPS\n', 'Lock\n']
Demo Output:
['Caps', 'Lock\n']
Note:
none | ```python
s = input()
if len(s) == 1:
s = s.swapcase()
else:
if s[0].islower() and s[1:].isupper():
s = s.swapcase()
if s.isupper():
s = s.lower()
print(s)
``` | 3 | |
356 | A | Knight Tournament | PROGRAMMING | 1,500 | [
"data structures",
"dsu"
] | null | null | Hooray! Berl II, the king of Berland is making a knight tournament. The king has already sent the message to all knights in the kingdom and they in turn agreed to participate in this grand event.
As for you, you're just a simple peasant. There's no surprise that you slept in this morning and were late for the tournament (it was a weekend, after all). Now you are really curious about the results of the tournament. This time the tournament in Berland went as follows:
- There are *n* knights participating in the tournament. Each knight was assigned his unique number — an integer from 1 to *n*. - The tournament consisted of *m* fights, in the *i*-th fight the knights that were still in the game with numbers at least *l**i* and at most *r**i* have fought for the right to continue taking part in the tournament. - After the *i*-th fight among all participants of the fight only one knight won — the knight number *x**i*, he continued participating in the tournament. Other knights left the tournament. - The winner of the last (the *m*-th) fight (the knight number *x**m*) became the winner of the tournament.
You fished out all the information about the fights from your friends. Now for each knight you want to know the name of the knight he was conquered by. We think that the knight number *b* was conquered by the knight number *a*, if there was a fight with both of these knights present and the winner was the knight number *a*.
Write the code that calculates for each knight, the name of the knight that beat him. | The first line contains two integers *n*, *m* (2<=≤<=*n*<=≤<=3·105; 1<=≤<=*m*<=≤<=3·105) — the number of knights and the number of fights. Each of the following *m* lines contains three integers *l**i*,<=*r**i*,<=*x**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*; *l**i*<=≤<=*x**i*<=≤<=*r**i*) — the description of the *i*-th fight.
It is guaranteed that the input is correct and matches the problem statement. It is guaranteed that at least two knights took part in each battle. | Print *n* integers. If the *i*-th knight lost, then the *i*-th number should equal the number of the knight that beat the knight number *i*. If the *i*-th knight is the winner, then the *i*-th number must equal 0. | [
"4 3\n1 2 1\n1 3 3\n1 4 4\n",
"8 4\n3 5 4\n3 7 6\n2 8 8\n1 8 1\n"
] | [
"3 1 4 0 ",
"0 8 4 6 4 8 6 1 "
] | Consider the first test case. Knights 1 and 2 fought the first fight and knight 1 won. Knights 1 and 3 fought the second fight and knight 3 won. The last fight was between knights 3 and 4, knight 4 won. | 500 | [
{
"input": "4 3\n1 2 1\n1 3 3\n1 4 4",
"output": "3 1 4 0 "
},
{
"input": "8 4\n3 5 4\n3 7 6\n2 8 8\n1 8 1",
"output": "0 8 4 6 4 8 6 1 "
},
{
"input": "2 1\n1 2 1",
"output": "0 1 "
},
{
"input": "2 1\n1 2 2",
"output": "2 0 "
},
{
"input": "3 1\n1 3 1",
"output": "0 1 1 "
},
{
"input": "3 1\n1 3 2",
"output": "2 0 2 "
},
{
"input": "3 1\n1 3 3",
"output": "3 3 0 "
},
{
"input": "3 2\n1 2 1\n1 3 3",
"output": "3 1 0 "
},
{
"input": "3 2\n1 2 2\n1 3 2",
"output": "2 0 2 "
},
{
"input": "3 2\n2 3 3\n1 3 3",
"output": "3 3 0 "
},
{
"input": "11 6\n1 2 2\n7 8 7\n3 4 4\n6 9 6\n5 10 10\n2 11 11",
"output": "2 11 4 11 10 10 6 7 6 11 0 "
},
{
"input": "10 6\n9 10 10\n6 7 7\n2 4 2\n2 5 5\n1 7 5\n4 10 8",
"output": "5 5 2 2 8 7 5 0 10 8 "
},
{
"input": "11 8\n3 5 5\n8 9 9\n4 6 6\n8 10 10\n5 7 7\n2 7 2\n10 11 11\n1 11 1",
"output": "0 1 5 5 6 7 2 9 10 11 1 "
},
{
"input": "10 7\n7 8 7\n7 9 9\n5 9 5\n5 10 10\n1 2 2\n3 4 4\n2 10 4",
"output": "2 4 4 0 10 5 9 7 5 4 "
},
{
"input": "11 5\n8 10 9\n6 10 7\n6 11 11\n3 5 5\n1 11 1",
"output": "0 1 5 5 1 7 11 9 7 9 1 "
},
{
"input": "10 6\n6 7 6\n5 7 5\n3 7 4\n2 8 2\n2 10 10\n1 10 10",
"output": "10 10 4 2 4 5 6 2 10 0 "
},
{
"input": "11 7\n7 8 8\n5 6 5\n1 3 3\n7 9 9\n5 10 10\n10 11 11\n1 11 4",
"output": "3 3 4 0 10 5 8 9 10 11 4 "
},
{
"input": "10 7\n8 9 9\n3 4 3\n2 3 2\n1 5 2\n6 7 6\n6 10 10\n1 10 10",
"output": "2 10 2 3 2 10 6 9 10 0 "
},
{
"input": "11 6\n1 2 1\n8 9 9\n3 5 5\n3 6 6\n9 10 10\n1 11 10",
"output": "10 1 5 5 6 10 10 9 10 0 10 "
},
{
"input": "10 5\n1 2 1\n8 10 8\n3 6 4\n4 7 7\n1 8 7",
"output": "7 1 4 7 4 4 0 7 8 8 "
},
{
"input": "4 3\n1 2 2\n1 3 3\n1 4 4",
"output": "2 3 4 0 "
}
] | 1,649,460,478 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 3,000 | 9,318,400 | from bisect import bisect_left
knights, battles = [int(x) for x in input().strip().split()]
res: list = [0 for _ in range(knights+1)]
alive_knights: list = [i for i in range(knights+1)][1:]
old_nal = knights
n_alive: int = knights
for _ in range(battles):
li, ri, xi = [int(x) for x in input().strip().split()]
l_idx = bisect_left(alive_knights, li)
idx = l_idx
while idx < len(alive_knights) and alive_knights[idx] <= ri:
k = alive_knights[idx]
if k != xi and res[k] == 0:
res[k] = xi
n_alive -= 1
idx += 1
if n_alive < old_nal/2:
alive_knights = [knight for knight in alive_knights if res[knight] == 0]
old_nal = n_alive
print(' '.join([str(x) for x in res[1:]]))
| Title: Knight Tournament
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hooray! Berl II, the king of Berland is making a knight tournament. The king has already sent the message to all knights in the kingdom and they in turn agreed to participate in this grand event.
As for you, you're just a simple peasant. There's no surprise that you slept in this morning and were late for the tournament (it was a weekend, after all). Now you are really curious about the results of the tournament. This time the tournament in Berland went as follows:
- There are *n* knights participating in the tournament. Each knight was assigned his unique number — an integer from 1 to *n*. - The tournament consisted of *m* fights, in the *i*-th fight the knights that were still in the game with numbers at least *l**i* and at most *r**i* have fought for the right to continue taking part in the tournament. - After the *i*-th fight among all participants of the fight only one knight won — the knight number *x**i*, he continued participating in the tournament. Other knights left the tournament. - The winner of the last (the *m*-th) fight (the knight number *x**m*) became the winner of the tournament.
You fished out all the information about the fights from your friends. Now for each knight you want to know the name of the knight he was conquered by. We think that the knight number *b* was conquered by the knight number *a*, if there was a fight with both of these knights present and the winner was the knight number *a*.
Write the code that calculates for each knight, the name of the knight that beat him.
Input Specification:
The first line contains two integers *n*, *m* (2<=≤<=*n*<=≤<=3·105; 1<=≤<=*m*<=≤<=3·105) — the number of knights and the number of fights. Each of the following *m* lines contains three integers *l**i*,<=*r**i*,<=*x**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*; *l**i*<=≤<=*x**i*<=≤<=*r**i*) — the description of the *i*-th fight.
It is guaranteed that the input is correct and matches the problem statement. It is guaranteed that at least two knights took part in each battle.
Output Specification:
Print *n* integers. If the *i*-th knight lost, then the *i*-th number should equal the number of the knight that beat the knight number *i*. If the *i*-th knight is the winner, then the *i*-th number must equal 0.
Demo Input:
['4 3\n1 2 1\n1 3 3\n1 4 4\n', '8 4\n3 5 4\n3 7 6\n2 8 8\n1 8 1\n']
Demo Output:
['3 1 4 0 ', '0 8 4 6 4 8 6 1 ']
Note:
Consider the first test case. Knights 1 and 2 fought the first fight and knight 1 won. Knights 1 and 3 fought the second fight and knight 3 won. The last fight was between knights 3 and 4, knight 4 won. | ```python
from bisect import bisect_left
knights, battles = [int(x) for x in input().strip().split()]
res: list = [0 for _ in range(knights+1)]
alive_knights: list = [i for i in range(knights+1)][1:]
old_nal = knights
n_alive: int = knights
for _ in range(battles):
li, ri, xi = [int(x) for x in input().strip().split()]
l_idx = bisect_left(alive_knights, li)
idx = l_idx
while idx < len(alive_knights) and alive_knights[idx] <= ri:
k = alive_knights[idx]
if k != xi and res[k] == 0:
res[k] = xi
n_alive -= 1
idx += 1
if n_alive < old_nal/2:
alive_knights = [knight for knight in alive_knights if res[knight] == 0]
old_nal = n_alive
print(' '.join([str(x) for x in res[1:]]))
``` | 0 | |
669 | A | Little Artem and Presents | PROGRAMMING | 800 | [
"math"
] | null | null | Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha? | The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday. | Print the maximum possible number of times Artem can give presents to Masha. | [
"1\n",
"2\n",
"3\n",
"4\n"
] | [
"1\n",
"1\n",
"2\n",
"3\n"
] | In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again. | 500 | [
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "3"
},
{
"input": "100",
"output": "67"
},
{
"input": "101",
"output": "67"
},
{
"input": "102",
"output": "68"
},
{
"input": "1000000000",
"output": "666666667"
},
{
"input": "5",
"output": "3"
},
{
"input": "6",
"output": "4"
},
{
"input": "999999999",
"output": "666666666"
},
{
"input": "999999998",
"output": "666666665"
},
{
"input": "999999997",
"output": "666666665"
},
{
"input": "999999996",
"output": "666666664"
},
{
"input": "999999995",
"output": "666666663"
},
{
"input": "999999994",
"output": "666666663"
},
{
"input": "999999993",
"output": "666666662"
},
{
"input": "999999992",
"output": "666666661"
},
{
"input": "999999991",
"output": "666666661"
},
{
"input": "1000",
"output": "667"
},
{
"input": "10000",
"output": "6667"
},
{
"input": "100000",
"output": "66667"
},
{
"input": "1000000",
"output": "666667"
},
{
"input": "10000000",
"output": "6666667"
},
{
"input": "100000000",
"output": "66666667"
},
{
"input": "7",
"output": "5"
}
] | 1,611,253,511 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 2,000 | 0 | count = 0
n = int(input())
choice = 1
while True :
if n >= 2 and choice == 2:
n-=2
count+=1
choice = 1
elif n >= 1 and choice == 1:
n-=1
count+=1
choice = 2
else :
break
print(count) | Title: Little Artem and Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.
Output Specification:
Print the maximum possible number of times Artem can give presents to Masha.
Demo Input:
['1\n', '2\n', '3\n', '4\n']
Demo Output:
['1\n', '1\n', '2\n', '3\n']
Note:
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again. | ```python
count = 0
n = int(input())
choice = 1
while True :
if n >= 2 and choice == 2:
n-=2
count+=1
choice = 1
elif n >= 1 and choice == 1:
n-=1
count+=1
choice = 2
else :
break
print(count)
``` | 0 | |
159 | A | Friends or Not | PROGRAMMING | 1,400 | [
"*special",
"greedy",
"implementation"
] | null | null | Polycarpus has a hobby — he develops an unusual social network. His work is almost completed, and there is only one more module to implement — the module which determines friends. Oh yes, in this social network one won't have to add friends manually! Pairs of friends are deduced in the following way. Let's assume that user *A* sent user *B* a message at time *t*1, and user *B* sent user *A* a message at time *t*2. If 0<=<<=*t*2<=-<=*t*1<=≤<=*d*, then user *B*'s message was an answer to user *A*'s one. Users *A* and *B* are considered to be friends if *A* answered at least one *B*'s message or *B* answered at least one *A*'s message.
You are given the log of messages in chronological order and a number *d*. Find all pairs of users who will be considered to be friends. | The first line of the input contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=1000). The next *n* lines contain the messages log. The *i*-th line contains one line of the log formatted as "*A**i* *B**i* *t**i*" (without the quotes), which means that user *A**i* sent a message to user *B**i* at time *t**i* (1<=≤<=*i*<=≤<=*n*). *A**i* and *B**i* are non-empty strings at most 20 characters long, consisting of lowercase letters ('a' ... 'z'), and *t**i* is an integer (0<=≤<=*t**i*<=≤<=10000). It is guaranteed that the lines are given in non-decreasing order of *t**i*'s and that no user sent a message to himself. The elements in the lines are separated by single spaces. | In the first line print integer *k* — the number of pairs of friends. In the next *k* lines print pairs of friends as "*A**i* *B**i*" (without the quotes). You can print users in pairs and the pairs themselves in any order. Each pair must be printed exactly once. | [
"4 1\nvasya petya 1\npetya vasya 2\nanya ivan 2\nivan anya 4\n",
"1 1000\na b 0\n"
] | [
"1\npetya vasya\n",
"0\n"
] | In the first sample test case Vasya and Petya are friends because their messages' sending times are one second apart. Anya and Ivan are not, because their messages' sending times differ by more than one second. | 500 | [
{
"input": "4 1\nvasya petya 1\npetya vasya 2\nanya ivan 2\nivan anya 4",
"output": "1\npetya vasya"
},
{
"input": "1 1000\na b 0",
"output": "0"
},
{
"input": "2 1\na b 0\nb a 0",
"output": "0"
},
{
"input": "3 1\na b 1\nb c 2\nc d 3",
"output": "0"
},
{
"input": "10 2\nlutdc xfavzancwrokyzzkpco 0\nxfavzancwrokyzzkpco lutdc 1\nlutdc vydvatbnibttqgn 2\nxfavzancwrokyzzkpco vydvatbnibttqgn 2\nvydvatbnibttqgn lutdc 3\nlutdc xfavzancwrokyzzkpco 4\nlutdc vydvatbnibttqgn 5\nlutdc vydvatbnibttqgn 6\nlutdc xfavzancwrokyzzkpco 6\nvydvatbnibttqgn xfavzancwrokyzzkpco 6",
"output": "2\nlutdc vydvatbnibttqgn\nlutdc xfavzancwrokyzzkpco"
},
{
"input": "10 2\nrvmykneiddpqyf jdhmt 0\nwcsjvh jdhmt 0\njdhmt rvmykneiddpqyf 1\nrvmykneiddpqyf jdhmt 1\nwcsjvh rvmykneiddpqyf 2\nrvmykneiddpqyf jdhmt 2\njdhmt rvmykneiddpqyf 3\njdhmt wcsjvh 5\njdhmt wcsjvh 5\nrvmykneiddpqyf jdhmt 6",
"output": "1\njdhmt rvmykneiddpqyf"
},
{
"input": "10 2\nliazxawm spxwktiqjgs 0\nnolq liazxawm 1\nliazxawm nolq 2\nliazxawm spxwktiqjgs 2\nnolq liazxawm 3\nspxwktiqjgs liazxawm 3\nspxwktiqjgs liazxawm 3\nspxwktiqjgs liazxawm 3\nspxwktiqjgs nolq 3\nnolq spxwktiqjgs 4",
"output": "3\nliazxawm nolq\nliazxawm spxwktiqjgs\nnolq spxwktiqjgs"
},
{
"input": "10 2\nfxn ipntr 0\nipntr fxn 1\nfxn ipntr 1\npfvpfteadph ipntr 2\nfxn pfvpfteadph 4\nipntr fxn 4\npfvpfteadph fxn 5\nfxn pfvpfteadph 5\npfvpfteadph ipntr 6\nipntr pfvpfteadph 6",
"output": "2\nfxn ipntr\nfxn pfvpfteadph"
},
{
"input": "10 2\nyltec xnzdtcgzxqqltvpfr 0\nfxxhcmbzzg xnzdtcgzxqqltvpfr 0\nfxxhcmbzzg xnzdtcgzxqqltvpfr 0\nfxxhcmbzzg yltec 1\nfxxhcmbzzg xnzdtcgzxqqltvpfr 2\nfxxhcmbzzg yltec 2\nyltec fxxhcmbzzg 3\nyltec xnzdtcgzxqqltvpfr 3\nyltec xnzdtcgzxqqltvpfr 5\nfxxhcmbzzg yltec 6",
"output": "1\nfxxhcmbzzg yltec"
},
{
"input": "10 2\nrclgdpxdefqu abrfhwigaihoqq 0\nabrfhwigaihoqq rclgdpxdefqu 1\nrclgdpxdefqu bvkfwutdtvxgvx 1\nrclgdpxdefqu abrfhwigaihoqq 1\nabrfhwigaihoqq bvkfwutdtvxgvx 2\nbvkfwutdtvxgvx abrfhwigaihoqq 2\nbvkfwutdtvxgvx abrfhwigaihoqq 3\nabrfhwigaihoqq rclgdpxdefqu 5\nabrfhwigaihoqq rclgdpxdefqu 6\nrclgdpxdefqu bvkfwutdtvxgvx 6",
"output": "2\nabrfhwigaihoqq bvkfwutdtvxgvx\nabrfhwigaihoqq rclgdpxdefqu"
},
{
"input": "3 1\na b 1\na b 2\nb a 2",
"output": "1\na b"
}
] | 1,643,352,333 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 13 | 62 | 0 | from math import sqrt,log,ceil,gcd
import sys
# sys.stdin=open('input.txt','r')
# sys.stdout=open('output.txt','w')
from random import randint as r
def solve():
n,limit=map(int,input().split())
d={}
st=set()
for i in range(n):
a,b,t=map(str,input().split())
t=int(t)
if (b,a) in d:
# print(i+1)
temp=d[(b,a)]
if t-temp>0 and t-temp<=limit:
if ((a,b) in st) or ((b,a) in st):continue
st.add((a,b))
d[(a,b)]=t
print(len(st))
for i in st:
print(i[0],i[1])
solve()
| Title: Friends or Not
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has a hobby — he develops an unusual social network. His work is almost completed, and there is only one more module to implement — the module which determines friends. Oh yes, in this social network one won't have to add friends manually! Pairs of friends are deduced in the following way. Let's assume that user *A* sent user *B* a message at time *t*1, and user *B* sent user *A* a message at time *t*2. If 0<=<<=*t*2<=-<=*t*1<=≤<=*d*, then user *B*'s message was an answer to user *A*'s one. Users *A* and *B* are considered to be friends if *A* answered at least one *B*'s message or *B* answered at least one *A*'s message.
You are given the log of messages in chronological order and a number *d*. Find all pairs of users who will be considered to be friends.
Input Specification:
The first line of the input contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=1000). The next *n* lines contain the messages log. The *i*-th line contains one line of the log formatted as "*A**i* *B**i* *t**i*" (without the quotes), which means that user *A**i* sent a message to user *B**i* at time *t**i* (1<=≤<=*i*<=≤<=*n*). *A**i* and *B**i* are non-empty strings at most 20 characters long, consisting of lowercase letters ('a' ... 'z'), and *t**i* is an integer (0<=≤<=*t**i*<=≤<=10000). It is guaranteed that the lines are given in non-decreasing order of *t**i*'s and that no user sent a message to himself. The elements in the lines are separated by single spaces.
Output Specification:
In the first line print integer *k* — the number of pairs of friends. In the next *k* lines print pairs of friends as "*A**i* *B**i*" (without the quotes). You can print users in pairs and the pairs themselves in any order. Each pair must be printed exactly once.
Demo Input:
['4 1\nvasya petya 1\npetya vasya 2\nanya ivan 2\nivan anya 4\n', '1 1000\na b 0\n']
Demo Output:
['1\npetya vasya\n', '0\n']
Note:
In the first sample test case Vasya and Petya are friends because their messages' sending times are one second apart. Anya and Ivan are not, because their messages' sending times differ by more than one second. | ```python
from math import sqrt,log,ceil,gcd
import sys
# sys.stdin=open('input.txt','r')
# sys.stdout=open('output.txt','w')
from random import randint as r
def solve():
n,limit=map(int,input().split())
d={}
st=set()
for i in range(n):
a,b,t=map(str,input().split())
t=int(t)
if (b,a) in d:
# print(i+1)
temp=d[(b,a)]
if t-temp>0 and t-temp<=limit:
if ((a,b) in st) or ((b,a) in st):continue
st.add((a,b))
d[(a,b)]=t
print(len(st))
for i in st:
print(i[0],i[1])
solve()
``` | 0 | |
659 | C | Tanya and Toys | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | null | null | In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collection of the *i*-th type costs *i* bourles.
Tania has managed to collect *n* different types of toys *a*1,<=*a*2,<=...,<=*a**n* from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than *m* bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has.
Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this. | The first line contains two integers *n* (1<=≤<=*n*<=≤<=100<=000) and *m* (1<=≤<=*m*<=≤<=109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys.
The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the types of toys that Tanya already has. | In the first line print a single integer *k* — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed *m*.
In the second line print *k* distinct space-separated integers *t*1,<=*t*2,<=...,<=*t**k* (1<=≤<=*t**i*<=≤<=109) — the types of toys that Tanya should choose.
If there are multiple answers, you may print any of them. Values of *t**i* can be printed in any order. | [
"3 7\n1 3 4\n",
"4 14\n4 6 12 8\n"
] | [
"2\n2 5 \n",
"4\n7 2 3 1\n"
] | In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys. | 1,000 | [
{
"input": "3 7\n1 3 4",
"output": "2\n2 5 "
},
{
"input": "4 14\n4 6 12 8",
"output": "4\n1 2 3 5 "
},
{
"input": "5 6\n97746 64770 31551 96547 65684",
"output": "3\n1 2 3 "
},
{
"input": "10 10\n94125 56116 29758 94024 29289 31663 99794 35076 25328 58656",
"output": "4\n1 2 3 4 "
},
{
"input": "30 38\n9560 64176 75619 53112 54160 68775 12655 13118 99502 89757 78434 42521 19210 1927 34097 5416 56110 44786 59126 44266 79240 65567 54602 25325 37171 2879 89291 89121 39568 28162",
"output": "8\n1 2 3 4 5 6 7 8 "
},
{
"input": "1 999999298\n85187",
"output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "1 999999119\n34421",
"output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "1 1000000000\n1",
"output": "44719\n2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "1 1000000000\n44720",
"output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "1 1000000000\n44719",
"output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "1 1000000000\n44721",
"output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "3 1000000000\n123456789 234567891 345678912",
"output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "2 5\n999999999 1000000000",
"output": "2\n1 2 "
},
{
"input": "2 1000000000\n1 1000000000",
"output": "44719\n2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "3 100000\n1000000000 100000000 1",
"output": "445\n2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 ..."
},
{
"input": "5 5\n100000000 200000000 300000000 400000000 1000000000",
"output": "2\n1 2 "
},
{
"input": "6 3\n1 2 3 4 5 6",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "1 1000000000\n1000000000",
"output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "5 1000000\n1000000000 100000000 10000000 99999999 123456789",
"output": "1413\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "2 10000000\n1234567 123456",
"output": "4471\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "1 1\n1000000000",
"output": "1\n1 "
},
{
"input": "1 1000000000\n9999999",
"output": "44720\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 ..."
},
{
"input": "5 10000\n1000000000 888888888 777777777 666666666 959595959",
"output": "140\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 "
},
{
"input": "3 1\n1000000000 999999999 999999998",
"output": "1\n1 "
},
{
"input": "5 100000000\n100000000 999999999 1 2 3",
"output": "14138\n4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 15..."
},
{
"input": "3 55\n100000000 1000000000 999999999",
"output": "10\n1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "2 10\n5 10000009",
"output": "4\n1 2 3 4 "
},
{
"input": "3 10000000\n999999999 999999998 999999997",
"output": "4471\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "1 1100\n1000000000",
"output": "46\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 "
},
{
"input": "1 40\n1000000000",
"output": "8\n1 2 3 4 5 6 7 8 "
}
] | 1,587,302,718 | 2,147,483,647 | Python 3 | OK | TESTS | 61 | 233 | 12,492,800 | n, m = map(int, input().split())
d = {}
for x in input().split():
d[int(x)] = 1
num = 1
count = 0
toys = []
while(m>0):
if m>=num and not d.get(num):
toys.append(num)
m -= num
count += 1
elif m<num:
break
num += 1
print(count)
for x in toys:
print(x, end = " ")
| Title: Tanya and Toys
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collection of the *i*-th type costs *i* bourles.
Tania has managed to collect *n* different types of toys *a*1,<=*a*2,<=...,<=*a**n* from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than *m* bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has.
Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this.
Input Specification:
The first line contains two integers *n* (1<=≤<=*n*<=≤<=100<=000) and *m* (1<=≤<=*m*<=≤<=109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys.
The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the types of toys that Tanya already has.
Output Specification:
In the first line print a single integer *k* — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed *m*.
In the second line print *k* distinct space-separated integers *t*1,<=*t*2,<=...,<=*t**k* (1<=≤<=*t**i*<=≤<=109) — the types of toys that Tanya should choose.
If there are multiple answers, you may print any of them. Values of *t**i* can be printed in any order.
Demo Input:
['3 7\n1 3 4\n', '4 14\n4 6 12 8\n']
Demo Output:
['2\n2 5 \n', '4\n7 2 3 1\n']
Note:
In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys. | ```python
n, m = map(int, input().split())
d = {}
for x in input().split():
d[int(x)] = 1
num = 1
count = 0
toys = []
while(m>0):
if m>=num and not d.get(num):
toys.append(num)
m -= num
count += 1
elif m<num:
break
num += 1
print(count)
for x in toys:
print(x, end = " ")
``` | 3 | |
96 | A | Football | PROGRAMMING | 900 | [
"implementation",
"strings"
] | A. Football | 2 | 256 | Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. | The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. | Print "YES" if the situation is dangerous. Otherwise, print "NO". | [
"001001\n",
"1000000001\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"output": "NO"
},
{
"input": "1010010100000000010",
"output": "YES"
},
{
"input": "101010101",
"output": "NO"
},
{
"input": "000000000100000000000110101100000",
"output": "YES"
},
{
"input": "100001000000110101100000",
"output": "NO"
},
{
"input": "100001000011010110000",
"output": "NO"
},
{
"input": "010",
"output": "NO"
},
{
"input": "10101011111111111111111111111100",
"output": "YES"
},
{
"input": "1001101100",
"output": "NO"
},
{
"input": "1001101010",
"output": "NO"
},
{
"input": "1111100111",
"output": "NO"
},
{
"input": "00110110001110001111",
"output": "NO"
},
{
"input": "11110001001111110001",
"output": "NO"
},
{
"input": "10001111001011111101",
"output": "NO"
},
{
"input": "10000010100000001000110001010100001001001010011",
"output": "YES"
},
{
"input": "01111011111010111100101100001011001010111110000010",
"output": "NO"
},
{
"input": "00100000100100101110011001011011101110110110010100",
"output": "NO"
},
{
"input": "10110100110001001011110101110010100010000000000100101010111110111110100011",
"output": "YES"
},
{
"input": "00011101010101111001011011001101101011111101000010100000111000011100101011",
"output": "NO"
},
{
"input": "01110000110100110101110100111000101101011101011110110100100111100001110111",
"output": "NO"
},
{
"input": "11110110011000100111100111101101011111110100010101011011111101110110110111",
"output": "YES"
},
{
"input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100",
"output": "NO"
},
{
"input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010",
"output": "NO"
},
{
"input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110",
"output": "NO"
},
{
"input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000",
"output": "NO"
},
{
"input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110",
"output": "YES"
},
{
"input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000",
"output": "NO"
},
{
"input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101",
"output": "YES"
},
{
"input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101",
"output": "YES"
},
{
"input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000",
"output": "NO"
},
{
"input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000",
"output": "YES"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "YES"
},
{
"input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111",
"output": "YES"
},
{
"input": "10100101000",
"output": "NO"
},
{
"input": "11110111011101",
"output": "NO"
},
{
"input": "10000000",
"output": "YES"
},
{
"input": "00000001",
"output": "YES"
},
{
"input": "01111111",
"output": "YES"
},
{
"input": "11111110",
"output": "YES"
}
] | 1,670,223,515 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | d = input()
print("YES" if (d.__contains__("1111111") or d.__contains__("0000000")) else "NO"):
quit()
| Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
d = input()
print("YES" if (d.__contains__("1111111") or d.__contains__("0000000")) else "NO"):
quit()
``` | -1 |
387 | B | George and Round | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"two pointers"
] | null | null | George decided to prepare a Codesecrof round, so he has prepared *m* problems for the round. Let's number the problems with integers 1 through *m*. George estimates the *i*-th problem's complexity by integer *b**i*.
To make the round good, he needs to put at least *n* problems there. Besides, he needs to have at least one problem with complexity exactly *a*1, at least one with complexity exactly *a*2, ..., and at least one with complexity exactly *a**n*. Of course, the round can also have problems with other complexities.
George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity *c* to any positive integer complexity *d* (*c*<=≥<=*d*), by changing limits on the input data.
However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of problems he needs to come up with in addition to the *m* he's prepared in order to make a good round. Note that George can come up with a new problem of any complexity. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3000) — the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*<=≤<=106) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b*1<=≤<=*b*2...<=≤<=*b**m*<=≤<=106) — the complexities of the problems prepared by George. | Print a single integer — the answer to the problem. | [
"3 5\n1 2 3\n1 2 2 3 3\n",
"3 5\n1 2 3\n1 1 1 1 1\n",
"3 1\n2 3 4\n1\n"
] | [
"0\n",
"2\n",
"3\n"
] | In the first sample the set of the prepared problems meets the requirements for a good round.
In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to get a good round.
In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4. | 1,000 | [
{
"input": "3 5\n1 2 3\n1 2 2 3 3",
"output": "0"
},
{
"input": "3 5\n1 2 3\n1 1 1 1 1",
"output": "2"
},
{
"input": "3 1\n2 3 4\n1",
"output": "3"
},
{
"input": "29 100\n20 32 41 67 72 155 331 382 399 412 465 470 484 511 515 529 616 637 679 715 733 763 826 843 862 903 925 979 989\n15 15 15 17 18 19 19 20 21 21 22 24 25 26 26 27 28 31 32 32 37 38 38 39 39 40 41 42 43 43 45 45 46 47 49 49 50 50 50 51 52 53 53 55 56 57 59 59 59 60 60 62 62 63 63 64 64 64 66 67 69 69 70 70 72 72 73 74 75 76 77 78 80 80 81 81 83 83 83 84 86 86 86 86 87 88 89 91 91 91 92 93 94 94 96 97 97 97 98 98",
"output": "24"
}
] | 1,624,434,107 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 77 | 307,200 | n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
j,cnt = 0,0
#print(a[1],b[4])
for i in range(n):
#print(i,j)
while j<m and b[j]<a[i]:
j+=1
if j==m:
cnt = n - i
break
j+=1
print(cnt) | Title: George and Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George decided to prepare a Codesecrof round, so he has prepared *m* problems for the round. Let's number the problems with integers 1 through *m*. George estimates the *i*-th problem's complexity by integer *b**i*.
To make the round good, he needs to put at least *n* problems there. Besides, he needs to have at least one problem with complexity exactly *a*1, at least one with complexity exactly *a*2, ..., and at least one with complexity exactly *a**n*. Of course, the round can also have problems with other complexities.
George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity *c* to any positive integer complexity *d* (*c*<=≥<=*d*), by changing limits on the input data.
However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of problems he needs to come up with in addition to the *m* he's prepared in order to make a good round. Note that George can come up with a new problem of any complexity.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3000) — the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*<=≤<=106) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b*1<=≤<=*b*2...<=≤<=*b**m*<=≤<=106) — the complexities of the problems prepared by George.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['3 5\n1 2 3\n1 2 2 3 3\n', '3 5\n1 2 3\n1 1 1 1 1\n', '3 1\n2 3 4\n1\n']
Demo Output:
['0\n', '2\n', '3\n']
Note:
In the first sample the set of the prepared problems meets the requirements for a good round.
In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to get a good round.
In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4. | ```python
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
j,cnt = 0,0
#print(a[1],b[4])
for i in range(n):
#print(i,j)
while j<m and b[j]<a[i]:
j+=1
if j==m:
cnt = n - i
break
j+=1
print(cnt)
``` | 3 | |
231 | A | Team | PROGRAMMING | 800 | [
"brute force",
"greedy"
] | null | null | One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. | Print a single integer — the number of problems the friends will implement on the contest. | [
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] | [
"2\n",
"1\n"
] | In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | 500 | [
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,698,438,379 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | n = input()
num_solve = 0;
for i in range(0,n):
count = 0
arr = [int(x) for x in split(input)]
for num in arr:
if num == 1:
count++;
if count > 1:
num_solve++
print(num_solve) | Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | ```python
n = input()
num_solve = 0;
for i in range(0,n):
count = 0
arr = [int(x) for x in split(input)]
for num in arr:
if num == 1:
count++;
if count > 1:
num_solve++
print(num_solve)
``` | -1 | |
200 | B | Drinks | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space. | Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4. | [
"3\n50 50 100\n",
"4\n0 25 50 75\n"
] | [
"66.666666666667\n",
"37.500000000000\n"
] | Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | 500 | [
{
"input": "3\n50 50 100",
"output": "66.666666666667"
},
{
"input": "4\n0 25 50 75",
"output": "37.500000000000"
},
{
"input": "3\n0 1 8",
"output": "3.000000000000"
},
{
"input": "5\n96 89 93 95 70",
"output": "88.600000000000"
},
{
"input": "7\n62 41 78 4 38 39 75",
"output": "48.142857142857"
},
{
"input": "13\n2 22 7 0 1 17 3 17 11 2 21 26 22",
"output": "11.615384615385"
},
{
"input": "21\n5 4 11 7 0 5 45 21 0 14 51 6 0 16 10 19 8 9 7 12 18",
"output": "12.761904761905"
},
{
"input": "26\n95 70 93 74 94 70 91 70 39 79 80 57 87 75 37 93 48 67 51 90 85 26 23 64 66 84",
"output": "69.538461538462"
},
{
"input": "29\n84 99 72 96 83 92 95 98 97 93 76 84 99 93 81 76 93 99 99 100 95 100 96 95 97 100 71 98 94",
"output": "91.551724137931"
},
{
"input": "33\n100 99 100 100 99 99 99 100 100 100 99 99 99 100 100 100 100 99 100 99 100 100 97 100 100 100 100 100 100 100 98 98 100",
"output": "99.515151515152"
},
{
"input": "34\n14 9 10 5 4 26 18 23 0 1 0 20 18 15 2 2 3 5 14 1 9 4 2 15 7 1 7 19 10 0 0 11 0 2",
"output": "8.147058823529"
},
{
"input": "38\n99 98 100 100 99 92 99 99 98 84 88 94 86 99 93 100 98 99 65 98 85 84 64 97 96 89 79 96 91 84 99 93 72 96 94 97 96 93",
"output": "91.921052631579"
},
{
"input": "52\n100 94 99 98 99 99 99 95 97 97 98 100 100 98 97 100 98 90 100 99 97 94 90 98 100 100 90 99 100 95 98 95 94 85 97 94 96 94 99 99 99 98 100 100 94 99 99 100 98 87 100 100",
"output": "97.019230769231"
},
{
"input": "58\n10 70 12 89 1 82 100 53 40 100 21 69 92 91 67 66 99 77 25 48 8 63 93 39 46 79 82 14 44 42 1 79 0 69 56 73 67 17 59 4 65 80 20 60 77 52 3 61 16 76 33 18 46 100 28 59 9 6",
"output": "50.965517241379"
},
{
"input": "85\n7 8 1 16 0 15 1 7 0 11 15 6 2 12 2 8 9 8 2 0 3 7 15 7 1 8 5 7 2 26 0 3 11 1 8 10 31 0 7 6 1 8 1 0 9 14 4 8 7 16 9 1 0 16 10 9 6 1 1 4 2 7 4 5 4 1 20 6 16 16 1 1 10 17 8 12 14 19 3 8 1 7 10 23 10",
"output": "7.505882352941"
},
{
"input": "74\n5 3 0 7 13 10 12 10 18 5 0 18 2 13 7 17 2 7 5 2 40 19 0 2 2 3 0 45 4 20 0 4 2 8 1 19 3 9 17 1 15 0 16 1 9 4 0 9 32 2 6 18 11 18 1 15 16 12 7 19 5 3 9 28 26 8 3 10 33 29 4 13 28 6",
"output": "10.418918918919"
},
{
"input": "98\n42 9 21 11 9 11 22 12 52 20 10 6 56 9 26 27 1 29 29 14 38 17 41 21 7 45 15 5 29 4 51 20 6 8 34 17 13 53 30 45 0 10 16 41 4 5 6 4 14 2 31 6 0 11 13 3 3 43 13 36 51 0 7 16 28 23 8 36 30 22 8 54 21 45 39 4 50 15 1 30 17 8 18 10 2 20 16 50 6 68 15 6 38 7 28 8 29 41",
"output": "20.928571428571"
},
{
"input": "99\n60 65 40 63 57 44 30 84 3 10 39 53 40 45 72 20 76 11 61 32 4 26 97 55 14 57 86 96 34 69 52 22 26 79 31 4 21 35 82 47 81 28 72 70 93 84 40 4 69 39 83 58 30 7 32 73 74 12 92 23 61 88 9 58 70 32 75 40 63 71 46 55 39 36 14 97 32 16 95 41 28 20 85 40 5 50 50 50 75 6 10 64 38 19 77 91 50 72 96",
"output": "49.191919191919"
},
{
"input": "99\n100 88 40 30 81 80 91 98 69 73 88 96 79 58 14 100 87 84 52 91 83 88 72 83 99 35 54 80 46 79 52 72 85 32 99 39 79 79 45 83 88 50 75 75 50 59 65 75 97 63 92 58 89 46 93 80 89 33 69 86 99 99 66 85 72 74 79 98 85 95 46 63 77 97 49 81 89 39 70 76 68 91 90 56 31 93 51 87 73 95 74 69 87 95 57 68 49 95 92",
"output": "73.484848484848"
},
{
"input": "100\n18 15 17 0 3 3 0 4 1 8 2 22 7 21 5 0 0 8 3 16 1 0 2 9 9 3 10 8 17 20 5 4 8 12 2 3 1 1 3 2 23 0 1 0 5 7 4 0 1 3 3 4 25 2 2 14 8 4 9 3 0 11 0 3 12 3 14 16 7 7 14 1 17 9 0 35 42 12 3 1 25 9 3 8 5 3 2 8 22 14 11 6 3 9 6 8 7 7 4 6",
"output": "7.640000000000"
},
{
"input": "100\n88 77 65 87 100 63 91 96 92 89 77 95 76 80 84 83 100 71 85 98 26 54 74 78 69 59 96 86 88 91 95 26 52 88 64 70 84 81 76 84 94 82 100 66 97 98 43 94 59 94 100 80 98 73 69 83 94 70 74 79 91 31 62 88 69 55 62 97 40 64 62 83 87 85 50 90 69 72 67 49 100 51 69 96 81 90 83 91 86 34 79 69 100 66 97 98 47 97 74 100",
"output": "77.660000000000"
},
{
"input": "100\n91 92 90 91 98 84 85 96 83 98 99 87 94 70 87 75 86 90 89 88 82 83 91 94 88 86 90 99 100 98 97 75 95 99 95 100 91 92 76 93 95 97 88 93 95 81 96 89 88 100 98 87 90 96 100 99 58 90 96 77 92 82 100 100 93 93 98 99 79 88 97 95 98 66 96 83 96 100 99 92 98 98 92 93 100 97 98 100 98 97 100 100 94 90 99 100 98 79 80 81",
"output": "91.480000000000"
},
{
"input": "1\n0",
"output": "0.000000000000"
},
{
"input": "1\n100",
"output": "100.000000000000"
},
{
"input": "1\n78",
"output": "78.000000000000"
},
{
"input": "2\n0 100",
"output": "50.000000000000"
},
{
"input": "2\n100 100",
"output": "100.000000000000"
},
{
"input": "5\n0 0 0 0 1",
"output": "0.200000000000"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99",
"output": "99.990000000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "0.010000000000"
},
{
"input": "5\n100 100 100 100 100",
"output": "100.000000000000"
}
] | 1,684,769,686 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 92 | 0 | n = int(input())
volume_fractions = list(map(int, input().split()))
sum_volume_fractions = sum(volume_fractions)
average_volume_fraction = sum_volume_fractions / n
print(average_volume_fraction) | Title: Drinks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
Output Specification:
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
Demo Input:
['3\n50 50 100\n', '4\n0 25 50 75\n']
Demo Output:
['66.666666666667\n', '37.500000000000\n']
Note:
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | ```python
n = int(input())
volume_fractions = list(map(int, input().split()))
sum_volume_fractions = sum(volume_fractions)
average_volume_fraction = sum_volume_fractions / n
print(average_volume_fraction)
``` | 3 | |
214 | A | System of Equations | PROGRAMMING | 800 | [
"brute force"
] | null | null | Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system. | A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space. | On a single line print the answer to the problem. | [
"9 3\n",
"14 28\n",
"4 20\n"
] | [
"1\n",
"1\n",
"0\n"
] | In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | 500 | [
{
"input": "9 3",
"output": "1"
},
{
"input": "14 28",
"output": "1"
},
{
"input": "4 20",
"output": "0"
},
{
"input": "18 198",
"output": "1"
},
{
"input": "22 326",
"output": "1"
},
{
"input": "26 104",
"output": "1"
},
{
"input": "14 10",
"output": "0"
},
{
"input": "8 20",
"output": "0"
},
{
"input": "2 8",
"output": "0"
},
{
"input": "20 11",
"output": "0"
},
{
"input": "57 447",
"output": "1"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "66 296",
"output": "1"
},
{
"input": "75 683",
"output": "1"
},
{
"input": "227 975",
"output": "1"
},
{
"input": "247 499",
"output": "1"
},
{
"input": "266 116",
"output": "1"
},
{
"input": "286 916",
"output": "1"
},
{
"input": "307 341",
"output": "1"
},
{
"input": "451 121",
"output": "1"
},
{
"input": "471 921",
"output": "1"
},
{
"input": "502 346",
"output": "1"
},
{
"input": "535 59",
"output": "1"
},
{
"input": "555 699",
"output": "1"
},
{
"input": "747 351",
"output": "1"
},
{
"input": "790 64",
"output": "1"
},
{
"input": "810 704",
"output": "1"
},
{
"input": "855 225",
"output": "1"
},
{
"input": "902 34",
"output": "1"
},
{
"input": "922 514",
"output": "1"
},
{
"input": "971 131",
"output": "1"
},
{
"input": "991 931",
"output": "1"
},
{
"input": "840 780",
"output": "0"
},
{
"input": "102 595",
"output": "0"
},
{
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"output": "0"
},
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"output": "0"
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"output": "0"
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"output": "0"
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"output": "0"
},
{
"input": "205 210",
"output": "0"
},
{
"input": "1 11",
"output": "0"
},
{
"input": "1000 1000",
"output": "0"
},
{
"input": "3 3",
"output": "0"
},
{
"input": "11 99",
"output": "0"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "11 1",
"output": "0"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "100 452",
"output": "0"
},
{
"input": "420 380",
"output": "0"
},
{
"input": "31 31",
"output": "0"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "112 134",
"output": "0"
},
{
"input": "13 13",
"output": "0"
},
{
"input": "1 571",
"output": "0"
}
] | 1,685,621,871 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | a=input()
b1=''
b2=''
if len(a)%2==0:
for i in range(0,int(len(a)/2)):
b1+=a[i]
for i in range(len(a)-1,int(len(a)/2)-1,-1):
b2+=a[i]
if b1==b2:
print('YES')
else:
print('NO')
else:
for i in range(0,int(len(a)/2)):
b1+=a[i]
for i in range(len(a)-1,int(len(a)/2),-1):
b2+=a[i]
if b1==b2:
print('YES')
else:
print('NO') | Title: System of Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system.
Input Specification:
A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space.
Output Specification:
On a single line print the answer to the problem.
Demo Input:
['9 3\n', '14 28\n', '4 20\n']
Demo Output:
['1\n', '1\n', '0\n']
Note:
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | ```python
a=input()
b1=''
b2=''
if len(a)%2==0:
for i in range(0,int(len(a)/2)):
b1+=a[i]
for i in range(len(a)-1,int(len(a)/2)-1,-1):
b2+=a[i]
if b1==b2:
print('YES')
else:
print('NO')
else:
for i in range(0,int(len(a)/2)):
b1+=a[i]
for i in range(len(a)-1,int(len(a)/2),-1):
b2+=a[i]
if b1==b2:
print('YES')
else:
print('NO')
``` | 0 | |
78 | B | Easter Eggs | PROGRAMMING | 1,200 | [
"constructive algorithms",
"implementation"
] | B. Easter Eggs | 2 | 256 | The Easter Rabbit laid *n* eggs in a circle and is about to paint them.
Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied:
- Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors.
Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible. | The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100). | Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet.
If there are several answers, print any of them. | [
"8\n",
"13\n"
] | [
"ROYGRBIV\n",
"ROYGBIVGBIVYG\n"
] | The way the eggs will be painted in the first sample is shown on the picture: | 1,000 | [
{
"input": "8",
"output": "ROYGBIVG"
},
{
"input": "13",
"output": "ROYGBIVOYGBIV"
},
{
"input": "7",
"output": "ROYGBIV"
},
{
"input": "10",
"output": "ROYGBIVYGB"
},
{
"input": "14",
"output": "ROYGBIVROYGBIV"
},
{
"input": "50",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "9",
"output": "ROYGBIVGB"
},
{
"input": "11",
"output": "ROYGBIVYGBI"
},
{
"input": "12",
"output": "ROYGBIVOYGBI"
},
{
"input": "15",
"output": "ROYGBIVROYGBIVG"
},
{
"input": "16",
"output": "ROYGBIVROYGBIVGB"
},
{
"input": "17",
"output": "ROYGBIVROYGBIVYGB"
},
{
"input": "18",
"output": "ROYGBIVROYGBIVYGBI"
},
{
"input": "19",
"output": "ROYGBIVROYGBIVOYGBI"
},
{
"input": "20",
"output": "ROYGBIVROYGBIVOYGBIV"
},
{
"input": "21",
"output": "ROYGBIVROYGBIVROYGBIV"
},
{
"input": "22",
"output": "ROYGBIVROYGBIVROYGBIVG"
},
{
"input": "23",
"output": "ROYGBIVROYGBIVROYGBIVGB"
},
{
"input": "24",
"output": "ROYGBIVROYGBIVROYGBIVYGB"
},
{
"input": "25",
"output": "ROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "28",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIV"
},
{
"input": "29",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "34",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVOYGBIV"
},
{
"input": "43",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "61",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBI"
},
{
"input": "79",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVGB"
},
{
"input": "81",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "92",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "95",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "96",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBI"
},
{
"input": "97",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBIV"
},
{
"input": "98",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIV"
},
{
"input": "99",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "100",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVGB"
}
] | 1,614,622,438 | 2,147,483,647 | PyPy 3 | OK | TESTS | 34 | 184 | 0 | n = int(input())
s = 'VIBGYOR'
times = n//7
remains = n%7
if remains > 4:
print(f'{s*times}{s[0:remains]}')
else:
print(f'{s*times}{s[3:3+remains]}')
| Title: Easter Eggs
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Easter Rabbit laid *n* eggs in a circle and is about to paint them.
Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied:
- Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors.
Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible.
Input Specification:
The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100).
Output Specification:
Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet.
If there are several answers, print any of them.
Demo Input:
['8\n', '13\n']
Demo Output:
['ROYGRBIV\n', 'ROYGBIVGBIVYG\n']
Note:
The way the eggs will be painted in the first sample is shown on the picture: | ```python
n = int(input())
s = 'VIBGYOR'
times = n//7
remains = n%7
if remains > 4:
print(f'{s*times}{s[0:remains]}')
else:
print(f'{s*times}{s[3:3+remains]}')
``` | 3.954 |
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