contestId int64 0 1.01k | index stringclasses 57 values | name stringlengths 2 58 | type stringclasses 2 values | rating int64 0 3.5k | tags listlengths 0 11 | title stringclasses 522 values | time-limit stringclasses 8 values | memory-limit stringclasses 8 values | problem-description stringlengths 0 7.15k | input-specification stringlengths 0 2.05k | output-specification stringlengths 0 1.5k | demo-input listlengths 0 7 | demo-output listlengths 0 7 | note stringlengths 0 5.24k | points float64 0 425k | test_cases listlengths 0 402 | creationTimeSeconds int64 1.37B 1.7B | relativeTimeSeconds int64 8 2.15B | programmingLanguage stringclasses 3 values | verdict stringclasses 14 values | testset stringclasses 12 values | passedTestCount int64 0 1k | timeConsumedMillis int64 0 15k | memoryConsumedBytes int64 0 805M | code stringlengths 3 65.5k | prompt stringlengths 262 8.2k | response stringlengths 17 65.5k | score float64 -1 3.99 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
678 | A | Johny Likes Numbers | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*. | The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109). | Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*. | [
"5 3\n",
"25 13\n",
"26 13\n"
] | [
"6\n",
"26\n",
"39\n"
] | none | 0 | [
{
"input": "5 3",
"output": "6"
},
{
"input": "25 13",
"output": "26"
},
{
"input": "26 13",
"output": "39"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "8 8",
"output": "16"
},
{
"input": "14 15",
"output": "15"
},
{
"input": "197 894",
"output": "894"
},
{
"input": "6058 8581",
"output": "8581"
},
{
"input": "97259 41764",
"output": "125292"
},
{
"input": "453145 333625",
"output": "667250"
},
{
"input": "2233224 4394826",
"output": "4394826"
},
{
"input": "76770926 13350712",
"output": "80104272"
},
{
"input": "687355301 142098087",
"output": "710490435"
},
{
"input": "1000000000 999999999",
"output": "1999999998"
},
{
"input": "1000000000 1000000000",
"output": "2000000000"
},
{
"input": "999999999 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 1",
"output": "1000000001"
},
{
"input": "1000000000 2",
"output": "1000000002"
},
{
"input": "999999999 1",
"output": "1000000000"
},
{
"input": "100000000 1",
"output": "100000001"
},
{
"input": "999999999 500000000",
"output": "1000000000"
},
{
"input": "999999990 10",
"output": "1000000000"
},
{
"input": "1000000000 999999997",
"output": "1999999994"
},
{
"input": "999999999 2",
"output": "1000000000"
},
{
"input": "999999984 1",
"output": "999999985"
},
{
"input": "999999983 1",
"output": "999999984"
},
{
"input": "666666666 1",
"output": "666666667"
},
{
"input": "1000000000 990000000",
"output": "1980000000"
},
{
"input": "41 48",
"output": "48"
},
{
"input": "123456 2",
"output": "123458"
},
{
"input": "111 111",
"output": "222"
},
{
"input": "878787 1",
"output": "878788"
},
{
"input": "121 1",
"output": "122"
},
{
"input": "114514 114514",
"output": "229028"
},
{
"input": "500000001 1000000000",
"output": "1000000000"
},
{
"input": "999999997 1",
"output": "999999998"
},
{
"input": "100000000 10",
"output": "100000010"
}
] | 1,500,005,385 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 62 | 5,529,600 | import math
string = input()
numbers = string.split()
a, b = int(numbers[0]), int(numbers[1])
print((math.floor(a / b) + 1) * b) | Title: Johny Likes Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
Output Specification:
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
Demo Input:
['5 3\n', '25 13\n', '26 13\n']
Demo Output:
['6\n', '26\n', '39\n']
Note:
none | ```python
import math
string = input()
numbers = string.split()
a, b = int(numbers[0]), int(numbers[1])
print((math.floor(a / b) + 1) * b)
``` | 3 | |
344 | A | Magnets | PROGRAMMING | 800 | [
"implementation"
] | null | null | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | On the single line of the output print the number of groups of magnets. | [
"6\n10\n10\n10\n01\n10\n10\n",
"4\n01\n01\n10\n10\n"
] | [
"3\n",
"2\n"
] | The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets. | 500 | [
{
"input": "6\n10\n10\n10\n01\n10\n10",
"output": "3"
},
{
"input": "4\n01\n01\n10\n10",
"output": "2"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n01\n10",
"output": "2"
},
{
"input": "2\n10\n10",
"output": "1"
},
{
"input": "3\n10\n01\n10",
"output": "3"
},
{
"input": "1\n01",
"output": "1"
},
{
"input": "2\n01\n01",
"output": "1"
},
{
"input": "2\n10\n01",
"output": "2"
},
{
"input": "3\n01\n01\n01",
"output": "1"
},
{
"input": "3\n10\n10\n01",
"output": "2"
},
{
"input": "3\n01\n10\n10",
"output": "2"
},
{
"input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10",
"output": "55"
}
] | 1,694,571,475 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 92 | 0 | n= int(input())
mag_seq = []
for i in range(n):
mag_seq.append(input())
point = 0
for j in range(n - 1):
if mag_seq[j] == mag_seq[j + 1]:
point += 1
print(point + 1) | Title: Magnets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output Specification:
On the single line of the output print the number of groups of magnets.
Demo Input:
['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n']
Demo Output:
['3\n', '2\n']
Note:
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets. | ```python
n= int(input())
mag_seq = []
for i in range(n):
mag_seq.append(input())
point = 0
for j in range(n - 1):
if mag_seq[j] == mag_seq[j + 1]:
point += 1
print(point + 1)
``` | 0 | |
749 | A | Bachgold Problem | PROGRAMMING | 800 | [
"greedy",
"implementation",
"math",
"number theory"
] | null | null | Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*. | The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000). | The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them. | [
"5\n",
"6\n"
] | [
"2\n2 3\n",
"3\n2 2 2\n"
] | none | 500 | [
{
"input": "5",
"output": "2\n2 3"
},
{
"input": "6",
"output": "3\n2 2 2"
},
{
"input": "2",
"output": "1\n2"
},
{
"input": "3",
"output": "1\n3"
},
{
"input": "99999",
"output": "49999\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 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 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 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 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 ..."
},
{
"input": "100000",
"output": "50000\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 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 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 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 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 ..."
},
{
"input": "7",
"output": "3\n2 2 3"
},
{
"input": "4",
"output": "2\n2 2"
},
{
"input": "8",
"output": "4\n2 2 2 2"
},
{
"input": "9",
"output": "4\n2 2 2 3"
},
{
"input": "99995",
"output": "49997\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 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 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 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 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 ..."
},
{
"input": "99996",
"output": "49998\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 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 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 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 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 ..."
},
{
"input": "10",
"output": "5\n2 2 2 2 2"
},
{
"input": "11",
"output": "5\n2 2 2 2 3"
},
{
"input": "99997",
"output": "49998\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 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 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 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 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 ..."
},
{
"input": "99998",
"output": "49999\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 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 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 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 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 ..."
},
{
"input": "12",
"output": "6\n2 2 2 2 2 2"
},
{
"input": "13",
"output": "6\n2 2 2 2 2 3"
},
{
"input": "99993",
"output": "49996\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 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 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 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 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 ..."
},
{
"input": "99994",
"output": "49997\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 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 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 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 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 ..."
},
{
"input": "14",
"output": "7\n2 2 2 2 2 2 2"
},
{
"input": "15",
"output": "7\n2 2 2 2 2 2 3"
},
{
"input": "53",
"output": "26\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 3"
},
{
"input": "57",
"output": "28\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 3"
},
{
"input": "61",
"output": "30\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 3"
},
{
"input": "774",
"output": "387\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 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 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 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 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 ..."
},
{
"input": "202",
"output": "101\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 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 2 2"
},
{
"input": "530",
"output": "265\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 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 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 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 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 ..."
},
{
"input": "7166",
"output": "3583\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 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 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 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 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..."
},
{
"input": "9294",
"output": "4647\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 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 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 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 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..."
},
{
"input": "2422",
"output": "1211\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 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 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 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 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..."
},
{
"input": "15326",
"output": "7663\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 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 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 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 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..."
},
{
"input": "11454",
"output": "5727\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 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 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 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 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..."
},
{
"input": "14878",
"output": "7439\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 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 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 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 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..."
},
{
"input": "90672",
"output": "45336\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 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 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 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 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 ..."
},
{
"input": "99544",
"output": "49772\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 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 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 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 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 ..."
},
{
"input": "90472",
"output": "45236\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 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 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 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 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 ..."
},
{
"input": "23",
"output": "11\n2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "93",
"output": "46\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 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "19",
"output": "9\n2 2 2 2 2 2 2 2 3"
},
{
"input": "11110",
"output": "5555\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 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 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 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 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..."
}
] | 1,675,780,687 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<bits/stdc++.h>
using namespace std;
int main()
{
int k;
cin>>k;
cout<<k/2<<endl;
if(k%2==0)
{
for(int i=0;i<(k/2);i++)
{
cout<<"2"<<" ";
}
}
else
{
for(int i=0;i<((k/2)-1);i++){
cout<<"2"<<" ";
}
cout<<"3"<<" ";
}
cout<<endl;
}
| Title: Bachgold Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
Output Specification:
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
Demo Input:
['5\n', '6\n']
Demo Output:
['2\n2 3\n', '3\n2 2 2\n']
Note:
none | ```python
#include<bits/stdc++.h>
using namespace std;
int main()
{
int k;
cin>>k;
cout<<k/2<<endl;
if(k%2==0)
{
for(int i=0;i<(k/2);i++)
{
cout<<"2"<<" ";
}
}
else
{
for(int i=0;i<((k/2)-1);i++){
cout<<"2"<<" ";
}
cout<<"3"<<" ";
}
cout<<endl;
}
``` | -1 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,611,783,328 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 122 | 307,200 | n = int(input())
lstx = []
lsty = []
lstz = []
for i in range(n) :
x, y, z = map(int, input().strip().split())
lstx.append(x)
lsty.append(y)
lstz.append(z)
u = eval('+'.join(lstx))
v = eval('+'.join(lsty))
w = eval('+'.join(lstz))
if u == 0 and v == 0 and w == 0 :
print("YES")
else :
print("NO")
exit() | Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
n = int(input())
lstx = []
lsty = []
lstz = []
for i in range(n) :
x, y, z = map(int, input().strip().split())
lstx.append(x)
lsty.append(y)
lstz.append(z)
u = eval('+'.join(lstx))
v = eval('+'.join(lsty))
w = eval('+'.join(lstz))
if u == 0 and v == 0 and w == 0 :
print("YES")
else :
print("NO")
exit()
``` | -1 |
967 | A | Mind the Gap | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | These days Arkady works as an air traffic controller at a large airport. He controls a runway which is usually used for landings only. Thus, he has a schedule of planes that are landing in the nearest future, each landing lasts $1$ minute.
He was asked to insert one takeoff in the schedule. The takeoff takes $1$ minute itself, but for safety reasons there should be a time space between the takeoff and any landing of at least $s$ minutes from both sides.
Find the earliest time when Arkady can insert the takeoff. | The first line of input contains two integers $n$ and $s$ ($1 \le n \le 100$, $1 \le s \le 60$) — the number of landings on the schedule and the minimum allowed time (in minutes) between a landing and a takeoff.
Each of next $n$ lines contains two integers $h$ and $m$ ($0 \le h \le 23$, $0 \le m \le 59$) — the time, in hours and minutes, when a plane will land, starting from current moment (i. e. the current time is $0$ $0$). These times are given in increasing order. | Print two integers $h$ and $m$ — the hour and the minute from the current moment of the earliest time Arkady can insert the takeoff. | [
"6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40\n",
"16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59\n",
"3 17\n0 30\n1 0\n12 0\n"
] | [
"6 1\n",
"24 50\n",
"0 0\n"
] | In the first example note that there is not enough time between 1:20 and 3:21, because each landing and the takeoff take one minute.
In the second example there is no gaps in the schedule, so Arkady can only add takeoff after all landings. Note that it is possible that one should wait more than $24$ hours to insert the takeoff.
In the third example Arkady can insert the takeoff even between the first landing. | 500 | [
{
"input": "6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40",
"output": "6 1"
},
{
"input": "16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59",
"output": "24 50"
},
{
"input": "3 17\n0 30\n1 0\n12 0",
"output": "0 0"
},
{
"input": "24 60\n0 21\n2 21\n2 46\n3 17\n4 15\n5 43\n6 41\n7 50\n8 21\n9 8\n10 31\n10 45\n12 30\n14 8\n14 29\n14 32\n14 52\n15 16\n16 7\n16 52\n18 44\n20 25\n21 13\n22 7",
"output": "23 8"
},
{
"input": "20 60\n0 9\n0 19\n0 57\n2 42\n3 46\n3 47\n5 46\n8 1\n9 28\n9 41\n10 54\n12 52\n13 0\n14 49\n17 28\n17 39\n19 34\n20 52\n21 35\n23 22",
"output": "6 47"
},
{
"input": "57 20\n0 2\n0 31\n1 9\n1 42\n1 58\n2 4\n2 35\n2 49\n3 20\n3 46\n4 23\n4 52\n5 5\n5 39\n6 7\n6 48\n6 59\n7 8\n7 35\n8 10\n8 46\n8 53\n9 19\n9 33\n9 43\n10 18\n10 42\n11 0\n11 26\n12 3\n12 5\n12 30\n13 1\n13 38\n14 13\n14 54\n15 31\n16 5\n16 44\n17 18\n17 30\n17 58\n18 10\n18 34\n19 13\n19 49\n19 50\n19 59\n20 17\n20 23\n20 40\n21 18\n21 57\n22 31\n22 42\n22 56\n23 37",
"output": "23 58"
},
{
"input": "66 20\n0 16\n0 45\n0 58\n1 6\n1 19\n2 7\n2 9\n3 9\n3 25\n3 57\n4 38\n4 58\n5 21\n5 40\n6 16\n6 19\n6 58\n7 6\n7 26\n7 51\n8 13\n8 36\n8 55\n9 1\n9 15\n9 33\n10 12\n10 37\n11 15\n11 34\n12 8\n12 37\n12 55\n13 26\n14 0\n14 34\n14 36\n14 48\n15 23\n15 29\n15 43\n16 8\n16 41\n16 45\n17 5\n17 7\n17 15\n17 29\n17 46\n18 12\n18 19\n18 38\n18 57\n19 32\n19 58\n20 5\n20 40\n20 44\n20 50\n21 18\n21 49\n22 18\n22 47\n23 1\n23 38\n23 50",
"output": "1 40"
},
{
"input": "1 1\n0 0",
"output": "0 2"
},
{
"input": "10 1\n0 2\n0 4\n0 5\n0 8\n0 9\n0 11\n0 13\n0 16\n0 19\n0 21",
"output": "0 0"
},
{
"input": "10 1\n0 2\n0 5\n0 8\n0 11\n0 15\n0 17\n0 25\n0 28\n0 29\n0 32",
"output": "0 0"
},
{
"input": "15 20\n0 47\n2 24\n4 19\n4 34\n5 46\n8 15\n9 8\n10 28\n17 47\n17 52\n18 32\n19 50\n20 46\n20 50\n23 21",
"output": "0 0"
},
{
"input": "1 5\n1 0",
"output": "0 0"
},
{
"input": "24 60\n1 0\n2 0\n3 0\n4 0\n5 0\n6 0\n7 0\n8 0\n9 0\n10 0\n11 0\n12 0\n13 0\n14 0\n15 0\n16 0\n17 0\n18 0\n19 0\n20 0\n21 0\n22 0\n23 0\n23 59",
"output": "25 0"
},
{
"input": "1 30\n0 29",
"output": "1 0"
},
{
"input": "1 2\n3 0",
"output": "0 0"
},
{
"input": "16 60\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59",
"output": "25 0"
},
{
"input": "1 5\n0 6",
"output": "0 0"
},
{
"input": "2 60\n0 59\n23 59",
"output": "2 0"
},
{
"input": "1 58\n0 1",
"output": "1 0"
},
{
"input": "25 60\n0 0\n1 0\n2 0\n3 0\n4 0\n5 0\n6 0\n7 0\n8 0\n9 0\n10 0\n11 0\n12 0\n13 0\n14 0\n15 0\n16 0\n17 0\n18 0\n19 0\n20 0\n21 0\n22 0\n23 0\n23 59",
"output": "25 0"
},
{
"input": "2 3\n0 3\n0 30",
"output": "0 7"
},
{
"input": "16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 9",
"output": "24 0"
},
{
"input": "1 60\n2 0",
"output": "0 0"
},
{
"input": "2 60\n0 0\n5 0",
"output": "1 1"
},
{
"input": "1 30\n0 31",
"output": "0 0"
},
{
"input": "2 60\n0 59\n3 1",
"output": "2 0"
},
{
"input": "2 60\n0 59\n5 0",
"output": "2 0"
},
{
"input": "1 59\n0 0",
"output": "1 0"
},
{
"input": "3 25\n0 0\n1 0\n2 0",
"output": "0 26"
},
{
"input": "1 2\n2 3",
"output": "0 0"
}
] | 1,525,348,542 | 642 | PyPy 3 | OK | TESTS | 34 | 124 | 21,401,600 | n,s=map(int,input().split())
v=[tuple(map(int,input().split())) for _ in range(n)]
v.sort()
cur=0
for a,b in v:
t=a*60+b
if t>=cur+1+s: print(cur//60,cur%60);break
cur=t+1+s
if cur>v[-1][0]*60+v[-1][1]: print(cur//60,cur%60) | Title: Mind the Gap
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
These days Arkady works as an air traffic controller at a large airport. He controls a runway which is usually used for landings only. Thus, he has a schedule of planes that are landing in the nearest future, each landing lasts $1$ minute.
He was asked to insert one takeoff in the schedule. The takeoff takes $1$ minute itself, but for safety reasons there should be a time space between the takeoff and any landing of at least $s$ minutes from both sides.
Find the earliest time when Arkady can insert the takeoff.
Input Specification:
The first line of input contains two integers $n$ and $s$ ($1 \le n \le 100$, $1 \le s \le 60$) — the number of landings on the schedule and the minimum allowed time (in minutes) between a landing and a takeoff.
Each of next $n$ lines contains two integers $h$ and $m$ ($0 \le h \le 23$, $0 \le m \le 59$) — the time, in hours and minutes, when a plane will land, starting from current moment (i. e. the current time is $0$ $0$). These times are given in increasing order.
Output Specification:
Print two integers $h$ and $m$ — the hour and the minute from the current moment of the earliest time Arkady can insert the takeoff.
Demo Input:
['6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40\n', '16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59\n', '3 17\n0 30\n1 0\n12 0\n']
Demo Output:
['6 1\n', '24 50\n', '0 0\n']
Note:
In the first example note that there is not enough time between 1:20 and 3:21, because each landing and the takeoff take one minute.
In the second example there is no gaps in the schedule, so Arkady can only add takeoff after all landings. Note that it is possible that one should wait more than $24$ hours to insert the takeoff.
In the third example Arkady can insert the takeoff even between the first landing. | ```python
n,s=map(int,input().split())
v=[tuple(map(int,input().split())) for _ in range(n)]
v.sort()
cur=0
for a,b in v:
t=a*60+b
if t>=cur+1+s: print(cur//60,cur%60);break
cur=t+1+s
if cur>v[-1][0]*60+v[-1][1]: print(cur//60,cur%60)
``` | 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"
}
] | 1,649,540,492 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | x=input()
y=input()
l1=[]
l2=[]
for i in x:
l1.append(i)
for j in y:
l2.append(j)
l1.reverse()
if l1==l2:
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
x=input()
y=input()
l1=[]
l2=[]
for i in x:
l1.append(i)
for j in y:
l2.append(j)
l1.reverse()
if l1==l2:
print("Yes")
else:
print('No')
``` | 0 |
939 | A | Love Triangle | PROGRAMMING | 800 | [
"graphs"
] | null | null | As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th. | Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case. | [
"5\n2 4 5 1 3\n",
"5\n5 5 5 5 1\n"
] | [
"YES\n",
"NO\n"
] | In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | 500 | [
{
"input": "5\n2 4 5 1 3",
"output": "YES"
},
{
"input": "5\n5 5 5 5 1",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "10\n4 10 9 5 3 1 5 10 6 4",
"output": "NO"
},
{
"input": "10\n5 5 4 9 10 9 9 5 3 1",
"output": "YES"
},
{
"input": "100\n50 40 60 87 39 58 44 84 46 68 16 57 77 87 92 95 42 31 74 15 36 84 30 3 47 15 87 90 76 66 6 63 74 19 40 49 6 84 41 9 77 34 7 12 11 73 58 24 81 14 81 29 65 100 1 85 64 32 38 4 54 67 32 81 80 7 100 71 29 80 4 52 47 7 78 56 52 75 81 37 16 41 27 28 58 60 62 47 29 40 37 14 59 91 12 54 25 58 12 43",
"output": "NO"
},
{
"input": "100\n25 6 46 37 87 99 70 31 46 12 94 40 87 56 28 8 94 39 13 12 67 13 71 39 83 48 40 14 62 41 16 71 20 41 83 41 68 98 23 82 62 83 62 35 49 22 31 21 66 98 54 39 34 52 11 28 47 89 25 44 68 36 91 46 82 86 88 48 27 93 7 9 53 36 16 100 84 84 44 25 58 66 16 46 72 21 91 78 4 17 44 17 47 67 93 89 75 44 56 50",
"output": "YES"
},
{
"input": "2\n2 1",
"output": "NO"
},
{
"input": "3\n2 3 1",
"output": "YES"
},
{
"input": "5\n2 1 4 5 3",
"output": "YES"
},
{
"input": "5\n5 4 5 5 2",
"output": "YES"
}
] | 1,657,857,135 | 735 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 |
n = int( input() )
p = list( map(int, input().split()) ) # Convierte texto con espacios a lista de enteros
def nxt(p, i):
n = p[i]
if n == i:
return -1
return n
for i in range(n):
fin = nxt(p, nxt(p, nxt(p, i)))
if fin == i:
print("YES")
break
else:
print("NO")
| Title: Love Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th.
Output Specification:
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Demo Input:
['5\n2 4 5 1 3\n', '5\n5 5 5 5 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | ```python
n = int( input() )
p = list( map(int, input().split()) ) # Convierte texto con espacios a lista de enteros
def nxt(p, i):
n = p[i]
if n == i:
return -1
return n
for i in range(n):
fin = nxt(p, nxt(p, nxt(p, i)))
if fin == i:
print("YES")
break
else:
print("NO")
``` | -1 | |
235 | A | LCM Challenge | PROGRAMMING | 1,600 | [
"number theory"
] | null | null | Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than *n*. Can you help me to find the maximum possible least common multiple of these three integers? | The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the *n* mentioned in the statement. | Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than *n*. | [
"9\n",
"7\n"
] | [
"504\n",
"210\n"
] | The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | 500 | [
{
"input": "9",
"output": "504"
},
{
"input": "7",
"output": "210"
},
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "60"
},
{
"input": "6",
"output": "60"
},
{
"input": "33",
"output": "32736"
},
{
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"output": "999996000003000000"
},
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}
] | 1,695,420,753 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 31 | 124 | 0 | from math import lcm
n = int(input())
if n<=2:
print(n)
else:
print(lcm(n,n-1,n-2))
| Title: LCM Challenge
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than *n*. Can you help me to find the maximum possible least common multiple of these three integers?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the *n* mentioned in the statement.
Output Specification:
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than *n*.
Demo Input:
['9\n', '7\n']
Demo Output:
['504\n', '210\n']
Note:
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | ```python
from math import lcm
n = int(input())
if n<=2:
print(n)
else:
print(lcm(n,n-1,n-2))
``` | 0 | |
803 | C | Maximal GCD | PROGRAMMING | 1,900 | [
"constructive algorithms",
"greedy",
"math"
] | null | null | You are given positive integer number *n*. You should create such strictly increasing sequence of *k* positive numbers *a*1,<=*a*2,<=...,<=*a**k*, that their sum is equal to *n* and greatest common divisor is maximal.
Greatest common divisor of sequence is maximum of such numbers that every element of sequence is divisible by them.
If there is no possible sequence then output -1. | The first line consists of two numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1010). | If the answer exists then output *k* numbers — resulting sequence. Otherwise output -1. If there are multiple answers, print any of them. | [
"6 3\n",
"8 2\n",
"5 3\n"
] | [
"1 2 3\n",
"2 6\n",
"-1\n"
] | none | 0 | [
{
"input": "6 3",
"output": "1 2 3"
},
{
"input": "8 2",
"output": "2 6"
},
{
"input": "5 3",
"output": "-1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "2 10000000000",
"output": "-1"
},
{
"input": "5 1",
"output": "5"
},
{
"input": "6 2",
"output": "2 4"
},
{
"input": "24 2",
"output": "8 16"
},
{
"input": "24 3",
"output": "4 8 12"
},
{
"input": "24 4",
"output": "2 4 6 12"
},
{
"input": "24 5",
"output": "1 2 3 4 14"
},
{
"input": "479001600 2",
"output": "159667200 319334400"
},
{
"input": "479001600 3",
"output": "79833600 159667200 239500800"
},
{
"input": "479001600 4",
"output": "47900160 95800320 143700480 191600640"
},
{
"input": "479001600 5",
"output": "31933440 63866880 95800320 127733760 159667200"
},
{
"input": "479001600 6",
"output": "22809600 45619200 68428800 91238400 114048000 136857600"
},
{
"input": "3000000021 1",
"output": "3000000021"
},
{
"input": "3000000021 2",
"output": "1000000007 2000000014"
},
{
"input": "3000000021 3",
"output": "3 6 3000000012"
},
{
"input": "3000000021 4",
"output": "3 6 9 3000000003"
},
{
"input": "3000000021 50000",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "3000000021 100000",
"output": "-1"
},
{
"input": "10000000000 100",
"output": "1953125 3906250 5859375 7812500 9765625 11718750 13671875 15625000 17578125 19531250 21484375 23437500 25390625 27343750 29296875 31250000 33203125 35156250 37109375 39062500 41015625 42968750 44921875 46875000 48828125 50781250 52734375 54687500 56640625 58593750 60546875 62500000 64453125 66406250 68359375 70312500 72265625 74218750 76171875 78125000 80078125 82031250 83984375 85937500 87890625 89843750 91796875 93750000 95703125 97656250 99609375 101562500 103515625 105468750 107421875 109375000 1113281..."
},
{
"input": "10000000000 2000",
"output": "4000 8000 12000 16000 20000 24000 28000 32000 36000 40000 44000 48000 52000 56000 60000 64000 68000 72000 76000 80000 84000 88000 92000 96000 100000 104000 108000 112000 116000 120000 124000 128000 132000 136000 140000 144000 148000 152000 156000 160000 164000 168000 172000 176000 180000 184000 188000 192000 196000 200000 204000 208000 212000 216000 220000 224000 228000 232000 236000 240000 244000 248000 252000 256000 260000 264000 268000 272000 276000 280000 284000 288000 292000 296000 300000 304000 30800..."
},
{
"input": "10000000000 5000",
"output": "640 1280 1920 2560 3200 3840 4480 5120 5760 6400 7040 7680 8320 8960 9600 10240 10880 11520 12160 12800 13440 14080 14720 15360 16000 16640 17280 17920 18560 19200 19840 20480 21120 21760 22400 23040 23680 24320 24960 25600 26240 26880 27520 28160 28800 29440 30080 30720 31360 32000 32640 33280 33920 34560 35200 35840 36480 37120 37760 38400 39040 39680 40320 40960 41600 42240 42880 43520 44160 44800 45440 46080 46720 47360 48000 48640 49280 49920 50560 51200 51840 52480 53120 53760 54400 55040 55680 56320..."
},
{
"input": "10000000000 100000",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "10000000000 100000000",
"output": "-1"
},
{
"input": "10000000000 10000000000",
"output": "-1"
},
{
"input": "10000000000 100001",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "1 4000000000",
"output": "-1"
},
{
"input": "4294967296 4294967296",
"output": "-1"
},
{
"input": "71227122 9603838834",
"output": "-1"
},
{
"input": "10000000000 9603838835",
"output": "-1"
},
{
"input": "5 5999999999",
"output": "-1"
},
{
"input": "2 9324327498",
"output": "-1"
},
{
"input": "9 2",
"output": "3 6"
},
{
"input": "10000000000 4294967296",
"output": "-1"
},
{
"input": "1 3500000000",
"output": "-1"
},
{
"input": "10000000000 4000000000",
"output": "-1"
},
{
"input": "2000 9324327498",
"output": "-1"
},
{
"input": "10000000000 8589934592",
"output": "-1"
},
{
"input": "5000150001 100001",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "10000000000 3037000500",
"output": "-1"
},
{
"input": "9400000000 9324327498",
"output": "-1"
},
{
"input": "10000000000 3307000500",
"output": "-1"
},
{
"input": "2 4000000000",
"output": "-1"
},
{
"input": "1000 4294967295",
"output": "-1"
},
{
"input": "36 3",
"output": "6 12 18"
},
{
"input": "2147483648 4294967296",
"output": "-1"
},
{
"input": "999 4294967295",
"output": "-1"
},
{
"input": "10000000000 130000",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "10000000000 140000",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "10000000000 6074001000",
"output": "-1"
},
{
"input": "12344321 1",
"output": "12344321"
},
{
"input": "2 2",
"output": "-1"
},
{
"input": "28 7",
"output": "1 2 3 4 5 6 7"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "1 3",
"output": "-1"
},
{
"input": "1 4",
"output": "-1"
},
{
"input": "1 5",
"output": "-1"
},
{
"input": "1 6",
"output": "-1"
},
{
"input": "1 7",
"output": "-1"
},
{
"input": "1 8",
"output": "-1"
},
{
"input": "1 9",
"output": "-1"
},
{
"input": "1 10",
"output": "-1"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "2 2",
"output": "-1"
},
{
"input": "2 3",
"output": "-1"
},
{
"input": "2 4",
"output": "-1"
},
{
"input": "2 5",
"output": "-1"
},
{
"input": "2 6",
"output": "-1"
},
{
"input": "2 7",
"output": "-1"
},
{
"input": "2 8",
"output": "-1"
},
{
"input": "2 9",
"output": "-1"
},
{
"input": "2 10",
"output": "-1"
},
{
"input": "3 1",
"output": "3"
},
{
"input": "3 2",
"output": "1 2"
},
{
"input": "3 3",
"output": "-1"
},
{
"input": "3 4",
"output": "-1"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "3 6",
"output": "-1"
},
{
"input": "3 7",
"output": "-1"
},
{
"input": "3 8",
"output": "-1"
},
{
"input": "3 9",
"output": "-1"
},
{
"input": "3 10",
"output": "-1"
},
{
"input": "4 1",
"output": "4"
},
{
"input": "4 2",
"output": "1 3"
},
{
"input": "4 3",
"output": "-1"
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "4 5",
"output": "-1"
},
{
"input": "4 6",
"output": "-1"
},
{
"input": "4 7",
"output": "-1"
},
{
"input": "4 8",
"output": "-1"
},
{
"input": "4 9",
"output": "-1"
},
{
"input": "4 10",
"output": "-1"
},
{
"input": "5 1",
"output": "5"
},
{
"input": "5 2",
"output": "1 4"
},
{
"input": "5 3",
"output": "-1"
},
{
"input": "5 4",
"output": "-1"
},
{
"input": "5 5",
"output": "-1"
},
{
"input": "5 6",
"output": "-1"
},
{
"input": "5 7",
"output": "-1"
},
{
"input": "5 8",
"output": "-1"
},
{
"input": "5 9",
"output": "-1"
},
{
"input": "5 10",
"output": "-1"
},
{
"input": "6 1",
"output": "6"
},
{
"input": "6 2",
"output": "2 4"
},
{
"input": "6 3",
"output": "1 2 3"
},
{
"input": "6 4",
"output": "-1"
},
{
"input": "6 5",
"output": "-1"
},
{
"input": "6 6",
"output": "-1"
},
{
"input": "6 7",
"output": "-1"
},
{
"input": "6 8",
"output": "-1"
},
{
"input": "6 9",
"output": "-1"
},
{
"input": "6 10",
"output": "-1"
},
{
"input": "7 1",
"output": "7"
},
{
"input": "7 2",
"output": "1 6"
},
{
"input": "7 3",
"output": "1 2 4"
},
{
"input": "7 4",
"output": "-1"
},
{
"input": "7 5",
"output": "-1"
},
{
"input": "7 6",
"output": "-1"
},
{
"input": "7 7",
"output": "-1"
},
{
"input": "7 8",
"output": "-1"
},
{
"input": "7 9",
"output": "-1"
},
{
"input": "7 10",
"output": "-1"
},
{
"input": "8 1",
"output": "8"
},
{
"input": "8 2",
"output": "2 6"
},
{
"input": "8 3",
"output": "1 2 5"
},
{
"input": "8 4",
"output": "-1"
},
{
"input": "8 5",
"output": "-1"
},
{
"input": "8 6",
"output": "-1"
},
{
"input": "8 7",
"output": "-1"
},
{
"input": "8 8",
"output": "-1"
},
{
"input": "8 9",
"output": "-1"
},
{
"input": "8 10",
"output": "-1"
}
] | 1,498,840,442 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 20 | 1,000 | 5,529,600 | n,k=list(map(int,input().strip().split(' ')))
L=n//k
if k*(k+1)//2>n:
print(-1)
else:
temp=(k*(k+1))//2
L=n//temp
while(n%L!=0):
L-=1
T=(n-((temp)*L))//L
ans=[]
for i in range(1,k+1):
if i!=k:
ans+=[i*L]
else:
ans+=[(i+T)*L]
print(*ans)
| Title: Maximal GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given positive integer number *n*. You should create such strictly increasing sequence of *k* positive numbers *a*1,<=*a*2,<=...,<=*a**k*, that their sum is equal to *n* and greatest common divisor is maximal.
Greatest common divisor of sequence is maximum of such numbers that every element of sequence is divisible by them.
If there is no possible sequence then output -1.
Input Specification:
The first line consists of two numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1010).
Output Specification:
If the answer exists then output *k* numbers — resulting sequence. Otherwise output -1. If there are multiple answers, print any of them.
Demo Input:
['6 3\n', '8 2\n', '5 3\n']
Demo Output:
['1 2 3\n', '2 6\n', '-1\n']
Note:
none | ```python
n,k=list(map(int,input().strip().split(' ')))
L=n//k
if k*(k+1)//2>n:
print(-1)
else:
temp=(k*(k+1))//2
L=n//temp
while(n%L!=0):
L-=1
T=(n-((temp)*L))//L
ans=[]
for i in range(1,k+1):
if i!=k:
ans+=[i*L]
else:
ans+=[(i+T)*L]
print(*ans)
``` | 0 | |
287 | A | IQ Test | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color. | Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black. | Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise. | [
"####\n.#..\n####\n....\n",
"####\n....\n####\n....\n"
] | [
"YES\n",
"NO\n"
] | In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column. | 500 | [
{
"input": "###.\n...#\n###.\n...#",
"output": "NO"
},
{
"input": ".##.\n#..#\n.##.\n#..#",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n.#.#\n#.#.",
"output": "NO"
},
{
"input": "##..\n..##\n##..\n..##",
"output": "NO"
},
{
"input": "#.#.\n#.#.\n.#.#\n.#.#",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n#.#.\n#.#.",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n#.#.\n.#.#",
"output": "NO"
},
{
"input": "#.#.\n#.#.\n#.#.\n#.#.",
"output": "NO"
},
{
"input": ".#.#\n.#.#\n.#.#\n.#.#",
"output": "NO"
},
{
"input": "#..#\n.##.\n####\n####",
"output": "YES"
},
{
"input": "#.#.\n.###\n#.#.\n.###",
"output": "YES"
},
{
"input": "#..#\n.##.\n.##.\n#..#",
"output": "YES"
},
{
"input": ".##.\n.#..\n##.#\n#..#",
"output": "YES"
},
{
"input": ".##.\n##..\n#..#\n..##",
"output": "YES"
},
{
"input": "##..\n##..\n..##\n..##",
"output": "YES"
},
{
"input": ".#..\n###.\n.#.#\n..#.",
"output": "YES"
},
{
"input": "###.\n###.\n...#\n...#",
"output": "YES"
},
{
"input": "#.##\n##.#\n#.##\n##.#",
"output": "YES"
},
{
"input": ".#.#\n#.#.\n.#.#\n#.##",
"output": "YES"
},
{
"input": "##..\n..##\n##..\n...#",
"output": "YES"
},
{
"input": ".#..\n..##\n##..\n..##",
"output": "YES"
},
{
"input": "##..\n...#\n##..\n...#",
"output": "YES"
},
{
"input": ".#..\n..#.\n.#..\n..#.",
"output": "YES"
},
{
"input": "....\n....\n....\n.#.#",
"output": "YES"
},
{
"input": "....\n....\n....\n...#",
"output": "YES"
}
] | 1,587,036,627 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 16 | 155 | 0 | l=[]
for i in range(4):
l.append(input())
b=1
for i in range(2):
k=0
for j in range(2):
if l[i][j]==l[i+1][j]:
k+=1
if l[i][j]==l[i][j+1]:
k+=1
if l[i][j]==l[i+1][j+1]:
k+=1
if k>2:
print("YES")
b=0
break
if b==1:
print("NO")
| Title: IQ Test
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color.
Input Specification:
Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black.
Output Specification:
Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise.
Demo Input:
['####\n.#..\n####\n....\n', '####\n....\n####\n....\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column. | ```python
l=[]
for i in range(4):
l.append(input())
b=1
for i in range(2):
k=0
for j in range(2):
if l[i][j]==l[i+1][j]:
k+=1
if l[i][j]==l[i][j+1]:
k+=1
if l[i][j]==l[i+1][j+1]:
k+=1
if k>2:
print("YES")
b=0
break
if b==1:
print("NO")
``` | 0 | |
900 | B | Position in Fraction | PROGRAMMING | 1,300 | [
"math",
"number theory"
] | null | null | You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point. | The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9). | Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1. | [
"1 2 0\n",
"2 3 7\n"
] | [
"2",
"-1"
] | The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction. | 1,000 | [
{
"input": "1 2 0",
"output": "2"
},
{
"input": "2 3 7",
"output": "-1"
},
{
"input": "1 100000 1",
"output": "5"
},
{
"input": "1 7 7",
"output": "6"
},
{
"input": "99999 100000 8",
"output": "-1"
},
{
"input": "44102 73848 2",
"output": "132"
},
{
"input": "7 31 3",
"output": "15"
},
{
"input": "8880 81608 9",
"output": "161"
},
{
"input": "4942 62768 5",
"output": "122"
},
{
"input": "69168 84860 4",
"output": "107"
},
{
"input": "971 1883 3",
"output": "130"
},
{
"input": "1636 3269 6",
"output": "150"
},
{
"input": "6873 7769 3",
"output": "163"
},
{
"input": "13805 15538 3",
"output": "164"
},
{
"input": "10958 21926 3",
"output": "117"
},
{
"input": "8 51 0",
"output": "10"
},
{
"input": "1 10 1",
"output": "1"
},
{
"input": "1 9 0",
"output": "-1"
},
{
"input": "53 101 6",
"output": "-1"
},
{
"input": "1 10001 9",
"output": "5"
},
{
"input": "25102 31579 2",
"output": "174"
},
{
"input": "38790 39359 0",
"output": "212"
},
{
"input": "47117 78718 0",
"output": "213"
},
{
"input": "1 57 0",
"output": "1"
},
{
"input": "1 3 0",
"output": "-1"
},
{
"input": "1 100 0",
"output": "1"
},
{
"input": "2 3 0",
"output": "-1"
},
{
"input": "99971 99989 0",
"output": "9"
},
{
"input": "567 1580 0",
"output": "6"
},
{
"input": "45 97 0",
"output": "39"
},
{
"input": "35 111 4",
"output": "-1"
},
{
"input": "1 2 5",
"output": "1"
},
{
"input": "1 7 0",
"output": "-1"
}
] | 1,634,453,086 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 46 | 0 | a,b,c=map(int,input().split())
t="{:.15f}".format(a/b)
print(t[t.find("."):13].find(str(c)))
| Title: Position in Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point.
Input Specification:
The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9).
Output Specification:
Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Demo Input:
['1 2 0\n', '2 3 7\n']
Demo Output:
['2', '-1']
Note:
The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction. | ```python
a,b,c=map(int,input().split())
t="{:.15f}".format(a/b)
print(t[t.find("."):13].find(str(c)))
``` | 0 | |
2 | A | Winner | PROGRAMMING | 1,500 | [
"hashing",
"implementation"
] | A. Winner | 1 | 64 | The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points. | The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive. | Print the name of the winner. | [
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] | [
"andrew\n",
"andrew\n"
] | none | 0 | [
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},
{
"input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734",
"output": "ksjuuerbnlklcfdjeyq"
},
{
"input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879",
"output": "fcgslzkicjrpbqaifgweyzreajjfdo"
},
{
"input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735",
"output": "aawtvezfntstrcpgbzjbf"
},
{
"input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158",
"output": "ivhgbxiv"
}
] | 1,673,421,634 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | # LUOGU_RID: 99361177
def input_mz_fs():
num = 0
mz,fs=input().split()
if abs(int(fs))<=1000 and len(mz)<=32:
li_mz.append(mz)
li_fs.append(int(fs))
for i in li_fs:
if mz==i:
li_mz.insert(0,mz)
li_mz.remove(i)
if li_fs(li_fs.index(i))>0:
# num += 1
li_fs.insert(0,fs+ li_fs(li_fs.index(i)) )#分数加减>0
else:
li_fs.insert(0, 0)#分数加减<=0
# num -= 1
# return num
def output_result():
max_fs=0
for i in li_fs:
if i>max_fs:
max_fs=i
max_fs_index=li_fs.index(max_fs)#最大值元素索引
print(f"{li_mz[max_fs_index]}")
li_mz = []
li_fs = []
def main():
n=int(input())
while n>0:
n -= 1
input_mz_fs()
num=0
for i in li_fs:
if i>0:
num += 1
else:
num -= 1
if num>0:
output_result()
if __name__ == '__main__':
main() | Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none | ```python
# LUOGU_RID: 99361177
def input_mz_fs():
num = 0
mz,fs=input().split()
if abs(int(fs))<=1000 and len(mz)<=32:
li_mz.append(mz)
li_fs.append(int(fs))
for i in li_fs:
if mz==i:
li_mz.insert(0,mz)
li_mz.remove(i)
if li_fs(li_fs.index(i))>0:
# num += 1
li_fs.insert(0,fs+ li_fs(li_fs.index(i)) )#分数加减>0
else:
li_fs.insert(0, 0)#分数加减<=0
# num -= 1
# return num
def output_result():
max_fs=0
for i in li_fs:
if i>max_fs:
max_fs=i
max_fs_index=li_fs.index(max_fs)#最大值元素索引
print(f"{li_mz[max_fs_index]}")
li_mz = []
li_fs = []
def main():
n=int(input())
while n>0:
n -= 1
input_mz_fs()
num=0
for i in li_fs:
if i>0:
num += 1
else:
num -= 1
if num>0:
output_result()
if __name__ == '__main__':
main()
``` | 0 |
825 | C | Multi-judge Solving | PROGRAMMING | 1,600 | [
"greedy",
"implementation"
] | null | null | Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty *d* on Decoforces is as hard as the problem with difficulty *d* on any other judge).
Makes has chosen *n* problems to solve on Decoforces with difficulties *a*1,<=*a*2,<=...,<=*a**n*. He can solve these problems in arbitrary order. Though he can solve problem *i* with difficulty *a**i* only if he had already solved some problem with difficulty (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty *k*.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer *y* there exist some problem with difficulty *y* on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces. | The first line contains two integer numbers *n*, *k* (1<=≤<=*n*<=≤<=103, 1<=≤<=*k*<=≤<=109).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). | Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces. | [
"3 3\n2 1 9\n",
"4 20\n10 3 6 3\n"
] | [
"1\n",
"0\n"
] | In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start. | 0 | [
{
"input": "3 3\n2 1 9",
"output": "1"
},
{
"input": "4 20\n10 3 6 3",
"output": "0"
},
{
"input": "1 1000000000\n1",
"output": "0"
},
{
"input": "1 1\n3",
"output": "1"
},
{
"input": "50 100\n74 55 33 5 83 24 75 59 30 36 13 4 62 28 96 17 6 35 45 53 33 11 37 93 34 79 61 72 13 31 44 75 7 3 63 46 18 16 44 89 62 25 32 12 38 55 75 56 61 82",
"output": "0"
},
{
"input": "100 10\n246 286 693 607 87 612 909 312 621 37 801 558 504 914 416 762 187 974 976 123 635 488 416 659 988 998 93 662 92 749 889 78 214 786 735 625 921 372 713 617 975 119 402 411 878 138 548 905 802 762 940 336 529 373 745 835 805 880 816 94 166 114 475 699 974 462 61 337 555 805 968 815 392 746 591 558 740 380 668 29 881 151 387 986 174 923 541 520 998 947 535 651 103 584 664 854 180 852 726 93",
"output": "1"
},
{
"input": "2 1\n1 1000000000",
"output": "29"
},
{
"input": "29 2\n1 3 7 15 31 63 127 255 511 1023 2047 4095 8191 16383 32767 65535 131071 262143 524287 1048575 2097151 4194303 8388607 16777215 33554431 67108863 134217727 268435455 536870911",
"output": "27"
},
{
"input": "1 1\n1000000000",
"output": "29"
},
{
"input": "7 6\n4 20 16 14 3 17 4",
"output": "1"
},
{
"input": "2 1\n3 6",
"output": "1"
},
{
"input": "1 1\n20",
"output": "4"
},
{
"input": "5 2\n86 81 53 25 18",
"output": "4"
},
{
"input": "4 1\n88 55 14 39",
"output": "4"
},
{
"input": "3 1\n2 3 6",
"output": "0"
},
{
"input": "3 2\n4 9 18",
"output": "1"
},
{
"input": "5 3\n6 6 6 13 27",
"output": "2"
},
{
"input": "5 1\n23 8 83 26 18",
"output": "4"
},
{
"input": "3 1\n4 5 6",
"output": "1"
},
{
"input": "3 1\n1 3 6",
"output": "1"
},
{
"input": "1 1\n2",
"output": "0"
},
{
"input": "3 2\n4 5 6",
"output": "0"
},
{
"input": "5 1\n100 200 400 1000 2000",
"output": "7"
},
{
"input": "2 1\n1 4",
"output": "1"
},
{
"input": "4 1\n2 4 8 32",
"output": "1"
},
{
"input": "2 10\n21 42",
"output": "1"
},
{
"input": "3 3\n1 7 13",
"output": "1"
},
{
"input": "3 1\n1 4 6",
"output": "1"
},
{
"input": "2 2\n2 8",
"output": "1"
},
{
"input": "1 1\n4",
"output": "1"
},
{
"input": "2 2\n8 16",
"output": "1"
},
{
"input": "3 1\n4 8 16",
"output": "1"
},
{
"input": "3 1\n3 6 9",
"output": "1"
},
{
"input": "2 1\n4 8",
"output": "1"
},
{
"input": "2 2\n7 14",
"output": "1"
},
{
"input": "1 4\n9",
"output": "1"
},
{
"input": "5 3\n1024 4096 16384 65536 536870913",
"output": "24"
},
{
"input": "2 5\n10 11",
"output": "0"
},
{
"input": "2 2\n3 6",
"output": "0"
},
{
"input": "2 2\n8 11",
"output": "1"
},
{
"input": "3 19905705\n263637263 417905394 108361057",
"output": "3"
},
{
"input": "4 25\n100 11 1 13",
"output": "1"
},
{
"input": "10 295206008\n67980321 440051990 883040288 135744260 96431758 242465794 576630162 972797687 356406646 547451696",
"output": "0"
},
{
"input": "4 2\n45 44 35 38",
"output": "4"
},
{
"input": "1 2\n9",
"output": "2"
},
{
"input": "3 6\n13 26 52",
"output": "1"
},
{
"input": "9 30111088\n824713578 11195876 458715185 731769293 680826358 189542586 550198537 860586039 101083021",
"output": "2"
},
{
"input": "3 72014068\n430005292 807436976 828082746",
"output": "2"
},
{
"input": "3 165219745\n737649884 652879952 506420386",
"output": "1"
},
{
"input": "2 60669400\n95037700 337255240",
"output": "1"
},
{
"input": "4 28\n34 1 86 90",
"output": "1"
},
{
"input": "2 1\n5 10",
"output": "2"
},
{
"input": "2 1\n4 1000000000",
"output": "28"
},
{
"input": "2 1\n2 3",
"output": "0"
},
{
"input": "2 1\n3 5",
"output": "1"
},
{
"input": "3 3\n1 5 20",
"output": "1"
},
{
"input": "9 1\n1 2 4 9 15 32 64 128 1024",
"output": "4"
}
] | 1,634,831,430 | 2,130 | PyPy 3 | WRONG_ANSWER | TESTS | 6 | 77 | 19,763,200 | n,k = map(int, input().split())
A = list(map(int, input().split()))
q1 = []
q2 = []
for a in A:
if 2*k >= a:
q1.append(-a)
else:
q2.append(-a)
import heapq
heapq.heapify(q1)
heapq.heapify(q2)
cur = k
ans = 0
for i in range(n):
if q1:
a = -heapq.heappop(q1)
cur = max(cur, a)
while q2 and 2*cur >= -q2[0]:
a = -heapq.heappop(q2)
heapq.heappush(q1, -a)
else:
a = -heapq.heappop(q2)
if 2*cur >=a:
continue
else:
ans += 1
cur = (a+1)//2
while q2 and 2*cur >= -q2[0]:
a = -heapq.heappop(q2)
heapq.heappush(q1, -a)
print(ans)
| Title: Multi-judge Solving
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty *d* on Decoforces is as hard as the problem with difficulty *d* on any other judge).
Makes has chosen *n* problems to solve on Decoforces with difficulties *a*1,<=*a*2,<=...,<=*a**n*. He can solve these problems in arbitrary order. Though he can solve problem *i* with difficulty *a**i* only if he had already solved some problem with difficulty (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty *k*.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer *y* there exist some problem with difficulty *y* on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input Specification:
The first line contains two integer numbers *n*, *k* (1<=≤<=*n*<=≤<=103, 1<=≤<=*k*<=≤<=109).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Demo Input:
['3 3\n2 1 9\n', '4 20\n10 3 6 3\n']
Demo Output:
['1\n', '0\n']
Note:
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start. | ```python
n,k = map(int, input().split())
A = list(map(int, input().split()))
q1 = []
q2 = []
for a in A:
if 2*k >= a:
q1.append(-a)
else:
q2.append(-a)
import heapq
heapq.heapify(q1)
heapq.heapify(q2)
cur = k
ans = 0
for i in range(n):
if q1:
a = -heapq.heappop(q1)
cur = max(cur, a)
while q2 and 2*cur >= -q2[0]:
a = -heapq.heappop(q2)
heapq.heappush(q1, -a)
else:
a = -heapq.heappop(q2)
if 2*cur >=a:
continue
else:
ans += 1
cur = (a+1)//2
while q2 and 2*cur >= -q2[0]:
a = -heapq.heappop(q2)
heapq.heappush(q1, -a)
print(ans)
``` | 0 | |
950 | B | Intercepted Message | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of *k* files of sizes *l*1,<=*l*2,<=...,<=*l**k* bytes, then the *i*-th file is split to one or more blocks *b**i*,<=1,<=*b**i*,<=2,<=...,<=*b**i*,<=*m**i* (here the total length of the blocks *b**i*,<=1<=+<=*b**i*,<=2<=+<=...<=+<=*b**i*,<=*m**i* is equal to the length of the file *l**i*), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct. | The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of blocks in the first and in the second messages.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=106) — the length of the blocks that form the first message.
The third line contains *m* integers *y*1,<=*y*2,<=...,<=*y**m* (1<=≤<=*y**i*<=≤<=106) — the length of the blocks that form the second message.
It is guaranteed that *x*1<=+<=...<=+<=*x**n*<==<=*y*1<=+<=...<=+<=*y**m*. Also, it is guaranteed that *x*1<=+<=...<=+<=*x**n*<=≤<=106. | Print the maximum number of files the intercepted array could consist of. | [
"7 6\n2 5 3 1 11 4 4\n7 8 2 4 1 8\n",
"3 3\n1 10 100\n1 100 10\n",
"1 4\n4\n1 1 1 1\n"
] | [
"3\n",
"2\n",
"1\n"
] | In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4. | 1,000 | [
{
"input": "7 6\n2 5 3 1 11 4 4\n7 8 2 4 1 8",
"output": "3"
},
{
"input": "3 3\n1 10 100\n1 100 10",
"output": "2"
},
{
"input": "1 4\n4\n1 1 1 1",
"output": "1"
},
{
"input": "1 1\n1000000\n1000000",
"output": "1"
},
{
"input": "3 5\n2 2 9\n2 1 4 2 4",
"output": "2"
},
{
"input": "5 3\n1 1 4 1 2\n1 4 4",
"output": "2"
},
{
"input": "30 50\n3 3 1 3 1 2 4 3 4 1 3 2 3 3 2 3 2 1 3 4 2 1 1 3 2 2 1 3 1 60\n4 4 1 2 2 2 3 1 3 2 1 2 4 4 2 1 2 3 1 3 4 4 3 3 4 4 4 1 2 1 3 3 1 1 3 3 4 3 2 3 2 4 1 4 2 3 2 2 3 1",
"output": "12"
},
{
"input": "50 50\n5733 740 547 3647 5382 5109 6842 7102 5879 1502 3574 1628 7905 4357 8569 9564 8268 3542 2487 8532 425 7713 2585 925 6458 2697 2844 69 324 9030 495 4428 6724 3524 3304 4874 1303 2098 1136 1048 2464 7316 274 9586 534 2450 2368 8060 7795 70692\n1918 4122 6806 4914 6517 6278 9842 9480 6609 4221 9373 1728 9508 9778 8578 5589 2673 6618 6031 9016 4017 6671 6008 2268 5154 9614 6834 9512 9618 6424 1736 1464 6520 9812 1722 9197 2412 2699 73 968 2906 2715 6573 8675 548 7061 5455 88 5565 2544",
"output": "1"
},
{
"input": "1 2\n2\n1 1",
"output": "1"
},
{
"input": "1 2\n1000000\n999999 1",
"output": "1"
},
{
"input": "2 2\n1 1\n1 1",
"output": "2"
},
{
"input": "2 2\n500000 500000\n1 999999",
"output": "1"
},
{
"input": "2 2\n2 3\n4 1",
"output": "1"
},
{
"input": "2 2\n2 3\n3 2",
"output": "1"
},
{
"input": "2 2\n2 3\n2 3",
"output": "2"
},
{
"input": "2 3\n2 2\n1 1 2",
"output": "2"
},
{
"input": "1 1\n1\n1",
"output": "1"
},
{
"input": "2 3\n3 2\n2 1 2",
"output": "2"
},
{
"input": "2 3\n2 3\n2 1 2",
"output": "2"
},
{
"input": "50 30\n2 3 1 2 2 4 3 4 3 2 1 4 2 3 1 3 1 2 2 3 1 1 1 2 3 1 4 3 1 2 1 2 2 1 2 4 4 3 3 2 2 1 1 1 2 2 2 4 3 3\n3 3 3 4 1 4 1 4 4 1 3 4 3 1 2 4 2 1 4 2 3 1 1 2 2 1 2 4 1 41",
"output": "12"
},
{
"input": "50 50\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\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",
"output": "50"
},
{
"input": "31 31\n5745 258 5486 13779 20931 407 1478 49032 30787 4957 36603 1034 5011 22319 50560 34419 22036 18235 62551 89259 36093 126169 106027 1673 52983 50127 640 30714 54574 20129 45984\n5745 258 5486 13779 20931 407 1478 49032 30787 4957 36603 1034 5011 22319 50560 34419 22036 18235 62551 89259 36093 126169 106027 1673 52983 50127 640 30714 54574 20129 45984",
"output": "31"
},
{
"input": "3 6\n8 4 1\n1 8 1 1 1 1",
"output": "2"
}
] | 1,521,119,885 | 2,147,483,647 | Python 3 | OK | TESTS | 59 | 187 | 9,113,600 | n, m = (int(x) for x in input().split())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
s1=0
s2=0
j=0
cnt=0
for i in range(0,n):
s1+=a[i]
while(s1>s2):
s2+=b[j]
j+=1
if (s1==s2):
cnt+=1
s1=0
s2=0
print(cnt)
| Title: Intercepted Message
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of *k* files of sizes *l*1,<=*l*2,<=...,<=*l**k* bytes, then the *i*-th file is split to one or more blocks *b**i*,<=1,<=*b**i*,<=2,<=...,<=*b**i*,<=*m**i* (here the total length of the blocks *b**i*,<=1<=+<=*b**i*,<=2<=+<=...<=+<=*b**i*,<=*m**i* is equal to the length of the file *l**i*), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of blocks in the first and in the second messages.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=106) — the length of the blocks that form the first message.
The third line contains *m* integers *y*1,<=*y*2,<=...,<=*y**m* (1<=≤<=*y**i*<=≤<=106) — the length of the blocks that form the second message.
It is guaranteed that *x*1<=+<=...<=+<=*x**n*<==<=*y*1<=+<=...<=+<=*y**m*. Also, it is guaranteed that *x*1<=+<=...<=+<=*x**n*<=≤<=106.
Output Specification:
Print the maximum number of files the intercepted array could consist of.
Demo Input:
['7 6\n2 5 3 1 11 4 4\n7 8 2 4 1 8\n', '3 3\n1 10 100\n1 100 10\n', '1 4\n4\n1 1 1 1\n']
Demo Output:
['3\n', '2\n', '1\n']
Note:
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4. | ```python
n, m = (int(x) for x in input().split())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
s1=0
s2=0
j=0
cnt=0
for i in range(0,n):
s1+=a[i]
while(s1>s2):
s2+=b[j]
j+=1
if (s1==s2):
cnt+=1
s1=0
s2=0
print(cnt)
``` | 3 | |
760 | B | Frodo and pillows | PROGRAMMING | 1,500 | [
"binary search",
"greedy"
] | null | null | *n* hobbits are planning to spend the night at Frodo's house. Frodo has *n* beds standing in a row and *m* pillows (*n*<=≤<=*m*). Each hobbit needs a bed and at least one pillow to sleep, however, everyone wants as many pillows as possible. Of course, it's not always possible to share pillows equally, but any hobbit gets hurt if he has at least two pillows less than some of his neighbors have.
Frodo will sleep on the *k*-th bed in the row. What is the maximum number of pillows he can have so that every hobbit has at least one pillow, every pillow is given to some hobbit and no one is hurt? | The only line contain three integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=*m*<=≤<=109, 1<=≤<=*k*<=≤<=*n*) — the number of hobbits, the number of pillows and the number of Frodo's bed. | Print single integer — the maximum number of pillows Frodo can have so that no one is hurt. | [
"4 6 2\n",
"3 10 3\n",
"3 6 1\n"
] | [
"2\n",
"4\n",
"3\n"
] | In the first example Frodo can have at most two pillows. In this case, he can give two pillows to the hobbit on the first bed, and one pillow to each of the hobbits on the third and the fourth beds.
In the second example Frodo can take at most four pillows, giving three pillows to each of the others.
In the third example Frodo can take three pillows, giving two pillows to the hobbit in the middle and one pillow to the hobbit on the third bed. | 1,000 | [
{
"input": "4 6 2",
"output": "2"
},
{
"input": "3 10 3",
"output": "4"
},
{
"input": "3 6 1",
"output": "3"
},
{
"input": "3 3 3",
"output": "1"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "1 1000000000 1",
"output": "1000000000"
},
{
"input": "100 1000000000 20",
"output": "10000034"
},
{
"input": "1000 1000 994",
"output": "1"
},
{
"input": "100000000 200000000 54345",
"output": "10001"
},
{
"input": "1000000000 1000000000 1",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 500000000",
"output": "1"
},
{
"input": "1000 1000 3",
"output": "1"
},
{
"input": "100000000 200020000 54345",
"output": "10001"
},
{
"input": "100 108037 18",
"output": "1115"
},
{
"input": "100000000 200020001 54345",
"output": "10002"
},
{
"input": "200 6585 2",
"output": "112"
},
{
"input": "30000 30593 5980",
"output": "25"
},
{
"input": "40000 42107 10555",
"output": "46"
},
{
"input": "50003 50921 192",
"output": "31"
},
{
"input": "100000 113611 24910",
"output": "117"
},
{
"input": "1000000 483447163 83104",
"output": "21965"
},
{
"input": "10000000 10021505 600076",
"output": "147"
},
{
"input": "100000000 102144805 2091145",
"output": "1465"
},
{
"input": "1000000000 1000000000 481982093",
"output": "1"
},
{
"input": "100 999973325 5",
"output": "9999778"
},
{
"input": "200 999999109 61",
"output": "5000053"
},
{
"input": "30000 999999384 5488",
"output": "43849"
},
{
"input": "40000 999997662 8976",
"output": "38038"
},
{
"input": "50003 999999649 405",
"output": "44320"
},
{
"input": "100000 999899822 30885",
"output": "31624"
},
{
"input": "1000000 914032367 528790",
"output": "30217"
},
{
"input": "10000000 999617465 673112",
"output": "31459"
},
{
"input": "100000000 993180275 362942",
"output": "29887"
},
{
"input": "1000000000 1000000000 331431458",
"output": "1"
},
{
"input": "100 10466 89",
"output": "144"
},
{
"input": "200 5701 172",
"output": "84"
},
{
"input": "30000 36932 29126",
"output": "84"
},
{
"input": "40000 40771 22564",
"output": "28"
},
{
"input": "50003 51705 49898",
"output": "42"
},
{
"input": "100000 149408 74707",
"output": "223"
},
{
"input": "1000000 194818222 998601",
"output": "18389"
},
{
"input": "10000000 10748901 8882081",
"output": "866"
},
{
"input": "100000000 106296029 98572386",
"output": "2510"
},
{
"input": "1000000000 1000000000 193988157",
"output": "1"
},
{
"input": "100 999981057 92",
"output": "9999852"
},
{
"input": "200 999989691 199",
"output": "5000046"
},
{
"input": "30000 999995411 24509",
"output": "43846"
},
{
"input": "40000 999998466 30827",
"output": "37930"
},
{
"input": "50003 999997857 48387",
"output": "43163"
},
{
"input": "100000 999731886 98615",
"output": "43371"
},
{
"input": "1000000 523220797 654341",
"output": "22853"
},
{
"input": "10000000 999922591 8157724",
"output": "31464"
},
{
"input": "100000000 999834114 93836827",
"output": "29998"
},
{
"input": "1000000000 1000000000 912549504",
"output": "1"
},
{
"input": "1000 97654978 234",
"output": "97976"
},
{
"input": "1000 97654977 234",
"output": "97975"
},
{
"input": "1000234 97653889 1",
"output": "13903"
},
{
"input": "1000234 97653890 1",
"output": "13904"
},
{
"input": "3450234 97656670 3000000",
"output": "9707"
},
{
"input": "3450234 97656669 3000000",
"output": "9706"
},
{
"input": "3 1000000000 2",
"output": "333333334"
},
{
"input": "2 1000000000 1",
"output": "500000000"
},
{
"input": "2 1000000000 2",
"output": "500000000"
},
{
"input": "3 1000000000 1",
"output": "333333334"
},
{
"input": "3 1000000000 3",
"output": "333333334"
},
{
"input": "2 999999999 1",
"output": "500000000"
},
{
"input": "2 999999999 2",
"output": "500000000"
},
{
"input": "1 999999999 1",
"output": "999999999"
}
] | 1,485,281,847 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 16 | 62 | 4,608,000 | def codeforces(n, m, k):
m -= n
l = k - 1
r = n - k
l_empty = l * (l + 1) // 2
r_empty = r * (r + 1) // 2
total = (m + l_empty + r_empty) // (l + r + 1)
x = total + 1
if l > x - 1:
tl = (x - 1) * x // 2 + l - x + 1
else:
tl = (x - 1 + x - l) * l // 2
if r > x - 1:
tr = (x - 1) * x // 2 + r - x + 1
else:
tr = (x - 1 + x - r) * r // 2
t = tl + tr + x
if t <= m + n:
return total + 1
while True:
a = m ** 0.5
if int(a) == a:
return int(min(a, total + 1)) + 1
m -= 1
n, m, k = map(int, input().split())
print(codeforces(n, m, k))
| Title: Frodo and pillows
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* hobbits are planning to spend the night at Frodo's house. Frodo has *n* beds standing in a row and *m* pillows (*n*<=≤<=*m*). Each hobbit needs a bed and at least one pillow to sleep, however, everyone wants as many pillows as possible. Of course, it's not always possible to share pillows equally, but any hobbit gets hurt if he has at least two pillows less than some of his neighbors have.
Frodo will sleep on the *k*-th bed in the row. What is the maximum number of pillows he can have so that every hobbit has at least one pillow, every pillow is given to some hobbit and no one is hurt?
Input Specification:
The only line contain three integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=*m*<=≤<=109, 1<=≤<=*k*<=≤<=*n*) — the number of hobbits, the number of pillows and the number of Frodo's bed.
Output Specification:
Print single integer — the maximum number of pillows Frodo can have so that no one is hurt.
Demo Input:
['4 6 2\n', '3 10 3\n', '3 6 1\n']
Demo Output:
['2\n', '4\n', '3\n']
Note:
In the first example Frodo can have at most two pillows. In this case, he can give two pillows to the hobbit on the first bed, and one pillow to each of the hobbits on the third and the fourth beds.
In the second example Frodo can take at most four pillows, giving three pillows to each of the others.
In the third example Frodo can take three pillows, giving two pillows to the hobbit in the middle and one pillow to the hobbit on the third bed. | ```python
def codeforces(n, m, k):
m -= n
l = k - 1
r = n - k
l_empty = l * (l + 1) // 2
r_empty = r * (r + 1) // 2
total = (m + l_empty + r_empty) // (l + r + 1)
x = total + 1
if l > x - 1:
tl = (x - 1) * x // 2 + l - x + 1
else:
tl = (x - 1 + x - l) * l // 2
if r > x - 1:
tr = (x - 1) * x // 2 + r - x + 1
else:
tr = (x - 1 + x - r) * r // 2
t = tl + tr + x
if t <= m + n:
return total + 1
while True:
a = m ** 0.5
if int(a) == a:
return int(min(a, total + 1)) + 1
m -= 1
n, m, k = map(int, input().split())
print(codeforces(n, m, k))
``` | 0 | |
868 | C | Qualification Rounds | PROGRAMMING | 1,500 | [
"bitmasks",
"brute force",
"constructive algorithms",
"dp"
] | null | null | Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of *n* problems, and they want to select any non-empty subset of it as a problemset.
*k* experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.
Determine if Snark and Philip can make an interesting problemset! | The first line contains two integers *n*, *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=4) — the number of problems and the number of experienced teams.
Each of the next *n* lines contains *k* integers, each equal to 0 or 1. The *j*-th number in the *i*-th line is 1 if *j*-th team knows *i*-th problem and 0 otherwise. | Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES"). | [
"5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0\n",
"3 2\n1 0\n1 1\n0 1\n"
] | [
"NO\n",
"YES\n"
] | In the first example you can't make any interesting problemset, because the first team knows all problems.
In the second example you can choose the first and the third problems. | 1,000 | [
{
"input": "5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0",
"output": "NO"
},
{
"input": "3 2\n1 0\n1 1\n0 1",
"output": "YES"
},
{
"input": "10 2\n1 0\n1 0\n0 0\n1 1\n0 0\n1 1\n0 0\n1 1\n0 1\n0 1",
"output": "YES"
},
{
"input": "10 3\n1 0 0\n0 1 1\n1 0 0\n0 1 0\n0 0 1\n1 0 1\n0 1 1\n1 0 0\n1 1 0\n0 0 0",
"output": "YES"
},
{
"input": "10 4\n1 0 1 0\n1 0 0 1\n1 1 0 1\n1 0 1 1\n1 1 0 1\n1 0 1 0\n0 0 0 0\n0 0 1 0\n1 0 1 0\n0 0 1 1",
"output": "YES"
},
{
"input": "2 2\n0 0\n1 0",
"output": "YES"
},
{
"input": "3 3\n1 0 1\n1 0 0\n1 1 1",
"output": "NO"
},
{
"input": "4 4\n0 0 0 0\n1 1 0 0\n1 1 1 1\n1 0 1 1",
"output": "YES"
},
{
"input": "4 1\n1\n1\n0\n0",
"output": "YES"
},
{
"input": "1 4\n0 0 0 0",
"output": "YES"
},
{
"input": "3 3\n0 0 1\n0 1 1\n1 0 0",
"output": "YES"
},
{
"input": "2 3\n0 0 1\n1 0 0",
"output": "YES"
},
{
"input": "1 1\n0",
"output": "YES"
},
{
"input": "2 4\n0 1 1 1\n1 0 0 0",
"output": "YES"
},
{
"input": "2 4\n1 0 1 0\n0 1 0 1",
"output": "YES"
},
{
"input": "2 4\n1 0 0 0\n0 0 0 1",
"output": "YES"
},
{
"input": "2 3\n0 1 0\n0 0 1",
"output": "YES"
},
{
"input": "3 4\n1 0 1 0\n0 1 0 1\n1 1 1 1",
"output": "YES"
},
{
"input": "3 4\n0 0 1 1\n1 1 1 0\n1 1 0 1",
"output": "NO"
},
{
"input": "4 4\n0 0 0 1\n0 0 0 1\n0 0 1 0\n0 0 1 0",
"output": "YES"
},
{
"input": "2 4\n1 1 0 0\n0 0 1 1",
"output": "YES"
},
{
"input": "2 4\n1 0 0 0\n0 1 0 0",
"output": "YES"
},
{
"input": "2 3\n1 0 0\n0 0 1",
"output": "YES"
},
{
"input": "3 4\n1 0 1 0\n0 1 1 1\n1 0 0 0",
"output": "YES"
},
{
"input": "1 2\n0 0",
"output": "YES"
},
{
"input": "6 3\n0 1 1\n1 0 1\n1 1 1\n0 1 0\n1 0 1\n1 1 0",
"output": "YES"
},
{
"input": "1 4\n0 0 1 1",
"output": "NO"
},
{
"input": "3 3\n1 0 0\n0 1 0\n0 0 1",
"output": "YES"
},
{
"input": "3 4\n1 0 0 0\n1 1 0 0\n0 1 1 1",
"output": "YES"
},
{
"input": "3 2\n0 0\n0 0\n0 0",
"output": "YES"
},
{
"input": "2 4\n1 0 0 0\n1 0 1 1",
"output": "NO"
},
{
"input": "2 4\n0 0 0 1\n1 0 0 0",
"output": "YES"
},
{
"input": "2 4\n1 0 0 0\n0 1 1 1",
"output": "YES"
},
{
"input": "4 4\n1 1 1 1\n0 0 0 1\n0 0 1 1\n1 0 1 1",
"output": "NO"
},
{
"input": "6 3\n1 0 0\n1 1 1\n1 1 1\n0 1 0\n0 1 0\n1 0 0",
"output": "YES"
},
{
"input": "4 4\n0 1 0 0\n1 1 1 1\n1 1 1 1\n1 0 1 1",
"output": "YES"
},
{
"input": "1 3\n0 0 0",
"output": "YES"
},
{
"input": "3 3\n1 0 0\n0 1 0\n0 0 0",
"output": "YES"
},
{
"input": "2 4\n0 1 1 0\n0 0 0 0",
"output": "YES"
},
{
"input": "1 4\n0 0 0 1",
"output": "NO"
},
{
"input": "4 4\n0 0 0 1\n0 0 0 1\n0 0 1 1\n1 1 1 0",
"output": "YES"
},
{
"input": "2 3\n1 0 0\n0 1 1",
"output": "YES"
},
{
"input": "3 2\n0 1\n0 1\n1 0",
"output": "YES"
},
{
"input": "4 3\n1 1 0\n1 1 1\n0 0 1\n0 0 1",
"output": "YES"
},
{
"input": "2 1\n0\n0",
"output": "YES"
},
{
"input": "2 4\n1 1 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "5 4\n1 1 1 0\n1 1 0 1\n1 0 1 1\n0 1 1 1\n1 1 0 0",
"output": "NO"
},
{
"input": "3 4\n0 1 1 0\n0 1 0 1\n0 0 1 1",
"output": "NO"
},
{
"input": "1 1\n1",
"output": "NO"
},
{
"input": "3 4\n1 0 0 0\n1 0 0 0\n0 1 1 1",
"output": "YES"
},
{
"input": "2 3\n1 1 0\n0 0 1",
"output": "YES"
},
{
"input": "3 3\n0 0 1\n1 1 1\n1 1 0",
"output": "YES"
},
{
"input": "4 4\n0 1 1 1\n1 0 1 0\n1 1 0 1\n1 0 1 0",
"output": "NO"
},
{
"input": "3 3\n1 0 0\n0 0 0\n1 0 0",
"output": "YES"
},
{
"input": "3 4\n1 1 0 0\n1 1 0 0\n0 0 1 1",
"output": "YES"
},
{
"input": "2 4\n1 0 0 1\n0 0 1 0",
"output": "YES"
},
{
"input": "2 4\n0 0 1 1\n1 1 0 0",
"output": "YES"
},
{
"input": "2 3\n0 0 1\n0 1 0",
"output": "YES"
},
{
"input": "2 3\n1 0 0\n0 1 0",
"output": "YES"
},
{
"input": "3 2\n1 0\n0 1\n0 1",
"output": "YES"
},
{
"input": "3 4\n1 1 0 1\n0 0 1 1\n1 0 1 0",
"output": "NO"
},
{
"input": "3 4\n0 0 1 1\n0 1 1 0\n1 1 0 0",
"output": "YES"
},
{
"input": "3 4\n0 0 0 1\n0 0 0 1\n1 1 1 0",
"output": "YES"
},
{
"input": "3 4\n1 1 1 0\n1 1 0 1\n0 0 1 0",
"output": "YES"
},
{
"input": "8 4\n0 0 0 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n1 1 1 0",
"output": "YES"
},
{
"input": "3 4\n1 0 1 1\n1 1 1 0\n0 1 0 1",
"output": "NO"
},
{
"input": "2 4\n1 1 0 0\n0 0 0 1",
"output": "YES"
},
{
"input": "10 4\n1 0 1 0\n1 0 1 0\n0 1 1 1\n1 0 1 1\n1 1 0 1\n1 0 0 1\n0 1 1 1\n0 0 0 1\n1 1 1 1\n1 0 1 0",
"output": "YES"
},
{
"input": "2 4\n0 1 0 0\n0 0 1 1",
"output": "YES"
},
{
"input": "3 3\n1 1 0\n1 0 1\n0 1 1",
"output": "NO"
},
{
"input": "3 3\n1 1 0\n0 0 1\n1 1 1",
"output": "YES"
},
{
"input": "4 4\n1 1 0 0\n1 0 1 0\n0 1 1 0\n0 0 1 1",
"output": "YES"
},
{
"input": "4 4\n1 0 0 0\n1 0 0 1\n1 0 0 1\n0 1 1 1",
"output": "YES"
},
{
"input": "4 3\n1 0 0\n1 0 0\n1 0 0\n0 1 1",
"output": "YES"
},
{
"input": "2 4\n0 0 1 0\n0 1 0 0",
"output": "YES"
},
{
"input": "1 2\n0 1",
"output": "NO"
},
{
"input": "3 4\n1 1 1 0\n0 0 1 1\n1 1 0 0",
"output": "YES"
},
{
"input": "3 4\n0 0 1 1\n0 1 0 1\n1 0 1 0",
"output": "YES"
},
{
"input": "2 3\n1 0 1\n0 1 0",
"output": "YES"
},
{
"input": "2 4\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "3 4\n0 1 0 1\n0 1 1 0\n1 0 0 1",
"output": "YES"
},
{
"input": "3 4\n0 1 1 0\n1 1 0 0\n0 0 1 1",
"output": "YES"
},
{
"input": "2 4\n1 0 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "2 3\n0 1 0\n1 0 0",
"output": "YES"
},
{
"input": "4 4\n0 0 1 1\n0 1 1 0\n1 1 0 0\n1 0 0 1",
"output": "YES"
},
{
"input": "10 4\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n1 1 0 0",
"output": "YES"
},
{
"input": "3 3\n1 1 0\n0 1 1\n1 0 1",
"output": "NO"
},
{
"input": "2 3\n0 0 1\n1 1 0",
"output": "YES"
},
{
"input": "4 4\n0 0 0 1\n0 0 1 1\n1 1 0 1\n1 1 1 0",
"output": "YES"
},
{
"input": "3 4\n0 0 1 1\n1 0 1 0\n0 1 0 1",
"output": "YES"
},
{
"input": "5 4\n1 1 1 0\n1 1 0 1\n1 0 1 1\n0 1 1 1\n0 0 1 1",
"output": "NO"
},
{
"input": "3 4\n1 0 0 0\n1 1 0 0\n0 1 1 0",
"output": "YES"
}
] | 1,651,665,814 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 2 | 77 | 2,969,600 | n,k = map(int,input().split())
a = set()
dit = {}
for i in range(1,15):
dit[i] = []
for j in range(1,15):
f = True
for p in range(3):
if (i>>p)&(j>>p)!=0:
f = False
break
if f:
dit[i].append(j)
f = False
for i in range(n):
p = int(''.join(input().split()),2)
if not f:
for e in dit[p]:
if e in a:
f = True
break
a.add(p)
if f:
print("YES")
else:
print("NO")
| Title: Qualification Rounds
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of *n* problems, and they want to select any non-empty subset of it as a problemset.
*k* experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.
Determine if Snark and Philip can make an interesting problemset!
Input Specification:
The first line contains two integers *n*, *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=4) — the number of problems and the number of experienced teams.
Each of the next *n* lines contains *k* integers, each equal to 0 or 1. The *j*-th number in the *i*-th line is 1 if *j*-th team knows *i*-th problem and 0 otherwise.
Output Specification:
Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0\n', '3 2\n1 0\n1 1\n0 1\n']
Demo Output:
['NO\n', 'YES\n']
Note:
In the first example you can't make any interesting problemset, because the first team knows all problems.
In the second example you can choose the first and the third problems. | ```python
n,k = map(int,input().split())
a = set()
dit = {}
for i in range(1,15):
dit[i] = []
for j in range(1,15):
f = True
for p in range(3):
if (i>>p)&(j>>p)!=0:
f = False
break
if f:
dit[i].append(j)
f = False
for i in range(n):
p = int(''.join(input().split()),2)
if not f:
for e in dit[p]:
if e in a:
f = True
break
a.add(p)
if f:
print("YES")
else:
print("NO")
``` | -1 | |
822 | A | I'm bored with life | PROGRAMMING | 800 | [
"implementation",
"math",
"number theory"
] | null | null | Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you? | The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12). | Print a single integer denoting the greatest common divisor of integers *A*! and *B*!. | [
"4 3\n"
] | [
"6\n"
] | Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | 500 | [
{
"input": "4 3",
"output": "6"
},
{
"input": "10 399603090",
"output": "3628800"
},
{
"input": "6 973151934",
"output": "720"
},
{
"input": "2 841668075",
"output": "2"
},
{
"input": "7 415216919",
"output": "5040"
},
{
"input": "3 283733059",
"output": "6"
},
{
"input": "11 562314608",
"output": "39916800"
},
{
"input": "3 990639260",
"output": "6"
},
{
"input": "11 859155400",
"output": "39916800"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "5 3",
"output": "6"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "5 4",
"output": "24"
},
{
"input": "1 12",
"output": "1"
},
{
"input": "9 7",
"output": "5040"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "6 11",
"output": "720"
},
{
"input": "6 7",
"output": "720"
},
{
"input": "11 11",
"output": "39916800"
},
{
"input": "4 999832660",
"output": "24"
},
{
"input": "7 999228288",
"output": "5040"
},
{
"input": "11 999257105",
"output": "39916800"
},
{
"input": "11 999286606",
"output": "39916800"
},
{
"input": "3 999279109",
"output": "6"
},
{
"input": "999632727 11",
"output": "39916800"
},
{
"input": "999625230 7",
"output": "5040"
},
{
"input": "999617047 3",
"output": "6"
},
{
"input": "999646548 7",
"output": "5040"
},
{
"input": "999639051 3",
"output": "6"
},
{
"input": "12 12",
"output": "479001600"
},
{
"input": "12 1",
"output": "1"
},
{
"input": "1213 5",
"output": "120"
},
{
"input": "8 9",
"output": "40320"
},
{
"input": "12 9",
"output": "362880"
},
{
"input": "12 1000000000",
"output": "479001600"
},
{
"input": "1000000000 1",
"output": "1"
},
{
"input": "12 13",
"output": "479001600"
},
{
"input": "2 29845",
"output": "2"
},
{
"input": "10 21",
"output": "3628800"
},
{
"input": "12 20",
"output": "479001600"
},
{
"input": "15 12",
"output": "479001600"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000 12",
"output": "479001600"
},
{
"input": "11 30",
"output": "39916800"
},
{
"input": "17 12",
"output": "479001600"
},
{
"input": "4 19",
"output": "24"
},
{
"input": "12 15",
"output": "479001600"
},
{
"input": "20 6",
"output": "720"
},
{
"input": "10 20",
"output": "3628800"
},
{
"input": "10 10",
"output": "3628800"
},
{
"input": "22 12",
"output": "479001600"
},
{
"input": "20 12",
"output": "479001600"
},
{
"input": "12 23",
"output": "479001600"
},
{
"input": "12 22",
"output": "479001600"
},
{
"input": "18 3",
"output": "6"
},
{
"input": "14 10",
"output": "3628800"
},
{
"input": "14 12",
"output": "479001600"
},
{
"input": "8 3",
"output": "6"
},
{
"input": "5 5",
"output": "120"
}
] | 1,652,892,126 | 2,147,483,647 | Python 3 | OK | TESTS | 59 | 46 | 4,300,800 | import math
m,n = map(int,input().split())
print(math.factorial(min(m,n))) | Title: I'm bored with life
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input Specification:
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
Output Specification:
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
Demo Input:
['4 3\n']
Demo Output:
['6\n']
Note:
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | ```python
import math
m,n = map(int,input().split())
print(math.factorial(min(m,n)))
``` | 3 | |
855 | A | Tom Riddle's Diary | PROGRAMMING | 800 | [
"brute force",
"implementation",
"strings"
] | null | null | Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence.
He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not.
Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=<<=*i*, otherwise, output "NO" (without quotes). | First line of input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of names in the list.
Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100. | Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not.
You can print each letter in any case (upper or lower). | [
"6\ntom\nlucius\nginny\nharry\nginny\nharry\n",
"3\na\na\na\n"
] | [
"NO\nNO\nNO\nNO\nYES\nYES\n",
"NO\nYES\nYES\n"
] | In test case 1, for *i* = 5 there exists *j* = 3 such that *s*<sub class="lower-index">*i*</sub> = *s*<sub class="lower-index">*j*</sub> and *j* < *i*, which means that answer for *i* = 5 is "YES". | 500 | [
{
"input": "6\ntom\nlucius\nginny\nharry\nginny\nharry",
"output": "NO\nNO\nNO\nNO\nYES\nYES"
},
{
"input": "3\na\na\na",
"output": "NO\nYES\nYES"
},
{
"input": "1\nzn",
"output": "NO"
},
{
"input": "9\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nhrtm\nssjqvixduertmotgagizamvfucfwtxqnhuowbqbzctgznivehelpcyigwrbbdsxnewfqvcf\nhyrtxvozpbveexfkgalmguozzakitjiwsduqxonb\nwcyxteiwtcyuztaguilqpbiwcwjaiq\nwcyxteiwtcyuztaguilqpbiwcwjaiq\nbdbivqzvhggth",
"output": "NO\nYES\nYES\nNO\nNO\nNO\nNO\nYES\nNO"
},
{
"input": "10\nkkiubdktydpdcbbttwpfdplhhjhrpqmpg\nkkiubdktydpdcbbttwpfdplhhjhrpqmpg\nmvutw\nqooeqoxzxwetlpecqiwgdbogiqqulttysyohwhzxzphvsfmnplizxoebzcvvfyppqbhxjksuzepuezqqzxlfmdanoeaoqmor\nmvutw\nvchawxjoreboqzuklifv\nvchawxjoreboqzuklifv\nnivijte\nrflybruq\nvchawxjoreboqzuklifv",
"output": "NO\nYES\nNO\nNO\nYES\nNO\nYES\nNO\nNO\nYES"
},
{
"input": "1\nz",
"output": "NO"
},
{
"input": "9\nl\ny\nm\nj\nn\nr\nj\nk\nf",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO"
},
{
"input": "14\nw\na\nh\np\nk\nw\ny\nv\ns\nf\nx\nd\nk\nr",
"output": "NO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO"
},
{
"input": "25\np\nk\nu\nl\nf\nt\nc\ns\nq\nd\nb\nq\no\ni\ni\nd\ni\nw\nn\ng\nw\nt\na\ne\ni",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nYES\nYES\nNO\nNO\nNO\nYES\nYES\nNO\nNO\nYES"
},
{
"input": "20\nd\nh\ng\no\np\ne\nt\nj\nv\ni\nt\nh\ns\ni\nw\nf\nx\na\nl\ni",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nYES"
},
{
"input": "3\nbbbbbbb\nbbbbbbbbb\nbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "NO\nNO\nNO"
},
{
"input": "2\nab\nba",
"output": "NO\nNO"
},
{
"input": "6\ntom\nlucius\nginnys\nharpy\nginny\nharry",
"output": "NO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "2\nabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcde\nabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcde",
"output": "NO\nYES"
},
{
"input": "42\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na",
"output": "NO\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
}
] | 1,691,166,233 | 2,147,483,647 | Python 3 | OK | TESTS | 55 | 46 | 0 | n=int(input())
name_set=set()
for i in range(n):
name=input().strip()
if name in name_set:
print("YES")
else:
print("NO")
name_set.add(name)
| Title: Tom Riddle's Diary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence.
He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not.
Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=<<=*i*, otherwise, output "NO" (without quotes).
Input Specification:
First line of input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of names in the list.
Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100.
Output Specification:
Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not.
You can print each letter in any case (upper or lower).
Demo Input:
['6\ntom\nlucius\nginny\nharry\nginny\nharry\n', '3\na\na\na\n']
Demo Output:
['NO\nNO\nNO\nNO\nYES\nYES\n', 'NO\nYES\nYES\n']
Note:
In test case 1, for *i* = 5 there exists *j* = 3 such that *s*<sub class="lower-index">*i*</sub> = *s*<sub class="lower-index">*j*</sub> and *j* < *i*, which means that answer for *i* = 5 is "YES". | ```python
n=int(input())
name_set=set()
for i in range(n):
name=input().strip()
if name in name_set:
print("YES")
else:
print("NO")
name_set.add(name)
``` | 3 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,661,976,071 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | word = input()
i = 0
l = 0
u = 0
while i<len(word):
if word[i].islower():
l+=1
else:
u+=1
i+=1
if u>l:
word = word.upper()
else:
word = word.lower()
print(word) | Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
word = input()
i = 0
l = 0
u = 0
while i<len(word):
if word[i].islower():
l+=1
else:
u+=1
i+=1
if u>l:
word = word.upper()
else:
word = word.lower()
print(word)
``` | 3.977 |
486 | A | Calculating Function | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*. | The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015). | Print *f*(*n*) in a single line. | [
"4\n",
"5\n"
] | [
"2\n",
"-3\n"
] | *f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3 | 500 | [
{
"input": "4",
"output": "2"
},
{
"input": "5",
"output": "-3"
},
{
"input": "1000000000",
"output": "500000000"
},
{
"input": "1000000001",
"output": "-500000001"
},
{
"input": "1000000000000000",
"output": "500000000000000"
},
{
"input": "100",
"output": "50"
},
{
"input": "101",
"output": "-51"
},
{
"input": "102",
"output": "51"
},
{
"input": "103",
"output": "-52"
},
{
"input": "104",
"output": "52"
},
{
"input": "105",
"output": "-53"
},
{
"input": "106",
"output": "53"
},
{
"input": "107",
"output": "-54"
},
{
"input": "108",
"output": "54"
},
{
"input": "109",
"output": "-55"
},
{
"input": "208170109961052",
"output": "104085054980526"
},
{
"input": "46017661651072",
"output": "23008830825536"
},
{
"input": "4018154546667",
"output": "-2009077273334"
},
{
"input": "288565475053",
"output": "-144282737527"
},
{
"input": "3052460231",
"output": "-1526230116"
},
{
"input": "29906716",
"output": "14953358"
},
{
"input": "87897701693326",
"output": "43948850846663"
},
{
"input": "8240",
"output": "4120"
},
{
"input": "577935",
"output": "-288968"
},
{
"input": "62",
"output": "31"
},
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "1"
},
{
"input": "9999999999999",
"output": "-5000000000000"
},
{
"input": "1000000000000",
"output": "500000000000"
},
{
"input": "99999999999999",
"output": "-50000000000000"
},
{
"input": "999999999999999",
"output": "-500000000000000"
},
{
"input": "42191359342",
"output": "21095679671"
},
{
"input": "100000000000000",
"output": "50000000000000"
},
{
"input": "145645214654154",
"output": "72822607327077"
},
{
"input": "4294967296",
"output": "2147483648"
},
{
"input": "3037000499",
"output": "-1518500250"
},
{
"input": "10000000000001",
"output": "-5000000000001"
},
{
"input": "100000017040846",
"output": "50000008520423"
},
{
"input": "98979894985999",
"output": "-49489947493000"
}
] | 1,693,745,348 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 46 | 0 | # import sys
# sys.stdin = open("/Users/swasti/Desktop/coding/cp/codeforces/input.txt", "r")
# sys.stdout = open("/Users/swasti/Desktop/coding/cp/codeforces/output.txt", "w")
n = int(input())
if n%2==0:
print( n//2)
else:
prev = (n-1)//2
print(prev-n)
| Title: Calculating Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*.
Input Specification:
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
Output Specification:
Print *f*(*n*) in a single line.
Demo Input:
['4\n', '5\n']
Demo Output:
['2\n', '-3\n']
Note:
*f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3 | ```python
# import sys
# sys.stdin = open("/Users/swasti/Desktop/coding/cp/codeforces/input.txt", "r")
# sys.stdout = open("/Users/swasti/Desktop/coding/cp/codeforces/output.txt", "w")
n = int(input())
if n%2==0:
print( n//2)
else:
prev = (n-1)//2
print(prev-n)
``` | 3 | |
496 | A | Minimum Difficulty | PROGRAMMING | 900 | [
"brute force",
"implementation",
"math"
] | null | null | Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold. | The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one). | Print a single number — the minimum difficulty of the track after removing a single hold. | [
"3\n1 4 6\n",
"5\n1 2 3 4 5\n",
"5\n1 2 3 7 8\n"
] | [
"5\n",
"2\n",
"4\n"
] | In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4. | 500 | [
{
"input": "3\n1 4 6",
"output": "5"
},
{
"input": "5\n1 2 3 4 5",
"output": "2"
},
{
"input": "5\n1 2 3 7 8",
"output": "4"
},
{
"input": "3\n1 500 1000",
"output": "999"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "2"
},
{
"input": "10\n1 4 9 16 25 36 49 64 81 100",
"output": "19"
},
{
"input": "10\n300 315 325 338 350 365 379 391 404 416",
"output": "23"
},
{
"input": "15\n87 89 91 92 93 95 97 99 101 103 105 107 109 111 112",
"output": "2"
},
{
"input": "60\n3 5 7 8 15 16 18 21 24 26 40 41 43 47 48 49 50 51 52 54 55 60 62 71 74 84 85 89 91 96 406 407 409 412 417 420 423 424 428 431 432 433 436 441 445 446 447 455 458 467 469 471 472 475 480 485 492 493 497 500",
"output": "310"
},
{
"input": "3\n159 282 405",
"output": "246"
},
{
"input": "81\n6 7 22 23 27 38 40 56 59 71 72 78 80 83 86 92 95 96 101 122 125 127 130 134 154 169 170 171 172 174 177 182 184 187 195 197 210 211 217 223 241 249 252 253 256 261 265 269 274 277 291 292 297 298 299 300 302 318 338 348 351 353 381 386 387 397 409 410 419 420 428 430 453 460 461 473 478 493 494 500 741",
"output": "241"
},
{
"input": "10\n218 300 388 448 535 629 680 740 836 925",
"output": "111"
},
{
"input": "100\n6 16 26 36 46 56 66 76 86 96 106 116 126 136 146 156 166 176 186 196 206 216 226 236 246 256 266 276 286 296 306 316 326 336 346 356 366 376 386 396 406 416 426 436 446 456 466 476 486 496 506 516 526 536 546 556 566 576 586 596 606 616 626 636 646 656 666 676 686 696 706 716 726 736 746 756 766 776 786 796 806 816 826 836 846 856 866 876 886 896 906 916 926 936 946 956 966 976 986 996",
"output": "20"
},
{
"input": "100\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 50 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000",
"output": "901"
},
{
"input": "100\n1 9 15 17 28 29 30 31 32 46 48 49 52 56 62 77 82 85 90 91 94 101 102 109 111 113 116 118 124 125 131 132 136 138 139 143 145 158 161 162 165 167 171 173 175 177 179 183 189 196 801 802 804 806 817 819 827 830 837 840 842 846 850 855 858 862 863 866 869 870 878 881 883 884 896 898 899 901 904 906 908 909 910 911 912 917 923 924 925 935 939 943 945 956 963 964 965 972 976 978",
"output": "605"
},
{
"input": "100\n2 43 47 49 50 57 59 67 74 98 901 903 904 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 938 939 940 942 943 944 945 946 947 948 949 950 952 953 954 956 957 958 959 960 961 962 963 965 966 967 968 969 970 971 972 973 974 975 976 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 998 999",
"output": "803"
},
{
"input": "72\n178 186 196 209 217 226 236 248 260 273 281 291 300 309 322 331 343 357 366 377 389 399 409 419 429 442 450 459 469 477 491 501 512 524 534 548 557 568 582 593 602 616 630 643 652 660 670 679 693 707 715 728 737 750 759 768 776 789 797 807 815 827 837 849 863 873 881 890 901 910 920 932",
"output": "17"
},
{
"input": "38\n1 28 55 82 109 136 163 190 217 244 271 298 325 352 379 406 433 460 487 514 541 568 595 622 649 676 703 730 757 784 811 838 865 892 919 946 973 1000",
"output": "54"
},
{
"input": "28\n1 38 75 112 149 186 223 260 297 334 371 408 445 482 519 556 593 630 667 704 741 778 815 852 889 926 963 1000",
"output": "74"
}
] | 1,634,520,558 | 2,147,483,647 | PyPy 3 | OK | TESTS | 19 | 93 | 0 | n, lst, res, bd = int(input()), list(map(int, input().split())), 10000000, 0
for x in range(n - 1): bd = max(bd, lst[x + 1] - lst[x])
for x in range(1, n - 1): res = min(res, max(bd, lst[x + 1] - lst[x - 1]))
print(res) | Title: Minimum Difficulty
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike is trying rock climbing but he is awful at it.
There are *n* holds on the wall, *i*-th hold is at height *a**i* off the ground. Besides, let the sequence *a**i* increase, that is, *a**i*<=<<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1; we will call such sequence a track. Mike thinks that the track *a*1, ..., *a**n* has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.
Today Mike decided to cover the track with holds hanging on heights *a*1, ..., *a**n*. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1,<=2,<=3,<=4,<=5) and remove the third element from it, we obtain the sequence (1,<=2,<=4,<=5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.
Help Mike determine the minimum difficulty of the track after removing one hold.
Input Specification:
The first line contains a single integer *n* (3<=≤<=*n*<=≤<=100) — the number of holds.
The next line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000), where *a**i* is the height where the hold number *i* hangs. The sequence *a**i* is increasing (i.e. each element except for the first one is strictly larger than the previous one).
Output Specification:
Print a single number — the minimum difficulty of the track after removing a single hold.
Demo Input:
['3\n1 4 6\n', '5\n1 2 3 4 5\n', '5\n1 2 3 7 8\n']
Demo Output:
['5\n', '2\n', '4\n']
Note:
In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.
In the second test after removing every hold the difficulty equals 2.
In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4. | ```python
n, lst, res, bd = int(input()), list(map(int, input().split())), 10000000, 0
for x in range(n - 1): bd = max(bd, lst[x + 1] - lst[x])
for x in range(1, n - 1): res = min(res, max(bd, lst[x + 1] - lst[x - 1]))
print(res)
``` | 3 | |
116 | A | Tram | PROGRAMMING | 800 | [
"implementation"
] | null | null | Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram. | The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0. | Print a single integer denoting the minimum possible capacity of the tram (0 is allowed). | [
"4\n0 3\n2 5\n4 2\n4 0\n"
] | [
"6\n"
] | For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | 500 | [
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\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",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\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 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,687,043,612 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 71 | 218 | 1,945,600 | s=int(input())
c=0
v=[]
for i in range(s):
b=list(map(int,input().split()))
c+=b[1]-b[0]
v+=[c]
print(max(v)) | Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | ```python
s=int(input())
c=0
v=[]
for i in range(s):
b=list(map(int,input().split()))
c+=b[1]-b[0]
v+=[c]
print(max(v))
``` | 3 | |
137 | A | Postcards and photos | PROGRAMMING | 900 | [
"implementation"
] | null | null | Polycarpus has postcards and photos hung in a row on the wall. He decided to put them away to the closet and hang on the wall a famous painter's picture. Polycarpus does it like that: he goes from the left to the right and removes the objects consecutively. As Polycarpus doesn't want any mix-ups to happen, he will not carry in his hands objects of two different types. In other words, Polycarpus can't carry both postcards and photos simultaneously. Sometimes he goes to the closet and puts the objects there, thus leaving his hands free. Polycarpus must put all the postcards and photos to the closet. He cannot skip objects. What minimum number of times he should visit the closet if he cannot carry more than 5 items? | The only line of the input data contains a non-empty string consisting of letters "С" and "P" whose length does not exceed 100 characters. If the *i*-th character in the string is the letter "С", that means that the *i*-th object (the numbering goes from the left to the right) on Polycarpus' wall is a postcard. And if the *i*-th character is the letter "P", than the *i*-th object on the wall is a photo. | Print the only number — the minimum number of times Polycarpus has to visit the closet. | [
"CPCPCPC\n",
"CCCCCCPPPPPP\n",
"CCCCCCPPCPPPPPPPPPP\n",
"CCCCCCCCCC\n"
] | [
"7\n",
"4\n",
"6\n",
"2\n"
] | In the first sample Polycarpus needs to take one item to the closet 7 times.
In the second sample Polycarpus can first take 3 postcards to the closet; then 3 more. He can take the 6 photos that are left in the similar way, going to the closet twice.
In the third sample Polycarpus can visit the closet twice, both times carrying 3 postcards. Then he can take there 2 photos at once, then one postcard and finally, he can carry the last 10 photos if he visits the closet twice.
In the fourth sample Polycarpus can visit the closet twice and take there all 10 postcards (5 items during each go). | 500 | [
{
"input": "CPCPCPC",
"output": "7"
},
{
"input": "CCCCCCPPPPPP",
"output": "4"
},
{
"input": "CCCCCCPPCPPPPPPPPPP",
"output": "6"
},
{
"input": "CCCCCCCCCC",
"output": "2"
},
{
"input": "CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC",
"output": "20"
},
{
"input": "CPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCPCP",
"output": "100"
},
{
"input": "CCCCCCPPPPPPCCCCCCPPPPPPCCCCCCPPPPPPCCCCCCPPPPPPCCCCCCPPPPPPCCCCCCPPPPPPCCCCCCPPPPPP",
"output": "28"
},
{
"input": "P",
"output": "1"
},
{
"input": "C",
"output": "1"
},
{
"input": "PC",
"output": "2"
},
{
"input": "PPPPP",
"output": "1"
},
{
"input": "PPPP",
"output": "1"
},
{
"input": "CCCCCCCCCC",
"output": "2"
},
{
"input": "CP",
"output": "2"
},
{
"input": "CPCCPCPPPC",
"output": "7"
},
{
"input": "PPCPCCPCPPCCPPPPPPCP",
"output": "12"
},
{
"input": "PCPCCPCPPCCPCPCCPPPPPCPCPCPCCC",
"output": "20"
},
{
"input": "CCPPPPPCPCCPPPCCPPCPCCPCPPCPPCCCPPCPPPCC",
"output": "21"
},
{
"input": "CPPCCCCCCPCCCCPCCPCPPPCPCCCCCCCPCCPPCCCPCCCCCPPCCC",
"output": "23"
},
{
"input": "PPCCCCPPCCPPPCCCCPPPPPCPPPCPPPCCCPCCCPCPPPCPCCCPCCPPCCPPPPPC",
"output": "26"
},
{
"input": "PPCPPCCCCCPCCCPCCPCCCCPPPCCCCPCPCCPCPCPCPPPPCCPPPPPPPCPCPPPCPCPCPCPPPC",
"output": "39"
},
{
"input": "CCPCPPPPCPPPPCCCCPCCPCPCCPPCPCCCPPCCCCPCCCPCPCCPPPCPPPCPCPPPPPCPCCPCCPPCCCPCPPPC",
"output": "43"
},
{
"input": "CCPPCPCPCPPCCCPCPPPCCCCCPCPPCCCPPCPCPPPPCPPCPPPPCCCPCCPCPPPCPCPPCCCPCCCCCCPCCCCPCCPPPPCCPP",
"output": "47"
},
{
"input": "PPCPPPPCCCCPPPPCPPPPPPPPCPCPPCCPPPPPPPPCPPPPCCCCPPPPCPPCPCPPPCCPPCPPCCCPCPPCCCCCCPCPCPCPPCPCPCPPPCCC",
"output": "49"
},
{
"input": "CCPCCCPPCPPCPCCCPCPPCPPCPPCCCCCCCPCPPCPCCPCCPCPCPCCCPCCCPPPCCPCCPPCCCCCPPPPCPCPPCPCPCCPCPPP",
"output": "53"
},
{
"input": "PCPCPPPPCPCPPPCPPCCCPCPCPCPPCPPPPCCPPPCPPPCPPPPCCPPCCCPCCPCCCCPCCPCPPCPCCCPCPPCP",
"output": "47"
},
{
"input": "PCCPPCCCPPCPPCC",
"output": "8"
},
{
"input": "CCCPPPPPPCCCCPCCPCCCCCCPCCCPPPCPC",
"output": "15"
},
{
"input": "CPPCCPPCCPPPCCCPPPPCPPPPPPPCCPCPCCPPPPCCCPPCCPCCPPCCCPCCPCPPPPCCPP",
"output": "31"
},
{
"input": "CCCCCPPPCCPCPCCPPPPCPCCCPCPPCPCPPPPPCCPCPCPC",
"output": "25"
},
{
"input": "PPPPPPPPPCPCP",
"output": "6"
},
{
"input": "PPPCPCPCCCPPCPCCPPPPCCCPCCP",
"output": "15"
},
{
"input": "PCPCCPCPPPPPPCPCCPCPCPCCPPPCPCPCPPCPPCCPCPCCCPCCCPPCPCPCCPCPPPPCCCCCCPPCCPCCCCCPCCCCPPPCPCCCCCPCPCP",
"output": "59"
},
{
"input": "PCCPCPPCCCCCPCCCPCCCPPCCCCCPPPCCPPPPPPPPCPPPCCPPCPPCPCP",
"output": "26"
},
{
"input": "CPCPCCPPPPCCPPCPPCPPCCCCCCPCCPPPCPPCPCCCCCCPCPCCCCCPCCCCCCPCCPPCCP",
"output": "35"
},
{
"input": "PPCCCCCCPP",
"output": "4"
},
{
"input": "CCCCCCCCCCCCPPCCCCPP",
"output": "6"
},
{
"input": "PPPPPPPPPPPCCCCCCCCCCCCCCCCCCP",
"output": "8"
},
{
"input": "PPPPPPPPPPPPPPPPPPPPPCCCCCCCCCCCPPPPCCCC",
"output": "10"
},
{
"input": "PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPCCCCCCCCCPPPC",
"output": "12"
},
{
"input": "CCCCCCCCCPPPPPPPPPPPPPPPPPPPPCCCCCCCCCCCCCCCCCCCCCCCCPPPPPCC",
"output": "13"
},
{
"input": "CCCCCCCCCCCCCCCCCCCCCCCCCPPPCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC",
"output": "15"
},
{
"input": "CCCCCPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPCCCCCCCCCCCCCPPPPPPPCCPPP",
"output": "18"
},
{
"input": "PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPCCCCCCCCCCCCCCCCCCCCCCCCCCCPPPPPPPPPPPPPPPPPPP",
"output": "19"
},
{
"input": "PPPPPPPPPPPPPPPPPPPPPPPCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCPPPPPCCCPPPPCCCCCPCC",
"output": "23"
}
] | 1,596,038,216 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 280 | 20,172,800 | s=input()
a=len(s)
s=s+'0'
p=0
c=s[0]
i=1
cnt = 0
while p < a:
p+=1#pointer of string
i+=1
if s[p] != c:
cnt += 1
i=0#number of stuff on his hands
c=s[p]
if i==5:
i=0
cnt+=1
c=s[p+1]
if i != 0:
cnt+=1
print(cnt) | Title: Postcards and photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has postcards and photos hung in a row on the wall. He decided to put them away to the closet and hang on the wall a famous painter's picture. Polycarpus does it like that: he goes from the left to the right and removes the objects consecutively. As Polycarpus doesn't want any mix-ups to happen, he will not carry in his hands objects of two different types. In other words, Polycarpus can't carry both postcards and photos simultaneously. Sometimes he goes to the closet and puts the objects there, thus leaving his hands free. Polycarpus must put all the postcards and photos to the closet. He cannot skip objects. What minimum number of times he should visit the closet if he cannot carry more than 5 items?
Input Specification:
The only line of the input data contains a non-empty string consisting of letters "С" and "P" whose length does not exceed 100 characters. If the *i*-th character in the string is the letter "С", that means that the *i*-th object (the numbering goes from the left to the right) on Polycarpus' wall is a postcard. And if the *i*-th character is the letter "P", than the *i*-th object on the wall is a photo.
Output Specification:
Print the only number — the minimum number of times Polycarpus has to visit the closet.
Demo Input:
['CPCPCPC\n', 'CCCCCCPPPPPP\n', 'CCCCCCPPCPPPPPPPPPP\n', 'CCCCCCCCCC\n']
Demo Output:
['7\n', '4\n', '6\n', '2\n']
Note:
In the first sample Polycarpus needs to take one item to the closet 7 times.
In the second sample Polycarpus can first take 3 postcards to the closet; then 3 more. He can take the 6 photos that are left in the similar way, going to the closet twice.
In the third sample Polycarpus can visit the closet twice, both times carrying 3 postcards. Then he can take there 2 photos at once, then one postcard and finally, he can carry the last 10 photos if he visits the closet twice.
In the fourth sample Polycarpus can visit the closet twice and take there all 10 postcards (5 items during each go). | ```python
s=input()
a=len(s)
s=s+'0'
p=0
c=s[0]
i=1
cnt = 0
while p < a:
p+=1#pointer of string
i+=1
if s[p] != c:
cnt += 1
i=0#number of stuff on his hands
c=s[p]
if i==5:
i=0
cnt+=1
c=s[p+1]
if i != 0:
cnt+=1
print(cnt)
``` | 0 | |
239 | A | Two Bags of Potatoes | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"math"
] | null | null | Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order. | The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105). | Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1. | [
"10 1 10\n",
"10 6 40\n"
] | [
"-1\n",
"2 8 14 20 26 \n"
] | none | 500 | [
{
"input": "10 1 10",
"output": "-1"
},
{
"input": "10 6 40",
"output": "2 8 14 20 26 "
},
{
"input": "10 1 20",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "1 10000 1000000000",
"output": "9999 19999 29999 39999 49999 59999 69999 79999 89999 99999 109999 119999 129999 139999 149999 159999 169999 179999 189999 199999 209999 219999 229999 239999 249999 259999 269999 279999 289999 299999 309999 319999 329999 339999 349999 359999 369999 379999 389999 399999 409999 419999 429999 439999 449999 459999 469999 479999 489999 499999 509999 519999 529999 539999 549999 559999 569999 579999 589999 599999 609999 619999 629999 639999 649999 659999 669999 679999 689999 699999 709999 719999 729999 739999 7499..."
},
{
"input": "84817 1 33457",
"output": "-1"
},
{
"input": "21 37 99",
"output": "16 53 "
},
{
"input": "78 7 15",
"output": "-1"
},
{
"input": "74 17 27",
"output": "-1"
},
{
"input": "79 23 43",
"output": "-1"
},
{
"input": "32 33 3",
"output": "-1"
},
{
"input": "55 49 44",
"output": "-1"
},
{
"input": "64 59 404",
"output": "54 113 172 231 290 "
},
{
"input": "61 69 820",
"output": "8 77 146 215 284 353 422 491 560 629 698 "
},
{
"input": "17 28 532",
"output": "11 39 67 95 123 151 179 207 235 263 291 319 347 375 403 431 459 487 515 "
},
{
"input": "46592 52 232",
"output": "-1"
},
{
"input": "1541 58 648",
"output": "-1"
},
{
"input": "15946 76 360",
"output": "-1"
},
{
"input": "30351 86 424",
"output": "-1"
},
{
"input": "1 2 37493",
"output": "1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 103 105 107 109 111 113 115 117 119 121 123 125 127 129 131 133 135 137 139 141 143 145 147 149 151 153 155 157 159 161 163 165 167 169 171 173 175 177 179 181 183 185 187 189 191 193 195 197 199 201 203 205 207 209 211 213 215 217 219 221 223 225 227 229 231 233 235 237 239 241 243 245 247 249 251 253 255 257 259 261 263 265 267 269 271 273 275 277 279 281 28..."
},
{
"input": "1 3 27764",
"output": "2 5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50 53 56 59 62 65 68 71 74 77 80 83 86 89 92 95 98 101 104 107 110 113 116 119 122 125 128 131 134 137 140 143 146 149 152 155 158 161 164 167 170 173 176 179 182 185 188 191 194 197 200 203 206 209 212 215 218 221 224 227 230 233 236 239 242 245 248 251 254 257 260 263 266 269 272 275 278 281 284 287 290 293 296 299 302 305 308 311 314 317 320 323 326 329 332 335 338 341 344 347 350 353 356 359 362 365 368 371 374 377 380 383 386 389 392 395 398 401 404 407 410..."
},
{
"input": "10 4 9174",
"output": "2 6 10 14 18 22 26 30 34 38 42 46 50 54 58 62 66 70 74 78 82 86 90 94 98 102 106 110 114 118 122 126 130 134 138 142 146 150 154 158 162 166 170 174 178 182 186 190 194 198 202 206 210 214 218 222 226 230 234 238 242 246 250 254 258 262 266 270 274 278 282 286 290 294 298 302 306 310 314 318 322 326 330 334 338 342 346 350 354 358 362 366 370 374 378 382 386 390 394 398 402 406 410 414 418 422 426 430 434 438 442 446 450 454 458 462 466 470 474 478 482 486 490 494 498 502 506 510 514 518 522 526 530 534 53..."
},
{
"input": "33 7 4971",
"output": "2 9 16 23 30 37 44 51 58 65 72 79 86 93 100 107 114 121 128 135 142 149 156 163 170 177 184 191 198 205 212 219 226 233 240 247 254 261 268 275 282 289 296 303 310 317 324 331 338 345 352 359 366 373 380 387 394 401 408 415 422 429 436 443 450 457 464 471 478 485 492 499 506 513 520 527 534 541 548 555 562 569 576 583 590 597 604 611 618 625 632 639 646 653 660 667 674 681 688 695 702 709 716 723 730 737 744 751 758 765 772 779 786 793 800 807 814 821 828 835 842 849 856 863 870 877 884 891 898 905 912 919..."
},
{
"input": "981 1 3387",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "386 1 2747",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "123 2 50000",
"output": "1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 103 105 107 109 111 113 115 117 119 121 123 125 127 129 131 133 135 137 139 141 143 145 147 149 151 153 155 157 159 161 163 165 167 169 171 173 175 177 179 181 183 185 187 189 191 193 195 197 199 201 203 205 207 209 211 213 215 217 219 221 223 225 227 229 231 233 235 237 239 241 243 245 247 249 251 253 255 257 259 261 263 265 267 269 271 273 275 277 279 281 28..."
},
{
"input": "3123 100 10000000",
"output": "77 177 277 377 477 577 677 777 877 977 1077 1177 1277 1377 1477 1577 1677 1777 1877 1977 2077 2177 2277 2377 2477 2577 2677 2777 2877 2977 3077 3177 3277 3377 3477 3577 3677 3777 3877 3977 4077 4177 4277 4377 4477 4577 4677 4777 4877 4977 5077 5177 5277 5377 5477 5577 5677 5777 5877 5977 6077 6177 6277 6377 6477 6577 6677 6777 6877 6977 7077 7177 7277 7377 7477 7577 7677 7777 7877 7977 8077 8177 8277 8377 8477 8577 8677 8777 8877 8977 9077 9177 9277 9377 9477 9577 9677 9777 9877 9977 10077 10177 10277 1037..."
},
{
"input": "2 10000 1000000000",
"output": "9998 19998 29998 39998 49998 59998 69998 79998 89998 99998 109998 119998 129998 139998 149998 159998 169998 179998 189998 199998 209998 219998 229998 239998 249998 259998 269998 279998 289998 299998 309998 319998 329998 339998 349998 359998 369998 379998 389998 399998 409998 419998 429998 439998 449998 459998 469998 479998 489998 499998 509998 519998 529998 539998 549998 559998 569998 579998 589998 599998 609998 619998 629998 639998 649998 659998 669998 679998 689998 699998 709998 719998 729998 739998 7499..."
},
{
"input": "3 10000 1000000000",
"output": "9997 19997 29997 39997 49997 59997 69997 79997 89997 99997 109997 119997 129997 139997 149997 159997 169997 179997 189997 199997 209997 219997 229997 239997 249997 259997 269997 279997 289997 299997 309997 319997 329997 339997 349997 359997 369997 379997 389997 399997 409997 419997 429997 439997 449997 459997 469997 479997 489997 499997 509997 519997 529997 539997 549997 559997 569997 579997 589997 599997 609997 619997 629997 639997 649997 659997 669997 679997 689997 699997 709997 719997 729997 739997 7499..."
},
{
"input": "12312223 10000 1000000000",
"output": "7777 17777 27777 37777 47777 57777 67777 77777 87777 97777 107777 117777 127777 137777 147777 157777 167777 177777 187777 197777 207777 217777 227777 237777 247777 257777 267777 277777 287777 297777 307777 317777 327777 337777 347777 357777 367777 377777 387777 397777 407777 417777 427777 437777 447777 457777 467777 477777 487777 497777 507777 517777 527777 537777 547777 557777 567777 577777 587777 597777 607777 617777 627777 637777 647777 657777 667777 677777 687777 697777 707777 717777 727777 737777 7477..."
},
{
"input": "500000000 1000000000 1000000000",
"output": "500000000 "
},
{
"input": "1 1000000000 1000000000",
"output": "999999999 "
},
{
"input": "10 6 11",
"output": "-1"
},
{
"input": "2 100 10",
"output": "-1"
},
{
"input": "1 100000007 1000000000",
"output": "100000006 200000013 300000020 400000027 500000034 600000041 700000048 800000055 900000062 "
},
{
"input": "1 999999999 1000000000",
"output": "999999998 "
},
{
"input": "100000000 1000000000 1000000000",
"output": "900000000 "
},
{
"input": "11 2 12",
"output": "1 "
},
{
"input": "31 10 39",
"output": "-1"
},
{
"input": "48 6 50",
"output": "-1"
},
{
"input": "500000000 500000000 1000000000",
"output": "500000000 "
},
{
"input": "1 1000000000 999999999",
"output": "-1"
},
{
"input": "4 2 10",
"output": "2 4 6 "
},
{
"input": "1000000000 1 1",
"output": "-1"
},
{
"input": "1000000000 1 100000",
"output": "-1"
},
{
"input": "1000000000 1 10",
"output": "-1"
},
{
"input": "10 5 14",
"output": "-1"
},
{
"input": "500000000 499999999 1000000000",
"output": "499999998 "
},
{
"input": "1 999999997 1000000000",
"output": "999999996 "
}
] | 1,541,151,343 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 686 | 3,276,800 | y, k, n = map(int,input().split())
s = 1
l = []
while k*s <= n :
if k*s-y > 0 :
l.append(k*s-y)
s += 1
if l :
print(*l)
else :
print(-1) | Title: Two Bags of Potatoes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.
Input Specification:
The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105).
Output Specification:
Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1.
Demo Input:
['10 1 10\n', '10 6 40\n']
Demo Output:
['-1\n', '2 8 14 20 26 \n']
Note:
none | ```python
y, k, n = map(int,input().split())
s = 1
l = []
while k*s <= n :
if k*s-y > 0 :
l.append(k*s-y)
s += 1
if l :
print(*l)
else :
print(-1)
``` | 3 | |
492 | B | Vanya and Lanterns | PROGRAMMING | 1,200 | [
"binary search",
"implementation",
"math",
"sortings"
] | null | null | Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns.
Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street? | The first line contains two integers *n*, *l* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*l*<=≤<=109) — the number of lanterns and the length of the street respectively.
The next line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street. | Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9. | [
"7 15\n15 5 3 7 9 14 0\n",
"2 5\n2 5\n"
] | [
"2.5000000000\n",
"2.0000000000\n"
] | Consider the second sample. At *d* = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment [3, 5]. Thus, the whole street will be lit. | 1,000 | [
{
"input": "7 15\n15 5 3 7 9 14 0",
"output": "2.5000000000"
},
{
"input": "2 5\n2 5",
"output": "2.0000000000"
},
{
"input": "46 615683844\n431749087 271781274 274974690 324606253 480870261 401650581 13285442 478090364 266585394 425024433 588791449 492057200 391293435 563090494 317950 173675329 473068378 356306865 311731938 192959832 321180686 141984626 578985584 512026637 175885185 590844074 47103801 212211134 330150 509886963 565955809 315640375 612907074 500474373 524310737 568681652 315339618 478782781 518873818 271322031 74600969 539099112 85129347 222068995 106014720 77282307",
"output": "22258199.5000000000"
},
{
"input": "2 1000000000\n0 1000000000",
"output": "500000000.0000000000"
},
{
"input": "2 555\n200 300",
"output": "255.0000000000"
},
{
"input": "1 1\n1",
"output": "1.0000000000"
},
{
"input": "1 1\n0",
"output": "1.0000000000"
},
{
"input": "1 1000000000\n0",
"output": "1000000000.0000000000"
},
{
"input": "1 1000000000\n1000000000",
"output": "1000000000.0000000000"
},
{
"input": "1 999999999\n499999999",
"output": "500000000.0000000000"
},
{
"input": "4 700\n0 250 475 700",
"output": "125.0000000000"
},
{
"input": "4 700\n0 225 450 700",
"output": "125.0000000000"
},
{
"input": "5 1000\n0 250 500 750 1000",
"output": "125.0000000000"
},
{
"input": "5 2000\n1001 1001 1001 1001 1001",
"output": "1001.0000000000"
},
{
"input": "1 999\n501",
"output": "501.0000000000"
},
{
"input": "2 1000\n400 600",
"output": "400.0000000000"
},
{
"input": "4 15\n1 3 9 11",
"output": "4.0000000000"
},
{
"input": "1 5\n2",
"output": "3.0000000000"
}
] | 1,699,894,288 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 8 | 62 | 0 | import math
import sys
n,l = map(int,input().split())
ls = list(map(int,input().split()))
ls.sort()
maxi = 0
for i in range(n-1):
maxi = max(maxi,ls[i+1]-ls[i])
maxi=maxi/2
if ls[0]==0:
print(maxi)
else:
a = ls[0]
print(max(maxi,a))
| Title: Vanya and Lanterns
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns.
Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street?
Input Specification:
The first line contains two integers *n*, *l* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*l*<=≤<=109) — the number of lanterns and the length of the street respectively.
The next line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.
Output Specification:
Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9.
Demo Input:
['7 15\n15 5 3 7 9 14 0\n', '2 5\n2 5\n']
Demo Output:
['2.5000000000\n', '2.0000000000\n']
Note:
Consider the second sample. At *d* = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment [3, 5]. Thus, the whole street will be lit. | ```python
import math
import sys
n,l = map(int,input().split())
ls = list(map(int,input().split()))
ls.sort()
maxi = 0
for i in range(n-1):
maxi = max(maxi,ls[i+1]-ls[i])
maxi=maxi/2
if ls[0]==0:
print(maxi)
else:
a = ls[0]
print(max(maxi,a))
``` | 0 | |
724 | A | Checking the Calendar | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | You are given names of two days of the week.
Please, determine whether it is possible that during some non-leap year the first day of some month was equal to the first day of the week you are given, while the first day of the next month was equal to the second day of the week you are given. Both months should belong to one year.
In this problem, we consider the Gregorian calendar to be used. The number of months in this calendar is equal to 12. The number of days in months during any non-leap year is: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31.
Names of the days of the week are given with lowercase English letters: "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday". | The input consists of two lines, each of them containing the name of exactly one day of the week. It's guaranteed that each string in the input is from the set "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday". | Print "YES" (without quotes) if such situation is possible during some non-leap year. Otherwise, print "NO" (without quotes). | [
"monday\ntuesday\n",
"sunday\nsunday\n",
"saturday\ntuesday\n"
] | [
"NO\n",
"YES\n",
"YES\n"
] | In the second sample, one can consider February 1 and March 1 of year 2015. Both these days were Sundays.
In the third sample, one can consider July 1 and August 1 of year 2017. First of these two days is Saturday, while the second one is Tuesday. | 500 | [
{
"input": "monday\ntuesday",
"output": "NO"
},
{
"input": "sunday\nsunday",
"output": "YES"
},
{
"input": "saturday\ntuesday",
"output": "YES"
},
{
"input": "tuesday\nthursday",
"output": "YES"
},
{
"input": "friday\nwednesday",
"output": "NO"
},
{
"input": "sunday\nsaturday",
"output": "NO"
},
{
"input": "monday\nmonday",
"output": "YES"
},
{
"input": "monday\nwednesday",
"output": "YES"
},
{
"input": "monday\nthursday",
"output": "YES"
},
{
"input": "monday\nfriday",
"output": "NO"
},
{
"input": "monday\nsaturday",
"output": "NO"
},
{
"input": "monday\nsunday",
"output": "NO"
},
{
"input": "tuesday\nmonday",
"output": "NO"
},
{
"input": "tuesday\ntuesday",
"output": "YES"
},
{
"input": "tuesday\nwednesday",
"output": "NO"
},
{
"input": "tuesday\nfriday",
"output": "YES"
},
{
"input": "tuesday\nsaturday",
"output": "NO"
},
{
"input": "tuesday\nsunday",
"output": "NO"
},
{
"input": "wednesday\nmonday",
"output": "NO"
},
{
"input": "wednesday\ntuesday",
"output": "NO"
},
{
"input": "wednesday\nwednesday",
"output": "YES"
},
{
"input": "wednesday\nthursday",
"output": "NO"
},
{
"input": "wednesday\nfriday",
"output": "YES"
},
{
"input": "wednesday\nsaturday",
"output": "YES"
},
{
"input": "wednesday\nsunday",
"output": "NO"
},
{
"input": "thursday\nmonday",
"output": "NO"
},
{
"input": "thursday\ntuesday",
"output": "NO"
},
{
"input": "thursday\nwednesday",
"output": "NO"
},
{
"input": "thursday\nthursday",
"output": "YES"
},
{
"input": "thursday\nfriday",
"output": "NO"
},
{
"input": "thursday\nsaturday",
"output": "YES"
},
{
"input": "thursday\nsunday",
"output": "YES"
},
{
"input": "friday\nmonday",
"output": "YES"
},
{
"input": "friday\ntuesday",
"output": "NO"
},
{
"input": "friday\nthursday",
"output": "NO"
},
{
"input": "friday\nsaturday",
"output": "NO"
},
{
"input": "friday\nsunday",
"output": "YES"
},
{
"input": "saturday\nmonday",
"output": "YES"
},
{
"input": "saturday\nwednesday",
"output": "NO"
},
{
"input": "saturday\nthursday",
"output": "NO"
},
{
"input": "saturday\nfriday",
"output": "NO"
},
{
"input": "saturday\nsaturday",
"output": "YES"
},
{
"input": "saturday\nsunday",
"output": "NO"
},
{
"input": "sunday\nmonday",
"output": "NO"
},
{
"input": "sunday\ntuesday",
"output": "YES"
},
{
"input": "sunday\nwednesday",
"output": "YES"
},
{
"input": "sunday\nthursday",
"output": "NO"
},
{
"input": "sunday\nfriday",
"output": "NO"
},
{
"input": "friday\nfriday",
"output": "YES"
},
{
"input": "friday\nsunday",
"output": "YES"
},
{
"input": "monday\nmonday",
"output": "YES"
},
{
"input": "friday\ntuesday",
"output": "NO"
},
{
"input": "thursday\nsaturday",
"output": "YES"
},
{
"input": "tuesday\nfriday",
"output": "YES"
},
{
"input": "sunday\nwednesday",
"output": "YES"
},
{
"input": "monday\nthursday",
"output": "YES"
},
{
"input": "saturday\nsunday",
"output": "NO"
},
{
"input": "friday\nmonday",
"output": "YES"
},
{
"input": "thursday\nthursday",
"output": "YES"
},
{
"input": "wednesday\nfriday",
"output": "YES"
},
{
"input": "thursday\nmonday",
"output": "NO"
},
{
"input": "wednesday\nsunday",
"output": "NO"
},
{
"input": "thursday\nfriday",
"output": "NO"
},
{
"input": "monday\nfriday",
"output": "NO"
},
{
"input": "wednesday\nsaturday",
"output": "YES"
},
{
"input": "thursday\nsunday",
"output": "YES"
},
{
"input": "saturday\nfriday",
"output": "NO"
},
{
"input": "saturday\nmonday",
"output": "YES"
}
] | 1,628,799,208 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 6,758,400 | days = ["monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday"]
d1 = input()
d2 = input()
d = days.index(d2) - days.index(d1)
print(d)
if d < 0:
d = d + 7
if d == 0 or d == 2 or d == 3:
print('YES')
else:
print('NO') | Title: Checking the Calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given names of two days of the week.
Please, determine whether it is possible that during some non-leap year the first day of some month was equal to the first day of the week you are given, while the first day of the next month was equal to the second day of the week you are given. Both months should belong to one year.
In this problem, we consider the Gregorian calendar to be used. The number of months in this calendar is equal to 12. The number of days in months during any non-leap year is: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31.
Names of the days of the week are given with lowercase English letters: "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".
Input Specification:
The input consists of two lines, each of them containing the name of exactly one day of the week. It's guaranteed that each string in the input is from the set "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".
Output Specification:
Print "YES" (without quotes) if such situation is possible during some non-leap year. Otherwise, print "NO" (without quotes).
Demo Input:
['monday\ntuesday\n', 'sunday\nsunday\n', 'saturday\ntuesday\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n']
Note:
In the second sample, one can consider February 1 and March 1 of year 2015. Both these days were Sundays.
In the third sample, one can consider July 1 and August 1 of year 2017. First of these two days is Saturday, while the second one is Tuesday. | ```python
days = ["monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday"]
d1 = input()
d2 = input()
d = days.index(d2) - days.index(d1)
print(d)
if d < 0:
d = d + 7
if d == 0 or d == 2 or d == 3:
print('YES')
else:
print('NO')
``` | 0 | |
892 | A | Greed | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not! | The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans.
The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans. | Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower). | [
"2\n3 5\n3 6\n",
"3\n6 8 9\n6 10 12\n",
"5\n0 0 5 0 0\n1 1 8 10 5\n",
"4\n4 1 0 3\n5 2 2 3\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"YES\n"
] | In the first sample, there are already 2 cans, so the answer is "YES". | 500 | [
{
"input": "2\n3 5\n3 6",
"output": "YES"
},
{
"input": "3\n6 8 9\n6 10 12",
"output": "NO"
},
{
"input": "5\n0 0 5 0 0\n1 1 8 10 5",
"output": "YES"
},
{
"input": "4\n4 1 0 3\n5 2 2 3",
"output": "YES"
},
{
"input": "10\n9 10 24 11 1 7 8 3 28 14\n86 20 34 11 22 94 8 16 73 85",
"output": "YES"
},
{
"input": "4\n25 35 7 31\n70 37 43 35",
"output": "YES"
},
{
"input": "10\n15 26 15 14 14 39 40 4 25 39\n27 72 16 44 69 48 53 17 63 42",
"output": "NO"
},
{
"input": "5\n22 5 19 16 32\n26 10 43 38 37",
"output": "NO"
},
{
"input": "5\n32 4 22 40 26\n39 20 36 98 44",
"output": "YES"
},
{
"input": "6\n18 25 3 10 13 37\n38 73 19 35 24 37",
"output": "YES"
},
{
"input": "2\n2 2\n2 2",
"output": "YES"
},
{
"input": "2\n2 5\n2 5",
"output": "YES"
},
{
"input": "2\n1000 1008\n10000 2352",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5\n1 2 3 4 11",
"output": "YES"
},
{
"input": "4\n1 0 0 0\n2 0 0 0",
"output": "YES"
},
{
"input": "2\n0 0\n1 2",
"output": "YES"
},
{
"input": "3\n9 13 4\n10 14 5",
"output": "NO"
},
{
"input": "2\n0 0\n1 1",
"output": "YES"
},
{
"input": "5\n1 1 2 3 1\n1 1 2 3 4",
"output": "NO"
},
{
"input": "2\n0 0\n0 0",
"output": "YES"
},
{
"input": "3\n5 1 1\n5 5 5",
"output": "YES"
}
] | 1,510,934,187 | 4,887 | Python 3 | OK | TESTS | 39 | 171 | 7,168,000 | n=int(input())
v=0
for i in input().split(' '):
v+=int(i)
a,b,c=0,0,0
for i in input().split(' '):
c=int(i)
if c>=a:
b=a
a=c
elif c>b:
b=c
if v <= a+b:
print("YES")
else:
print("NO") | Title: Greed
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input Specification:
The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans.
The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans.
Output Specification:
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Demo Input:
['2\n3 5\n3 6\n', '3\n6 8 9\n6 10 12\n', '5\n0 0 5 0 0\n1 1 8 10 5\n', '4\n4 1 0 3\n5 2 2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'YES\n']
Note:
In the first sample, there are already 2 cans, so the answer is "YES". | ```python
n=int(input())
v=0
for i in input().split(' '):
v+=int(i)
a,b,c=0,0,0
for i in input().split(' '):
c=int(i)
if c>=a:
b=a
a=c
elif c>b:
b=c
if v <= a+b:
print("YES")
else:
print("NO")
``` | 3 | |
294 | A | Shaass and Oskols | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots. | The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment. | On the *i*-th line of the output print the number of birds on the *i*-th wire. | [
"5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n",
"3\n2 4 1\n1\n2 2\n"
] | [
"0\n12\n5\n0\n16\n",
"3\n0\n3\n"
] | none | 500 | [
{
"input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6",
"output": "0\n12\n5\n0\n16"
},
{
"input": "3\n2 4 1\n1\n2 2",
"output": "3\n0\n3"
},
{
"input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43",
"output": "0\n66\n57\n7\n0"
},
{
"input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51",
"output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52"
},
{
"input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1",
"output": "0\n0"
},
{
"input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5",
"output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21"
},
{
"input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0",
"output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95"
},
{
"input": "1\n100\n1\n1 100",
"output": "0"
},
{
"input": "1\n100\n1\n1 1",
"output": "0"
},
{
"input": "1\n50\n1\n1 25",
"output": "0"
},
{
"input": "2\n50 0\n1\n1 1",
"output": "0\n49"
},
{
"input": "1\n10\n0",
"output": "10"
}
] | 1,683,456,932 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 92 | 0 | n=int(input())
a=list(map(int,input().split()))
m=int(input())
w=[[0]*2 for i in range(m)]
for i in range(m):
w[i][0],w[i][1]=map(int,input().split())
for i in range(m):
left=w[i][1]-1
right=a[w[i][0]-1]-w[i][1]
if w[i][0]!=1:
a[w[i][0]-2]+=left
if w[i][0]!=n:
a[w[i][0]]+=right
a[w[i][0]-1]=0
for i in range(n):
print(a[i]) | Title: Shaass and Oskols
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
Input Specification:
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
Output Specification:
On the *i*-th line of the output print the number of birds on the *i*-th wire.
Demo Input:
['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n']
Demo Output:
['0\n12\n5\n0\n16\n', '3\n0\n3\n']
Note:
none | ```python
n=int(input())
a=list(map(int,input().split()))
m=int(input())
w=[[0]*2 for i in range(m)]
for i in range(m):
w[i][0],w[i][1]=map(int,input().split())
for i in range(m):
left=w[i][1]-1
right=a[w[i][0]-1]-w[i][1]
if w[i][0]!=1:
a[w[i][0]-2]+=left
if w[i][0]!=n:
a[w[i][0]]+=right
a[w[i][0]-1]=0
for i in range(n):
print(a[i])
``` | 3 | |
946 | A | Partition | PROGRAMMING | 800 | [
"greedy"
] | null | null | You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences.
Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*? | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100) — the elements of sequence *a*. | Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*. | [
"3\n1 -2 0\n",
"6\n16 23 16 15 42 8\n"
] | [
"3\n",
"120\n"
] | In the first example we may choose *b* = {1, 0}, *c* = { - 2}. Then *B* = 1, *C* = - 2, *B* - *C* = 3.
In the second example we choose *b* = {16, 23, 16, 15, 42, 8}, *c* = {} (an empty sequence). Then *B* = 120, *C* = 0, *B* - *C* = 120. | 0 | [
{
"input": "3\n1 -2 0",
"output": "3"
},
{
"input": "6\n16 23 16 15 42 8",
"output": "120"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "100\n-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 -100",
"output": "10000"
},
{
"input": "2\n-1 5",
"output": "6"
},
{
"input": "3\n-2 0 1",
"output": "3"
},
{
"input": "12\n-1 -2 -3 4 4 -6 -6 56 3 3 -3 3",
"output": "94"
},
{
"input": "4\n1 -1 1 -1",
"output": "4"
},
{
"input": "4\n100 -100 100 -100",
"output": "400"
},
{
"input": "3\n-2 -5 10",
"output": "17"
},
{
"input": "5\n1 -2 3 -4 5",
"output": "15"
},
{
"input": "3\n-100 100 -100",
"output": "300"
},
{
"input": "6\n1 -1 1 -1 1 -1",
"output": "6"
},
{
"input": "6\n2 -2 2 -2 2 -2",
"output": "12"
},
{
"input": "9\n12 93 -2 0 0 0 3 -3 -9",
"output": "122"
},
{
"input": "6\n-1 2 4 -5 -3 55",
"output": "70"
},
{
"input": "6\n-12 8 68 -53 1 -15",
"output": "157"
},
{
"input": "2\n-2 1",
"output": "3"
},
{
"input": "3\n100 -100 100",
"output": "300"
},
{
"input": "5\n100 100 -1 -100 2",
"output": "303"
},
{
"input": "6\n-5 -4 -3 -2 -1 0",
"output": "15"
},
{
"input": "6\n4 4 4 -3 -3 2",
"output": "20"
},
{
"input": "2\n-1 2",
"output": "3"
},
{
"input": "1\n100",
"output": "100"
},
{
"input": "5\n-1 -2 3 1 2",
"output": "9"
},
{
"input": "5\n100 -100 100 -100 100",
"output": "500"
},
{
"input": "5\n1 -1 1 -1 1",
"output": "5"
},
{
"input": "4\n0 0 0 -1",
"output": "1"
},
{
"input": "5\n100 -100 -1 2 100",
"output": "303"
},
{
"input": "2\n75 0",
"output": "75"
},
{
"input": "4\n55 56 -59 -58",
"output": "228"
},
{
"input": "2\n9 71",
"output": "80"
},
{
"input": "2\n9 70",
"output": "79"
},
{
"input": "2\n9 69",
"output": "78"
},
{
"input": "2\n100 -100",
"output": "200"
},
{
"input": "4\n-9 4 -9 5",
"output": "27"
},
{
"input": "42\n91 -27 -79 -56 80 -93 -23 10 80 94 61 -89 -64 81 34 99 31 -32 -69 92 79 -9 73 66 -8 64 99 99 58 -19 -40 21 1 -33 93 -23 -62 27 55 41 57 36",
"output": "2348"
},
{
"input": "7\n-1 2 2 2 -1 2 -1",
"output": "11"
},
{
"input": "6\n-12 8 17 -69 7 -88",
"output": "201"
},
{
"input": "3\n1 -2 5",
"output": "8"
},
{
"input": "6\n-2 3 -4 5 6 -1",
"output": "21"
},
{
"input": "2\n-5 1",
"output": "6"
},
{
"input": "4\n2 2 -2 4",
"output": "10"
},
{
"input": "68\n21 47 -75 -25 64 83 83 -21 89 24 43 44 -35 34 -42 92 -96 -52 -66 64 14 -87 25 -61 -78 83 -96 -18 95 83 -93 -28 75 49 87 65 -93 -69 -2 95 -24 -36 -61 -71 88 -53 -93 -51 -81 -65 -53 -46 -56 6 65 58 19 100 57 61 -53 44 -58 48 -8 80 -88 72",
"output": "3991"
},
{
"input": "5\n5 5 -10 -1 1",
"output": "22"
},
{
"input": "3\n-1 2 3",
"output": "6"
},
{
"input": "76\n57 -38 -48 -81 93 -32 96 55 -44 2 38 -46 42 64 71 -73 95 31 -39 -62 -1 75 -17 57 28 52 12 -11 82 -84 59 -86 73 -97 34 97 -57 -85 -6 39 -5 -54 95 24 -44 35 -18 9 91 7 -22 -61 -80 54 -40 74 -90 15 -97 66 -52 -49 -24 65 21 -93 -29 -24 -4 -1 76 -93 7 -55 -53 1",
"output": "3787"
},
{
"input": "5\n-1 -2 1 2 3",
"output": "9"
},
{
"input": "4\n2 2 -2 -2",
"output": "8"
},
{
"input": "6\n100 -100 100 -100 100 -100",
"output": "600"
},
{
"input": "100\n-59 -33 34 0 69 24 -22 58 62 -36 5 45 -19 -73 61 -9 95 42 -73 -64 91 -96 2 53 -8 82 -79 16 18 -5 -53 26 71 38 -31 12 -33 -1 -65 -6 3 -89 22 33 -27 -36 41 11 -47 -32 47 -56 -38 57 -63 -41 23 41 29 78 16 -65 90 -58 -12 6 -60 42 -36 -52 -54 -95 -10 29 70 50 -94 1 93 48 -71 -77 -16 54 56 -60 66 76 31 8 44 -61 -74 23 37 38 18 -18 29 41",
"output": "4362"
},
{
"input": "2\n-1 1",
"output": "2"
},
{
"input": "3\n1 -2 100",
"output": "103"
},
{
"input": "5\n1 -2 3 1 2",
"output": "9"
},
{
"input": "10\n100 -10 -100 10 10 10 10 10 10 10",
"output": "280"
},
{
"input": "4\n2 0 -2 4",
"output": "8"
},
{
"input": "4\n3 -3 1 -1",
"output": "8"
},
{
"input": "3\n1 -1 1",
"output": "3"
},
{
"input": "4\n2 5 -2 4",
"output": "13"
},
{
"input": "2\n-2 2",
"output": "4"
},
{
"input": "3\n1 -2 1",
"output": "4"
},
{
"input": "5\n-1 -2 1 1 -1",
"output": "6"
},
{
"input": "4\n-2 0 2 4",
"output": "8"
},
{
"input": "8\n-42 7 87 -16 -5 65 -88 1",
"output": "311"
},
{
"input": "3\n1 -3 4",
"output": "8"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "3\n-1 2 -1",
"output": "4"
},
{
"input": "18\n-21 12 65 66 -24 62 82 35 -45 -47 28 37 5 -32 22 -14 -69 -95",
"output": "761"
},
{
"input": "4\n-1 1 -1 1",
"output": "4"
},
{
"input": "5\n-1 2 1 1 1",
"output": "6"
},
{
"input": "3\n1 1 1",
"output": "3"
}
] | 1,657,284,820 | 2,147,483,647 | PyPy 3 | OK | TESTS | 72 | 77 | 0 | # problem D
n = int(input())
elements = list(map(int, input().split()))
a = []
b = []
for x in elements:
if x < 0:
a.append(x)
else:
b.append(x)
A = sum(a)
B = sum(b)
print(B - A)
| Title: Partition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences.
Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*?
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100) — the elements of sequence *a*.
Output Specification:
Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*.
Demo Input:
['3\n1 -2 0\n', '6\n16 23 16 15 42 8\n']
Demo Output:
['3\n', '120\n']
Note:
In the first example we may choose *b* = {1, 0}, *c* = { - 2}. Then *B* = 1, *C* = - 2, *B* - *C* = 3.
In the second example we choose *b* = {16, 23, 16, 15, 42, 8}, *c* = {} (an empty sequence). Then *B* = 120, *C* = 0, *B* - *C* = 120. | ```python
# problem D
n = int(input())
elements = list(map(int, input().split()))
a = []
b = []
for x in elements:
if x < 0:
a.append(x)
else:
b.append(x)
A = sum(a)
B = sum(b)
print(B - A)
``` | 3 | |
1,006 | D | Two Strings Swaps | PROGRAMMING | 1,700 | [
"implementation"
] | null | null | You are given two strings $a$ and $b$ consisting of lowercase English letters, both of length $n$. The characters of both strings have indices from $1$ to $n$, inclusive.
You are allowed to do the following changes:
- Choose any index $i$ ($1 \le i \le n$) and swap characters $a_i$ and $b_i$; - Choose any index $i$ ($1 \le i \le n$) and swap characters $a_i$ and $a_{n - i + 1}$; - Choose any index $i$ ($1 \le i \le n$) and swap characters $b_i$ and $b_{n - i + 1}$.
Note that if $n$ is odd, you are formally allowed to swap $a_{\lceil\frac{n}{2}\rceil}$ with $a_{\lceil\frac{n}{2}\rceil}$ (and the same with the string $b$) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in $a$ with another character. In other words, in a single preprocess move you can choose any index $i$ ($1 \le i \le n$), any character $c$ and set $a_i := c$.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings $a$ and $b$ equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string $b$ or make any preprocess moves after the first change is made. | The first line of the input contains one integer $n$ ($1 \le n \le 10^5$) — the length of strings $a$ and $b$.
The second line contains the string $a$ consisting of exactly $n$ lowercase English letters.
The third line contains the string $b$ consisting of exactly $n$ lowercase English letters. | Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string $a$ equal to string $b$ with a sequence of changes from the list above. | [
"7\nabacaba\nbacabaa\n",
"5\nzcabd\ndbacz\n"
] | [
"4\n",
"0\n"
] | In the first example preprocess moves are as follows: $a_1 := $'b', $a_3 := $'c', $a_4 := $'a' and $a_5:=$'b'. Afterwards, $a = $"bbcabba". Then we can obtain equal strings by the following sequence of changes: $swap(a_2, b_2)$ and $swap(a_2, a_6)$. There is no way to use fewer than $4$ preprocess moves before a sequence of changes to make string equal, so the answer in this example is $4$.
In the second example no preprocess moves are required. We can use the following sequence of changes to make $a$ and $b$ equal: $swap(b_1, b_5)$, $swap(a_2, a_4)$. | 0 | [
{
"input": "7\nabacaba\nbacabaa",
"output": "4"
},
{
"input": "5\nzcabd\ndbacz",
"output": "0"
},
{
"input": "1\na\nb",
"output": "1"
},
{
"input": "5\nahmad\nyogaa",
"output": "3"
}
] | 1,650,454,682 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 23 | 78 | 3,584,000 | n=int(input())
a=input()
b=input()
ans=0
for i in range(n//2):
seen=set()
if b[i]==b[n-i-1]:
if a[i]==a[n-i-1]:
ans+=0
else:
ans+=1
else :
seen.add(b[i])
if b[n-i-1] in seen:
seen.remove(b[n-i-1])
else :
seen.add(b[n-i-1])
if a[i] in seen:
seen.remove(a[i])
else :
ans+=1
if a[n-1-i] in seen:
seen.remove(a[n-i-1])
else :
ans+=1
if n%2:
if a[n//2]!=b[n//2]:
ans+=1
print(ans) | Title: Two Strings Swaps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two strings $a$ and $b$ consisting of lowercase English letters, both of length $n$. The characters of both strings have indices from $1$ to $n$, inclusive.
You are allowed to do the following changes:
- Choose any index $i$ ($1 \le i \le n$) and swap characters $a_i$ and $b_i$; - Choose any index $i$ ($1 \le i \le n$) and swap characters $a_i$ and $a_{n - i + 1}$; - Choose any index $i$ ($1 \le i \le n$) and swap characters $b_i$ and $b_{n - i + 1}$.
Note that if $n$ is odd, you are formally allowed to swap $a_{\lceil\frac{n}{2}\rceil}$ with $a_{\lceil\frac{n}{2}\rceil}$ (and the same with the string $b$) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in $a$ with another character. In other words, in a single preprocess move you can choose any index $i$ ($1 \le i \le n$), any character $c$ and set $a_i := c$.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings $a$ and $b$ equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string $b$ or make any preprocess moves after the first change is made.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 10^5$) — the length of strings $a$ and $b$.
The second line contains the string $a$ consisting of exactly $n$ lowercase English letters.
The third line contains the string $b$ consisting of exactly $n$ lowercase English letters.
Output Specification:
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string $a$ equal to string $b$ with a sequence of changes from the list above.
Demo Input:
['7\nabacaba\nbacabaa\n', '5\nzcabd\ndbacz\n']
Demo Output:
['4\n', '0\n']
Note:
In the first example preprocess moves are as follows: $a_1 := $'b', $a_3 := $'c', $a_4 := $'a' and $a_5:=$'b'. Afterwards, $a = $"bbcabba". Then we can obtain equal strings by the following sequence of changes: $swap(a_2, b_2)$ and $swap(a_2, a_6)$. There is no way to use fewer than $4$ preprocess moves before a sequence of changes to make string equal, so the answer in this example is $4$.
In the second example no preprocess moves are required. We can use the following sequence of changes to make $a$ and $b$ equal: $swap(b_1, b_5)$, $swap(a_2, a_4)$. | ```python
n=int(input())
a=input()
b=input()
ans=0
for i in range(n//2):
seen=set()
if b[i]==b[n-i-1]:
if a[i]==a[n-i-1]:
ans+=0
else:
ans+=1
else :
seen.add(b[i])
if b[n-i-1] in seen:
seen.remove(b[n-i-1])
else :
seen.add(b[n-i-1])
if a[i] in seen:
seen.remove(a[i])
else :
ans+=1
if a[n-1-i] in seen:
seen.remove(a[n-i-1])
else :
ans+=1
if n%2:
if a[n//2]!=b[n//2]:
ans+=1
print(ans)
``` | 3 | |
230 | B | T-primes | PROGRAMMING | 1,300 | [
"binary search",
"implementation",
"math",
"number theory"
] | null | null | We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not. | The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier. | Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't. | [
"3\n4 5 6\n"
] | [
"YES\nNO\nNO\n"
] | The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO". | 500 | [
{
"input": "3\n4 5 6",
"output": "YES\nNO\nNO"
},
{
"input": "2\n48 49",
"output": "NO\nYES"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "NO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO"
},
{
"input": "1\n36",
"output": "NO"
},
{
"input": "1\n999966000289",
"output": "YES"
},
{
"input": "1\n999993399999",
"output": "NO"
},
{
"input": "9\n111 121 131 111 121 131 111 121 131",
"output": "NO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO"
},
{
"input": "1\n1",
"output": "NO"
},
{
"input": "1\n10",
"output": "NO"
},
{
"input": "1\n976197352729",
"output": "NO"
},
{
"input": "1\n1000000000000",
"output": "NO"
},
{
"input": "1\n9",
"output": "YES"
},
{
"input": "6\n549755813888 847288609443 762939453125 678223072849 285311670611 137858491849",
"output": "NO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "3\n223092870 6469693230 200560490130",
"output": "NO\nNO\nNO"
},
{
"input": "2\n81 25",
"output": "NO\nYES"
},
{
"input": "1\n16",
"output": "NO"
},
{
"input": "22\n1 2 3 4 5 6 7 8 9 10 12752041 64 121 144 27550356289 124 24657 23756 135153365 25235235235 42351351 81",
"output": "NO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "1\n225",
"output": "NO"
},
{
"input": "1\n1521",
"output": "NO"
},
{
"input": "1\n81",
"output": "NO"
}
] | 1,697,040,065 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 0 | 2,000 | 1,331,200 | import math
def judge(a):#生成10^6以内质数表
b=False
for i in range(2,math.ceil(math.sqrt(a))+1):#平方数是上限
if a%i==0:
b=True
break
if b==False:
l.append(a)
l=[]
l.append(2)
for a in range(3,1000001):
judge(a)
def ch(i):
if math.sqrt(i)%1==0 and i!=1 :
q=int(math.sqrt(i))
if q in l:
print('YES')
else:
print('NO')
else:
print('NO')
n=int(input())
m=list(map(int,input().split()))
for i in m:
ch(i) | Title: T-primes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
Input Specification:
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
Demo Input:
['3\n4 5 6\n']
Demo Output:
['YES\nNO\nNO\n']
Note:
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO". | ```python
import math
def judge(a):#生成10^6以内质数表
b=False
for i in range(2,math.ceil(math.sqrt(a))+1):#平方数是上限
if a%i==0:
b=True
break
if b==False:
l.append(a)
l=[]
l.append(2)
for a in range(3,1000001):
judge(a)
def ch(i):
if math.sqrt(i)%1==0 and i!=1 :
q=int(math.sqrt(i))
if q in l:
print('YES')
else:
print('NO')
else:
print('NO')
n=int(input())
m=list(map(int,input().split()))
for i in m:
ch(i)
``` | 0 | |
920 | C | Swap Adjacent Elements | PROGRAMMING | 1,400 | [
"dfs and similar",
"greedy",
"math",
"sortings",
"two pointers"
] | null | null | You have an array *a* consisting of *n* integers. Each integer from 1 to *n* appears exactly once in this array.
For some indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1) it is possible to swap *i*-th element with (*i*<=+<=1)-th, for other indices it is not possible. You may perform any number of swapping operations any order. There is no limit on the number of times you swap *i*-th element with (*i*<=+<=1)-th (if the position is not forbidden).
Can you make this array sorted in ascending order performing some sequence of swapping operations? | The first line contains one integer *n* (2<=≤<=*n*<=≤<=200000) — the number of elements in the array.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=200000) — the elements of the array. Each integer from 1 to *n* appears exactly once.
The third line contains a string of *n*<=-<=1 characters, each character is either 0 or 1. If *i*-th character is 1, then you can swap *i*-th element with (*i*<=+<=1)-th any number of times, otherwise it is forbidden to swap *i*-th element with (*i*<=+<=1)-th. | If it is possible to sort the array in ascending order using any sequence of swaps you are allowed to make, print YES. Otherwise, print NO. | [
"6\n1 2 5 3 4 6\n01110\n",
"6\n1 2 5 3 4 6\n01010\n"
] | [
"YES\n",
"NO\n"
] | In the first example you may swap *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, and then swap *a*<sub class="lower-index">4</sub> and *a*<sub class="lower-index">5</sub>. | 0 | [
{
"input": "6\n1 2 5 3 4 6\n01110",
"output": "YES"
},
{
"input": "6\n1 2 5 3 4 6\n01010",
"output": "NO"
},
{
"input": "6\n1 6 3 4 5 2\n01101",
"output": "NO"
},
{
"input": "6\n2 3 1 4 5 6\n01111",
"output": "NO"
},
{
"input": "4\n2 3 1 4\n011",
"output": "NO"
},
{
"input": "2\n2 1\n0",
"output": "NO"
},
{
"input": "5\n1 2 4 5 3\n0101",
"output": "NO"
},
{
"input": "5\n1 2 4 5 3\n0001",
"output": "NO"
},
{
"input": "5\n1 4 5 2 3\n0110",
"output": "NO"
},
{
"input": "5\n4 5 1 2 3\n0111",
"output": "NO"
},
{
"input": "3\n3 1 2\n10",
"output": "NO"
},
{
"input": "5\n2 3 4 5 1\n0011",
"output": "NO"
},
{
"input": "16\n3 4 14 16 11 7 13 9 10 8 6 5 15 12 1 2\n111111101111111",
"output": "NO"
},
{
"input": "5\n1 5 3 4 2\n1101",
"output": "NO"
},
{
"input": "6\n6 1 2 3 4 5\n11101",
"output": "NO"
},
{
"input": "3\n2 3 1\n01",
"output": "NO"
},
{
"input": "6\n1 6 3 4 5 2\n01110",
"output": "NO"
},
{
"input": "7\n1 7 3 4 5 6 2\n010001",
"output": "NO"
},
{
"input": "5\n5 2 3 4 1\n1001",
"output": "NO"
},
{
"input": "4\n1 3 4 2\n001",
"output": "NO"
},
{
"input": "5\n4 5 1 2 3\n1011",
"output": "NO"
},
{
"input": "6\n1 5 3 4 2 6\n11011",
"output": "NO"
},
{
"input": "5\n1 4 2 5 3\n1101",
"output": "NO"
},
{
"input": "5\n3 2 4 1 5\n1010",
"output": "NO"
},
{
"input": "6\n1 4 3 5 6 2\n01101",
"output": "NO"
},
{
"input": "6\n2 3 4 5 1 6\n00010",
"output": "NO"
},
{
"input": "10\n5 2 7 9 1 10 3 4 6 8\n111101000",
"output": "NO"
},
{
"input": "5\n2 4 3 1 5\n0110",
"output": "NO"
},
{
"input": "4\n3 1 2 4\n100",
"output": "NO"
},
{
"input": "6\n1 5 3 4 2 6\n01010",
"output": "NO"
},
{
"input": "4\n3 1 2 4\n101",
"output": "NO"
},
{
"input": "4\n2 4 3 1\n011",
"output": "NO"
},
{
"input": "4\n2 3 4 1\n001",
"output": "NO"
},
{
"input": "4\n3 4 1 2\n011",
"output": "NO"
},
{
"input": "5\n2 4 1 3 5\n0110",
"output": "NO"
},
{
"input": "4\n1 3 4 2\n101",
"output": "NO"
},
{
"input": "20\n20 19 18 17 16 15 1 2 3 4 5 14 13 12 11 10 9 8 7 6\n1111111011111111111",
"output": "NO"
},
{
"input": "6\n6 5 4 1 2 3\n11100",
"output": "NO"
},
{
"input": "5\n2 3 5 1 4\n0011",
"output": "NO"
},
{
"input": "4\n1 4 2 3\n010",
"output": "NO"
},
{
"input": "6\n1 6 3 4 5 2\n01001",
"output": "NO"
},
{
"input": "7\n1 7 2 4 3 5 6\n011110",
"output": "NO"
},
{
"input": "5\n1 3 4 2 5\n0010",
"output": "NO"
},
{
"input": "5\n5 4 3 1 2\n1110",
"output": "NO"
},
{
"input": "5\n2 5 4 3 1\n0111",
"output": "NO"
},
{
"input": "4\n2 3 4 1\n101",
"output": "NO"
},
{
"input": "5\n1 4 5 2 3\n1011",
"output": "NO"
},
{
"input": "5\n1 3 2 5 4\n1110",
"output": "NO"
},
{
"input": "6\n3 2 4 1 5 6\n10111",
"output": "NO"
},
{
"input": "7\n3 1 7 4 5 2 6\n101110",
"output": "NO"
},
{
"input": "10\n5 4 10 9 2 1 6 7 3 8\n011111111",
"output": "NO"
},
{
"input": "5\n1 5 3 2 4\n1110",
"output": "NO"
},
{
"input": "4\n2 3 4 1\n011",
"output": "NO"
},
{
"input": "5\n5 4 3 2 1\n0000",
"output": "NO"
},
{
"input": "12\n6 9 11 1 12 7 5 8 10 4 3 2\n11111111110",
"output": "NO"
},
{
"input": "5\n3 1 5 2 4\n1011",
"output": "NO"
},
{
"input": "5\n4 5 1 2 3\n1110",
"output": "NO"
},
{
"input": "10\n1 2 3 4 5 6 8 9 7 10\n000000000",
"output": "NO"
},
{
"input": "6\n5 6 3 2 4 1\n01111",
"output": "NO"
},
{
"input": "5\n1 3 4 2 5\n0100",
"output": "NO"
},
{
"input": "4\n2 1 4 3\n100",
"output": "NO"
},
{
"input": "6\n1 2 3 4 6 5\n00000",
"output": "NO"
},
{
"input": "6\n4 6 5 3 2 1\n01111",
"output": "NO"
},
{
"input": "5\n3 1 4 5 2\n1001",
"output": "NO"
},
{
"input": "5\n5 2 3 1 4\n1011",
"output": "NO"
},
{
"input": "3\n2 3 1\n10",
"output": "NO"
},
{
"input": "10\n6 5 9 4 3 2 8 10 7 1\n111111110",
"output": "NO"
},
{
"input": "7\n1 2 7 3 4 5 6\n111101",
"output": "NO"
},
{
"input": "6\n5 6 1 2 4 3\n11101",
"output": "NO"
},
{
"input": "6\n4 6 3 5 2 1\n11110",
"output": "NO"
},
{
"input": "5\n5 4 2 3 1\n1110",
"output": "NO"
},
{
"input": "2\n2 1\n1",
"output": "YES"
},
{
"input": "3\n1 3 2\n10",
"output": "NO"
},
{
"input": "5\n3 4 5 1 2\n1110",
"output": "NO"
},
{
"input": "5\n3 4 2 1 5\n0110",
"output": "NO"
},
{
"input": "6\n6 1 2 3 4 5\n10001",
"output": "NO"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10\n000000000",
"output": "YES"
},
{
"input": "3\n3 2 1\n00",
"output": "NO"
},
{
"input": "5\n5 4 3 2 1\n1110",
"output": "NO"
},
{
"input": "6\n3 1 2 5 6 4\n10011",
"output": "NO"
},
{
"input": "6\n3 2 1 6 5 4\n11000",
"output": "NO"
},
{
"input": "2\n1 2\n0",
"output": "YES"
},
{
"input": "2\n1 2\n1",
"output": "YES"
},
{
"input": "11\n1 2 3 4 5 6 7 8 9 10 11\n0000000000",
"output": "YES"
},
{
"input": "4\n2 4 3 1\n101",
"output": "NO"
},
{
"input": "4\n3 4 1 2\n101",
"output": "NO"
},
{
"input": "3\n1 3 2\n01",
"output": "YES"
},
{
"input": "6\n6 2 3 1 4 5\n11110",
"output": "NO"
},
{
"input": "3\n2 1 3\n01",
"output": "NO"
},
{
"input": "5\n1 5 4 3 2\n0111",
"output": "YES"
},
{
"input": "6\n1 2 6 3 4 5\n11110",
"output": "NO"
},
{
"input": "7\n2 3 1 7 6 5 4\n011111",
"output": "NO"
},
{
"input": "6\n5 6 1 2 3 4\n01111",
"output": "NO"
},
{
"input": "4\n1 2 4 3\n001",
"output": "YES"
},
{
"input": "6\n1 2 3 6 4 5\n11001",
"output": "NO"
},
{
"input": "11\n9 8 10 11 1 2 3 4 5 6 7\n1101111111",
"output": "NO"
},
{
"input": "5\n1 5 3 4 2\n0101",
"output": "NO"
},
{
"input": "10\n9 1 2 3 7 8 5 6 4 10\n110111100",
"output": "NO"
},
{
"input": "7\n1 2 7 3 4 5 6\n111011",
"output": "NO"
},
{
"input": "10\n3 10 1 2 6 4 5 7 8 9\n111111001",
"output": "NO"
},
{
"input": "10\n1 3 6 5 2 9 7 8 4 10\n001101111",
"output": "NO"
},
{
"input": "10\n1 8 9 7 6 10 4 2 3 5\n111111101",
"output": "NO"
},
{
"input": "7\n1 2 5 3 6 4 7\n111011",
"output": "NO"
},
{
"input": "4\n2 4 3 1\n100",
"output": "NO"
},
{
"input": "6\n1 2 3 4 6 5\n00001",
"output": "YES"
},
{
"input": "6\n2 1 3 4 5 6\n10000",
"output": "YES"
},
{
"input": "5\n3 2 1 5 4\n1100",
"output": "NO"
},
{
"input": "9\n2 1 3 6 5 4 7 9 8\n10011001",
"output": "YES"
},
{
"input": "8\n2 6 4 1 5 7 3 8\n1010010",
"output": "NO"
},
{
"input": "5\n1 2 4 5 3\n1101",
"output": "NO"
},
{
"input": "6\n1 3 5 2 4 6\n00110",
"output": "NO"
},
{
"input": "6\n1 3 6 2 4 5\n10111",
"output": "NO"
},
{
"input": "9\n9 8 7 6 5 4 3 1 2\n11111110",
"output": "NO"
},
{
"input": "10\n6 7 8 9 10 1 2 3 4 5\n111111110",
"output": "NO"
},
{
"input": "8\n6 1 7 8 3 2 5 4\n1011111",
"output": "NO"
},
{
"input": "70\n4 65 66 30 67 16 39 35 57 14 42 51 5 21 61 53 63 13 60 29 68 70 69 46 20 2 43 47 49 52 26 44 54 62 25 19 12 28 27 24 18 36 6 33 7 8 11 1 45 32 64 38 23 22 56 59 15 9 41 37 40 55 3 31 34 48 50 10 17 58\n111111101101111111111110101111111111111101101111010010110011011110010",
"output": "NO"
},
{
"input": "5\n5 3 2 4 1\n0100",
"output": "NO"
},
{
"input": "6\n3 2 6 5 1 4\n11011",
"output": "NO"
},
{
"input": "6\n1 2 4 5 6 3\n10011",
"output": "NO"
},
{
"input": "7\n1 7 3 2 5 6 4\n111001",
"output": "NO"
}
] | 1,517,589,288 | 7,188 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 46 | 5,632,000 | n = int(input())
l = [int(x) for x in input().split()]
a = input()
p = [0] * n
s = set()
k = 0
for i in range(n-1):
if a[i] == '1':
s.add(l[i])
s.add(l[i+1])
else:
if k < l[i]:
if k < min(s) and max(s) < l[i]:
k = l[i]
else:
print('NO')
break
else:
print('YES') | Title: Swap Adjacent Elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have an array *a* consisting of *n* integers. Each integer from 1 to *n* appears exactly once in this array.
For some indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1) it is possible to swap *i*-th element with (*i*<=+<=1)-th, for other indices it is not possible. You may perform any number of swapping operations any order. There is no limit on the number of times you swap *i*-th element with (*i*<=+<=1)-th (if the position is not forbidden).
Can you make this array sorted in ascending order performing some sequence of swapping operations?
Input Specification:
The first line contains one integer *n* (2<=≤<=*n*<=≤<=200000) — the number of elements in the array.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=200000) — the elements of the array. Each integer from 1 to *n* appears exactly once.
The third line contains a string of *n*<=-<=1 characters, each character is either 0 or 1. If *i*-th character is 1, then you can swap *i*-th element with (*i*<=+<=1)-th any number of times, otherwise it is forbidden to swap *i*-th element with (*i*<=+<=1)-th.
Output Specification:
If it is possible to sort the array in ascending order using any sequence of swaps you are allowed to make, print YES. Otherwise, print NO.
Demo Input:
['6\n1 2 5 3 4 6\n01110\n', '6\n1 2 5 3 4 6\n01010\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example you may swap *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, and then swap *a*<sub class="lower-index">4</sub> and *a*<sub class="lower-index">5</sub>. | ```python
n = int(input())
l = [int(x) for x in input().split()]
a = input()
p = [0] * n
s = set()
k = 0
for i in range(n-1):
if a[i] == '1':
s.add(l[i])
s.add(l[i+1])
else:
if k < l[i]:
if k < min(s) and max(s) < l[i]:
k = l[i]
else:
print('NO')
break
else:
print('YES')
``` | -1 | |
960 | A | Check the string | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes). | The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'. | Print "YES" or "NO", according to the condition. | [
"aaabccc\n",
"bbacc\n",
"aabc\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | 500 | [
{
"input": "aaabccc",
"output": "YES"
},
{
"input": "bbacc",
"output": "NO"
},
{
"input": "aabc",
"output": "YES"
},
{
"input": "aabbcc",
"output": "YES"
},
{
"input": "aaacccbb",
"output": "NO"
},
{
"input": "abc",
"output": "YES"
},
{
"input": "acba",
"output": "NO"
},
{
"input": "bbabbc",
"output": "NO"
},
{
"input": "bbbabacca",
"output": "NO"
},
{
"input": "aabcbcaca",
"output": "NO"
},
{
"input": "aaaaabbbbbb",
"output": "NO"
},
{
"input": "c",
"output": "NO"
},
{
"input": "cc",
"output": "NO"
},
{
"input": "bbb",
"output": "NO"
},
{
"input": "bc",
"output": "NO"
},
{
"input": "ccbcc",
"output": "NO"
},
{
"input": "aaa",
"output": "NO"
},
{
"input": "aaccaa",
"output": "NO"
},
{
"input": "a",
"output": "NO"
},
{
"input": "b",
"output": "NO"
},
{
"input": "abca",
"output": "NO"
},
{
"input": "aabbcccc",
"output": "NO"
},
{
"input": "abac",
"output": "NO"
},
{
"input": "abcc",
"output": "NO"
},
{
"input": "abcb",
"output": "NO"
},
{
"input": "aacc",
"output": "NO"
},
{
"input": "aabbaacccc",
"output": "NO"
},
{
"input": "aabb",
"output": "NO"
},
{
"input": "ac",
"output": "NO"
},
{
"input": "abbacc",
"output": "NO"
},
{
"input": "abacc",
"output": "NO"
},
{
"input": "ababc",
"output": "NO"
},
{
"input": "aa",
"output": "NO"
},
{
"input": "aabaccc",
"output": "NO"
},
{
"input": "bbcc",
"output": "NO"
},
{
"input": "aaabcbc",
"output": "NO"
},
{
"input": "acbbc",
"output": "NO"
},
{
"input": "babc",
"output": "NO"
},
{
"input": "bbbcc",
"output": "NO"
},
{
"input": "bbc",
"output": "NO"
},
{
"input": "abababccc",
"output": "NO"
},
{
"input": "ccbbaa",
"output": "NO"
}
] | 1,661,967,931 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 33 | 46 | 0 | s = input()
def func(s):
ca = 0
cb = 0
cc = 0
for i in range(1, len(s)):
if s[i-1] > s[i]:
return "NO"
for i in range(0, len(s)):
if s[i]=="a":
ca += 1
if s[i]=="b":
cb += 1
if s[i]=="c":
cc += 1
if (cc == ca or cc == cb or cc == ca + cb):
if ca <= 0 or cb <= 0:
return "NO"
return "YES"
else:
return 'NO'
print(func(s)) | Title: Check the string
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes).
Input Specification:
The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'.
Output Specification:
Print "YES" or "NO", according to the condition.
Demo Input:
['aaabccc\n', 'bbacc\n', 'aabc\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | ```python
s = input()
def func(s):
ca = 0
cb = 0
cc = 0
for i in range(1, len(s)):
if s[i-1] > s[i]:
return "NO"
for i in range(0, len(s)):
if s[i]=="a":
ca += 1
if s[i]=="b":
cb += 1
if s[i]=="c":
cc += 1
if (cc == ca or cc == cb or cc == ca + cb):
if ca <= 0 or cb <= 0:
return "NO"
return "YES"
else:
return 'NO'
print(func(s))
``` | 0 | |
505 | B | Mr. Kitayuta's Colorful Graph | PROGRAMMING | 1,400 | [
"dfs and similar",
"dp",
"dsu",
"graphs"
] | null | null | Mr. Kitayuta has just bought an undirected graph consisting of *n* vertices and *m* edges. The vertices of the graph are numbered from 1 to *n*. Each edge, namely edge *i*, has a color *c**i*, connecting vertex *a**i* and *b**i*.
Mr. Kitayuta wants you to process the following *q* queries.
In the *i*-th query, he gives you two integers — *u**i* and *v**i*.
Find the number of the colors that satisfy the following condition: the edges of that color connect vertex *u**i* and vertex *v**i* directly or indirectly. | The first line of the input contains space-separated two integers — *n* and *m* (2<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100), denoting the number of the vertices and the number of the edges, respectively.
The next *m* lines contain space-separated three integers — *a**i*, *b**i* (1<=≤<=*a**i*<=<<=*b**i*<=≤<=*n*) and *c**i* (1<=≤<=*c**i*<=≤<=*m*). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if *i*<=≠<=*j*, (*a**i*,<=*b**i*,<=*c**i*)<=≠<=(*a**j*,<=*b**j*,<=*c**j*).
The next line contains a integer — *q* (1<=≤<=*q*<=≤<=100), denoting the number of the queries.
Then follows *q* lines, containing space-separated two integers — *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*). It is guaranteed that *u**i*<=≠<=*v**i*. | For each query, print the answer in a separate line. | [
"4 5\n1 2 1\n1 2 2\n2 3 1\n2 3 3\n2 4 3\n3\n1 2\n3 4\n1 4\n",
"5 7\n1 5 1\n2 5 1\n3 5 1\n4 5 1\n1 2 2\n2 3 2\n3 4 2\n5\n1 5\n5 1\n2 5\n1 5\n1 4\n"
] | [
"2\n1\n0\n",
"1\n1\n1\n1\n2\n"
] | Let's consider the first sample.
- Vertex 1 and vertex 2 are connected by color 1 and 2. - Vertex 3 and vertex 4 are connected by color 3. - Vertex 1 and vertex 4 are not connected by any single color. | 1,000 | [
{
"input": "4 5\n1 2 1\n1 2 2\n2 3 1\n2 3 3\n2 4 3\n3\n1 2\n3 4\n1 4",
"output": "2\n1\n0"
},
{
"input": "5 7\n1 5 1\n2 5 1\n3 5 1\n4 5 1\n1 2 2\n2 3 2\n3 4 2\n5\n1 5\n5 1\n2 5\n1 5\n1 4",
"output": "1\n1\n1\n1\n2"
},
{
"input": "2 1\n1 2 1\n1\n1 2",
"output": "1"
},
{
"input": "2 3\n1 2 3\n1 2 2\n1 2 1\n1\n1 2",
"output": "3"
},
{
"input": "2 5\n1 2 1\n1 2 2\n1 2 3\n1 2 4\n1 2 5\n1\n1 2",
"output": "5"
}
] | 1,655,623,951 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 29 | 78 | 3,993,600 | def make_set(color, v):
if (parent[color][v] != -1):
return
parent[color][v] = v
rank[color][v] = 0
def find_set(color, v):
if (parent[color][v] == -1):
return -1
if (v == parent[color][v]):
return v
parent[color][v] = find_set(color, parent[color][v])
return parent[color][v]
def union_set(color, a, b):
a = find_set(color, a)
b = find_set(color, b)
if (a != b):
if (rank[color][a] < rank[color][b]):
a, b = b, a
parent[color][b] = a
if (rank[color][a] == rank[color][b]):
rank[color][a] += 1
n, m = list(map(int, input().split()))
parent = [[-1 for _ in range(n+1)] for _ in range(m+1)]
rank = [[-1 for _ in range(n+1)] for _ in range(m+1)]
for i in range(m):
a, b, c = list(map(int, input().split()))
make_set(c, a)
make_set(c, b)
union_set(c, a, b)
q = int(input())
for i in range(q):
u, v = list(map(int, input().split()))
result = 0
for j in range(m+1):
parent_u = find_set(j, u)
parent_v = find_set(j, v)
if (parent_u == -1 or parent_v == -1):
continue
if (parent_u == parent_v):
result += 1
print(result)
| Title: Mr. Kitayuta's Colorful Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mr. Kitayuta has just bought an undirected graph consisting of *n* vertices and *m* edges. The vertices of the graph are numbered from 1 to *n*. Each edge, namely edge *i*, has a color *c**i*, connecting vertex *a**i* and *b**i*.
Mr. Kitayuta wants you to process the following *q* queries.
In the *i*-th query, he gives you two integers — *u**i* and *v**i*.
Find the number of the colors that satisfy the following condition: the edges of that color connect vertex *u**i* and vertex *v**i* directly or indirectly.
Input Specification:
The first line of the input contains space-separated two integers — *n* and *m* (2<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100), denoting the number of the vertices and the number of the edges, respectively.
The next *m* lines contain space-separated three integers — *a**i*, *b**i* (1<=≤<=*a**i*<=<<=*b**i*<=≤<=*n*) and *c**i* (1<=≤<=*c**i*<=≤<=*m*). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if *i*<=≠<=*j*, (*a**i*,<=*b**i*,<=*c**i*)<=≠<=(*a**j*,<=*b**j*,<=*c**j*).
The next line contains a integer — *q* (1<=≤<=*q*<=≤<=100), denoting the number of the queries.
Then follows *q* lines, containing space-separated two integers — *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*). It is guaranteed that *u**i*<=≠<=*v**i*.
Output Specification:
For each query, print the answer in a separate line.
Demo Input:
['4 5\n1 2 1\n1 2 2\n2 3 1\n2 3 3\n2 4 3\n3\n1 2\n3 4\n1 4\n', '5 7\n1 5 1\n2 5 1\n3 5 1\n4 5 1\n1 2 2\n2 3 2\n3 4 2\n5\n1 5\n5 1\n2 5\n1 5\n1 4\n']
Demo Output:
['2\n1\n0\n', '1\n1\n1\n1\n2\n']
Note:
Let's consider the first sample.
- Vertex 1 and vertex 2 are connected by color 1 and 2. - Vertex 3 and vertex 4 are connected by color 3. - Vertex 1 and vertex 4 are not connected by any single color. | ```python
def make_set(color, v):
if (parent[color][v] != -1):
return
parent[color][v] = v
rank[color][v] = 0
def find_set(color, v):
if (parent[color][v] == -1):
return -1
if (v == parent[color][v]):
return v
parent[color][v] = find_set(color, parent[color][v])
return parent[color][v]
def union_set(color, a, b):
a = find_set(color, a)
b = find_set(color, b)
if (a != b):
if (rank[color][a] < rank[color][b]):
a, b = b, a
parent[color][b] = a
if (rank[color][a] == rank[color][b]):
rank[color][a] += 1
n, m = list(map(int, input().split()))
parent = [[-1 for _ in range(n+1)] for _ in range(m+1)]
rank = [[-1 for _ in range(n+1)] for _ in range(m+1)]
for i in range(m):
a, b, c = list(map(int, input().split()))
make_set(c, a)
make_set(c, b)
union_set(c, a, b)
q = int(input())
for i in range(q):
u, v = list(map(int, input().split()))
result = 0
for j in range(m+1):
parent_u = find_set(j, u)
parent_v = find_set(j, v)
if (parent_u == -1 or parent_v == -1):
continue
if (parent_u == parent_v):
result += 1
print(result)
``` | 3 | |
385 | B | Bear and Strings | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"implementation",
"math",
"strings"
] | null | null | The bear has a string *s*<==<=*s*1*s*2... *s*|*s*| (record |*s*| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices *i*,<=*j* (1<=≤<=*i*<=≤<=*j*<=≤<=|*s*|), that string *x*(*i*,<=*j*)<==<=*s**i**s**i*<=+<=1... *s**j* contains at least one string "bear" as a substring.
String *x*(*i*,<=*j*) contains string "bear", if there is such index *k* (*i*<=≤<=*k*<=≤<=*j*<=-<=3), that *s**k*<==<=*b*, *s**k*<=+<=1<==<=*e*, *s**k*<=+<=2<==<=*a*, *s**k*<=+<=3<==<=*r*.
Help the bear cope with the given problem. | The first line contains a non-empty string *s* (1<=≤<=|*s*|<=≤<=5000). It is guaranteed that the string only consists of lowercase English letters. | Print a single number — the answer to the problem. | [
"bearbtear\n",
"bearaabearc\n"
] | [
"6\n",
"20\n"
] | In the first sample, the following pairs (*i*, *j*) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (*i*, *j*) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11). | 1,000 | [
{
"input": "bearbtear",
"output": "6"
},
{
"input": "bearaabearc",
"output": "20"
},
{
"input": "pbearbearhbearzqbearjkterasjhy",
"output": "291"
},
{
"input": "pbearjbearbebearnbabcffbearbearwubearjezpiorrbearbearjbdlbearbearqbearjbearwipmsbearoaftrsebearzsnqb",
"output": "4419"
},
{
"input": "bear",
"output": "1"
},
{
"input": "a",
"output": "0"
},
{
"input": "be",
"output": "0"
}
] | 1,393,943,163 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 62 | 307,200 | # input part
string = input()
# solution
result = 0
pos = string.find("bear")
while pos >= 0:
result += len(string) - (pos + 3)
string = string[1:]
pos = string.find("bear")
print(result) | Title: Bear and Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The bear has a string *s*<==<=*s*1*s*2... *s*|*s*| (record |*s*| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices *i*,<=*j* (1<=≤<=*i*<=≤<=*j*<=≤<=|*s*|), that string *x*(*i*,<=*j*)<==<=*s**i**s**i*<=+<=1... *s**j* contains at least one string "bear" as a substring.
String *x*(*i*,<=*j*) contains string "bear", if there is such index *k* (*i*<=≤<=*k*<=≤<=*j*<=-<=3), that *s**k*<==<=*b*, *s**k*<=+<=1<==<=*e*, *s**k*<=+<=2<==<=*a*, *s**k*<=+<=3<==<=*r*.
Help the bear cope with the given problem.
Input Specification:
The first line contains a non-empty string *s* (1<=≤<=|*s*|<=≤<=5000). It is guaranteed that the string only consists of lowercase English letters.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['bearbtear\n', 'bearaabearc\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, the following pairs (*i*, *j*) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (*i*, *j*) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11). | ```python
# input part
string = input()
# solution
result = 0
pos = string.find("bear")
while pos >= 0:
result += len(string) - (pos + 3)
string = string[1:]
pos = string.find("bear")
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"
}
] | 1,577,713,285 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 124 | 307,200 | word = input()
try:
h_index = word.index("h")
e_index = word[h_index + 1:].index("e") + len(word) - len(word[h_index + 1:])
l1_index = word[e_index + 1:].index("l") + len(word) - len(word[e_index + 1:])
l2_index = word[l1_index + 1:].index("l") + len(word) - len(word[l1_index + 1:])
o_index = word[l2_index + 1:].index("o") + len(word) - len(word[l2_index + 1:])
print("YES")
except Exception:
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
word = input()
try:
h_index = word.index("h")
e_index = word[h_index + 1:].index("e") + len(word) - len(word[h_index + 1:])
l1_index = word[e_index + 1:].index("l") + len(word) - len(word[e_index + 1:])
l2_index = word[l1_index + 1:].index("l") + len(word) - len(word[l1_index + 1:])
o_index = word[l2_index + 1:].index("o") + len(word) - len(word[l2_index + 1:])
print("YES")
except Exception:
print("NO")
``` | 3.937428 |
277 | A | Learning Languages | PROGRAMMING | 1,400 | [
"dfs and similar",
"dsu"
] | null | null | The "BerCorp" company has got *n* employees. These employees can use *m* approved official languages for the formal correspondence. The languages are numbered with integers from 1 to *m*. For each employee we have the list of languages, which he knows. This list could be empty, i. e. an employee may know no official languages. But the employees are willing to learn any number of official languages, as long as the company pays their lessons. A study course in one language for one employee costs 1 berdollar.
Find the minimum sum of money the company needs to spend so as any employee could correspond to any other one (their correspondence can be indirect, i. e. other employees can help out translating). | The first line contains two integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of employees and the number of languages.
Then *n* lines follow — each employee's language list. At the beginning of the *i*-th line is integer *k**i* (0<=≤<=*k**i*<=≤<=*m*) — the number of languages the *i*-th employee knows. Next, the *i*-th line contains *k**i* integers — *a**ij* (1<=≤<=*a**ij*<=≤<=*m*) — the identifiers of languages the *i*-th employee knows. It is guaranteed that all the identifiers in one list are distinct. Note that an employee may know zero languages.
The numbers in the lines are separated by single spaces. | Print a single integer — the minimum amount of money to pay so that in the end every employee could write a letter to every other one (other employees can help out translating). | [
"5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5\n",
"8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1\n",
"2 2\n1 2\n0\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the second sample the employee 1 can learn language 2, and employee 8 can learn language 4.
In the third sample employee 2 must learn language 2. | 500 | [
{
"input": "5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5",
"output": "0"
},
{
"input": "8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1",
"output": "2"
},
{
"input": "2 2\n1 2\n0",
"output": "1"
},
{
"input": "2 2\n0\n0",
"output": "2"
},
{
"input": "5 5\n1 3\n0\n0\n2 4 1\n0",
"output": "4"
},
{
"input": "6 2\n0\n0\n2 1 2\n1 1\n1 1\n0",
"output": "3"
},
{
"input": "7 3\n3 1 3 2\n3 2 1 3\n2 2 3\n1 1\n2 2 3\n3 3 2 1\n3 2 3 1",
"output": "0"
},
{
"input": "8 4\n0\n0\n4 2 3 1 4\n4 2 1 4 3\n3 4 3 1\n1 2\n2 4 1\n2 4 2",
"output": "2"
},
{
"input": "10 10\n5 7 5 2 8 1\n7 10 6 9 5 8 2 4\n2 2 7\n5 8 6 9 10 1\n2 9 5\n3 6 5 2\n6 5 8 7 9 10 4\n0\n1 1\n2 8 6",
"output": "1"
},
{
"input": "11 42\n4 20 26 9 24\n14 34 7 28 32 12 15 26 4 10 38 21 20 8 11\n4 21 8 36 6\n11 32 1 39 11 21 10 25 17 26 15 4\n2 8 12\n2 21 31\n8 17 10 3 39 32 30 5 15\n20 24 20 38 17 4 7 21 19 32 28 31 22 30 37 10 5 33 2 13 9\n7 38 34 42 27 20 11 6\n3 40 3 39\n14 39 40 4 30 33 8 36 28 14 23 16 7 25 9",
"output": "0"
},
{
"input": "100 100\n1 33\n0\n2 35 4\n2 40 78\n1 69\n0\n1 2\n0\n0\n2 81 34\n1 45\n0\n1 87\n1 50\n1 97\n0\n2 100 59\n0\n0\n0\n1 45\n0\n0\n0\n1 6\n1 54\n0\n0\n0\n4 79 96 52 84\n1 86\n0\n0\n0\n0\n0\n0\n0\n0\n3 90 2 80\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 2\n0\n1 64\n0\n1 78\n1 82\n0\n0\n0\n0\n0\n1 6\n0\n0\n2 47 57\n1 95\n0\n2 91 79\n0\n1 27\n0\n1 74\n0\n0\n1 14\n0\n0\n2 90 19\n0\n1 10\n0\n0\n0\n0\n0\n1 57\n0\n2 28 50\n0\n0\n0\n1 47\n0\n0\n1 14\n0\n1 84\n1 1\n0\n0",
"output": "87"
},
{
"input": "2 2\n2 1 2\n2 1 2",
"output": "0"
},
{
"input": "2 2\n2 1 2\n1 1",
"output": "0"
},
{
"input": "2 2\n1 2\n1 1",
"output": "1"
},
{
"input": "3 100\n0\n0\n0",
"output": "3"
},
{
"input": "100 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "100"
},
{
"input": "3 3\n0\n0\n0",
"output": "3"
}
] | 1,664,008,718 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 19 | 248 | 3,584,000 | from collections import defaultdict
n, m = list(map(int, input().split()))
language_speakers = defaultdict(set)
total_languages_spoken = set()
zero_speakers = 0
for i in range(n):
li = input().split()
if li[0] == '0':
zero_speakers += 1
for j in li[1:]:
language_speakers[j].add(i) # ith person speakes this language
_max_language = 0
need = 0
groups_found = set()
for i in list(language_speakers.keys()):
for j in list(language_speakers.keys()):
if len(language_speakers[i].intersection(language_speakers[j])) > 0:
language_speakers[i] = language_speakers[i].union(language_speakers[j])
first = True
for person in language_speakers[i]:
if person not in groups_found:
groups_found.add(person)
if first:
first = False
need += 1
_max_language = max(_max_language, len(language_speakers[i]))
print(max(0, need - 1) + zero_speakers) | Title: Learning Languages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The "BerCorp" company has got *n* employees. These employees can use *m* approved official languages for the formal correspondence. The languages are numbered with integers from 1 to *m*. For each employee we have the list of languages, which he knows. This list could be empty, i. e. an employee may know no official languages. But the employees are willing to learn any number of official languages, as long as the company pays their lessons. A study course in one language for one employee costs 1 berdollar.
Find the minimum sum of money the company needs to spend so as any employee could correspond to any other one (their correspondence can be indirect, i. e. other employees can help out translating).
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of employees and the number of languages.
Then *n* lines follow — each employee's language list. At the beginning of the *i*-th line is integer *k**i* (0<=≤<=*k**i*<=≤<=*m*) — the number of languages the *i*-th employee knows. Next, the *i*-th line contains *k**i* integers — *a**ij* (1<=≤<=*a**ij*<=≤<=*m*) — the identifiers of languages the *i*-th employee knows. It is guaranteed that all the identifiers in one list are distinct. Note that an employee may know zero languages.
The numbers in the lines are separated by single spaces.
Output Specification:
Print a single integer — the minimum amount of money to pay so that in the end every employee could write a letter to every other one (other employees can help out translating).
Demo Input:
['5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5\n', '8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1\n', '2 2\n1 2\n0\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the second sample the employee 1 can learn language 2, and employee 8 can learn language 4.
In the third sample employee 2 must learn language 2. | ```python
from collections import defaultdict
n, m = list(map(int, input().split()))
language_speakers = defaultdict(set)
total_languages_spoken = set()
zero_speakers = 0
for i in range(n):
li = input().split()
if li[0] == '0':
zero_speakers += 1
for j in li[1:]:
language_speakers[j].add(i) # ith person speakes this language
_max_language = 0
need = 0
groups_found = set()
for i in list(language_speakers.keys()):
for j in list(language_speakers.keys()):
if len(language_speakers[i].intersection(language_speakers[j])) > 0:
language_speakers[i] = language_speakers[i].union(language_speakers[j])
first = True
for person in language_speakers[i]:
if person not in groups_found:
groups_found.add(person)
if first:
first = False
need += 1
_max_language = max(_max_language, len(language_speakers[i]))
print(max(0, need - 1) + zero_speakers)
``` | 0 | |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,684,774,852 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | s=input()
k=set()
for i in s:
k.add(i)
if len(k)%2!=0:
print("IGNORE HIM!")
else:
print("CHAT WITH HER!")
| Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
s=input()
k=set()
for i in s:
k.add(i)
if len(k)%2!=0:
print("IGNORE HIM!")
else:
print("CHAT WITH HER!")
``` | 0 |
931 | A | Friends Meeting | PROGRAMMING | 800 | [
"brute force",
"greedy",
"implementation",
"math"
] | null | null | Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*.
Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6.
The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point. | The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend.
The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend.
It is guaranteed that *a*<=≠<=*b*. | Print the minimum possible total tiredness if the friends meet in the same point. | [
"3\n4\n",
"101\n99\n",
"5\n10\n"
] | [
"1\n",
"2\n",
"9\n"
] | In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1.
In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2.
In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9. | 500 | [
{
"input": "3\n4",
"output": "1"
},
{
"input": "101\n99",
"output": "2"
},
{
"input": "5\n10",
"output": "9"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n1000",
"output": "250000"
},
{
"input": "999\n1000",
"output": "1"
},
{
"input": "1000\n999",
"output": "1"
},
{
"input": "1000\n1",
"output": "250000"
},
{
"input": "2\n1",
"output": "1"
},
{
"input": "2\n999",
"output": "249001"
},
{
"input": "2\n998",
"output": "248502"
},
{
"input": "999\n2",
"output": "249001"
},
{
"input": "998\n2",
"output": "248502"
},
{
"input": "2\n1000",
"output": "249500"
},
{
"input": "1000\n2",
"output": "249500"
},
{
"input": "1\n999",
"output": "249500"
},
{
"input": "999\n1",
"output": "249500"
},
{
"input": "188\n762",
"output": "82656"
},
{
"input": "596\n777",
"output": "8281"
},
{
"input": "773\n70",
"output": "123904"
},
{
"input": "825\n729",
"output": "2352"
},
{
"input": "944\n348",
"output": "89102"
},
{
"input": "352\n445",
"output": "2209"
},
{
"input": "529\n656",
"output": "4096"
},
{
"input": "19\n315",
"output": "22052"
},
{
"input": "138\n370",
"output": "13572"
},
{
"input": "546\n593",
"output": "576"
},
{
"input": "285\n242",
"output": "484"
},
{
"input": "773\n901",
"output": "4160"
},
{
"input": "892\n520",
"output": "34782"
},
{
"input": "864\n179",
"output": "117649"
},
{
"input": "479\n470",
"output": "25"
},
{
"input": "967\n487",
"output": "57840"
},
{
"input": "648\n106",
"output": "73712"
},
{
"input": "58\n765",
"output": "125316"
},
{
"input": "235\n56",
"output": "8100"
},
{
"input": "285\n153",
"output": "4422"
},
{
"input": "943\n13",
"output": "216690"
},
{
"input": "675\n541",
"output": "4556"
},
{
"input": "4\n912",
"output": "206570"
}
] | 1,525,461,682 | 142 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define inf 0x3f3f3f3f
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define ii pair<int, int>
#define all(x) (x).begin(), (x).end()
#define N 1000007 // 10e6 + 7
int main(int argc, char const *argv[]){
int a, b;
cin >> a >> b;
if(a>b)swap(a, b);
int c=1;
int tot=0;
while(a!=b){
a++;
tot+=c;
if(a!=b){
b--;
tot+=c;
}
c++;
}
cout << tot << endl;
return 0;
} | Title: Friends Meeting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*.
Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6.
The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point.
Input Specification:
The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend.
The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend.
It is guaranteed that *a*<=≠<=*b*.
Output Specification:
Print the minimum possible total tiredness if the friends meet in the same point.
Demo Input:
['3\n4\n', '101\n99\n', '5\n10\n']
Demo Output:
['1\n', '2\n', '9\n']
Note:
In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1.
In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2.
In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9. | ```python
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define inf 0x3f3f3f3f
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define ii pair<int, int>
#define all(x) (x).begin(), (x).end()
#define N 1000007 // 10e6 + 7
int main(int argc, char const *argv[]){
int a, b;
cin >> a >> b;
if(a>b)swap(a, b);
int c=1;
int tot=0;
while(a!=b){
a++;
tot+=c;
if(a!=b){
b--;
tot+=c;
}
c++;
}
cout << tot << endl;
return 0;
}
``` | -1 | |
479 | A | Expression | PROGRAMMING | 1,000 | [
"brute force",
"math"
] | null | null | Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get. | The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10). | Print the maximum value of the expression that you can obtain. | [
"1\n2\n3\n",
"2\n10\n3\n"
] | [
"9\n",
"60\n"
] | none | 500 | [
{
"input": "1\n2\n3",
"output": "9"
},
{
"input": "2\n10\n3",
"output": "60"
},
{
"input": "1\n1\n1",
"output": "3"
},
{
"input": "1\n2\n1",
"output": "4"
},
{
"input": "10\n10\n10",
"output": "1000"
},
{
"input": "5\n1\n3",
"output": "20"
},
{
"input": "3\n1\n5",
"output": "20"
},
{
"input": "6\n7\n1",
"output": "48"
},
{
"input": "1\n8\n3",
"output": "27"
},
{
"input": "9\n7\n2",
"output": "126"
},
{
"input": "1\n1\n10",
"output": "20"
},
{
"input": "9\n1\n1",
"output": "18"
},
{
"input": "10\n5\n6",
"output": "300"
},
{
"input": "8\n9\n7",
"output": "504"
},
{
"input": "4\n2\n10",
"output": "80"
},
{
"input": "2\n8\n3",
"output": "48"
},
{
"input": "3\n5\n7",
"output": "105"
},
{
"input": "1\n10\n1",
"output": "12"
},
{
"input": "2\n2\n2",
"output": "8"
},
{
"input": "5\n6\n1",
"output": "35"
},
{
"input": "10\n1\n1",
"output": "20"
},
{
"input": "1\n6\n1",
"output": "8"
},
{
"input": "1\n9\n1",
"output": "11"
},
{
"input": "2\n1\n2",
"output": "6"
},
{
"input": "2\n6\n1",
"output": "14"
},
{
"input": "9\n2\n1",
"output": "27"
},
{
"input": "1\n9\n2",
"output": "20"
},
{
"input": "1\n3\n1",
"output": "5"
},
{
"input": "2\n1\n1",
"output": "4"
}
] | 1,694,877,986 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 29 | 62 | 0 | a,b,c = int(input()),int(input()),int(input())
ans = a * b * c
ans = max(ans, a + b + c, (a + b) * c, a * (b + c), a * b + c, a + b * c)
print(ans)
| Title: Expression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
Input Specification:
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
Output Specification:
Print the maximum value of the expression that you can obtain.
Demo Input:
['1\n2\n3\n', '2\n10\n3\n']
Demo Output:
['9\n', '60\n']
Note:
none | ```python
a,b,c = int(input()),int(input()),int(input())
ans = a * b * c
ans = max(ans, a + b + c, (a + b) * c, a * (b + c), a * b + c, a + b * c)
print(ans)
``` | 3 | |
837 | A | Text Volume | PROGRAMMING | 800 | [
"implementation"
] | null | null | You are given a text of single-space separated words, consisting of small and capital Latin letters.
Volume of the word is number of capital letters in the word. Volume of the text is maximum volume of all words in the text.
Calculate the volume of the given text. | The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200) — length of the text.
The second line contains text of single-space separated words *s*1,<=*s*2,<=...,<=*s**i*, consisting only of small and capital Latin letters. | Print one integer number — volume of text. | [
"7\nNonZERO\n",
"24\nthis is zero answer text\n",
"24\nHarbour Space University\n"
] | [
"5\n",
"0\n",
"1\n"
] | In the first example there is only one word, there are 5 capital letters in it.
In the second example all of the words contain 0 capital letters. | 0 | [
{
"input": "7\nNonZERO",
"output": "5"
},
{
"input": "24\nthis is zero answer text",
"output": "0"
},
{
"input": "24\nHarbour Space University",
"output": "1"
},
{
"input": "2\nWM",
"output": "2"
},
{
"input": "200\nLBmJKQLCKUgtTxMoDsEerwvLOXsxASSydOqWyULsRcjMYDWdDCgaDvBfATIWPVSXlbcCLHPYahhxMEYUiaxoCebghJqvmRnaNHYTKLeOiaLDnATPZAOgSNfBzaxLymTGjfzvTegbXsAthTxyDTcmBUkqyGlVGZhoazQzVSoKbTFcCRvYsgSCwjGMxBfWEwMHuagTBxkz",
"output": "105"
},
{
"input": "199\no A r v H e J q k J k v w Q F p O R y R Z o a K R L Z E H t X y X N y y p b x B m r R S q i A x V S u i c L y M n N X c C W Z m S j e w C w T r I S X T D F l w o k f t X u n W w p Z r A k I Y E h s g",
"output": "1"
},
{
"input": "200\nhCyIdivIiISmmYIsCLbpKcTyHaOgTUQEwnQACXnrLdHAVFLtvliTEMlzBVzTesQbhXmcqvwPDeojglBMIjOXANfyQxCSjOJyO SIqOTnRzVzseGIDDYNtrwIusScWSuEhPyEmgQIVEzXofRptjeMzzhtUQxJgcUWILUhEaaRmYRBVsjoqgmyPIKwSajdlNPccOOtWrez",
"output": "50"
},
{
"input": "1\ne",
"output": "0"
},
{
"input": "1\nA",
"output": "1"
},
{
"input": "200\nABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ ABCDEFGHIJ KLMNOPRSTU KLMNOPRSTU KLMNOPRSTU VWXYZABCDE KLMNOPRSTU KLMNOPRSTU KLMNOPRSTU KLMNOPRSTU KZ",
"output": "10"
},
{
"input": "200\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "200"
},
{
"input": "200\nffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff",
"output": "0"
},
{
"input": "24\nHarbour Space UniversitY",
"output": "2"
},
{
"input": "5\naA AA",
"output": "2"
},
{
"input": "10\nas AS ASDA",
"output": "4"
},
{
"input": "10\nas AS ASDZ",
"output": "4"
},
{
"input": "3\na A",
"output": "1"
},
{
"input": "24\nHarbour space UniversitY",
"output": "2"
},
{
"input": "10\nas AS ASAa",
"output": "3"
},
{
"input": "15\naAb ABCDFGRHTJS",
"output": "11"
},
{
"input": "53\nsdfAZEZR AZE dfdf dsdRFGSDF ZZDZSD dfsd ERBGF dsfsdfR",
"output": "6"
},
{
"input": "10\nABC ABc AB",
"output": "3"
},
{
"input": "10\nA c de CDE",
"output": "3"
},
{
"input": "4\nA AB",
"output": "2"
},
{
"input": "18\nHARbour Space UNIV",
"output": "4"
},
{
"input": "13\na b c d e f A",
"output": "1"
},
{
"input": "6\nA B CA",
"output": "2"
},
{
"input": "4\naa A",
"output": "1"
},
{
"input": "3\nA a",
"output": "1"
}
] | 1,631,708,384 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 62 | 6,758,400 | x=input()
n=input().split(' ')
ans=0
for i in n:
cur=0
for j in i:
if j>='A' and j<='Z':
cur+=1
ans=max(ans,cur)
print(ans) | Title: Text Volume
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a text of single-space separated words, consisting of small and capital Latin letters.
Volume of the word is number of capital letters in the word. Volume of the text is maximum volume of all words in the text.
Calculate the volume of the given text.
Input Specification:
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200) — length of the text.
The second line contains text of single-space separated words *s*1,<=*s*2,<=...,<=*s**i*, consisting only of small and capital Latin letters.
Output Specification:
Print one integer number — volume of text.
Demo Input:
['7\nNonZERO\n', '24\nthis is zero answer text\n', '24\nHarbour Space University\n']
Demo Output:
['5\n', '0\n', '1\n']
Note:
In the first example there is only one word, there are 5 capital letters in it.
In the second example all of the words contain 0 capital letters. | ```python
x=input()
n=input().split(' ')
ans=0
for i in n:
cur=0
for j in i:
if j>='A' and j<='Z':
cur+=1
ans=max(ans,cur)
print(ans)
``` | 3 | |
489 | B | BerSU Ball | PROGRAMMING | 1,200 | [
"dfs and similar",
"dp",
"graph matchings",
"greedy",
"sortings",
"two pointers"
] | null | null | The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! *n* boys and *m* girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from *n* boys and *m* girls. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of boys. The second line contains sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is the *i*-th boy's dancing skill.
Similarly, the third line contains an integer *m* (1<=≤<=*m*<=≤<=100) — the number of girls. The fourth line contains sequence *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=100), where *b**j* is the *j*-th girl's dancing skill. | Print a single number — the required maximum possible number of pairs. | [
"4\n1 4 6 2\n5\n5 1 5 7 9\n",
"4\n1 2 3 4\n4\n10 11 12 13\n",
"5\n1 1 1 1 1\n3\n1 2 3\n"
] | [
"3\n",
"0\n",
"2\n"
] | none | 1,000 | [
{
"input": "4\n1 4 6 2\n5\n5 1 5 7 9",
"output": "3"
},
{
"input": "4\n1 2 3 4\n4\n10 11 12 13",
"output": "0"
},
{
"input": "5\n1 1 1 1 1\n3\n1 2 3",
"output": "2"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1 10\n1\n9",
"output": "1"
},
{
"input": "4\n4 5 4 4\n5\n5 3 4 2 4",
"output": "4"
},
{
"input": "1\n2\n1\n1",
"output": "1"
},
{
"input": "1\n3\n2\n3 2",
"output": "1"
},
{
"input": "1\n4\n3\n4 4 4",
"output": "1"
},
{
"input": "1\n2\n4\n3 1 4 2",
"output": "1"
},
{
"input": "1\n4\n5\n2 5 5 3 1",
"output": "1"
},
{
"input": "2\n2 2\n1\n2",
"output": "1"
},
{
"input": "2\n4 2\n2\n4 4",
"output": "1"
},
{
"input": "2\n4 1\n3\n2 3 2",
"output": "2"
},
{
"input": "2\n4 3\n4\n5 5 5 6",
"output": "1"
},
{
"input": "2\n5 7\n5\n4 6 7 2 5",
"output": "2"
},
{
"input": "3\n1 2 3\n1\n1",
"output": "1"
},
{
"input": "3\n5 4 5\n2\n2 1",
"output": "0"
},
{
"input": "3\n6 3 4\n3\n4 5 2",
"output": "3"
},
{
"input": "3\n7 7 7\n4\n2 7 2 4",
"output": "1"
},
{
"input": "3\n1 3 3\n5\n1 3 4 1 2",
"output": "3"
},
{
"input": "4\n1 2 1 3\n1\n4",
"output": "1"
},
{
"input": "4\n4 4 6 6\n2\n2 1",
"output": "0"
},
{
"input": "4\n3 1 1 1\n3\n1 6 7",
"output": "1"
},
{
"input": "4\n2 5 1 2\n4\n2 3 3 1",
"output": "3"
},
{
"input": "4\n9 1 7 1\n5\n9 9 9 8 4",
"output": "2"
},
{
"input": "5\n1 6 5 5 6\n1\n2",
"output": "1"
},
{
"input": "5\n5 2 4 5 6\n2\n7 4",
"output": "2"
},
{
"input": "5\n4 1 3 1 4\n3\n6 3 6",
"output": "1"
},
{
"input": "5\n5 2 3 1 4\n4\n1 3 1 7",
"output": "3"
},
{
"input": "5\n9 8 10 9 10\n5\n2 1 5 4 6",
"output": "0"
},
{
"input": "1\n48\n100\n76 90 78 44 29 30 35 85 98 38 27 71 51 100 15 98 78 45 85 26 48 66 98 71 45 85 83 77 92 17 23 95 98 43 11 15 39 53 71 25 74 53 77 41 39 35 66 4 92 44 44 55 35 87 91 6 44 46 57 24 46 82 15 44 81 40 65 17 64 24 42 52 13 12 64 82 26 7 66 85 93 89 58 92 92 77 37 91 47 73 35 69 31 22 60 60 97 21 52 6",
"output": "1"
},
{
"input": "100\n9 90 66 62 60 9 10 97 47 73 26 81 97 60 80 84 19 4 25 77 19 17 91 12 1 27 15 54 18 45 71 79 96 90 51 62 9 13 92 34 7 52 55 8 16 61 96 12 52 38 50 9 60 3 30 3 48 46 77 64 90 35 16 16 21 42 67 70 23 19 90 14 50 96 98 92 82 62 7 51 93 38 84 82 37 78 99 3 20 69 44 96 94 71 3 55 27 86 92 82\n1\n58",
"output": "0"
},
{
"input": "10\n20 87 3 39 20 20 8 40 70 51\n100\n69 84 81 84 35 97 69 68 63 97 85 80 95 58 70 91 100 65 72 80 41 87 87 87 22 49 96 96 78 96 97 56 90 31 62 98 89 74 100 86 95 88 66 54 93 62 41 60 95 79 29 69 63 70 52 63 87 58 54 52 48 57 26 75 39 61 98 78 52 73 99 49 74 50 59 90 31 97 16 85 63 72 81 68 75 59 70 67 73 92 10 88 57 95 3 71 80 95 84 96",
"output": "6"
},
{
"input": "100\n10 10 9 18 56 64 92 66 54 42 66 65 58 5 74 68 80 57 58 30 58 69 70 13 38 19 34 63 38 17 26 24 66 83 48 77 44 37 78 97 13 90 51 56 60 23 49 32 14 86 90 100 13 14 52 69 85 95 81 53 5 3 91 66 2 64 45 59 7 30 80 42 61 82 70 10 62 82 5 34 50 28 24 47 85 68 27 50 24 61 76 17 63 24 3 67 83 76 42 60\n10\n66 74 40 67 28 82 99 57 93 64",
"output": "9"
},
{
"input": "100\n4 1 1 1 3 3 2 5 1 2 1 2 1 1 1 6 1 3 1 1 1 1 2 4 1 1 4 2 2 8 2 2 1 8 2 4 3 3 8 1 3 2 3 2 1 3 8 2 2 3 1 1 2 2 5 1 4 3 1 1 3 1 3 1 7 1 1 1 3 2 1 2 2 3 7 2 1 4 3 2 1 1 3 4 1 1 3 5 1 8 4 1 1 1 3 10 2 2 1 2\n100\n1 1 5 2 13 2 2 3 6 12 1 13 8 1 1 16 1 1 5 6 2 4 6 4 2 7 4 1 7 3 3 9 5 3 1 7 4 1 6 6 8 2 2 5 2 3 16 3 6 3 8 6 1 8 1 2 6 5 3 4 11 3 4 8 2 13 2 5 2 7 3 3 1 8 1 4 4 2 4 7 7 1 5 7 6 3 6 9 1 1 1 3 1 11 5 2 5 11 13 1",
"output": "76"
},
{
"input": "4\n1 6 9 15\n2\n5 8",
"output": "2"
},
{
"input": "2\n2 4\n2\n3 1",
"output": "2"
},
{
"input": "3\n2 3 5\n3\n3 4 6",
"output": "3"
},
{
"input": "3\n1 3 4\n3\n2 1 5",
"output": "3"
},
{
"input": "2\n5 5\n4\n1 1 1 5",
"output": "1"
},
{
"input": "2\n3 2\n2\n3 4",
"output": "2"
},
{
"input": "2\n3 1\n2\n2 4",
"output": "2"
},
{
"input": "2\n2 3\n2\n2 1",
"output": "2"
},
{
"input": "2\n10 12\n2\n11 9",
"output": "2"
},
{
"input": "3\n1 2 3\n3\n3 2 1",
"output": "3"
},
{
"input": "2\n1 3\n2\n2 1",
"output": "2"
},
{
"input": "2\n4 5\n2\n5 3",
"output": "2"
},
{
"input": "2\n7 5\n2\n6 8",
"output": "2"
},
{
"input": "4\n4 3 2 1\n4\n1 2 3 4",
"output": "4"
},
{
"input": "2\n2 3\n2\n3 1",
"output": "2"
},
{
"input": "2\n2 4\n3\n3 1 8",
"output": "2"
},
{
"input": "3\n3 1 1\n3\n2 4 4",
"output": "2"
},
{
"input": "2\n5 3\n2\n4 6",
"output": "2"
},
{
"input": "4\n1 1 3 3\n4\n2 2 1 1",
"output": "4"
},
{
"input": "3\n3 2 1\n3\n2 4 3",
"output": "3"
},
{
"input": "5\n1 2 3 4 5\n5\n2 3 4 5 1",
"output": "5"
},
{
"input": "3\n3 2 1\n3\n1 2 3",
"output": "3"
},
{
"input": "2\n5 4\n2\n4 6",
"output": "2"
},
{
"input": "4\n3 3 5 5\n4\n4 4 2 2",
"output": "4"
},
{
"input": "3\n2 7 5\n3\n2 4 8",
"output": "3"
},
{
"input": "100\n2 3 3 4 2 1 4 4 5 5 2 1 5 2 3 3 5 4 3 2 4 2 3 3 2 2 3 4 2 2 2 3 1 2 3 2 2 3 5 3 3 3 3 4 5 2 2 1 1 1 3 1 2 2 3 5 5 2 5 1 3 4 5 3 5 4 1 1 2 3 4 4 5 3 2 4 5 5 5 2 1 4 2 4 5 4 4 5 5 3 2 5 1 4 4 2 2 2 5 3\n100\n4 5 3 3 2 2 4 3 1 5 4 3 3 2 2 4 5 2 5 2 1 4 3 4 2 3 5 3 4 4 1 2 3 5 2 2 1 5 4 2 4 3 4 3 4 2 3 1 3 3 4 1 1 1 4 4 5 3 1 4 2 3 2 1 3 3 2 3 2 1 1 2 3 2 1 3 3 4 3 3 1 1 3 3 3 1 1 3 5 3 3 3 3 4 4 5 2 5 4 5",
"output": "100"
},
{
"input": "1\n3\n2\n2 3",
"output": "1"
},
{
"input": "2\n5 6\n3\n1 5 100",
"output": "1"
},
{
"input": "2\n2 7\n2\n6 8",
"output": "1"
},
{
"input": "4\n4 10 15 17\n4\n3 12 16 16",
"output": "3"
},
{
"input": "2\n2 3\n2\n1 2",
"output": "2"
}
] | 1,693,323,376 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | import sys
def main():
n, m, pairs = int(sys.stdin.readline()), int(sys.stdin.readline()), 0
a = list(map(int, sys.stdin.readline().split()))
b = list(map(int, sys.stdin.readline().split()))
a.sort()
b.sort()
for i in range(n):
for j in range(m):
if abs(a[i] - b[j]) < 2:
b[j] = 1000
pairs += 1
break
print(pairs)
if __name__ == "__main__":
main()
| Title: BerSU Ball
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! *n* boys and *m* girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from *n* boys and *m* girls.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of boys. The second line contains sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is the *i*-th boy's dancing skill.
Similarly, the third line contains an integer *m* (1<=≤<=*m*<=≤<=100) — the number of girls. The fourth line contains sequence *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=100), where *b**j* is the *j*-th girl's dancing skill.
Output Specification:
Print a single number — the required maximum possible number of pairs.
Demo Input:
['4\n1 4 6 2\n5\n5 1 5 7 9\n', '4\n1 2 3 4\n4\n10 11 12 13\n', '5\n1 1 1 1 1\n3\n1 2 3\n']
Demo Output:
['3\n', '0\n', '2\n']
Note:
none | ```python
import sys
def main():
n, m, pairs = int(sys.stdin.readline()), int(sys.stdin.readline()), 0
a = list(map(int, sys.stdin.readline().split()))
b = list(map(int, sys.stdin.readline().split()))
a.sort()
b.sort()
for i in range(n):
for j in range(m):
if abs(a[i] - b[j]) < 2:
b[j] = 1000
pairs += 1
break
print(pairs)
if __name__ == "__main__":
main()
``` | -1 | |
160 | A | Twins | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | null | null | Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces. | In the single line print the single number — the minimum needed number of coins. | [
"2\n3 3\n",
"3\n2 1 2\n"
] | [
"2\n",
"2\n"
] | In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2. | 500 | [
{
"input": "2\n3 3",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "5\n4 2 2 2 2",
"output": "3"
},
{
"input": "7\n1 10 1 2 1 1 1",
"output": "1"
},
{
"input": "5\n3 2 3 3 1",
"output": "3"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "4"
},
{
"input": "7\n10 10 5 5 5 5 1",
"output": "3"
},
{
"input": "20\n2 1 2 2 2 1 1 2 1 2 2 1 1 1 1 2 1 1 1 1",
"output": "8"
},
{
"input": "20\n4 2 4 4 3 4 2 2 4 2 3 1 1 2 2 3 3 3 1 4",
"output": "8"
},
{
"input": "20\n35 26 41 40 45 46 22 26 39 23 11 15 47 42 18 15 27 10 45 40",
"output": "8"
},
{
"input": "20\n7 84 100 10 31 35 41 2 63 44 57 4 63 11 23 49 98 71 16 90",
"output": "6"
},
{
"input": "50\n19 2 12 26 17 27 10 26 17 17 5 24 11 15 3 9 16 18 19 1 25 23 18 6 2 7 25 7 21 25 13 29 16 9 25 3 14 30 18 4 10 28 6 10 8 2 2 4 8 28",
"output": "14"
},
{
"input": "70\n2 18 18 47 25 5 14 9 19 46 36 49 33 32 38 23 32 39 8 29 31 17 24 21 10 15 33 37 46 21 22 11 20 35 39 13 11 30 28 40 39 47 1 17 24 24 21 46 12 2 20 43 8 16 44 11 45 10 13 44 31 45 45 46 11 10 33 35 23 42",
"output": "22"
},
{
"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": "51"
},
{
"input": "100\n1 2 2 1 2 1 1 2 1 1 1 2 2 1 1 1 2 2 2 1 2 1 1 1 1 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 1 1 1 2 2 1 2 1 2 1 2 2 2 1 2 1 2 2 1 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 2 1 1 2 2 1 2 1 1 2 2 1 1 1 1 2 1 1 1 1 2 2 2 2",
"output": "37"
},
{
"input": "100\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2",
"output": "36"
},
{
"input": "100\n5 5 4 3 5 1 2 5 1 1 3 5 4 4 1 1 1 1 5 4 4 5 1 5 5 1 2 1 3 1 5 1 3 3 3 2 2 2 1 1 5 1 3 4 1 1 3 2 5 2 2 5 5 4 4 1 3 4 3 3 4 5 3 3 3 1 2 1 4 2 4 4 1 5 1 3 5 5 5 5 3 4 4 3 1 2 5 2 3 5 4 2 4 5 3 2 4 2 4 3",
"output": "33"
},
{
"input": "100\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2",
"output": "30"
},
{
"input": "100\n16 9 11 8 11 4 9 17 4 8 4 10 9 10 6 3 3 15 1 6 1 15 12 18 6 14 13 18 1 7 18 4 10 7 10 12 3 16 14 4 10 8 10 7 19 13 15 1 4 8 16 10 6 4 3 16 11 10 7 3 4 16 1 20 1 11 4 16 10 7 7 12 18 19 3 17 19 3 4 19 2 12 11 3 18 20 2 2 14 4 20 13 13 11 16 20 19 14 7 2",
"output": "29"
},
{
"input": "100\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1",
"output": "30"
},
{
"input": "100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52",
"output": "29"
},
{
"input": "100\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",
"output": "30"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "4\n4 1 2 3",
"output": "2"
},
{
"input": "5\n2 2 2 4 4",
"output": "2"
},
{
"input": "4\n100 99 99 100",
"output": "2"
}
] | 1,694,876,605 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 122 | 0 | n = int(input())
a = [int(i) for i in input().split()]
a.sort()
s = sum(a)
tek = 0
for i in range(len(a)):
el = a[i]
tek += el
s -= el
if tek > s:
print(i + 1)
break
| Title: Twins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
Output Specification:
In the single line print the single number — the minimum needed number of coins.
Demo Input:
['2\n3 3\n', '3\n2 1 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2. | ```python
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
s = sum(a)
tek = 0
for i in range(len(a)):
el = a[i]
tek += el
s -= el
if tek > s:
print(i + 1)
break
``` | 0 | |
961 | D | Pair Of Lines | PROGRAMMING | 2,000 | [
"geometry"
] | null | null | You are given *n* points on Cartesian plane. Every point is a lattice point (i.<=e. both of its coordinates are integers), and all points are distinct.
You may draw two straight lines (not necessarily distinct). Is it possible to do this in such a way that every point lies on at least one of these lines? | The first line contains one integer *n* (1<=≤<=*n*<=≤<=105) — the number of points you are given.
Then *n* lines follow, each line containing two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109)— coordinates of *i*-th point. All *n* points are distinct. | If it is possible to draw two straight lines in such a way that each of given points belongs to at least one of these lines, print YES. Otherwise, print NO. | [
"5\n0 0\n0 1\n1 1\n1 -1\n2 2\n",
"5\n0 0\n1 0\n2 1\n1 1\n2 3\n"
] | [
"YES\n",
"NO\n"
] | In the first example it is possible to draw two lines, the one containing the points 1, 3 and 5, and another one containing two remaining points. | 0 | [
{
"input": "5\n0 0\n0 1\n1 1\n1 -1\n2 2",
"output": "YES"
},
{
"input": "5\n0 0\n1 0\n2 1\n1 1\n2 3",
"output": "NO"
},
{
"input": "1\n-1000000000 1000000000",
"output": "YES"
},
{
"input": "5\n2 -1\n-4 1\n0 -9\n5 -9\n9 -10",
"output": "NO"
},
{
"input": "5\n6 1\n10 5\n10 -2\n-2 -10\n-4 -9",
"output": "YES"
},
{
"input": "5\n-10 3\n4 -5\n-9 5\n-5 -3\n-4 -6",
"output": "NO"
},
{
"input": "5\n2 9\n-1 -4\n-3 -8\n-4 8\n7 2",
"output": "NO"
},
{
"input": "10\n315 202\n315 203\n315 204\n-138 -298\n-136 -295\n-134 -292\n-132 -289\n-130 -286\n-128 -283\n-126 -280",
"output": "YES"
},
{
"input": "10\n416 -473\n-162 491\n-164 488\n-170 479\n-166 485\n-172 476\n416 -475\n416 -474\n-168 482\n-160 494",
"output": "YES"
},
{
"input": "6\n0 0\n1 1\n0 1\n1 0\n0 2\n2 0",
"output": "NO"
},
{
"input": "5\n3 3\n6 3\n0 0\n10 0\n-10 0",
"output": "YES"
},
{
"input": "1\n0 0",
"output": "YES"
},
{
"input": "10\n0 0\n1 0\n0 1\n0 2\n2 0\n3 0\n0 3\n0 4\n4 0\n0 -10000000",
"output": "YES"
},
{
"input": "6\n0 0\n0 1\n0 2\n1 1\n1 2\n2 1",
"output": "NO"
},
{
"input": "6\n0 -1\n1 -1\n3 3\n2 0\n-2 -2\n1 -2",
"output": "NO"
},
{
"input": "5\n1000000000 1000000000\n999999999 999999999\n999999999 999999998\n-1000000000 1000000000\n-1000000000 999999999",
"output": "NO"
},
{
"input": "5\n0 0\n1 0\n0 1\n1 1\n-1 1",
"output": "YES"
},
{
"input": "6\n0 0\n0 1\n0 -1\n1 1\n1 -1\n2 -1",
"output": "NO"
},
{
"input": "4\n0 0\n0 1\n1 0\n1 1",
"output": "YES"
},
{
"input": "6\n0 0\n1 0\n2 1\n1 1\n0 1\n6 0",
"output": "YES"
},
{
"input": "10\n536870912 536870912\n268435456 368435456\n268435456 168435456\n1 3\n2 4\n3 5\n4 6\n5 7\n6 8\n7 9",
"output": "NO"
},
{
"input": "5\n0 0\n0 1\n100 100\n100 99\n100 98",
"output": "YES"
},
{
"input": "8\n0 0\n1 0\n2 1\n1 1\n0 1\n6 0\n5 0\n7 0",
"output": "YES"
},
{
"input": "5\n0 0\n2 0\n1 1\n0 2\n5 1",
"output": "YES"
},
{
"input": "7\n0 0\n4 0\n1 1\n2 2\n3 1\n5 1\n6 2",
"output": "NO"
},
{
"input": "6\n1 1\n2 2\n3 2\n4 1\n5 2\n6 1",
"output": "YES"
},
{
"input": "8\n0 0\n1 0\n2 0\n3 0\n0 1\n1 1\n2 1\n3 1",
"output": "YES"
},
{
"input": "12\n0 0\n1 1\n2 2\n3 3\n10 11\n20 11\n30 11\n40 11\n-1 1\n-2 2\n-3 3\n-4 4",
"output": "NO"
},
{
"input": "6\n0 0\n165580141 267914296\n331160282 535828592\n267914296 433494437\n535828592 866988874\n433494437 701408733",
"output": "NO"
},
{
"input": "5\n-1000000000 -1000000000\n-588442013 -868997024\n-182303377 -739719081\n-999999999 -999999999\n229254610 -608716105",
"output": "NO"
},
{
"input": "5\n-1000000000 -1000000000\n229254610 -608716105\n-588442013 -868997024\n-182303377 -739719081\n-176884026 -737994048",
"output": "YES"
},
{
"input": "6\n0 0\n0 1\n0 2\n5 0\n5 1\n5 -1",
"output": "YES"
},
{
"input": "5\n-1 1\n1 0\n1 1\n1 -1\n-1 -1",
"output": "YES"
},
{
"input": "5\n-1000000000 -1000000000\n229254610 -608716105\n-588442013 -868997024\n-182303377 -739719081\n-999999999 -999999999",
"output": "NO"
},
{
"input": "6\n1 1\n0 0\n-1 -1\n1 0\n0 -1\n-1 -10",
"output": "NO"
},
{
"input": "5\n8 8\n3303829 10\n10 1308\n4 2\n6 3",
"output": "NO"
},
{
"input": "5\n0 0\n0 1\n0 2\n0 3\n1 0",
"output": "YES"
},
{
"input": "5\n0 0\n165580142 267914296\n331160283 535828592\n267914296 433494437\n535828592 866988874",
"output": "YES"
},
{
"input": "59\n1 0\n0 2\n0 3\n0 4\n0 5\n6 0\n7 0\n8 0\n9 0\n10 0\n0 11\n12 0\n13 0\n14 0\n15 0\n0 16\n0 17\n18 0\n19 0\n20 0\n21 0\n0 22\n23 0\n24 0\n0 25\n26 0\n27 0\n0 28\n0 29\n30 0\n31 0\n0 32\n33 0\n34 0\n0 35\n0 36\n37 0\n0 38\n39 0\n40 0\n0 41\n42 0\n0 43\n0 44\n0 45\n0 46\n47 0\n0 48\n0 49\n50 0\n0 51\n0 52\n53 0\n0 54\n55 0\n0 56\n57 0\n0 58\n59 0",
"output": "YES"
},
{
"input": "5\n10000000 40000100\n3 112\n2 400000100\n1 104\n1000000 701789036",
"output": "YES"
},
{
"input": "5\n514 2131\n312 52362\n1 1\n2 2\n3 3",
"output": "YES"
},
{
"input": "9\n-65536 65536\n0 65536\n65536 65536\n-65536 0\n0 0\n65536 0\n-65536 -65536\n0 -65536\n65536 -65536",
"output": "NO"
},
{
"input": "5\n0 -7\n0 10000\n1 1000000000\n100 0\n200 0",
"output": "NO"
},
{
"input": "7\n0 0\n2 2\n2 -2\n-2 2\n-2 -2\n0 1\n0 3",
"output": "NO"
},
{
"input": "5\n3 0\n4 1\n0 0\n1 1\n2 2",
"output": "YES"
},
{
"input": "5\n-65536 -65536\n65536 0\n131072 0\n0 65536\n0 131072",
"output": "NO"
},
{
"input": "4\n0 0\n1 0\n0 1\n1 1",
"output": "YES"
},
{
"input": "6\n0 0\n2 0\n0 2\n0 -2\n-2 1\n-4 2",
"output": "NO"
},
{
"input": "5\n-1000000000 -1000000000\n134903170 -298591267\n-566505563 -732085704\n-298591267 -566505563\n-133011126 -464171408",
"output": "YES"
},
{
"input": "5\n-1000000000 -1000000000\n134903170 -298591267\n-566505563 -732085704\n-298591267 -566505563\n-999999999 -999999999",
"output": "NO"
},
{
"input": "5\n1 1\n-1 0\n0 1\n-1 1\n0 0",
"output": "YES"
},
{
"input": "5\n0 0\n-1 -1\n0 -1\n-1 1\n-1 0",
"output": "YES"
},
{
"input": "5\n0 0\n-1 1\n-1 0\n0 -1\n-1 -1",
"output": "YES"
},
{
"input": "6\n0 0\n-1 1\n-1 0\n1 1\n-1 -1\n0 -1",
"output": "NO"
},
{
"input": "5\n-1 2\n-1 1\n2 1\n-2 2\n1 1",
"output": "YES"
},
{
"input": "6\n-1 -1\n-1 -2\n-1 -3\n1000000000 1\n-1000000000 0\n999999999 1",
"output": "NO"
},
{
"input": "6\n-1 -1\n-1 -2\n-1 -3\n0 0\n65536 65536\n65536 131072",
"output": "NO"
},
{
"input": "6\n-1 -1\n-1 -2\n-1 -3\n1000000000 1\n999999999 1\n-1000000000 0",
"output": "NO"
},
{
"input": "3\n-1 1\n-1 -1\n0 0",
"output": "YES"
},
{
"input": "7\n1 -1\n3 -3\n1 2\n0 -2\n1 -3\n0 1\n0 2",
"output": "NO"
},
{
"input": "4\n0 0\n-1 1\n-1 -1\n1 0",
"output": "YES"
},
{
"input": "6\n0 0\n0 1\n-1 1\n0 -1\n1 0\n-1 -1",
"output": "NO"
},
{
"input": "5\n1 1\n0 0\n-1 0\n0 1\n1 0",
"output": "YES"
},
{
"input": "11\n-2 -2\n2 3\n3 -2\n1 -2\n2 -2\n2 0\n2 2\n-3 -2\n-1 -2\n2 -3\n2 1",
"output": "YES"
},
{
"input": "5\n0 0\n-1 0\n-1 1\n1 0\n1 -1",
"output": "YES"
},
{
"input": "5\n1 -1\n0 0\n0 1\n-1 1\n1 1",
"output": "YES"
},
{
"input": "5\n0 0\n1 1\n0 -2\n1 -1\n1 2",
"output": "YES"
},
{
"input": "5\n-999999998 -999999998\n229254612 -608716103\n-588442011 -868997022\n-182303375 -739719079\n-176884024 -737994046",
"output": "YES"
}
] | 1,615,294,148 | 2,948 | PyPy 3 | WRONG_ANSWER | TESTS | 8 | 93 | 0 | import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n=int(input())
if n==1:
print('YES')
exit()
points=[]
for _ in range(n):
x,y=map(int,input().split())
points.append((x,y))
x1,y1=points[0]
x2,y2=points[1]
flag=0
for i in range(2,n):
xk,yk=points[i]
if (xk-x2)*(y2-y1)!=(yk-y2)*(x2-x1):
flag=1
x3=xk
y3=yk
thirdpoint=i
break
if flag==0:
print('YES')
exit()
flag=0
for i in range(2,n):
if i==thirdpoint:
continue
xk,yk=points[i]
if (xk-x2)*(y2-y1)!=(yk-y2)*(x2-x1) and (xk-x3)*(y3-y2)!=(yk-y3)*(x3-x2) and (xk-x1)*(y1-y3)!=(yk-y1)*(x1-x3):
flag=1
x4=xk
y4=yk
fourthpoint=i
break
if flag==0:
print('YES')
exit()
line1x1=x1
line1x2=x2
line1y1=y1
line1y2=y2
line2x1=x3
line2x2=x4
line2y1=y3
line2y2=y4
flag=0
for i in range(n):
xk,yk=points[i]
if (xk-line1x2)*(line1y2-line1y1)!=(yk-line1y2)*(line1x2-line1x1) and (xk-line2x2)*(line2y2-line2y1)!=(yk-line2y2)*(line2x2-line2x1):
flag=1
break
if flag==0:
print('YES')
exit()
line1x1=x1
line1x2=x3
line1y1=y1
line1y2=y3
line2x1=x2
line2x2=x4
line2y1=y2
line2y2=y4
flag=0
for i in range(n):
xk,yk=points[i]
if (xk-line1x2)*(line1y2-line1y1)!=(yk-line1y2)*(line1x2-line1x1) and (xk-line2x2)*(line2y2-line2y1)!=(yk-line2y2)*(line2x2-line2x1):
flag=1
break
if flag==0:
print('YES')
exit()
line1x1=x1
line1x2=x4
line1y1=y2
line1y2=y3
line2x1=x1
line2x2=x4
line2y1=y2
line2y2=y3
flag=0
for i in range(n):
xk,yk=points[i]
if (xk-line1x2)*(line1y2-line1y1)!=(yk-line1y2)*(line1x2-line1x1) and (xk-line2x2)*(line2y2-line2y1)!=(yk-line2y2)*(line2x2-line2x1):
flag=1
break
if flag==0:
print('YES')
exit()
print('NO')
| Title: Pair Of Lines
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* points on Cartesian plane. Every point is a lattice point (i.<=e. both of its coordinates are integers), and all points are distinct.
You may draw two straight lines (not necessarily distinct). Is it possible to do this in such a way that every point lies on at least one of these lines?
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=105) — the number of points you are given.
Then *n* lines follow, each line containing two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109)— coordinates of *i*-th point. All *n* points are distinct.
Output Specification:
If it is possible to draw two straight lines in such a way that each of given points belongs to at least one of these lines, print YES. Otherwise, print NO.
Demo Input:
['5\n0 0\n0 1\n1 1\n1 -1\n2 2\n', '5\n0 0\n1 0\n2 1\n1 1\n2 3\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example it is possible to draw two lines, the one containing the points 1, 3 and 5, and another one containing two remaining points. | ```python
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n=int(input())
if n==1:
print('YES')
exit()
points=[]
for _ in range(n):
x,y=map(int,input().split())
points.append((x,y))
x1,y1=points[0]
x2,y2=points[1]
flag=0
for i in range(2,n):
xk,yk=points[i]
if (xk-x2)*(y2-y1)!=(yk-y2)*(x2-x1):
flag=1
x3=xk
y3=yk
thirdpoint=i
break
if flag==0:
print('YES')
exit()
flag=0
for i in range(2,n):
if i==thirdpoint:
continue
xk,yk=points[i]
if (xk-x2)*(y2-y1)!=(yk-y2)*(x2-x1) and (xk-x3)*(y3-y2)!=(yk-y3)*(x3-x2) and (xk-x1)*(y1-y3)!=(yk-y1)*(x1-x3):
flag=1
x4=xk
y4=yk
fourthpoint=i
break
if flag==0:
print('YES')
exit()
line1x1=x1
line1x2=x2
line1y1=y1
line1y2=y2
line2x1=x3
line2x2=x4
line2y1=y3
line2y2=y4
flag=0
for i in range(n):
xk,yk=points[i]
if (xk-line1x2)*(line1y2-line1y1)!=(yk-line1y2)*(line1x2-line1x1) and (xk-line2x2)*(line2y2-line2y1)!=(yk-line2y2)*(line2x2-line2x1):
flag=1
break
if flag==0:
print('YES')
exit()
line1x1=x1
line1x2=x3
line1y1=y1
line1y2=y3
line2x1=x2
line2x2=x4
line2y1=y2
line2y2=y4
flag=0
for i in range(n):
xk,yk=points[i]
if (xk-line1x2)*(line1y2-line1y1)!=(yk-line1y2)*(line1x2-line1x1) and (xk-line2x2)*(line2y2-line2y1)!=(yk-line2y2)*(line2x2-line2x1):
flag=1
break
if flag==0:
print('YES')
exit()
line1x1=x1
line1x2=x4
line1y1=y2
line1y2=y3
line2x1=x1
line2x2=x4
line2y1=y2
line2y2=y3
flag=0
for i in range(n):
xk,yk=points[i]
if (xk-line1x2)*(line1y2-line1y1)!=(yk-line1y2)*(line1x2-line1x1) and (xk-line2x2)*(line2y2-line2y1)!=(yk-line2y2)*(line2x2-line2x1):
flag=1
break
if flag==0:
print('YES')
exit()
print('NO')
``` | 0 | |
295 | B | Greg and Graph | PROGRAMMING | 1,700 | [
"dp",
"graphs",
"shortest paths"
] | null | null | Greg has a weighed directed graph, consisting of *n* vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:
- The game consists of *n* steps. - On the *i*-th step Greg removes vertex number *x**i* from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex. - Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that *d*(*i*,<=*v*,<=*u*) is the shortest path between vertices *v* and *u* in the graph that formed before deleting vertex *x**i*, then Greg wants to know the value of the following sum: .
Help Greg, print the value of the required sum before each step. | The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the number of vertices in the graph.
Next *n* lines contain *n* integers each — the graph adjacency matrix: the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*a**ij*<=≤<=105,<=*a**ii*<==<=0) represents the weight of the edge that goes from vertex *i* to vertex *j*.
The next line contains *n* distinct integers: *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=*n*) — the vertices that Greg deletes. | Print *n* integers — the *i*-th number equals the required sum before the *i*-th step.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier. | [
"1\n0\n1\n",
"2\n0 5\n4 0\n1 2\n",
"4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3\n"
] | [
"0 ",
"9 0 ",
"17 23 404 0 "
] | none | 1,000 | [
{
"input": "1\n0\n1",
"output": "0 "
},
{
"input": "2\n0 5\n4 0\n1 2",
"output": "9 0 "
},
{
"input": "4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3",
"output": "17 23 404 0 "
},
{
"input": "4\n0 57148 51001 13357\n71125 0 98369 67226\n49388 90852 0 66291\n39573 38165 97007 0\n2 3 1 4",
"output": "723897 306638 52930 0 "
},
{
"input": "5\n0 27799 15529 16434 44291\n47134 0 90227 26873 52252\n41605 21269 0 9135 55784\n70744 17563 79061 0 73981\n70529 35681 91073 52031 0\n5 2 3 1 4",
"output": "896203 429762 232508 87178 0 "
},
{
"input": "6\n0 72137 71041 29217 96749 46417\n40199 0 55907 57677 68590 78796\n83463 50721 0 30963 31779 28646\n94529 47831 98222 0 61665 73941\n24397 66286 2971 81613 0 52501\n26285 3381 51438 45360 20160 0\n6 3 2 4 5 1",
"output": "1321441 1030477 698557 345837 121146 0 "
},
{
"input": "7\n0 34385 31901 51111 10191 14089 95685\n11396 0 8701 33277 1481 517 46253\n51313 2255 0 5948 66085 37201 65310\n21105 60985 10748 0 89271 42883 77345\n34686 29401 73565 47795 0 13793 66997\n70279 49576 62900 40002 70943 0 89601\n65045 1681 28239 12023 40414 89585 0\n3 5 7 6 1 2 4",
"output": "1108867 1016339 729930 407114 206764 94262 0 "
},
{
"input": "8\n0 74961 47889 4733 72876 21399 63105 48239\n15623 0 9680 89133 57989 63401 26001 29608\n42369 82390 0 32866 46171 11871 67489 54070\n23425 80027 18270 0 28105 42657 40876 29267\n78793 18701 7655 94798 0 88885 71424 86914\n44835 76636 11553 46031 13617 0 16971 51915\n33037 53719 43116 52806 56897 71241 0 11629\n2119 62373 93265 69513 5770 90751 36619 0\n3 7 6 5 8 1 2 4",
"output": "1450303 1188349 900316 531281 383344 219125 169160 0 "
},
{
"input": "9\n0 85236 27579 82251 69479 24737 87917 15149 52311\n59640 0 74687 34711 3685 30121 4961 7552 83399\n33376 68733 0 81357 18042 74297 15466 29476 5865\n7493 5601 3321 0 20263 55901 45756 55361 87633\n26751 17161 76681 40376 0 39745 50717 56887 90055\n18885 76353 47089 43601 21561 0 60571 33551 53753\n74595 877 71853 93156 97499 70876 0 22713 63961\n67725 25309 56358 92376 40641 35433 39781 0 97482\n81818 12561 85961 81445 3941 76799 31701 43725 0\n6 2 9 3 5 7 1 4 8",
"output": "2106523 1533575 1645151 1255230 946667 618567 287636 147737 0 "
}
] | 1,668,503,308 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 25 | 3,000 | 10,956,800 | #from pyrival import *
import math
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
n = int(input())
dist = [[0]*n for _ in range(n)]
for i in range(n):
dist[i] = list(map(int, input().split()))
x = list(map(int, input().split()))
x = [xi - 1 for xi in x]
x.reverse()
totals = []
for pos in range(n):
mid = x[pos]
totals.append(0)
# First fix the floyd invariant
for left in x[:pos]:
for right in x[:pos]:
dist[left][mid] = min(dist[left][mid], dist[left][right] + dist[right][mid])
dist[mid][right] = min(dist[mid][right], dist[mid][left] + dist[left][right])
# Now the new vert has the right distance
for left in x[:pos]:
for right in x[:pos]:
dist[left][right] = min(dist[left][right], dist[left][mid] + dist[mid][right])
for left in x[:pos + 1]:
for right in x[:pos + 1]:
totals[-1] += dist[left][right]
#[print(row) for row in dist]
totals.reverse()
print(*totals)
| Title: Greg and Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg has a weighed directed graph, consisting of *n* vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:
- The game consists of *n* steps. - On the *i*-th step Greg removes vertex number *x**i* from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex. - Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that *d*(*i*,<=*v*,<=*u*) is the shortest path between vertices *v* and *u* in the graph that formed before deleting vertex *x**i*, then Greg wants to know the value of the following sum: .
Help Greg, print the value of the required sum before each step.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the number of vertices in the graph.
Next *n* lines contain *n* integers each — the graph adjacency matrix: the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*a**ij*<=≤<=105,<=*a**ii*<==<=0) represents the weight of the edge that goes from vertex *i* to vertex *j*.
The next line contains *n* distinct integers: *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=*n*) — the vertices that Greg deletes.
Output Specification:
Print *n* integers — the *i*-th number equals the required sum before the *i*-th step.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Demo Input:
['1\n0\n1\n', '2\n0 5\n4 0\n1 2\n', '4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3\n']
Demo Output:
['0 ', '9 0 ', '17 23 404 0 ']
Note:
none | ```python
#from pyrival import *
import math
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
n = int(input())
dist = [[0]*n for _ in range(n)]
for i in range(n):
dist[i] = list(map(int, input().split()))
x = list(map(int, input().split()))
x = [xi - 1 for xi in x]
x.reverse()
totals = []
for pos in range(n):
mid = x[pos]
totals.append(0)
# First fix the floyd invariant
for left in x[:pos]:
for right in x[:pos]:
dist[left][mid] = min(dist[left][mid], dist[left][right] + dist[right][mid])
dist[mid][right] = min(dist[mid][right], dist[mid][left] + dist[left][right])
# Now the new vert has the right distance
for left in x[:pos]:
for right in x[:pos]:
dist[left][right] = min(dist[left][right], dist[left][mid] + dist[mid][right])
for left in x[:pos + 1]:
for right in x[:pos + 1]:
totals[-1] += dist[left][right]
#[print(row) for row in dist]
totals.reverse()
print(*totals)
``` | 0 | |
732 | A | Buy a Shovel | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null | Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel. | The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels. | Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change. | [
"117 3\n",
"237 7\n",
"15 2\n"
] | [
"9\n",
"1\n",
"2\n"
] | In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | 500 | [
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,673,939,067 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 0 | 1,000 | 1,331,200 | l=0
a,b=map(int,input().split())
while True:
l+=1
if a%10==b:
break
else:
a+=a
print(l) | Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | ```python
l=0
a,b=map(int,input().split())
while True:
l+=1
if a%10==b:
break
else:
a+=a
print(l)
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number *k* is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by *k*2, and the loser's score is multiplied by *k*. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all *n* games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not. | In the first string, the number of games *n* (1<=≤<=*n*<=≤<=350000) is given.
Each game is represented by a pair of scores *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) – the results of Slastyona and Pushok, correspondingly. | For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower). | [
"6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000\n"
] | [
"Yes\nYes\nYes\nNo\nNo\nYes\n"
] | First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3. | 0 | [
{
"input": "6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000",
"output": "Yes\nYes\nYes\nNo\nNo\nYes"
},
{
"input": "3\n1 1\n8 27\n1000 1331",
"output": "Yes\nNo\nNo"
},
{
"input": "1\n12004 18012002",
"output": "Yes"
},
{
"input": "1\n3331 11095561",
"output": "Yes"
},
{
"input": "1\n2 3",
"output": "No"
},
{
"input": "1\n1062961 1031",
"output": "Yes"
},
{
"input": "1\n6 12",
"output": "No"
},
{
"input": "1\n3 1",
"output": "No"
},
{
"input": "1\n3 10",
"output": "No"
},
{
"input": "1\n31159 970883281",
"output": "Yes"
},
{
"input": "1\n9907 98148649",
"output": "Yes"
},
{
"input": "1\n16 8",
"output": "No"
},
{
"input": "1\n90 72",
"output": "No"
}
] | 1,679,714,161 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 46 | 1,433,600 | for i in range(int(input())):
a,b=map(int,input().split())
c=2
while a>1:
if (a%(c**2)==0 and b%c==0):
a//=c**2
b//=c
elif (b%(c**2)==0 and a%c==0):
b//=c**2
a//=c
elif (a%c!=0 and b%c!=0):
c+=1
else:
a=0
break
if a==0 or b>1:
print("NO")
else:
print("YES") | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number *k* is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by *k*2, and the loser's score is multiplied by *k*. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all *n* games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input Specification:
In the first string, the number of games *n* (1<=≤<=*n*<=≤<=350000) is given.
Each game is represented by a pair of scores *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) – the results of Slastyona and Pushok, correspondingly.
Output Specification:
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Demo Input:
['6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000\n']
Demo Output:
['Yes\nYes\nYes\nNo\nNo\nYes\n']
Note:
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3. | ```python
for i in range(int(input())):
a,b=map(int,input().split())
c=2
while a>1:
if (a%(c**2)==0 and b%c==0):
a//=c**2
b//=c
elif (b%(c**2)==0 and a%c==0):
b//=c**2
a//=c
elif (a%c!=0 and b%c!=0):
c+=1
else:
a=0
break
if a==0 or b>1:
print("NO")
else:
print("YES")
``` | 0 | |
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"
}
] | 1,699,008,976 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 21 | 156 | 1,740,800 | t = int(input())
count = 0
for i in range(t):
a, b, c = map(int, input().split())
views = [a, b, c]
# Count the number of friends sure about the solution for each problem
sure_count = views.count(1)
# If at least two friends are sure, increment the total count of problems
if sure_count >= 2:
count += 1
print(count)
| 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
t = int(input())
count = 0
for i in range(t):
a, b, c = map(int, input().split())
views = [a, b, c]
# Count the number of friends sure about the solution for each problem
sure_count = views.count(1)
# If at least two friends are sure, increment the total count of problems
if sure_count >= 2:
count += 1
print(count)
``` | 3 | |
260 | A | Adding Digits | PROGRAMMING | 1,400 | [
"implementation",
"math"
] | null | null | Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times. | The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105). | In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them. | [
"5 4 5\n",
"12 11 1\n",
"260 150 10\n"
] | [
"524848\n",
"121\n",
"-1\n"
] | none | 500 | [
{
"input": "5 4 5",
"output": "524848"
},
{
"input": "12 11 1",
"output": "121"
},
{
"input": "260 150 10",
"output": "-1"
},
{
"input": "78843 5684 42717",
"output": "-1"
},
{
"input": "93248 91435 1133",
"output": "-1"
},
{
"input": "100000 10 64479",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 21 73839",
"output": "9999990000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99991 623 36438",
"output": "9999150000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 334 94854",
"output": "9999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99252 9827 84849",
"output": "9925270000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99313 9833 10561",
"output": "9931330000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "94885 55815 11417",
"output": "9488550000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99492 58525 53481",
"output": "9949250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99858 28531 79193",
"output": "9985850000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99136 47208 42607",
"output": "9913680000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "63270 19953 5555",
"output": "-1"
},
{
"input": "10240 128 100000",
"output": "1024000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "12 11 3",
"output": "12100"
},
{
"input": "14 12 99998",
"output": "1440000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "1 11 3",
"output": "1100"
},
{
"input": "3 40 1",
"output": "-1"
},
{
"input": "150 100 10",
"output": "1500000000000"
},
{
"input": "5 10 1",
"output": "50"
},
{
"input": "1 15 10",
"output": "15000000000"
},
{
"input": "3 13 2",
"output": "390"
}
] | 1,614,603,045 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 62 | 409,600 | a,b,n=map(int,input().split())
digits=[0,1,2,3,4,5,6,7,8,9]
length,prev=1,1
flag=False
if a%b==0:
print(str(a)+"0"*n)
for i in range(10):
cum=str(a)+str(digits[i])
if int(cum)%b==0:
a=cum
break
else:
print(-1)
flag=True
if flag==False:
print(str(a)+"0"*(n-1))
| Title: Adding Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
Input Specification:
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
Output Specification:
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
Demo Input:
['5 4 5\n', '12 11 1\n', '260 150 10\n']
Demo Output:
['524848\n', '121\n', '-1\n']
Note:
none | ```python
a,b,n=map(int,input().split())
digits=[0,1,2,3,4,5,6,7,8,9]
length,prev=1,1
flag=False
if a%b==0:
print(str(a)+"0"*n)
for i in range(10):
cum=str(a)+str(digits[i])
if int(cum)%b==0:
a=cum
break
else:
print(-1)
flag=True
if flag==False:
print(str(a)+"0"*(n-1))
``` | 0 | |
862 | B | Mahmoud and Ehab and the bipartiteness | PROGRAMMING | 1,300 | [
"dfs and similar",
"graphs",
"trees"
] | null | null | Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.
A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.
Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?
A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same . | The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105).
The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree.
It's guaranteed that the given graph is a tree. | Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions. | [
"3\n1 2\n1 3\n",
"5\n1 2\n2 3\n3 4\n4 5\n"
] | [
"0\n",
"2\n"
] | Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory))
Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph)
In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.
In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5). | 1,000 | [
{
"input": "3\n1 2\n1 3",
"output": "0"
},
{
"input": "5\n1 2\n2 3\n3 4\n4 5",
"output": "2"
},
{
"input": "10\n3 8\n6 2\n9 7\n10 1\n3 5\n1 3\n6 7\n5 4\n3 6",
"output": "16"
},
{
"input": "10\n7 6\n2 7\n4 1\n8 5\n9 4\n5 3\n8 7\n10 8\n10 4",
"output": "16"
},
{
"input": "10\n2 6\n3 7\n8 4\n4 10\n6 9\n9 7\n3 10\n1 2\n5 8",
"output": "16"
},
{
"input": "10\n6 9\n9 7\n9 4\n10 9\n9 1\n9 8\n9 2\n9 5\n3 9",
"output": "0"
},
{
"input": "2\n1 2",
"output": "0"
}
] | 1,695,281,631 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | n = int(input())
g = [[] for i in range(n+1)]
for _ in range(n-1):
u, v = map(int, input().split())
g[u].append(v)
g[v].append(u)
visited = [False for _ in range(n+1)]
def dfs(node, prev):
if prev:
visited[node] = 1
else:
visited[node] = 2
for w in g[node]:
if not visited[w]:
dfs(w, 3 - prev)
dfs(1, 1)
cnt = 0
for visit in visited:
if visit == 1:
cnt += visit
print(cnt*(n-cnt)-n+1) | Title: Mahmoud and Ehab and the bipartiteness
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.
A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.
Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?
A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same .
Input Specification:
The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105).
The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree.
It's guaranteed that the given graph is a tree.
Output Specification:
Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions.
Demo Input:
['3\n1 2\n1 3\n', '5\n1 2\n2 3\n3 4\n4 5\n']
Demo Output:
['0\n', '2\n']
Note:
Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory))
Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph)
In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.
In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5). | ```python
n = int(input())
g = [[] for i in range(n+1)]
for _ in range(n-1):
u, v = map(int, input().split())
g[u].append(v)
g[v].append(u)
visited = [False for _ in range(n+1)]
def dfs(node, prev):
if prev:
visited[node] = 1
else:
visited[node] = 2
for w in g[node]:
if not visited[w]:
dfs(w, 3 - prev)
dfs(1, 1)
cnt = 0
for visit in visited:
if visit == 1:
cnt += visit
print(cnt*(n-cnt)-n+1)
``` | 0 | |
262 | B | Roma and Changing Signs | PROGRAMMING | 1,200 | [
"greedy"
] | null | null | Roma works in a company that sells TVs. Now he has to prepare a report for the last year.
Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times.
The operation of changing a number's sign is the operation of multiplying this number by -1.
Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes. | The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made.
The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104).
The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order. | In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes. | [
"3 2\n-1 -1 1\n",
"3 1\n-1 -1 1\n"
] | [
"3\n",
"1\n"
] | In the first sample we can get sequence [1, 1, 1], thus the total income equals 3.
In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1. | 1,000 | [
{
"input": "3 2\n-1 -1 1",
"output": "3"
},
{
"input": "3 1\n-1 -1 1",
"output": "1"
},
{
"input": "17 27\n257 320 676 1136 2068 2505 2639 4225 4951 5786 7677 7697 7851 8337 8429 8469 9343",
"output": "81852"
},
{
"input": "69 28\n-9822 -9264 -9253 -9221 -9139 -9126 -9096 -8981 -8521 -8313 -8257 -8253 -7591 -7587 -7301 -7161 -7001 -6847 -6441 -6241 -5949 -5896 -5713 -5692 -5644 -5601 -5545 -5525 -5331 -5253 -5041 -5000 -4951 -4855 -4384 -4293 -4251 -4001 -3991 -3762 -3544 -3481 -3261 -2983 -2882 -2857 -2713 -2691 -2681 -2653 -2221 -2043 -2011 -1997 -1601 -1471 -1448 -1363 -1217 -1217 -1129 -961 -926 -801 -376 -327 -305 -174 -91",
"output": "102443"
},
{
"input": "12 28\n-6652 -6621 -6471 -5559 -5326 -4551 -4401 -4326 -3294 -1175 -1069 -43",
"output": "49488"
},
{
"input": "78 13\n-9961 -9922 -9817 -9813 -9521 -9368 -9361 -9207 -9153 -9124 -9008 -8981 -8951 -8911 -8551 -8479 -8245 -8216 -7988 -7841 -7748 -7741 -7734 -7101 -6846 -6804 -6651 -6526 -6519 -6463 -6297 -6148 -6090 -5845 -5209 -5201 -5161 -5061 -4537 -4529 -4433 -4370 -4266 -4189 -4125 -3945 -3843 -3777 -3751 -3476 -3461 -3279 -3205 -3001 -2889 -2761 -2661 -2521 -2481 -2305 -2278 -2269 -2225 -1648 -1524 -1476 -1353 -1097 -867 -785 -741 -711 -692 -440 -401 -225 -65 -41",
"output": "-147832"
},
{
"input": "4 1\n218 3441 4901 7601",
"output": "15725"
},
{
"input": "73 26\n-8497 -8363 -7603 -7388 -6830 -6827 -6685 -6389 -6237 -6099 -6013 -5565 -5465 -4965 -4947 -4201 -3851 -3793 -3421 -3410 -3201 -3169 -3156 -2976 -2701 -2623 -2321 -2169 -1469 -1221 -950 -926 -9 47 236 457 773 1321 1485 1545 1671 1736 2014 2137 2174 2301 2625 3181 3536 3851 4041 4685 4981 4987 5145 5163 5209 5249 6011 6337 6790 7254 7361 7407 7969 7982 8083 8251 8407 8735 9660 9855 9957",
"output": "315919"
},
{
"input": "53 5\n-9821 -9429 -9146 -8973 -8807 -8801 -8321 -7361 -7222 -7161 -6913 -5961 -4877 -4756 -4753 -4661 -3375 -3031 -2950 -2661 -2161 -2041 -1111 -1071 -905 -697 -397 323 772 1617 1752 2736 2737 3201 3465 4029 4121 4463 4561 4637 4814 6119 6610 6641 6961 7217 7523 8045 8610 8915 9004 9265 9576",
"output": "92703"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n10000",
"output": "-10000"
},
{
"input": "1 2\n-1",
"output": "-1"
},
{
"input": "2 1\n0 1",
"output": "1"
},
{
"input": "2 2\n-1 0",
"output": "1"
},
{
"input": "5 6\n-3 -2 -1 5 6",
"output": "15"
},
{
"input": "3 3\n-50 -10 30",
"output": "70"
},
{
"input": "4 4\n-100 -90 -80 1",
"output": "269"
},
{
"input": "4 3\n-3 -2 1 2",
"output": "6"
},
{
"input": "6 4\n-6 -3 -2 1 2 3",
"output": "15"
},
{
"input": "5 6\n-10 -9 -8 1 2",
"output": "28"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "2 2\n-1 3",
"output": "2"
},
{
"input": "4 7\n-3 -2 1 6",
"output": "10"
},
{
"input": "4 3\n-7 -6 1 5",
"output": "17"
},
{
"input": "4 4\n-5 -1 1 2",
"output": "9"
}
] | 1,637,566,061 | 2,147,483,647 | PyPy 3 | OK | TESTS | 56 | 248 | 8,294,400 | import sys,os,io
import math
from bisect import bisect_left as bl
from collections import defaultdict
def ii():
return int(input())
def li():
return list(map(int,input().split()))
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
else:
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n,k = li()
a = li()
k1 = k
# print(a)
for i in range(min(n,k1)):
if a[i]<0:
a[i]=-a[i]
# print(i,"f",a[i])
k-=1
if k>0:
if 0 in a:
k = 0
if k%2==0:
k = 0
else:
a.sort()
a[0] = -a[0]
print(sum(a))
| Title: Roma and Changing Signs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Roma works in a company that sells TVs. Now he has to prepare a report for the last year.
Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times.
The operation of changing a number's sign is the operation of multiplying this number by -1.
Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made.
The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104).
The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order.
Output Specification:
In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes.
Demo Input:
['3 2\n-1 -1 1\n', '3 1\n-1 -1 1\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample we can get sequence [1, 1, 1], thus the total income equals 3.
In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1. | ```python
import sys,os,io
import math
from bisect import bisect_left as bl
from collections import defaultdict
def ii():
return int(input())
def li():
return list(map(int,input().split()))
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
else:
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n,k = li()
a = li()
k1 = k
# print(a)
for i in range(min(n,k1)):
if a[i]<0:
a[i]=-a[i]
# print(i,"f",a[i])
k-=1
if k>0:
if 0 in a:
k = 0
if k%2==0:
k = 0
else:
a.sort()
a[0] = -a[0]
print(sum(a))
``` | 3 | |
462 | B | Appleman and Card Game | PROGRAMMING | 1,300 | [
"greedy"
] | null | null | Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get? | The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman. | Print a single integer – the answer to the problem. | [
"15 10\nDZFDFZDFDDDDDDF\n",
"6 4\nYJSNPI\n"
] | [
"82\n",
"4\n"
] | In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | 1,000 | [
{
"input": "15 10\nDZFDFZDFDDDDDDF",
"output": "82"
},
{
"input": "6 4\nYJSNPI",
"output": "4"
},
{
"input": "5 3\nAOWBY",
"output": "3"
},
{
"input": "1 1\nV",
"output": "1"
},
{
"input": "2 1\nWT",
"output": "1"
},
{
"input": "2 2\nBL",
"output": "2"
},
{
"input": "5 1\nFACJT",
"output": "1"
},
{
"input": "5 5\nMJDIJ",
"output": "7"
},
{
"input": "15 5\nAZBIPTOFTJCJJIK",
"output": "13"
},
{
"input": "100 1\nEVEEVEEEGGECFEHEFVFVFHVHEEEEEFCVEEEEEEVFVEEVEEHEEVEFEVVEFEEEFEVECEHGHEEFGEEVCEECCECEFHEVEEEEEEGEEHVH",
"output": "1"
},
{
"input": "100 15\nKKTFFUTFCKUIKKKKFIFFKTUKUUKUKKIKKKTIFKTKUCFFKKKIIKKKKKKTFKFKKIRKKKFKUUKIKUUUFFKKKKTUZKITUIKKIKUKKTIK",
"output": "225"
},
{
"input": "100 50\nYYIYYAAAIEAAYAYAEAIIIAAEAAYEAEYYYIAEYAYAYYAAAIAYAEAAYAYYIYAAYYAAAAAAIYYYAAYAAEAAYAIEIYIYAYAYAYIIAAEY",
"output": "1972"
},
{
"input": "100 90\nFAFAOOAOOAFAOTFAFAFFATAAAOFAAOAFBAAAFBOAOFFFOAOAFAPFOFAOFAAFOAAAAFAAFOFAAOFPPAAOOAAOOFFOFFFOFAOTOFAF",
"output": "2828"
},
{
"input": "100 99\nBFFBBFBFBQFFFFFQBFFBFFBQFBFQFBBFQFFFBFFFBFQFQFBFFBBFYQFBFFFFFFFBQQFQBFBQBQFFFBQQFFFBQFYFBFBFFFBBBQQY",
"output": "3713"
},
{
"input": "100 100\nMQSBDAJABILIBCUEOWGWCEXMUTEYQKAIWGINXVQEOFDUBSVULROQHQRZZAALVQFEFRAAAYUIMGCAFQGIAEFBETRECGSFQJNXHHDN",
"output": "514"
},
{
"input": "100 50\nBMYIXQSJNHGFVFPJBIOBXIKSFNUFPVODCUBQYSIIQNVNXXCWXWRHKFEUPPIIDDGRDBJLZDCBMNJMYRMWFIHOSTDJJHXHPNRKWNFD",
"output": "328"
},
{
"input": "100 50\nENFNEMLJEMDMFMNNGNIMNINALGLLLAEMENEMNLMMIEIJNAINBJEJMFJLLIMINELGFLAIAMJMHMGNLIEFJIEEFEFGLLLDLMEAEIMM",
"output": "748"
}
] | 1,620,149,153 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 77 | 409,600 | from collections import *
n,k=map(int,input().split())
s=input()
l=dict(Counter(s))
l=sorted(l.items(),reverse=True,key=lambda x:(x[1],x[0]))
c=0
for x,y in l:
if k>=y:
k=k-y
c=c+y*y
else:
c=c+k*k
break
print(c)
| Title: Appleman and Card Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
Output Specification:
Print a single integer – the answer to the problem.
Demo Input:
['15 10\nDZFDFZDFDDDDDDF\n', '6 4\nYJSNPI\n']
Demo Output:
['82\n', '4\n']
Note:
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | ```python
from collections import *
n,k=map(int,input().split())
s=input()
l=dict(Counter(s))
l=sorted(l.items(),reverse=True,key=lambda x:(x[1],x[0]))
c=0
for x,y in l:
if k>=y:
k=k-y
c=c+y*y
else:
c=c+k*k
break
print(c)
``` | 3 | |
383 | C | Propagating tree | PROGRAMMING | 2,000 | [
"data structures",
"dfs and similar",
"trees"
] | null | null | Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of *n* nodes numbered from 1 to *n*, each node *i* having an initial value *a**i*. The root of the tree is node 1.
This tree has a special property: when a value *val* is added to a value of node *i*, the value -*val* is added to values of all the children of node *i*. Note that when you add value -*val* to a child of node *i*, you also add -(-*val*) to all children of the child of node *i* and so on. Look an example explanation to understand better how it works.
This tree supports two types of queries:
- "1 *x* *val*" — *val* is added to the value of node *x*; - "2 *x*" — print the current value of node *x*.
In order to help Iahub understand the tree better, you must answer *m* queries of the preceding type. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=200000). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000). Each of the next *n*–1 lines contains two integers *v**i* and *u**i* (1<=≤<=*v**i*,<=*u**i*<=≤<=*n*), meaning that there is an edge between nodes *v**i* and *u**i*.
Each of the next *m* lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1<=≤<=*x*<=≤<=*n*,<=1<=≤<=*val*<=≤<=1000. | For each query of type two (print the value of node *x*) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input. | [
"5 5\n1 2 1 1 2\n1 2\n1 3\n2 4\n2 5\n1 2 3\n1 1 2\n2 1\n2 2\n2 4\n"
] | [
"3\n3\n0\n"
] | The values of the nodes are [1, 2, 1, 1, 2] at the beginning.
Then value 3 is added to node 2. It propagates and value -3 is added to it's sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1, - 2, - 1].
Then value 2 is added to node 1. It propagates and value -2 is added to it's sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it's sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3, - 1, 0, 1].
You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory) | 1,500 | [
{
"input": "5 5\n1 2 1 1 2\n1 2\n1 3\n2 4\n2 5\n1 2 3\n1 1 2\n2 1\n2 2\n2 4",
"output": "3\n3\n0"
},
{
"input": "10 10\n137 197 856 768 825 894 86 174 218 326\n7 8\n4 7\n8 9\n7 10\n1 2\n2 4\n3 6\n3 5\n2 3\n1 9 624\n2 1\n2 4\n1 6 505\n1 8 467\n1 3 643\n2 1\n1 8 631\n2 4\n1 7 244",
"output": "137\n768\n137\n768"
},
{
"input": "10 10\n418 45 865 869 745 901 177 773 854 462\n4 8\n1 4\n3 6\n1 5\n1 10\n5 9\n1 2\n4 7\n1 3\n2 2\n1 6 246\n1 4 296\n1 2 378\n1 8 648\n2 6\n1 5 288\n1 6 981\n1 2 868\n2 7",
"output": "45\n1147\n-119"
}
] | 1,675,839,000 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 54 | 1,107 | 130,252,800 | class BIT():
"""区間加算、一点取得クエリをそれぞれO(logN)で応えるデータ構造を構築する
add: 区間[begin, end)にvalを加える
get_val: i番目(0-indexed)の値を求める
"""
def __init__(self, n):
self.n = n
self.bit = [0] * (n + 1)
def get_val(self, i):
i = i + 1
s = 0
while i <= self.n:
s += self.bit[i]
i += i & -i
return s
def _add(self, i, val):
while i > 0:
self.bit[i] += val
i -= i & -i
def add(self, i, j, val):
self._add(j, val)
self._add(i, -val)
from collections import deque
import sys
input = sys.stdin.readline
def eular_tour(tree: list, root: int):
n = len(tree)
res = []
begin = [-1] * n
end = [-1] * n
visited = [False] * n
visited[root] = True
q = deque([root])
while q:
pos = q.pop()
res.append(pos)
end[pos] = len(res)
if begin[pos] == -1:
begin[pos] = len(res) - 1
for next_pos in tree[pos]:
if visited[next_pos]:
continue
else:
visited[next_pos] = True
q.append(pos)
q.append(next_pos)
return res, begin, end
n, q = map(int, input().split())
init_cost = list(map(int, input().split()))
info = [list(map(int, input().split())) for i in range(n-1)]
query = [list(map(int, input().split())) for i in range(q)]
tree = [[] for i in range(n)]
for i in range(n-1):
a, b = info[i]
a -= 1
b -= 1
tree[a].append(b)
tree[b].append(a)
res, begin, end = eular_tour(tree, 0)
even_res = []
odd_res = []
for i in range(len(res)):
if i % 2 == 0:
even_res.append(res[i])
else:
odd_res.append(res[i])
even_bit = BIT(len(even_res))
odd_bit = BIT(len(odd_res))
for i in range(q):
if query[i][0] == 1:
_, pos, cost = query[i]
pos -= 1
if begin[pos] % 2 == 0:
even_bit.add(begin[pos] // 2, (end[pos] + 1) // 2, cost)
odd_bit.add(begin[pos] // 2, end[pos] // 2, -cost)
else:
odd_bit.add(begin[pos] // 2, end[pos] // 2, cost)
even_bit.add((begin[pos] + 1) // 2, end[pos] // 2, -cost)
else:
_, pos = query[i]
pos -= 1
if begin[pos] % 2 == 0:
ans = even_bit.get_val(begin[pos] // 2)
else:
ans = odd_bit.get_val(begin[pos] // 2)
print(ans + init_cost[pos]) | Title: Propagating tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of *n* nodes numbered from 1 to *n*, each node *i* having an initial value *a**i*. The root of the tree is node 1.
This tree has a special property: when a value *val* is added to a value of node *i*, the value -*val* is added to values of all the children of node *i*. Note that when you add value -*val* to a child of node *i*, you also add -(-*val*) to all children of the child of node *i* and so on. Look an example explanation to understand better how it works.
This tree supports two types of queries:
- "1 *x* *val*" — *val* is added to the value of node *x*; - "2 *x*" — print the current value of node *x*.
In order to help Iahub understand the tree better, you must answer *m* queries of the preceding type.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=200000). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000). Each of the next *n*–1 lines contains two integers *v**i* and *u**i* (1<=≤<=*v**i*,<=*u**i*<=≤<=*n*), meaning that there is an edge between nodes *v**i* and *u**i*.
Each of the next *m* lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1<=≤<=*x*<=≤<=*n*,<=1<=≤<=*val*<=≤<=1000.
Output Specification:
For each query of type two (print the value of node *x*) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input.
Demo Input:
['5 5\n1 2 1 1 2\n1 2\n1 3\n2 4\n2 5\n1 2 3\n1 1 2\n2 1\n2 2\n2 4\n']
Demo Output:
['3\n3\n0\n']
Note:
The values of the nodes are [1, 2, 1, 1, 2] at the beginning.
Then value 3 is added to node 2. It propagates and value -3 is added to it's sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1, - 2, - 1].
Then value 2 is added to node 1. It propagates and value -2 is added to it's sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it's sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3, - 1, 0, 1].
You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory) | ```python
class BIT():
"""区間加算、一点取得クエリをそれぞれO(logN)で応えるデータ構造を構築する
add: 区間[begin, end)にvalを加える
get_val: i番目(0-indexed)の値を求める
"""
def __init__(self, n):
self.n = n
self.bit = [0] * (n + 1)
def get_val(self, i):
i = i + 1
s = 0
while i <= self.n:
s += self.bit[i]
i += i & -i
return s
def _add(self, i, val):
while i > 0:
self.bit[i] += val
i -= i & -i
def add(self, i, j, val):
self._add(j, val)
self._add(i, -val)
from collections import deque
import sys
input = sys.stdin.readline
def eular_tour(tree: list, root: int):
n = len(tree)
res = []
begin = [-1] * n
end = [-1] * n
visited = [False] * n
visited[root] = True
q = deque([root])
while q:
pos = q.pop()
res.append(pos)
end[pos] = len(res)
if begin[pos] == -1:
begin[pos] = len(res) - 1
for next_pos in tree[pos]:
if visited[next_pos]:
continue
else:
visited[next_pos] = True
q.append(pos)
q.append(next_pos)
return res, begin, end
n, q = map(int, input().split())
init_cost = list(map(int, input().split()))
info = [list(map(int, input().split())) for i in range(n-1)]
query = [list(map(int, input().split())) for i in range(q)]
tree = [[] for i in range(n)]
for i in range(n-1):
a, b = info[i]
a -= 1
b -= 1
tree[a].append(b)
tree[b].append(a)
res, begin, end = eular_tour(tree, 0)
even_res = []
odd_res = []
for i in range(len(res)):
if i % 2 == 0:
even_res.append(res[i])
else:
odd_res.append(res[i])
even_bit = BIT(len(even_res))
odd_bit = BIT(len(odd_res))
for i in range(q):
if query[i][0] == 1:
_, pos, cost = query[i]
pos -= 1
if begin[pos] % 2 == 0:
even_bit.add(begin[pos] // 2, (end[pos] + 1) // 2, cost)
odd_bit.add(begin[pos] // 2, end[pos] // 2, -cost)
else:
odd_bit.add(begin[pos] // 2, end[pos] // 2, cost)
even_bit.add((begin[pos] + 1) // 2, end[pos] // 2, -cost)
else:
_, pos = query[i]
pos -= 1
if begin[pos] % 2 == 0:
ans = even_bit.get_val(begin[pos] // 2)
else:
ans = odd_bit.get_val(begin[pos] // 2)
print(ans + init_cost[pos])
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Pupils decided to go to amusement park. Some of them were with parents. In total, *n* people came to the park and they all want to get to the most extreme attraction and roll on it exactly once.
Tickets for group of *x* people are sold on the attraction, there should be at least one adult in each group (it is possible that the group consists of one adult). The ticket price for such group is *c*1<=+<=*c*2·(*x*<=-<=1)2 (in particular, if the group consists of one person, then the price is *c*1).
All pupils who came to the park and their parents decided to split into groups in such a way that each visitor join exactly one group, and the total price of visiting the most extreme attraction is as low as possible. You are to determine this minimum possible total price. There should be at least one adult in each group. | The first line contains three integers *n*, *c*1 and *c*2 (1<=≤<=*n*<=≤<=200<=000, 1<=≤<=*c*1,<=*c*2<=≤<=107) — the number of visitors and parameters for determining the ticket prices for a group.
The second line contains the string of length *n*, which consists of zeros and ones. If the *i*-th symbol of the string is zero, then the *i*-th visitor is a pupil, otherwise the *i*-th person is an adult. It is guaranteed that there is at least one adult. It is possible that there are no pupils. | Print the minimum price of visiting the most extreme attraction for all pupils and their parents. Each of them should roll on the attraction exactly once. | [
"3 4 1\n011\n",
"4 7 2\n1101\n"
] | [
"8\n",
"18\n"
] | In the first test one group of three people should go to the attraction. Then they have to pay 4 + 1 * (3 - 1)<sup class="upper-index">2</sup> = 8.
In the second test it is better to go to the attraction in two groups. The first group should consist of two adults (for example, the first and the second person), the second should consist of one pupil and one adult (the third and the fourth person). Then each group will have a size of two and for each the price of ticket is 7 + 2 * (2 - 1)<sup class="upper-index">2</sup> = 9. Thus, the total price for two groups is 18. | 0 | [
{
"input": "3 4 1\n011",
"output": "8"
},
{
"input": "4 7 2\n1101",
"output": "18"
},
{
"input": "1 2 2\n1",
"output": "2"
},
{
"input": "2 3 10\n01",
"output": "13"
},
{
"input": "5 10 3\n11100",
"output": "35"
},
{
"input": "10 2 2\n1111101111",
"output": "20"
},
{
"input": "20 3 13\n01111110011111010101",
"output": "120"
},
{
"input": "50 13 44\n11101110100110111100010110001111001001110010111011",
"output": "1270"
},
{
"input": "100 1000 1000\n0000010100101100110100101111001111111111100101111100111011110001011110110111111010000000101000111000",
"output": "100000"
}
] | 1,491,502,834 | 2,147,483,647 | PyPy 3 | OK | TESTS | 31 | 202 | 27,340,800 | # import sys
i=lambda:map(int,input().split())
n,c1,c2=i()
s=input()
co=0
for j in range(0 , n):
if(s[j]=='1'):
co=co+1
# ans=sys.maxsize
ans = c1*1
ans += c2*(n-1)*(n-1)*1
for j in range(2,co+1):
a=int(c1*j)
b=int((n-j)/j)
c=int((n-j)%j)
a=int(a+(c2*(b*b)*(j-c))+(c2*(b+1)*(b+1)*c))
ans=min(a,ans)
print(ans) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pupils decided to go to amusement park. Some of them were with parents. In total, *n* people came to the park and they all want to get to the most extreme attraction and roll on it exactly once.
Tickets for group of *x* people are sold on the attraction, there should be at least one adult in each group (it is possible that the group consists of one adult). The ticket price for such group is *c*1<=+<=*c*2·(*x*<=-<=1)2 (in particular, if the group consists of one person, then the price is *c*1).
All pupils who came to the park and their parents decided to split into groups in such a way that each visitor join exactly one group, and the total price of visiting the most extreme attraction is as low as possible. You are to determine this minimum possible total price. There should be at least one adult in each group.
Input Specification:
The first line contains three integers *n*, *c*1 and *c*2 (1<=≤<=*n*<=≤<=200<=000, 1<=≤<=*c*1,<=*c*2<=≤<=107) — the number of visitors and parameters for determining the ticket prices for a group.
The second line contains the string of length *n*, which consists of zeros and ones. If the *i*-th symbol of the string is zero, then the *i*-th visitor is a pupil, otherwise the *i*-th person is an adult. It is guaranteed that there is at least one adult. It is possible that there are no pupils.
Output Specification:
Print the minimum price of visiting the most extreme attraction for all pupils and their parents. Each of them should roll on the attraction exactly once.
Demo Input:
['3 4 1\n011\n', '4 7 2\n1101\n']
Demo Output:
['8\n', '18\n']
Note:
In the first test one group of three people should go to the attraction. Then they have to pay 4 + 1 * (3 - 1)<sup class="upper-index">2</sup> = 8.
In the second test it is better to go to the attraction in two groups. The first group should consist of two adults (for example, the first and the second person), the second should consist of one pupil and one adult (the third and the fourth person). Then each group will have a size of two and for each the price of ticket is 7 + 2 * (2 - 1)<sup class="upper-index">2</sup> = 9. Thus, the total price for two groups is 18. | ```python
# import sys
i=lambda:map(int,input().split())
n,c1,c2=i()
s=input()
co=0
for j in range(0 , n):
if(s[j]=='1'):
co=co+1
# ans=sys.maxsize
ans = c1*1
ans += c2*(n-1)*(n-1)*1
for j in range(2,co+1):
a=int(c1*j)
b=int((n-j)/j)
c=int((n-j)%j)
a=int(a+(c2*(b*b)*(j-c))+(c2*(b+1)*(b+1)*c))
ans=min(a,ans)
print(ans)
``` | 3 | |
928 | C | Dependency management | PROGRAMMING | 1,900 | [
"*special",
"graphs",
"implementation"
] | null | null | Polycarp is currently developing a project in Vaja language and using a popular dependency management system called Vamen. From Vamen's point of view both Vaja project and libraries are treated projects for simplicity.
A project in Vaja has its own uniqie non-empty name consisting of lowercase latin letters with length not exceeding 10 and version — positive integer from 1 to 106. Each project (keep in mind that it is determined by both its name and version) might depend on other projects. For sure, there are no cyclic dependencies.
You're given a list of project descriptions. The first of the given projects is the one being developed by Polycarp at this moment. Help Polycarp determine all projects that his project depends on (directly or via a certain chain).
It's possible that Polycarp's project depends on two different versions of some project. In this case collision resolving is applied, i.e. for each such project the system chooses the version that minimizes the distance from it to Polycarp's project. If there are several options, the newer (with the maximum version) is preferred. This version is considered actual; other versions and their dependencies are ignored.
More formal, choose such a set of projects of minimum possible size that the following conditions hold:
- Polycarp's project is chosen; - Polycarp's project depends (directly or indirectly) on all other projects in the set; - no two projects share the name; - for each project *x* that some other project in the set depends on we have either *x* or some *y* with other version and shorter chain to Polycarp's project chosen. In case of ties the newer one is chosen.
Output all Polycarp's project's dependencies (Polycarp's project itself should't be printed) in lexicographical order. | The first line contains an only integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of projects in Vaja.
The following lines contain the project descriptions. Each project is described by a line consisting of its name and version separated by space. The next line gives the number of direct dependencies (from 0 to *n*<=-<=1) and the dependencies themselves (one in a line) in arbitrary order. Each dependency is specified by its name and version. The projects are also given in arbitrary order, but the first of them is always Polycarp's. Project descriptions are separated by one empty line. Refer to samples for better understanding.
It's guaranteed that there are no cyclic dependencies. | Output all Polycarp's project's dependencies in lexicographical order. | [
"4\na 3\n2\nb 1\nc 1\n \nb 2\n0\n \nb 1\n1\nb 2\n \nc 1\n1\nb 2\n",
"9\ncodehorses 5\n3\nwebfrmk 6\nmashadb 1\nmashadb 2\n \ncommons 2\n0\n \nmashadb 3\n0\n \nwebfrmk 6\n2\nmashadb 3\ncommons 2\n \nextra 4\n1\nextra 3\n \nextra 3\n0\n \nextra 1\n0\n \nmashadb 1\n1\nextra 3\n \nmashadb 2\n1\nextra 1\n",
"3\nabc 1\n2\nabc 3\ncba 2\n\nabc 3\n0\n\ncba 2\n0\n"
] | [
"2\nb 1\nc 1\n",
"4\ncommons 2\nextra 1\nmashadb 2\nwebfrmk 6\n",
"1\ncba 2\n"
] | The first sample is given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «a» (version 3) depends on are painted black.
The second sample is again given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «codehorses» (version 5) depends on are paint it black. Note that «extra 1» is chosen instead of «extra 3» since «mashadb 1» and all of its dependencies are ignored due to «mashadb 2». | 2,000 | [
{
"input": "4\na 3\n2\nb 1\nc 1\n\nb 2\n0\n\nb 1\n1\nb 2\n\nc 1\n1\nb 2",
"output": "2\nb 1\nc 1"
},
{
"input": "9\ncodehorses 5\n3\nwebfrmk 6\nmashadb 1\nmashadb 2\n\ncommons 2\n0\n\nmashadb 3\n0\n\nwebfrmk 6\n2\nmashadb 3\ncommons 2\n\nextra 4\n1\nextra 3\n\nextra 3\n0\n\nextra 1\n0\n\nmashadb 1\n1\nextra 3\n\nmashadb 2\n1\nextra 1",
"output": "4\ncommons 2\nextra 1\nmashadb 2\nwebfrmk 6"
},
{
"input": "3\nabc 1\n2\nabc 3\ncba 2\n\nabc 3\n0\n\ncba 2\n0",
"output": "1\ncba 2"
},
{
"input": "1\nabc 1000000\n0",
"output": "0"
},
{
"input": "3\nppdpd 283157\n1\npddpdpp 424025\n\nppdpd 529292\n1\nppdpd 283157\n\npddpdpp 424025\n0",
"output": "1\npddpdpp 424025"
},
{
"input": "5\nabbzzz 646068\n0\n\nzabza 468048\n2\nbb 902619\nzabza 550912\n\nzabza 217401\n2\nabbzzz 646068\nbb 902619\n\nzabza 550912\n1\nzabza 217401\n\nbb 902619\n1\nabbzzz 646068",
"output": "0"
},
{
"input": "5\nyyyy 223967\n1\nyyyyyyy 254197\n\nyyyyyyy 254197\n0\n\ny 442213\n0\n\ny 965022\n1\nyyyyyyy 254197\n\nyyyy 766922\n4\nyyyyyyy 254197\ny 442213\nyyyy 223967\ny 965022",
"output": "1\nyyyyyyy 254197"
},
{
"input": "3\nvvgvvgv 991444\n1\ngvgvgvgvgg 206648\n\nvvgvvgv 296188\n0\n\ngvgvgvgvgg 206648\n1\nvvgvvgv 296188",
"output": "1\ngvgvgvgvgg 206648"
},
{
"input": "5\ntctocototo 984516\n1\ncttocottt 486791\n\ntctocototo 688522\n2\ncttocottt 486791\ntctocototo 984516\n\ncttocottt 486791\n0\n\ntctocototo 676435\n1\ntctocototo 394244\n\ntctocototo 394244\n2\ntctocototo 688522\ncttocottt 486791",
"output": "1\ncttocottt 486791"
},
{
"input": "5\nggggggggg 202537\n4\ngggggggg 868552\ngggg 234633\ngggg 402994\ngggggggg 86863\n\ngggg 402994\n0\n\ngggggggg 868552\n1\ngggg 234633\n\ngggg 234633\n1\ngggg 402994\n\ngggggggg 86863\n0",
"output": "2\ngggg 402994\ngggggggg 868552"
},
{
"input": "4\ncf 1\n2\ndb 1\ndb 2\n\ndb 2\n0\n\ndb 1\n1\nold 1\n\nold 1\n0",
"output": "1\ndb 2"
},
{
"input": "5\ncf 1\n2\ndb 1\ndb 2\n\ndb 2\n1\nold 1\n\ndb 1\n1\nold 2\n\nold 1\n0\n\nold 2\n0",
"output": "2\ndb 2\nold 1"
}
] | 1,519,567,700 | 81,200 | PyPy 3 | WRONG_ANSWER | PRETESTS | 0 | 62 | 19,968,000 | # python3
# utf-8
projs_nr = int(input())
proj___deps = {}
start_proj = None
for proj_idx in range(projs_nr):
name, version = input().split()
version = int(version)
deps_nr = int(input())
proj = (name, version)
if proj_idx == 0:
start_proj = proj
proj___deps[proj] = []
for dep_idx in range(deps_nr):
dep_name, dep_version = input().split()
dep_version = int(dep_version)
dep_proj = (dep_name, dep_version)
proj___deps[proj].append(dep_proj)
if proj_idx != projs_nr - 1:
_ = input()
proj_name___version_used = {}
curr_unresolved_projs = [start_proj]
next_unresolved_projs = []
while curr_unresolved_projs:
for proj in curr_unresolved_projs:
proj_name___version_used[proj[0]] = proj[1]
for proj in curr_unresolved_projs:
for dep_proj in proj___deps[proj]:
if dep_proj[0] not in proj_name___version_used:
next_unresolved_projs.append(dep_proj)
unresolved_name___versions = {}
for unresolved_name, unresolved_version in next_unresolved_projs:
if unresolved_name not in unresolved_name___versions:
unresolved_name___versions[unresolved_name] = []
unresolved_name___versions[unresolved_name].append(unresolved_version)
unresolved_name___final_version = {}
for unresolved_name, versions in unresolved_name___versions.items():
unresolved_name___final_version[unresolved_name] = max(versions)
curr_unresolved_projs = \
[(name, version) for name, version in unresolved_name___final_version.items()]
next_unresolved_projs = []
del proj_name___version_used[start_proj[0]]
projects = [name + ' ' + str(version) for name, version in proj_name___version_used.items()]
projects = sorted(projects)
print(*projects, sep='\n')
| Title: Dependency management
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is currently developing a project in Vaja language and using a popular dependency management system called Vamen. From Vamen's point of view both Vaja project and libraries are treated projects for simplicity.
A project in Vaja has its own uniqie non-empty name consisting of lowercase latin letters with length not exceeding 10 and version — positive integer from 1 to 106. Each project (keep in mind that it is determined by both its name and version) might depend on other projects. For sure, there are no cyclic dependencies.
You're given a list of project descriptions. The first of the given projects is the one being developed by Polycarp at this moment. Help Polycarp determine all projects that his project depends on (directly or via a certain chain).
It's possible that Polycarp's project depends on two different versions of some project. In this case collision resolving is applied, i.e. for each such project the system chooses the version that minimizes the distance from it to Polycarp's project. If there are several options, the newer (with the maximum version) is preferred. This version is considered actual; other versions and their dependencies are ignored.
More formal, choose such a set of projects of minimum possible size that the following conditions hold:
- Polycarp's project is chosen; - Polycarp's project depends (directly or indirectly) on all other projects in the set; - no two projects share the name; - for each project *x* that some other project in the set depends on we have either *x* or some *y* with other version and shorter chain to Polycarp's project chosen. In case of ties the newer one is chosen.
Output all Polycarp's project's dependencies (Polycarp's project itself should't be printed) in lexicographical order.
Input Specification:
The first line contains an only integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of projects in Vaja.
The following lines contain the project descriptions. Each project is described by a line consisting of its name and version separated by space. The next line gives the number of direct dependencies (from 0 to *n*<=-<=1) and the dependencies themselves (one in a line) in arbitrary order. Each dependency is specified by its name and version. The projects are also given in arbitrary order, but the first of them is always Polycarp's. Project descriptions are separated by one empty line. Refer to samples for better understanding.
It's guaranteed that there are no cyclic dependencies.
Output Specification:
Output all Polycarp's project's dependencies in lexicographical order.
Demo Input:
['4\na 3\n2\nb 1\nc 1\n\xa0\nb 2\n0\n\xa0\nb 1\n1\nb 2\n\xa0\nc 1\n1\nb 2\n', '9\ncodehorses 5\n3\nwebfrmk 6\nmashadb 1\nmashadb 2\n\xa0\ncommons 2\n0\n\xa0\nmashadb 3\n0\n\xa0\nwebfrmk 6\n2\nmashadb 3\ncommons 2\n\xa0\nextra 4\n1\nextra 3\n\xa0\nextra 3\n0\n\xa0\nextra 1\n0\n\xa0\nmashadb 1\n1\nextra 3\n\xa0\nmashadb 2\n1\nextra 1\n', '3\nabc 1\n2\nabc 3\ncba 2\n\nabc 3\n0\n\ncba 2\n0\n']
Demo Output:
['2\nb 1\nc 1\n', '4\ncommons 2\nextra 1\nmashadb 2\nwebfrmk 6\n', '1\ncba 2\n']
Note:
The first sample is given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «a» (version 3) depends on are painted black.
The second sample is again given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «codehorses» (version 5) depends on are paint it black. Note that «extra 1» is chosen instead of «extra 3» since «mashadb 1» and all of its dependencies are ignored due to «mashadb 2». | ```python
# python3
# utf-8
projs_nr = int(input())
proj___deps = {}
start_proj = None
for proj_idx in range(projs_nr):
name, version = input().split()
version = int(version)
deps_nr = int(input())
proj = (name, version)
if proj_idx == 0:
start_proj = proj
proj___deps[proj] = []
for dep_idx in range(deps_nr):
dep_name, dep_version = input().split()
dep_version = int(dep_version)
dep_proj = (dep_name, dep_version)
proj___deps[proj].append(dep_proj)
if proj_idx != projs_nr - 1:
_ = input()
proj_name___version_used = {}
curr_unresolved_projs = [start_proj]
next_unresolved_projs = []
while curr_unresolved_projs:
for proj in curr_unresolved_projs:
proj_name___version_used[proj[0]] = proj[1]
for proj in curr_unresolved_projs:
for dep_proj in proj___deps[proj]:
if dep_proj[0] not in proj_name___version_used:
next_unresolved_projs.append(dep_proj)
unresolved_name___versions = {}
for unresolved_name, unresolved_version in next_unresolved_projs:
if unresolved_name not in unresolved_name___versions:
unresolved_name___versions[unresolved_name] = []
unresolved_name___versions[unresolved_name].append(unresolved_version)
unresolved_name___final_version = {}
for unresolved_name, versions in unresolved_name___versions.items():
unresolved_name___final_version[unresolved_name] = max(versions)
curr_unresolved_projs = \
[(name, version) for name, version in unresolved_name___final_version.items()]
next_unresolved_projs = []
del proj_name___version_used[start_proj[0]]
projects = [name + ' ' + str(version) for name, version in proj_name___version_used.items()]
projects = sorted(projects)
print(*projects, sep='\n')
``` | 0 | |
798 | A | Mike and palindrome | PROGRAMMING | 1,000 | [
"brute force",
"constructive algorithms",
"strings"
] | null | null | Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not. | The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15). | Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise. | [
"abccaa\n",
"abbcca\n",
"abcda\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "abccaa",
"output": "YES"
},
{
"input": "abbcca",
"output": "NO"
},
{
"input": "abcda",
"output": "YES"
},
{
"input": "kyw",
"output": "YES"
},
{
"input": "fccf",
"output": "NO"
},
{
"input": "mnlm",
"output": "YES"
},
{
"input": "gqrk",
"output": "NO"
},
{
"input": "glxlg",
"output": "YES"
},
{
"input": "czhfc",
"output": "YES"
},
{
"input": "broon",
"output": "NO"
},
{
"input": "rmggmr",
"output": "NO"
},
{
"input": "wvxxzw",
"output": "YES"
},
{
"input": "ukvciu",
"output": "NO"
},
{
"input": "vrnwnrv",
"output": "YES"
},
{
"input": "vlkjkav",
"output": "YES"
},
{
"input": "guayhmg",
"output": "NO"
},
{
"input": "lkvhhvkl",
"output": "NO"
},
{
"input": "ffdsslff",
"output": "YES"
},
{
"input": "galjjtyw",
"output": "NO"
},
{
"input": "uosgwgsou",
"output": "YES"
},
{
"input": "qjwmjmljq",
"output": "YES"
},
{
"input": "ustrvrodf",
"output": "NO"
},
{
"input": "a",
"output": "YES"
},
{
"input": "qjfyjjyfjq",
"output": "NO"
},
{
"input": "ysxibbixsq",
"output": "YES"
},
{
"input": "howfslfwmh",
"output": "NO"
},
{
"input": "ekhajrjahke",
"output": "YES"
},
{
"input": "ucnolsloncw",
"output": "YES"
},
{
"input": "jrzsfrrkrtj",
"output": "NO"
},
{
"input": "typayzzyapyt",
"output": "NO"
},
{
"input": "uwdhkzokhdwu",
"output": "YES"
},
{
"input": "xokxpyyuafij",
"output": "NO"
},
{
"input": "eusneioiensue",
"output": "YES"
},
{
"input": "fuxpuajabpxuf",
"output": "YES"
},
{
"input": "guvggtfhlgruy",
"output": "NO"
},
{
"input": "cojhkhxxhkhjoc",
"output": "NO"
},
{
"input": "mhifbmmmmbmihm",
"output": "YES"
},
{
"input": "kxfqqncnebpami",
"output": "NO"
},
{
"input": "scfwrjevejrwfcs",
"output": "YES"
},
{
"input": "thdaonpepdoadht",
"output": "YES"
},
{
"input": "jsfzcbnhsccuqsj",
"output": "NO"
},
{
"input": "nn",
"output": "NO"
},
{
"input": "nm",
"output": "YES"
},
{
"input": "jdj",
"output": "YES"
},
{
"input": "bbcaa",
"output": "NO"
},
{
"input": "abcde",
"output": "NO"
},
{
"input": "abcdf",
"output": "NO"
},
{
"input": "aa",
"output": "NO"
},
{
"input": "abecd",
"output": "NO"
},
{
"input": "abccacb",
"output": "NO"
},
{
"input": "aabc",
"output": "NO"
},
{
"input": "anpqb",
"output": "NO"
},
{
"input": "c",
"output": "YES"
},
{
"input": "abcdefg",
"output": "NO"
},
{
"input": "aanbb",
"output": "NO"
},
{
"input": "aabbb",
"output": "NO"
},
{
"input": "aaabbab",
"output": "NO"
},
{
"input": "ab",
"output": "YES"
},
{
"input": "aabbc",
"output": "NO"
},
{
"input": "ecabd",
"output": "NO"
},
{
"input": "abcdrty",
"output": "NO"
},
{
"input": "abcdmnp",
"output": "NO"
},
{
"input": "bbbbbb",
"output": "NO"
},
{
"input": "abcxuio",
"output": "NO"
},
{
"input": "abcdabcde",
"output": "NO"
},
{
"input": "abcxpoi",
"output": "NO"
},
{
"input": "aba",
"output": "YES"
},
{
"input": "aacbb",
"output": "NO"
},
{
"input": "abcedca",
"output": "NO"
},
{
"input": "abcdd",
"output": "NO"
},
{
"input": "abbcs",
"output": "NO"
},
{
"input": "aaabccc",
"output": "NO"
},
{
"input": "paxkxbq",
"output": "NO"
},
{
"input": "z",
"output": "YES"
},
{
"input": "b",
"output": "YES"
},
{
"input": "abcdefghi",
"output": "NO"
},
{
"input": "abcqr",
"output": "NO"
},
{
"input": "abcdc",
"output": "NO"
},
{
"input": "abcb",
"output": "NO"
},
{
"input": "aabcd",
"output": "NO"
},
{
"input": "abbba",
"output": "YES"
},
{
"input": "aaabbb",
"output": "NO"
},
{
"input": "bb",
"output": "NO"
},
{
"input": "aaacbbb",
"output": "NO"
},
{
"input": "abbzcca",
"output": "NO"
},
{
"input": "abxab",
"output": "NO"
},
{
"input": "bbb",
"output": "YES"
},
{
"input": "abcrtyu",
"output": "NO"
},
{
"input": "cbacb",
"output": "NO"
},
{
"input": "acbb",
"output": "NO"
},
{
"input": "ww",
"output": "NO"
},
{
"input": "aaaaaa",
"output": "NO"
},
{
"input": "jizzz",
"output": "NO"
},
{
"input": "aaakcba",
"output": "NO"
},
{
"input": "acbak",
"output": "NO"
},
{
"input": "bddeffd",
"output": "NO"
},
{
"input": "aaa",
"output": "YES"
},
{
"input": "afghqwe",
"output": "NO"
},
{
"input": "abcdfga",
"output": "NO"
}
] | 1,671,299,483 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | s = str(input())
l = len(s)
for i in range(l):
if s[i] != s[(l - 1) - i]:
print("no")
break
else:
print("yes")
| Title: Mike and palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.
Input Specification:
The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15).
Output Specification:
Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.
Demo Input:
['abccaa\n', 'abbcca\n', 'abcda\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none | ```python
s = str(input())
l = len(s)
for i in range(l):
if s[i] != s[(l - 1) - i]:
print("no")
break
else:
print("yes")
``` | 0 | |
770 | A | New Password | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions. | The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists. | Print any password which satisfies all conditions given by Innokentiy. | [
"4 3\n",
"6 6\n",
"5 2\n"
] | [
"java\n",
"python\n",
"phphp\n"
] | In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests. | 500 | [
{
"input": "4 3",
"output": "abca"
},
{
"input": "6 6",
"output": "abcdef"
},
{
"input": "5 2",
"output": "ababa"
},
{
"input": "3 2",
"output": "aba"
},
{
"input": "10 2",
"output": "ababababab"
},
{
"input": "26 13",
"output": "abcdefghijklmabcdefghijklm"
},
{
"input": "100 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "100 10",
"output": "abcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij"
},
{
"input": "3 3",
"output": "abc"
},
{
"input": "6 3",
"output": "abcabc"
},
{
"input": "10 3",
"output": "abcabcabca"
},
{
"input": "50 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab"
},
{
"input": "90 2",
"output": "ababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "6 2",
"output": "ababab"
},
{
"input": "99 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc"
},
{
"input": "4 2",
"output": "abab"
},
{
"input": "100 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "40 22",
"output": "abcdefghijklmnopqrstuvabcdefghijklmnopqr"
},
{
"input": "13 8",
"output": "abcdefghabcde"
},
{
"input": "16 15",
"output": "abcdefghijklmnoa"
},
{
"input": "17 17",
"output": "abcdefghijklmnopq"
},
{
"input": "19 4",
"output": "abcdabcdabcdabcdabc"
},
{
"input": "100 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv"
},
{
"input": "100 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxy"
},
{
"input": "26 26",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "27 26",
"output": "abcdefghijklmnopqrstuvwxyza"
},
{
"input": "2 2",
"output": "ab"
},
{
"input": "26 25",
"output": "abcdefghijklmnopqrstuvwxya"
},
{
"input": "99 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababa"
},
{
"input": "99 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstu"
},
{
"input": "4 4",
"output": "abcd"
},
{
"input": "5 3",
"output": "abcab"
},
{
"input": "5 4",
"output": "abcda"
},
{
"input": "5 5",
"output": "abcde"
},
{
"input": "24 22",
"output": "abcdefghijklmnopqrstuvab"
},
{
"input": "26 14",
"output": "abcdefghijklmnabcdefghijkl"
},
{
"input": "26 15",
"output": "abcdefghijklmnoabcdefghijk"
},
{
"input": "30 12",
"output": "abcdefghijklabcdefghijklabcdef"
},
{
"input": "35 4",
"output": "abcdabcdabcdabcdabcdabcdabcdabcdabc"
},
{
"input": "79 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "79 14",
"output": "abcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghi"
},
{
"input": "85 13",
"output": "abcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefg"
},
{
"input": "90 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmno"
},
{
"input": "90 19",
"output": "abcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmn"
},
{
"input": "26 24",
"output": "abcdefghijklmnopqrstuvwxab"
},
{
"input": "100 17",
"output": "abcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmno"
},
{
"input": "26 2",
"output": "ababababababababababababab"
}
] | 1,626,121,707 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 2 | 140 | 23,040,000 | from itertools import product
from math import ceil
import string
def binary_table(string_with_all_characters, length_to_make):
return [''.join(x) for x in product(string_with_all_characters, repeat=length_to_make)]
def all_possible_substrings(string):
return [int(string[i: j]) for i in range(len(string)) for j in range(i + 1, len(string) + 1)]
def number_of_substrings(length):
return int(length * (length + 1) / 2)
def is_prime(num):
for i in range(2, num):
if num / i == int(num / i) and num != i:
return False
return True
"""for enumeration in range(int(input())):
"""
alphabet = string.ascii_lowercase
used = []
build = ''
length, distinct = map(int, input().split())
for i in range(distinct):
build += alphabet[i]
used.append(alphabet[i])
for i in range(length - distinct):
build += used[i]
print(build) | Title: New Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
Input Specification:
The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
Output Specification:
Print any password which satisfies all conditions given by Innokentiy.
Demo Input:
['4 3\n', '6 6\n', '5 2\n']
Demo Output:
['java\n', 'python\n', 'phphp\n']
Note:
In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests. | ```python
from itertools import product
from math import ceil
import string
def binary_table(string_with_all_characters, length_to_make):
return [''.join(x) for x in product(string_with_all_characters, repeat=length_to_make)]
def all_possible_substrings(string):
return [int(string[i: j]) for i in range(len(string)) for j in range(i + 1, len(string) + 1)]
def number_of_substrings(length):
return int(length * (length + 1) / 2)
def is_prime(num):
for i in range(2, num):
if num / i == int(num / i) and num != i:
return False
return True
"""for enumeration in range(int(input())):
"""
alphabet = string.ascii_lowercase
used = []
build = ''
length, distinct = map(int, input().split())
for i in range(distinct):
build += alphabet[i]
used.append(alphabet[i])
for i in range(length - distinct):
build += used[i]
print(build)
``` | -1 | |
1,009 | B | Minimum Ternary String | PROGRAMMING | 1,400 | [
"greedy",
"implementation"
] | null | null | You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$. | The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive). | Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero). | [
"100210\n",
"11222121\n",
"20\n"
] | [
"001120\n",
"11112222\n",
"20\n"
] | none | 0 | [
{
"input": "100210",
"output": "001120"
},
{
"input": "11222121",
"output": "11112222"
},
{
"input": "20",
"output": "20"
},
{
"input": "1002",
"output": "0012"
},
{
"input": "10",
"output": "01"
},
{
"input": "000021",
"output": "000012"
},
{
"input": "021",
"output": "012"
},
{
"input": "2",
"output": "2"
},
{
"input": "201",
"output": "120"
},
{
"input": "2112120",
"output": "1112220"
},
{
"input": "102",
"output": "012"
},
{
"input": "202",
"output": "202"
},
{
"input": "220201",
"output": "122020"
},
{
"input": "12",
"output": "12"
},
{
"input": "100022202",
"output": "000122202"
},
{
"input": "01",
"output": "01"
},
{
"input": "1",
"output": "1"
}
] | 1,532,610,311 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 5 | 124 | 307,200 | s, t, k = input(), '', 0
for i in s:
if i != '1':
t += i
else:
k += 1
i = t.index('2')
print (t[:i] + '1' * k + t[i:]) | Title: Minimum Ternary String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$.
Input Specification:
The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive).
Output Specification:
Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).
Demo Input:
['100210\n', '11222121\n', '20\n']
Demo Output:
['001120\n', '11112222\n', '20\n']
Note:
none | ```python
s, t, k = input(), '', 0
for i in s:
if i != '1':
t += i
else:
k += 1
i = t.index('2')
print (t[:i] + '1' * k + t[i:])
``` | -1 | |
228 | A | Is your horseshoe on the other hoof? | PROGRAMMING | 800 | [
"implementation"
] | null | null | Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. | The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers. | Print a single integer — the minimum number of horseshoes Valera needs to buy. | [
"1 7 3 3\n",
"7 7 7 7\n"
] | [
"1\n",
"3\n"
] | none | 500 | [
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 586975278",
"output": "1"
},
{
"input": "259504825 377489979 588153796 377489979",
"output": "1"
},
{
"input": "652588203 931100304 931100304 652588203",
"output": "2"
},
{
"input": "391958720 651507265 391958720 651507265",
"output": "2"
},
{
"input": "90793237 90793237 90793237 90793237",
"output": "3"
},
{
"input": "551651653 551651653 551651653 551651653",
"output": "3"
},
{
"input": "156630260 609654355 668943582 973622757",
"output": "0"
},
{
"input": "17061017 110313588 434481173 796661222",
"output": "0"
},
{
"input": "24975422 256716298 337790533 690960249",
"output": "0"
},
{
"input": "255635360 732742923 798648949 883146723",
"output": "0"
},
{
"input": "133315691 265159773 734556507 265159773",
"output": "1"
},
{
"input": "28442865 741657755 978106882 978106882",
"output": "1"
},
{
"input": "131245479 174845575 497483467 131245479",
"output": "1"
},
{
"input": "139159884 616215581 958341883 616215581",
"output": "1"
},
{
"input": "147784432 947653080 947653080 947653080",
"output": "2"
},
{
"input": "94055790 756126496 756126496 94055790",
"output": "2"
},
{
"input": "240458500 511952208 240458500 511952208",
"output": "2"
},
{
"input": "681828506 972810624 972810624 681828506",
"output": "2"
},
{
"input": "454961014 454961014 454961014 454961014",
"output": "3"
},
{
"input": "915819430 915819430 915819430 915819430",
"output": "3"
},
{
"input": "671645142 671645142 671645142 671645142",
"output": "3"
},
{
"input": "132503558 132503558 132503558 132503558",
"output": "3"
},
{
"input": "5 5 999999 6",
"output": "1"
},
{
"input": "1 1 2 5",
"output": "1"
},
{
"input": "2 1 2 3",
"output": "1"
},
{
"input": "1 1 3 5",
"output": "1"
},
{
"input": "1 1 3 3",
"output": "2"
},
{
"input": "2 2 2 1",
"output": "2"
},
{
"input": "3 1 1 1",
"output": "2"
},
{
"input": "1 2 2 2",
"output": "2"
}
] | 1,696,801,733 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 154 | 2,764,800 | t=[0]*4
necessity=0
for i in range (4):
x=int(input())
if x in t:
necessity+=1
t[i]=x
print(necessity) | Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none | ```python
t=[0]*4
necessity=0
for i in range (4):
x=int(input())
if x in t:
necessity+=1
t[i]=x
print(necessity)
``` | -1 | |
108 | A | Palindromic Times | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | A. Palindromic Times | 2 | 256 | Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him. | The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits. | Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time. | [
"12:21\n",
"23:59\n"
] | [
"13:31\n",
"00:00\n"
] | none | 500 | [
{
"input": "12:21",
"output": "13:31"
},
{
"input": "23:59",
"output": "00:00"
},
{
"input": "15:51",
"output": "20:02"
},
{
"input": "10:44",
"output": "11:11"
},
{
"input": "04:02",
"output": "04:40"
},
{
"input": "02:11",
"output": "02:20"
},
{
"input": "12:15",
"output": "12:21"
},
{
"input": "07:07",
"output": "10:01"
},
{
"input": "00:17",
"output": "01:10"
},
{
"input": "04:55",
"output": "05:50"
},
{
"input": "02:17",
"output": "02:20"
},
{
"input": "07:56",
"output": "10:01"
},
{
"input": "00:29",
"output": "01:10"
},
{
"input": "23:31",
"output": "23:32"
},
{
"input": "19:30",
"output": "20:02"
},
{
"input": "12:14",
"output": "12:21"
},
{
"input": "17:32",
"output": "20:02"
},
{
"input": "03:44",
"output": "04:40"
},
{
"input": "07:15",
"output": "10:01"
},
{
"input": "18:42",
"output": "20:02"
},
{
"input": "08:56",
"output": "10:01"
},
{
"input": "04:50",
"output": "05:50"
},
{
"input": "14:32",
"output": "14:41"
},
{
"input": "23:23",
"output": "23:32"
},
{
"input": "08:35",
"output": "10:01"
},
{
"input": "03:32",
"output": "04:40"
},
{
"input": "07:59",
"output": "10:01"
},
{
"input": "14:12",
"output": "14:41"
},
{
"input": "23:52",
"output": "00:00"
},
{
"input": "16:36",
"output": "20:02"
},
{
"input": "17:50",
"output": "20:02"
},
{
"input": "06:59",
"output": "10:01"
},
{
"input": "16:50",
"output": "20:02"
},
{
"input": "00:00",
"output": "01:10"
},
{
"input": "23:59",
"output": "00:00"
},
{
"input": "23:33",
"output": "00:00"
}
] | 1,679,373,415 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 92 | 0 | h,m=map(int,input().split(':'))
while 1:
m+=1
if m==60:h+=1;m=0
if h==24:h=0
a=str(h);a='0'*(2-len(a))+a
b=str(m);a+=':'+'0'*(2-len(b))+b
if(a==a[::-1]):print(a);exit()
| Title: Palindromic Times
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him.
Input Specification:
The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits.
Output Specification:
Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time.
Demo Input:
['12:21\n', '23:59\n']
Demo Output:
['13:31\n', '00:00\n']
Note:
none | ```python
h,m=map(int,input().split(':'))
while 1:
m+=1
if m==60:h+=1;m=0
if h==24:h=0
a=str(h);a='0'*(2-len(a))+a
b=str(m);a+=':'+'0'*(2-len(b))+b
if(a==a[::-1]):print(a);exit()
``` | 3.977 |
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"
}
] | 1,645,457,989 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 60 | 0 | x=int(input())
y=int(input())
z=(x*y)/2
print(z) | 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
x=int(input())
y=int(input())
z=(x*y)/2
print(z)
``` | -1 |
907 | A | Masha and Bears | PROGRAMMING | 1,300 | [
"brute force",
"implementation"
] | null | null | A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars. | You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3. | Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes). | [
"50 30 10 10\n",
"100 50 10 21\n"
] | [
"50\n30\n10\n",
"-1\n"
] | In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20. | 500 | [
{
"input": "50 30 10 10",
"output": "50\n30\n10"
},
{
"input": "100 50 10 21",
"output": "-1"
},
{
"input": "100 50 19 10",
"output": "100\n50\n19"
},
{
"input": "99 50 25 49",
"output": "100\n99\n49"
},
{
"input": "3 2 1 1",
"output": "4\n3\n1"
},
{
"input": "100 99 98 100",
"output": "-1"
},
{
"input": "100 40 30 40",
"output": "-1"
},
{
"input": "100 50 19 25",
"output": "100\n51\n25"
},
{
"input": "100 50 19 30",
"output": "100\n61\n30"
},
{
"input": "49 48 25 49",
"output": "-1"
},
{
"input": "48 47 23 46",
"output": "94\n93\n46"
},
{
"input": "37 23 16 20",
"output": "42\n41\n20"
},
{
"input": "98 2 1 1",
"output": "98\n3\n1"
},
{
"input": "99 22 13 14",
"output": "99\n29\n14"
},
{
"input": "97 95 3 2",
"output": "97\n95\n3"
},
{
"input": "27 3 2 3",
"output": "-1"
},
{
"input": "13 7 6 2",
"output": "-1"
},
{
"input": "19 17 11 6",
"output": "19\n17\n11"
},
{
"input": "15 6 4 5",
"output": "15\n11\n5"
},
{
"input": "21 3 1 3",
"output": "-1"
},
{
"input": "100 99 98 97",
"output": "196\n195\n98"
},
{
"input": "99 32 13 5",
"output": "-1"
},
{
"input": "5 4 3 1",
"output": "-1"
},
{
"input": "50 20 4 8",
"output": "50\n20\n8"
},
{
"input": "60 30 10 1",
"output": "-1"
},
{
"input": "100 50 10 1",
"output": "-1"
},
{
"input": "40 30 20 10",
"output": "40\n30\n20"
},
{
"input": "50 49 5 10",
"output": "50\n49\n10"
},
{
"input": "99 98 50 97",
"output": "196\n195\n97"
},
{
"input": "50 30 10 5",
"output": "50\n30\n10"
},
{
"input": "4 3 2 1",
"output": "4\n3\n2"
},
{
"input": "100 50 3 1",
"output": "-1"
}
] | 1,612,132,770 | 570 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 93 | 0 | V = list(map(int, input().split()))
R = [0]*3
L = [0]*3
for i in range(3):
if i != 2:
R[i] = 2*V[i]
L[i] = max(V[i], 2*V[3]+1)
else:
R[i] = min(V[i], V[3])
L[i] = max(V[i], V[3])
#print(R)
#print(L)
ans = [-1]*3
for i in range(3):
if R[i] >= L[i]:
ans[i] = L[i]
else:
print(-1)
exit()
print(*ans, sep='\n')
| Title: Masha and Bears
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input Specification:
You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3.
Output Specification:
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Demo Input:
['50 30 10 10\n', '100 50 10 21\n']
Demo Output:
['50\n30\n10\n', '-1\n']
Note:
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20. | ```python
V = list(map(int, input().split()))
R = [0]*3
L = [0]*3
for i in range(3):
if i != 2:
R[i] = 2*V[i]
L[i] = max(V[i], 2*V[3]+1)
else:
R[i] = min(V[i], V[3])
L[i] = max(V[i], V[3])
#print(R)
#print(L)
ans = [-1]*3
for i in range(3):
if R[i] >= L[i]:
ans[i] = L[i]
else:
print(-1)
exit()
print(*ans, sep='\n')
``` | 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"
}
] | 1,651,276,535 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 12 | 46 | 0 | import math
n, m, x = map(int, input().split())
dec={}
S=[]
for i in range(n):
s=input()
for j in range(m):
if s[j]=='S':
S.append([i+1,j+1])
else:
dec[s[j]]=[i+1,j+1]
dec[s[j].upper()] = [i + 1, j + 1]
k=int(input())
tst=input()
f = 1
count = 0
for t in tst:
if t in dec:
if ord(t)<96:
if len(S)==0:
f=0
else:
ll=0
lll=0
for o in S:
lll+=1
i,j=o
I,J=dec[t]
dis=math.sqrt(abs(i-I)**2 +abs(j-J)**2)
if dis>x:
ll+=1
if ll==lll:
count+=1
else:
f=0
if f:
print(count)
else:
print(-1) | 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
import math
n, m, x = map(int, input().split())
dec={}
S=[]
for i in range(n):
s=input()
for j in range(m):
if s[j]=='S':
S.append([i+1,j+1])
else:
dec[s[j]]=[i+1,j+1]
dec[s[j].upper()] = [i + 1, j + 1]
k=int(input())
tst=input()
f = 1
count = 0
for t in tst:
if t in dec:
if ord(t)<96:
if len(S)==0:
f=0
else:
ll=0
lll=0
for o in S:
lll+=1
i,j=o
I,J=dec[t]
dis=math.sqrt(abs(i-I)**2 +abs(j-J)**2)
if dis>x:
ll+=1
if ll==lll:
count+=1
else:
f=0
if f:
print(count)
else:
print(-1)
``` | 0 |
424 | A | Squats | PROGRAMMING | 900 | [
"implementation"
] | null | null | Pasha has many hamsters and he makes them work out. Today, *n* hamsters (*n* is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.
For another exercise, Pasha needs exactly hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well? | The first line contains integer *n* (2<=≤<=*n*<=≤<=200; *n* is even). The next line contains *n* characters without spaces. These characters describe the hamsters' position: the *i*-th character equals 'X', if the *i*-th hamster in the row is standing, and 'x', if he is sitting. | In the first line, print a single integer — the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them. | [
"4\nxxXx\n",
"2\nXX\n",
"6\nxXXxXx\n"
] | [
"1\nXxXx\n",
"1\nxX\n",
"0\nxXXxXx\n"
] | none | 500 | [
{
"input": "4\nxxXx",
"output": "1\nXxXx"
},
{
"input": "2\nXX",
"output": "1\nxX"
},
{
"input": "6\nxXXxXx",
"output": "0\nxXXxXx"
},
{
"input": "4\nxXXX",
"output": "1\nxxXX"
},
{
"input": "2\nXx",
"output": "0\nXx"
},
{
"input": "22\nXXxXXxxXxXxXXXXXXXXXxx",
"output": "4\nxxxxxxxXxXxXXXXXXXXXxx"
},
{
"input": "30\nXXxXxxXXXXxxXXxxXXxxxxXxxXXXxx",
"output": "0\nXXxXxxXXXXxxXXxxXXxxxxXxxXXXxx"
},
{
"input": "104\nxxXxXxxXXXxxXxXxxXXXxxxXxxXXXxxXXXxXxXxXXxxXxxxxxXXXXxXXXXxXXXxxxXxxxxxxxXxxXxXXxxXXXXxXXXxxXXXXXXXXXxXX",
"output": "4\nxxxxxxxxxXxxXxXxxXXXxxxXxxXXXxxXXXxXxXxXXxxXxxxxxXXXXxXXXXxXXXxxxXxxxxxxxXxxXxXXxxXXXXxXXXxxXXXXXXXXXxXX"
},
{
"input": "78\nxxxXxxXxXxxXxxxxxXxXXXxXXXXxxxxxXxXXXxxXxXXXxxxxXxxXXXxxxxxxxxXXXXxXxXXxXXXxXX",
"output": "3\nXXXXxxXxXxxXxxxxxXxXXXxXXXXxxxxxXxXXXxxXxXXXxxxxXxxXXXxxxxxxxxXXXXxXxXXxXXXxXX"
},
{
"input": "200\nxxXXxxXXxXxxXxxXxXxxXxXxXxXxxxxxXXxXXxxXXXXxXXXxXXxXxXxxxxXxxXXXxxxXxXxxxXxxXXxXxXxxxxxxxXxxXxXxxXxXXXxxXxXXXXxxXxxxXxXXXXXXxXxXXxxxxXxxxXxxxXxXXXxXxXXXXxXXxxxXxXXxxXXxxxXxXxXXxXXXxXxXxxxXXxxxxXXxXXXX",
"output": "4\nXXXXXXXXxXxxXxxXxXxxXxXxXxXxxxxxXXxXXxxXXXXxXXXxXXxXxXxxxxXxxXXXxxxXxXxxxXxxXXxXxXxxxxxxxXxxXxXxxXxXXXxxXxXXXXxxXxxxXxXXXXXXxXxXXxxxxXxxxXxxxXxXXXxXxXXXXxXXxxxXxXXxxXXxxxXxXxXXxXXXxXxXxxxXXxxxxXXxXXXX"
},
{
"input": "198\nxXxxXxxXxxXXxXxXxXxxXXXxxXxxxxXXXXxxXxxxxXXXXxXxXXxxxXXXXXXXxXXXxxxxXXxXXxXxXXxxxxXxXXXXXXxXxxXxXxxxXxXXXXxxXXxxXxxxXXxXxXXxXxXXxXXXXxxxxxXxXXxxxXxXXXXxXxXXxxXxXXxXxXXxxxXxXXXXxXxxXxXXXxxxxXxXXXXxXx",
"output": "5\nxxxxxxxxxxxxxXxXxXxxXXXxxXxxxxXXXXxxXxxxxXXXXxXxXXxxxXXXXXXXxXXXxxxxXXxXXxXxXXxxxxXxXXXXXXxXxxXxXxxxXxXXXXxxXXxxXxxxXXxXxXXxXxXXxXXXXxxxxxXxXXxxxXxXXXXxXxXXxxXxXXxXxXXxxxXxXXXXxXxxXxXXXxxxxXxXXXXxXx"
},
{
"input": "200\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "100\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "198\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "99\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "200\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX",
"output": "100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX"
},
{
"input": "198\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX",
"output": "99\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX"
},
{
"input": "2\nxx",
"output": "1\nXx"
},
{
"input": "2\nXx",
"output": "0\nXx"
},
{
"input": "2\nxX",
"output": "0\nxX"
},
{
"input": "4\nXXXX",
"output": "2\nxxXX"
},
{
"input": "4\nxxxx",
"output": "2\nXXxx"
},
{
"input": "4\nxxXX",
"output": "0\nxxXX"
},
{
"input": "4\nXXxx",
"output": "0\nXXxx"
},
{
"input": "4\nxXxx",
"output": "1\nXXxx"
},
{
"input": "4\nXxxx",
"output": "1\nXXxx"
},
{
"input": "4\nxxxX",
"output": "1\nXxxX"
},
{
"input": "4\nxxXx",
"output": "1\nXxXx"
},
{
"input": "4\nXXXx",
"output": "1\nxXXx"
},
{
"input": "4\nxXXX",
"output": "1\nxxXX"
},
{
"input": "4\nXxXX",
"output": "1\nxxXX"
},
{
"input": "4\nXXxX",
"output": "1\nxXxX"
},
{
"input": "4\nXxXx",
"output": "0\nXxXx"
},
{
"input": "6\nxXXxXX",
"output": "1\nxxXxXX"
}
] | 1,501,350,965 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 62 | 4,608,000 | n = int(input())
global s
s = input()
a, b = s.count("x"), s.count("X")
p = abs(a - b) // 2
def change(a, b):
global s
q = 0
for x in range(n):
if q == p:
break
if s[x] == a:
s = s[:x] + b + s[x + 1:]
q += 1
print(p)
if a > b:
change("x", "X")
else:
change("X", "x")
print(s) | Title: Squats
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has many hamsters and he makes them work out. Today, *n* hamsters (*n* is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.
For another exercise, Pasha needs exactly hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well?
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=200; *n* is even). The next line contains *n* characters without spaces. These characters describe the hamsters' position: the *i*-th character equals 'X', if the *i*-th hamster in the row is standing, and 'x', if he is sitting.
Output Specification:
In the first line, print a single integer — the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them.
Demo Input:
['4\nxxXx\n', '2\nXX\n', '6\nxXXxXx\n']
Demo Output:
['1\nXxXx\n', '1\nxX\n', '0\nxXXxXx\n']
Note:
none | ```python
n = int(input())
global s
s = input()
a, b = s.count("x"), s.count("X")
p = abs(a - b) // 2
def change(a, b):
global s
q = 0
for x in range(n):
if q == p:
break
if s[x] == a:
s = s[:x] + b + s[x + 1:]
q += 1
print(p)
if a > b:
change("x", "X")
else:
change("X", "x")
print(s)
``` | 3 | |
699 | B | One Bomb | PROGRAMMING | 1,400 | [
"implementation"
] | null | null | You are given a description of a depot. It is a rectangular checkered field of *n*<=×<=*m* size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").
You have one bomb. If you lay the bomb at the cell (*x*,<=*y*), then after triggering it will wipe out all walls in the row *x* and all walls in the column *y*.
You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall. | The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the depot field.
The next *n* lines contain *m* symbols "." and "*" each — the description of the field. *j*-th symbol in *i*-th of them stands for cell (*i*,<=*j*). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall. | If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).
Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them. | [
"3 4\n.*..\n....\n.*..\n",
"3 3\n..*\n.*.\n*..\n",
"6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..\n"
] | [
"YES\n1 2\n",
"NO\n",
"YES\n3 3\n"
] | none | 1,000 | [
{
"input": "3 4\n.*..\n....\n.*..",
"output": "YES\n1 2"
},
{
"input": "3 3\n..*\n.*.\n*..",
"output": "NO"
},
{
"input": "6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..",
"output": "YES\n3 3"
},
{
"input": "1 10\n**********",
"output": "YES\n1 1"
},
{
"input": "10 1\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*",
"output": "YES\n1 1"
},
{
"input": "10 10\n.........*\n.........*\n........**\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*",
"output": "YES\n3 10"
},
{
"input": "10 10\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*\n.........*",
"output": "YES\n1 10"
},
{
"input": "2 2\n.*\n*.",
"output": "YES\n2 2"
},
{
"input": "4 4\n....\n...*\n....\n*..*",
"output": "YES\n4 4"
},
{
"input": "4 4\n*...\n*...\n....\n****",
"output": "YES\n4 1"
},
{
"input": "1 1\n*",
"output": "YES\n1 1"
},
{
"input": "1 1\n.",
"output": "YES\n1 1"
},
{
"input": "1 2\n.*",
"output": "YES\n1 2"
},
{
"input": "2 1\n.\n*",
"output": "YES\n1 1"
},
{
"input": "2 2\n**\n**",
"output": "NO"
},
{
"input": "3 1\n*\n*\n*",
"output": "YES\n1 1"
},
{
"input": "3 2\n*.\n.*\n.*",
"output": "YES\n1 2"
},
{
"input": "3 3\n***\n***\n***",
"output": "NO"
},
{
"input": "2 2\n..\n.*",
"output": "YES\n1 2"
},
{
"input": "6 5\n..*..\n..*..\n**.**\n..*..\n..*..\n..*..",
"output": "YES\n3 3"
},
{
"input": "3 3\n.*.\n*.*\n.*.",
"output": "YES\n2 2"
},
{
"input": "4 4\n*...\n....\n....\n...*",
"output": "YES\n4 1"
},
{
"input": "2 4\n...*\n...*",
"output": "YES\n1 4"
},
{
"input": "2 2\n..\n..",
"output": "YES\n1 1"
},
{
"input": "3 3\n..*\n.*.\n..*",
"output": "YES\n2 3"
},
{
"input": "2 2\n*.\n.*",
"output": "YES\n2 1"
},
{
"input": "3 2\n.*\n*.\n.*",
"output": "YES\n2 2"
},
{
"input": "3 3\n***\n.*.\n.*.",
"output": "YES\n1 2"
},
{
"input": "4 4\n*.*.\n..*.\n.***\n..*.",
"output": "NO"
},
{
"input": "2 3\n..*\n**.",
"output": "YES\n2 3"
},
{
"input": "3 2\n*.\n.*\n*.",
"output": "YES\n2 1"
},
{
"input": "4 4\n..*.\n**.*\n..*.\n..*.",
"output": "YES\n2 3"
},
{
"input": "3 3\n*..\n*..\n***",
"output": "YES\n3 1"
},
{
"input": "3 3\n...\n*.*\n.*.",
"output": "YES\n2 2"
},
{
"input": "3 2\n..\n..\n**",
"output": "YES\n3 1"
},
{
"input": "3 4\n...*\n...*\n...*",
"output": "YES\n1 4"
},
{
"input": "5 5\n..*..\n..*..\n**.**\n..*..\n..*..",
"output": "YES\n3 3"
},
{
"input": "6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*.*",
"output": "NO"
},
{
"input": "3 3\n...\n.*.\n..*",
"output": "YES\n3 2"
},
{
"input": "3 5\n....*\n....*\n....*",
"output": "YES\n1 5"
},
{
"input": "3 3\n...\n...\n.*.",
"output": "YES\n1 2"
},
{
"input": "3 3\n*..\n...\n..*",
"output": "YES\n3 1"
},
{
"input": "2 3\n..*\n..*",
"output": "YES\n1 3"
},
{
"input": "2 2\n**\n.*",
"output": "YES\n1 2"
},
{
"input": "3 3\n..*\n*..\n*..",
"output": "YES\n1 1"
},
{
"input": "5 4\n.*..\n*.**\n.*..\n.*..\n.*..",
"output": "YES\n2 2"
},
{
"input": "6 5\n*.*..\n..*..\n*****\n..*..\n..*..\n..*..",
"output": "NO"
},
{
"input": "4 4\n.*..\n*.**\n....\n.*..",
"output": "YES\n2 2"
},
{
"input": "3 5\n....*\n....*\n*****",
"output": "YES\n3 5"
},
{
"input": "3 3\n..*\n*..\n..*",
"output": "YES\n2 3"
},
{
"input": "6 6\n..*...\n......\n......\n......\n......\n*....*",
"output": "YES\n6 3"
},
{
"input": "4 4\n.*..\n*...\n.*..\n.*..",
"output": "YES\n2 2"
},
{
"input": "3 3\n...\n..*\n.*.",
"output": "YES\n3 3"
},
{
"input": "3 2\n.*\n*.\n*.",
"output": "YES\n1 1"
},
{
"input": "4 2\n**\n.*\n.*\n.*",
"output": "YES\n1 2"
},
{
"input": "5 5\n*...*\n.....\n.....\n.....\n..*..",
"output": "YES\n1 3"
},
{
"input": "3 3\n**.\n...\n..*",
"output": "YES\n1 3"
},
{
"input": "3 3\n*.*\n*..\n*.*",
"output": "NO"
},
{
"input": "5 4\n....\n....\n*..*\n....\n.*..",
"output": "YES\n3 2"
},
{
"input": "5 5\n...*.\n...*.\n...*.\n...*.\n***.*",
"output": "YES\n5 4"
},
{
"input": "5 5\n*****\n*****\n*****\n*****\n*****",
"output": "NO"
},
{
"input": "3 3\n.*.\n..*\n.*.",
"output": "YES\n2 2"
},
{
"input": "3 3\n*.*\n...\n*.*",
"output": "NO"
},
{
"input": "2 3\n.*.\n*.*",
"output": "YES\n2 2"
},
{
"input": "3 10\n.......*..\n........*.\n.........*",
"output": "NO"
},
{
"input": "3 3\n.*.\n.*.\n.**",
"output": "YES\n3 2"
},
{
"input": "4 4\n*...\n....\n....\n..**",
"output": "YES\n4 1"
},
{
"input": "4 4\n****\n****\n****\n****",
"output": "NO"
},
{
"input": "3 2\n.*\n.*\n*.",
"output": "YES\n3 2"
},
{
"input": "3 3\n..*\n..*\n**.",
"output": "YES\n3 3"
},
{
"input": "6 3\n...\n...\n...\n...\n**.\n.*.",
"output": "YES\n5 2"
},
{
"input": "3 4\n****\n..*.\n..*.",
"output": "YES\n1 3"
},
{
"input": "5 5\n*..*.\n.....\n.....\n.....\n...*.",
"output": "YES\n1 4"
},
{
"input": "6 5\n..*..\n..*..\n.*...\n..*..\n..*..\n..*..",
"output": "YES\n3 3"
}
] | 1,571,120,230 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 109 | 0 | n, m = map(int, input().split())
counter = 0
b = -1
for i in range(n):
a = input()
x, y = a.find("*"), a.count("*")
if b != x + 1 and y == 1 and b != -1:
exit(print("NO"))
if y > 1 and counter != 0:
exit(print("NO"))
if y > 1:
counter = i + 1
if y == 1:
b = x + 1
print("YES", (counter, 1)[counter == 0], (b, 1)[b == -1]) | Title: One Bomb
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a description of a depot. It is a rectangular checkered field of *n*<=×<=*m* size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").
You have one bomb. If you lay the bomb at the cell (*x*,<=*y*), then after triggering it will wipe out all walls in the row *x* and all walls in the column *y*.
You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.
Input Specification:
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the depot field.
The next *n* lines contain *m* symbols "." and "*" each — the description of the field. *j*-th symbol in *i*-th of them stands for cell (*i*,<=*j*). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall.
Output Specification:
If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).
Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.
Demo Input:
['3 4\n.*..\n....\n.*..\n', '3 3\n..*\n.*.\n*..\n', '6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..\n']
Demo Output:
['YES\n1 2\n', 'NO\n', 'YES\n3 3\n']
Note:
none | ```python
n, m = map(int, input().split())
counter = 0
b = -1
for i in range(n):
a = input()
x, y = a.find("*"), a.count("*")
if b != x + 1 and y == 1 and b != -1:
exit(print("NO"))
if y > 1 and counter != 0:
exit(print("NO"))
if y > 1:
counter = i + 1
if y == 1:
b = x + 1
print("YES", (counter, 1)[counter == 0], (b, 1)[b == -1])
``` | 0 | |
225 | B | Well-known Numbers | PROGRAMMING | 1,600 | [
"binary search",
"greedy",
"number theory"
] | null | null | Numbers *k*-bonacci (*k* is integer, *k*<=><=1) are a generalization of Fibonacci numbers and are determined as follows:
- *F*(*k*,<=*n*)<==<=0, for integer *n*, 1<=≤<=*n*<=<<=*k*; - *F*(*k*,<=*k*)<==<=1; - *F*(*k*,<=*n*)<==<=*F*(*k*,<=*n*<=-<=1)<=+<=*F*(*k*,<=*n*<=-<=2)<=+<=...<=+<=*F*(*k*,<=*n*<=-<=*k*), for integer *n*, *n*<=><=*k*.
Note that we determine the *k*-bonacci numbers, *F*(*k*,<=*n*), only for integer values of *n* and *k*.
You've got a number *s*, represent it as a sum of several (at least two) distinct *k*-bonacci numbers. | The first line contains two integers *s* and *k* (1<=≤<=*s*,<=*k*<=≤<=109; *k*<=><=1). | In the first line print an integer *m* (*m*<=≥<=2) that shows how many numbers are in the found representation. In the second line print *m* distinct integers *a*1,<=*a*2,<=...,<=*a**m*. Each printed integer should be a *k*-bonacci number. The sum of printed integers must equal *s*.
It is guaranteed that the answer exists. If there are several possible answers, print any of them. | [
"5 2\n",
"21 5\n"
] | [
"3\n0 2 3\n",
"3\n4 1 16\n"
] | none | 1,000 | [
{
"input": "5 2",
"output": "3\n0 2 3"
},
{
"input": "21 5",
"output": "3\n4 1 16"
},
{
"input": "1 1000",
"output": "2\n1 0 "
},
{
"input": "1000000000 1000000000",
"output": "14\n536870912 268435456 134217728 33554432 16777216 8388608 1048576 524288 131072 32768 16384 2048 512 0 "
},
{
"input": "122 7",
"output": "6\n64 32 16 8 2 0 "
},
{
"input": "4 3",
"output": "2\n4 0 "
},
{
"input": "321123 3211232",
"output": "11\n262144 32768 16384 8192 1024 512 64 32 2 1 0 "
},
{
"input": "1 2",
"output": "2\n1 0 "
},
{
"input": "2 2",
"output": "2\n2 0 "
},
{
"input": "3 2",
"output": "2\n3 0 "
},
{
"input": "8 2",
"output": "2\n8 0 "
},
{
"input": "17 2",
"output": "4\n13 3 1 0 "
},
{
"input": "137 2",
"output": "5\n89 34 13 1 0 "
},
{
"input": "7298 2",
"output": "7\n6765 377 144 8 3 1 0 "
},
{
"input": "76754 2",
"output": "7\n75025 1597 89 34 8 1 0 "
},
{
"input": "12345678 2",
"output": "8\n9227465 2178309 832040 75025 28657 4181 1 0 "
},
{
"input": "987654321 2",
"output": "16\n701408733 267914296 14930352 2178309 832040 317811 46368 17711 6765 1597 233 89 13 3 1 0 "
},
{
"input": "1000000000 2",
"output": "15\n701408733 267914296 24157817 5702887 514229 196418 75025 28657 1597 233 89 13 5 1 0 "
},
{
"input": "701408733 2",
"output": "2\n701408733 0 "
},
{
"input": "1 3",
"output": "2\n1 0 "
},
{
"input": "2 3",
"output": "2\n2 0 "
},
{
"input": "3 3",
"output": "3\n2 1 0 "
},
{
"input": "100 3",
"output": "5\n81 13 4 2 0 "
},
{
"input": "87783 3",
"output": "8\n66012 19513 1705 504 44 4 1 0 "
},
{
"input": "615693473 3",
"output": "23\n334745777 181997601 53798080 29249425 8646064 4700770 1389537 755476 223317 121415 35890 19513 5768 3136 927 504 149 81 24 13 4 2 0 "
},
{
"input": "615693474 3",
"output": "2\n615693474 0 "
},
{
"input": "1000000000 3",
"output": "15\n615693474 334745777 29249425 15902591 2555757 1389537 410744 35890 10609 5768 274 149 4 1 0 "
},
{
"input": "1 4",
"output": "2\n1 0 "
},
{
"input": "2 4",
"output": "2\n2 0 "
},
{
"input": "17 4",
"output": "3\n15 2 0 "
},
{
"input": "234 4",
"output": "6\n208 15 8 2 1 0 "
},
{
"input": "23435345 4",
"output": "13\n14564533 7555935 1055026 147312 76424 20569 10671 2872 1490 401 108 4 0 "
},
{
"input": "989464701 4",
"output": "18\n747044834 201061985 28074040 7555935 3919944 1055026 547337 147312 39648 10671 5536 1490 773 108 56 4 2 0 "
},
{
"input": "464 5",
"output": "2\n464 0 "
},
{
"input": "7647474 5",
"output": "8\n5976577 1546352 103519 13624 6930 464 8 0 "
},
{
"input": "457787655 5",
"output": "14\n345052351 89277256 23099186 203513 103519 26784 13624 6930 3525 912 31 16 8 0 "
},
{
"input": "764747 6",
"output": "13\n463968 233904 59448 3840 1936 976 492 125 32 16 8 2 0 "
},
{
"input": "980765665 7",
"output": "16\n971364608 7805695 987568 495776 62725 31489 15808 1004 504 253 127 64 32 8 4 0 "
},
{
"input": "877655444 8",
"output": "17\n512966536 256993248 64504063 32316160 8111200 2035872 510994 128257 64256 16128 8080 509 128 8 4 1 0 "
},
{
"input": "567886500 9",
"output": "11\n525375999 32965728 8257696 1035269 129792 64960 32512 16272 8144 128 0 "
},
{
"input": "656777660 10",
"output": "13\n531372800 66519472 33276064 16646200 8327186 521472 65280 32656 16336 128 64 2 0 "
},
{
"input": "197445609 11",
"output": "18\n133628064 33423378 16715781 8359937 4180992 1045760 65424 16364 8184 1024 512 128 32 16 8 4 1 0 "
},
{
"input": "647474474 12",
"output": "18\n535625888 66977797 33492993 8375296 2094336 523712 261888 65488 32748 16376 4095 2048 1024 512 256 16 1 0 "
},
{
"input": "856644446 14",
"output": "16\n536592385 268304384 33541120 16771072 1048320 262096 65528 32765 16383 8192 2048 128 16 8 1 0 "
},
{
"input": "980345678 19",
"output": "18\n536864768 268432640 134216448 33554176 4194284 2097144 524287 262144 131072 65536 2048 1024 64 32 8 2 1 0 "
},
{
"input": "561854567 23",
"output": "17\n536870656 16777213 4194304 2097152 1048576 524288 262144 65536 8192 4096 2048 256 64 32 8 2 0 "
},
{
"input": "987654321 27",
"output": "20\n536870904 268435453 134217727 33554432 8388608 4194304 1048576 524288 262144 131072 16384 8192 2048 128 32 16 8 4 1 0 "
},
{
"input": "780787655 29",
"output": "18\n536870911 134217728 67108864 33554432 8388608 524288 65536 32768 16384 4096 2048 1024 512 256 128 64 8 0 "
},
{
"input": "999999999 30",
"output": "22\n536870912 268435456 134217728 33554432 16777216 8388608 1048576 524288 131072 32768 16384 2048 256 128 64 32 16 8 4 2 1 0 "
},
{
"input": "1 50",
"output": "2\n1 0 "
},
{
"input": "5 54",
"output": "3\n4 1 0 "
},
{
"input": "378 83",
"output": "7\n256 64 32 16 8 2 0 "
},
{
"input": "283847 111",
"output": "10\n262144 16384 4096 1024 128 64 4 2 1 0 "
},
{
"input": "38746466 2847",
"output": "14\n33554432 4194304 524288 262144 131072 65536 8192 4096 2048 256 64 32 2 0 "
},
{
"input": "83768466 12345",
"output": "15\n67108864 8388608 4194304 2097152 1048576 524288 262144 131072 8192 4096 1024 128 16 2 0 "
},
{
"input": "987654321 7475657",
"output": "18\n536870912 268435456 134217728 33554432 8388608 4194304 1048576 524288 262144 131072 16384 8192 2048 128 32 16 1 0 "
},
{
"input": "10 174764570",
"output": "3\n8 2 0 "
},
{
"input": "967755664 974301345",
"output": "17\n536870912 268435456 134217728 16777216 8388608 2097152 524288 262144 131072 32768 16384 1024 512 256 128 16 0 "
},
{
"input": "76 758866446",
"output": "4\n64 8 4 0 "
},
{
"input": "1 1000000000",
"output": "2\n1 0 "
},
{
"input": "469766205 719342208",
"output": "10\n268435456 134217728 67108864 4096 32 16 8 4 1 0 "
},
{
"input": "918938066 77",
"output": "17\n536870912 268435456 67108864 33554432 8388608 4194304 262144 65536 32768 16384 8192 256 128 64 16 2 0 "
},
{
"input": "856089381 19",
"output": "15\n536864768 268432640 33554176 16777104 262144 131072 65536 1024 512 256 128 16 4 1 0 "
},
{
"input": "152235195 16",
"output": "16\n134204416 16775936 1048528 131069 65535 8192 1024 256 128 64 32 8 4 2 1 0 "
},
{
"input": "429960894 3101",
"output": "17\n268435456 134217728 16777216 8388608 2097152 32768 8192 2048 1024 512 128 32 16 8 4 2 0 "
},
{
"input": "450695564 7",
"output": "18\n244804400 122895984 61695880 15548665 3918592 987568 495776 248888 62725 31489 3984 1004 504 64 32 8 1 0 "
},
{
"input": "154517270 24",
"output": "18\n134217708 16777215 2097152 1048576 262144 65536 32768 8192 4096 2048 1024 512 256 32 8 2 1 0 "
},
{
"input": "300919980 24",
"output": "20\n268435408 16777215 8388608 4194304 2097152 524288 262144 131072 65536 32768 8192 2048 1024 128 64 16 8 4 1 0 "
},
{
"input": "900077555 2",
"output": "16\n701408733 165580141 24157817 5702887 2178309 832040 196418 17711 2584 610 233 55 13 3 1 0 "
},
{
"input": "172285923 26",
"output": "17\n134217725 33554432 4194304 262144 32768 16384 4096 2048 1024 512 256 128 64 32 4 2 0 "
}
] | 1,580,751,142 | 6,742 | PyPy 3 | OK | TESTS | 70 | 498 | 14,131,200 | from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
# M = mod = 998244353
# def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
l = []
s,k = li()
i = 0
sum = [0 for i in range(10**6)]
a = sum[:]
a[1]=1;
sum[1]=1;
for i in range(2,10**13):
if(i<=k):
a[i]=sum[i-1];
else:a[i]=sum[i-1]-sum[i-k-1];
if(a[i]>s):
n=i;break;
sum[i]=sum[i-1]+a[i];
l = sorted(list(set(l)))
ans = [0]
i = n-1
while s and i:
if s - a[i] >= 0:
ans.append(a[i])
s -= a[i]
i-=1
print(len(ans))
print(*ans) | Title: Well-known Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Numbers *k*-bonacci (*k* is integer, *k*<=><=1) are a generalization of Fibonacci numbers and are determined as follows:
- *F*(*k*,<=*n*)<==<=0, for integer *n*, 1<=≤<=*n*<=<<=*k*; - *F*(*k*,<=*k*)<==<=1; - *F*(*k*,<=*n*)<==<=*F*(*k*,<=*n*<=-<=1)<=+<=*F*(*k*,<=*n*<=-<=2)<=+<=...<=+<=*F*(*k*,<=*n*<=-<=*k*), for integer *n*, *n*<=><=*k*.
Note that we determine the *k*-bonacci numbers, *F*(*k*,<=*n*), only for integer values of *n* and *k*.
You've got a number *s*, represent it as a sum of several (at least two) distinct *k*-bonacci numbers.
Input Specification:
The first line contains two integers *s* and *k* (1<=≤<=*s*,<=*k*<=≤<=109; *k*<=><=1).
Output Specification:
In the first line print an integer *m* (*m*<=≥<=2) that shows how many numbers are in the found representation. In the second line print *m* distinct integers *a*1,<=*a*2,<=...,<=*a**m*. Each printed integer should be a *k*-bonacci number. The sum of printed integers must equal *s*.
It is guaranteed that the answer exists. If there are several possible answers, print any of them.
Demo Input:
['5 2\n', '21 5\n']
Demo Output:
['3\n0 2 3\n', '3\n4 1 16\n']
Note:
none | ```python
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
# M = mod = 998244353
# def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
l = []
s,k = li()
i = 0
sum = [0 for i in range(10**6)]
a = sum[:]
a[1]=1;
sum[1]=1;
for i in range(2,10**13):
if(i<=k):
a[i]=sum[i-1];
else:a[i]=sum[i-1]-sum[i-k-1];
if(a[i]>s):
n=i;break;
sum[i]=sum[i-1]+a[i];
l = sorted(list(set(l)))
ans = [0]
i = n-1
while s and i:
if s - a[i] >= 0:
ans.append(a[i])
s -= a[i]
i-=1
print(len(ans))
print(*ans)
``` | 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"
}
] | 1,602,956,622 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | /******** All Required Header Files ********/
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <string>
/******** For Loops ********/
#define rep(i,n) for(int i=0;i<(n);i++)
/******** Using ********/
using namespace std;
/******** Type Defs ********/
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<string> vs;
/******** Init ********/
void c_p_c()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
/***********************************************************/
/************************ Start ************************/
/***********************************************************/
int main()
{
c_p_c();
string word;
cin >> word;
string target = "hello";
int pos = 0;
for (char &c : word) {
if (c == target[pos]) {
pos++;
}
if (pos == target.size()) {
cout << "YES" << endl;
return 0;
}
}
cout << "NO" << endl;
return 0;
}
| 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
/******** All Required Header Files ********/
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <string>
/******** For Loops ********/
#define rep(i,n) for(int i=0;i<(n);i++)
/******** Using ********/
using namespace std;
/******** Type Defs ********/
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<string> vs;
/******** Init ********/
void c_p_c()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
/***********************************************************/
/************************ Start ************************/
/***********************************************************/
int main()
{
c_p_c();
string word;
cin >> word;
string target = "hello";
int pos = 0;
for (char &c : word) {
if (c == target[pos]) {
pos++;
}
if (pos == target.size()) {
cout << "YES" << endl;
return 0;
}
}
cout << "NO" << endl;
return 0;
}
``` | -1 |
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"
}
] | 1,537,293,229 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 19 | 124 | 0 | from math import log2
str = input('')
lst = [int(s) for s in str.split() if s.isdigit()]
n = lst[0]
print(2 ** (int(log2(n)) + 1) - 1)
| 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
from math import log2
str = input('')
lst = [int(s) for s in str.split() if s.isdigit()]
n = lst[0]
print(2 ** (int(log2(n)) + 1) - 1)
``` | 0 | |
466 | C | Number of Ways | PROGRAMMING | 1,700 | [
"binary search",
"brute force",
"data structures",
"dp",
"two pointers"
] | null | null | You've got array *a*[1],<=*a*[2],<=...,<=*a*[*n*], consisting of *n* integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices *i*,<=*j* (2<=≤<=*i*<=≤<=*j*<=≤<=*n*<=-<=1), that . | The first line contains integer *n* (1<=≤<=*n*<=≤<=5·105), showing how many numbers are in the array. The second line contains *n* integers *a*[1], *a*[2], ..., *a*[*n*] (|*a*[*i*]|<=≤<=<=109) — the elements of array *a*. | Print a single integer — the number of ways to split the array into three parts with the same sum. | [
"5\n1 2 3 0 3\n",
"4\n0 1 -1 0\n",
"2\n4 1\n"
] | [
"2\n",
"1\n",
"0\n"
] | none | 1,500 | [
{
"input": "5\n1 2 3 0 3",
"output": "2"
},
{
"input": "4\n0 1 -1 0",
"output": "1"
},
{
"input": "2\n4 1",
"output": "0"
},
{
"input": "9\n0 0 0 0 0 0 0 0 0",
"output": "28"
},
{
"input": "10\n2 5 -2 2 -3 -2 3 5 -5 -2",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "0"
},
{
"input": "100\n1 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 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": "2030"
},
{
"input": "6\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "10\n1 0 0 0 1 1 1 0 1 1",
"output": "2"
},
{
"input": "10\n-2 2 -2 0 -2 -1 1 -1 2 0",
"output": "0"
},
{
"input": "4\n0 2 -1 2",
"output": "0"
},
{
"input": "5\n3 3 -3 3 3",
"output": "3"
},
{
"input": "5\n1 1 1 1 1",
"output": "0"
},
{
"input": "8\n-1 -1 -1 -1 -1 -1 -1 -1",
"output": "0"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "4\n1 -1 0 0",
"output": "1"
},
{
"input": "3\n6 -3 6",
"output": "0"
}
] | 1,699,878,658 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 6 | 61 | 8,294,400 | n=int(input())
a=[int(i) for i in input().split()]
summ=0
x=0
y=0
sumn=sum(a)
z=int(sumn/3)
if(z!=0):
for i in range(n):
summ+=a[i]
if(summ==z):
x+=1
elif(summ==2*z):
y+=1
print(x*y)
else:
for i in range(n):
summ+=a[i]
if(summ==0):
x+=1
x-=1
print(int(x*(x-1)/2))
| Title: Number of Ways
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got array *a*[1],<=*a*[2],<=...,<=*a*[*n*], consisting of *n* integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices *i*,<=*j* (2<=≤<=*i*<=≤<=*j*<=≤<=*n*<=-<=1), that .
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5·105), showing how many numbers are in the array. The second line contains *n* integers *a*[1], *a*[2], ..., *a*[*n*] (|*a*[*i*]|<=≤<=<=109) — the elements of array *a*.
Output Specification:
Print a single integer — the number of ways to split the array into three parts with the same sum.
Demo Input:
['5\n1 2 3 0 3\n', '4\n0 1 -1 0\n', '2\n4 1\n']
Demo Output:
['2\n', '1\n', '0\n']
Note:
none | ```python
n=int(input())
a=[int(i) for i in input().split()]
summ=0
x=0
y=0
sumn=sum(a)
z=int(sumn/3)
if(z!=0):
for i in range(n):
summ+=a[i]
if(summ==z):
x+=1
elif(summ==2*z):
y+=1
print(x*y)
else:
for i in range(n):
summ+=a[i]
if(summ==0):
x+=1
x-=1
print(int(x*(x-1)/2))
``` | 0 | |
462 | B | Appleman and Card Game | PROGRAMMING | 1,300 | [
"greedy"
] | null | null | Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get? | The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman. | Print a single integer – the answer to the problem. | [
"15 10\nDZFDFZDFDDDDDDF\n",
"6 4\nYJSNPI\n"
] | [
"82\n",
"4\n"
] | In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | 1,000 | [
{
"input": "15 10\nDZFDFZDFDDDDDDF",
"output": "82"
},
{
"input": "6 4\nYJSNPI",
"output": "4"
},
{
"input": "5 3\nAOWBY",
"output": "3"
},
{
"input": "1 1\nV",
"output": "1"
},
{
"input": "2 1\nWT",
"output": "1"
},
{
"input": "2 2\nBL",
"output": "2"
},
{
"input": "5 1\nFACJT",
"output": "1"
},
{
"input": "5 5\nMJDIJ",
"output": "7"
},
{
"input": "15 5\nAZBIPTOFTJCJJIK",
"output": "13"
},
{
"input": "100 1\nEVEEVEEEGGECFEHEFVFVFHVHEEEEEFCVEEEEEEVFVEEVEEHEEVEFEVVEFEEEFEVECEHGHEEFGEEVCEECCECEFHEVEEEEEEGEEHVH",
"output": "1"
},
{
"input": "100 15\nKKTFFUTFCKUIKKKKFIFFKTUKUUKUKKIKKKTIFKTKUCFFKKKIIKKKKKKTFKFKKIRKKKFKUUKIKUUUFFKKKKTUZKITUIKKIKUKKTIK",
"output": "225"
},
{
"input": "100 50\nYYIYYAAAIEAAYAYAEAIIIAAEAAYEAEYYYIAEYAYAYYAAAIAYAEAAYAYYIYAAYYAAAAAAIYYYAAYAAEAAYAIEIYIYAYAYAYIIAAEY",
"output": "1972"
},
{
"input": "100 90\nFAFAOOAOOAFAOTFAFAFFATAAAOFAAOAFBAAAFBOAOFFFOAOAFAPFOFAOFAAFOAAAAFAAFOFAAOFPPAAOOAAOOFFOFFFOFAOTOFAF",
"output": "2828"
},
{
"input": "100 99\nBFFBBFBFBQFFFFFQBFFBFFBQFBFQFBBFQFFFBFFFBFQFQFBFFBBFYQFBFFFFFFFBQQFQBFBQBQFFFBQQFFFBQFYFBFBFFFBBBQQY",
"output": "3713"
},
{
"input": "100 100\nMQSBDAJABILIBCUEOWGWCEXMUTEYQKAIWGINXVQEOFDUBSVULROQHQRZZAALVQFEFRAAAYUIMGCAFQGIAEFBETRECGSFQJNXHHDN",
"output": "514"
},
{
"input": "100 50\nBMYIXQSJNHGFVFPJBIOBXIKSFNUFPVODCUBQYSIIQNVNXXCWXWRHKFEUPPIIDDGRDBJLZDCBMNJMYRMWFIHOSTDJJHXHPNRKWNFD",
"output": "328"
},
{
"input": "100 50\nENFNEMLJEMDMFMNNGNIMNINALGLLLAEMENEMNLMMIEIJNAINBJEJMFJLLIMINELGFLAIAMJMHMGNLIEFJIEEFEFGLLLDLMEAEIMM",
"output": "748"
}
] | 1,604,599,571 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 155 | 614,400 | def maxim():
m = 0
index = 0
for i in range(26):
if(not visited[i] and arr[i] > m):
m = arr[i]
index = i
visited[index] = True
return m
n, k = map(int, input().split())
s = input()
arr = [0]*26
for i in range(n):
arr[ord(s[i])-ord("A")] += 1
i = 0
visited = [False]*26
m = maxim()
count = 0
while i+m < k:
count += m*m
i+=m
m = maxim()
if(k-i > 0):
ex = k-i
count+=ex*ex
print(count) | Title: Appleman and Card Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
Output Specification:
Print a single integer – the answer to the problem.
Demo Input:
['15 10\nDZFDFZDFDDDDDDF\n', '6 4\nYJSNPI\n']
Demo Output:
['82\n', '4\n']
Note:
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | ```python
def maxim():
m = 0
index = 0
for i in range(26):
if(not visited[i] and arr[i] > m):
m = arr[i]
index = i
visited[index] = True
return m
n, k = map(int, input().split())
s = input()
arr = [0]*26
for i in range(n):
arr[ord(s[i])-ord("A")] += 1
i = 0
visited = [False]*26
m = maxim()
count = 0
while i+m < k:
count += m*m
i+=m
m = maxim()
if(k-i > 0):
ex = k-i
count+=ex*ex
print(count)
``` | 3 | |
959 | A | Mahmoud and Ehab and the even-odd game | PROGRAMMING | 800 | [
"games",
"math"
] | null | null | Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that:
- 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally? | The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game. | Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise. | [
"1\n",
"2\n"
] | [
"Ehab",
"Mahmoud"
] | In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins. | 500 | [
{
"input": "1",
"output": "Ehab"
},
{
"input": "2",
"output": "Mahmoud"
},
{
"input": "10000",
"output": "Mahmoud"
},
{
"input": "33333",
"output": "Ehab"
},
{
"input": "5",
"output": "Ehab"
},
{
"input": "1000000000",
"output": "Mahmoud"
},
{
"input": "999999999",
"output": "Ehab"
},
{
"input": "123123123",
"output": "Ehab"
},
{
"input": "22222221",
"output": "Ehab"
},
{
"input": "22222220",
"output": "Mahmoud"
},
{
"input": "3",
"output": "Ehab"
},
{
"input": "4",
"output": "Mahmoud"
},
{
"input": "6",
"output": "Mahmoud"
},
{
"input": "7",
"output": "Ehab"
},
{
"input": "8",
"output": "Mahmoud"
},
{
"input": "9",
"output": "Ehab"
},
{
"input": "10",
"output": "Mahmoud"
},
{
"input": "536870912",
"output": "Mahmoud"
},
{
"input": "536870913",
"output": "Ehab"
},
{
"input": "536870911",
"output": "Ehab"
}
] | 1,652,559,104 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 4,300,800 | n=input()
if n%2==0:
print('Mahmoud')
else:
print('Ehab')
# Sat May 14 2022 20:08:03 GMT+0000 (Coordinated Universal Time)
# Sat May 14 2022 20:08:09 GMT+0000 (Coordinated Universal Time)
# Sat May 14 2022 20:11:24 GMT+0000 (Coordinated Universal Time)
| Title: Mahmoud and Ehab and the even-odd game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer *n* and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer *a* and subtract it from *n* such that:
- 1<=≤<=*a*<=≤<=*n*. - If it's Mahmoud's turn, *a* has to be even, but if it's Ehab's turn, *a* has to be odd.
If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?
Input Specification:
The only line contains an integer *n* (1<=≤<=*n*<=≤<=109), the number at the beginning of the game.
Output Specification:
Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.
Demo Input:
['1\n', '2\n']
Demo Output:
['Ehab', 'Mahmoud']
Note:
In the first sample, Mahmoud can't choose any integer *a* initially because there is no positive even integer less than or equal to 1 so Ehab wins.
In the second sample, Mahmoud has to choose *a* = 2 and subtract it from *n*. It's Ehab's turn and *n* = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins. | ```python
n=input()
if n%2==0:
print('Mahmoud')
else:
print('Ehab')
# Sat May 14 2022 20:08:03 GMT+0000 (Coordinated Universal Time)
# Sat May 14 2022 20:08:09 GMT+0000 (Coordinated Universal Time)
# Sat May 14 2022 20:11:24 GMT+0000 (Coordinated Universal Time)
``` | -1 | |
583 | A | Asphalting Roads | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted. | The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct. | In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1. | [
"2\n1 1\n1 2\n2 1\n2 2\n",
"1\n1 1\n"
] | [
"1 4 \n",
"1 \n"
] | In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | 500 | [
{
"input": "2\n1 1\n1 2\n2 1\n2 2",
"output": "1 4 "
},
{
"input": "1\n1 1",
"output": "1 "
},
{
"input": "2\n1 1\n2 2\n1 2\n2 1",
"output": "1 2 "
},
{
"input": "2\n1 2\n2 2\n2 1\n1 1",
"output": "1 3 "
},
{
"input": "3\n2 2\n1 2\n3 2\n3 3\n1 1\n2 3\n1 3\n3 1\n2 1",
"output": "1 4 5 "
},
{
"input": "3\n1 3\n3 1\n2 1\n1 1\n1 2\n2 2\n3 2\n3 3\n2 3",
"output": "1 2 6 "
},
{
"input": "4\n1 3\n2 3\n2 4\n4 4\n3 1\n1 1\n3 4\n2 1\n1 4\n4 3\n4 1\n3 2\n1 2\n4 2\n2 2\n3 3",
"output": "1 3 5 14 "
},
{
"input": "4\n3 3\n4 2\n2 3\n3 4\n4 4\n1 2\n3 2\n2 2\n1 4\n3 1\n4 1\n2 1\n1 3\n1 1\n4 3\n2 4",
"output": "1 2 9 12 "
},
{
"input": "9\n4 5\n2 3\n8 3\n5 6\n9 3\n4 4\n5 4\n4 7\n1 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n9 9\n8 7\n7 5\n3 7\n6 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n1 1\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 2\n3 9\n2 9\n7 7\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n3 3\n3 8\n5 5\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n8 8\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8",
"output": "1 2 4 9 10 14 16 32 56 "
},
{
"input": "8\n1 1\n1 2\n1 3\n1 4\n1 5\n8 6\n1 7\n1 8\n2 1\n8 5\n2 3\n2 4\n2 5\n2 6\n4 3\n2 2\n3 1\n3 2\n3 3\n3 4\n3 5\n3 6\n5 6\n3 8\n4 1\n4 2\n2 7\n4 4\n8 8\n4 6\n4 7\n4 8\n5 1\n5 2\n5 3\n6 5\n5 5\n3 7\n5 7\n5 8\n6 1\n6 2\n6 3\n6 4\n5 4\n6 6\n6 7\n6 8\n7 1\n7 2\n7 3\n7 4\n7 5\n7 6\n7 7\n7 8\n8 1\n8 2\n8 3\n8 4\n2 8\n1 6\n8 7\n4 5",
"output": "1 6 11 18 28 36 39 56 "
},
{
"input": "9\n9 9\n5 5\n8 8\n3 3\n2 2\n6 6\n4 4\n1 1\n7 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n1 7\n8 7\n7 5\n3 7\n5 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n4 5\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 3\n3 9\n2 9\n5 4\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n8 3\n3 8\n9 3\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n4 7\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8",
"output": "1 2 3 4 5 6 7 8 9 "
}
] | 1,605,500,696 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 108 | 0 | n = int(input())
arr = []
for i in range(1,n**2+1):
h,v = map(int,input().split())
if h==v:
if h not in arr:
arr.append(h)
print(i,end=' ')
else:
if h not in arr and v not in arr:
arr.extend([h,v])
print(i,end= ' ')
| Title: Asphalting Roads
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.
Output Specification:
In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.
Demo Input:
['2\n1 1\n1 2\n2 1\n2 2\n', '1\n1 1\n']
Demo Output:
['1 4 \n', '1 \n']
Note:
In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | ```python
n = int(input())
arr = []
for i in range(1,n**2+1):
h,v = map(int,input().split())
if h==v:
if h not in arr:
arr.append(h)
print(i,end=' ')
else:
if h not in arr and v not in arr:
arr.extend([h,v])
print(i,end= ' ')
``` | 0 | |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,692,416,768 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 46 | 0 | for i in range(int(input())):
word = str(input())
length = len(word)
if (length > 10):
abbreviations = str(length-2)
print(word[0]+abbreviations+word[length-1])
else :
print(word) | Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
for i in range(int(input())):
word = str(input())
length = len(word)
if (length > 10):
abbreviations = str(length-2)
print(word[0]+abbreviations+word[length-1])
else :
print(word)
``` | 3.977 |
599 | A | Patrick and Shopping | PROGRAMMING | 800 | [
"implementation"
] | null | null | Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled. | The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops. | Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house. | [
"10 20 30\n",
"1 1 5\n"
] | [
"60\n",
"4\n"
] | The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. | 500 | [
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
{
"input": "789 101112 131415",
"output": "203802"
},
{
"input": "27485716 99999999 35182",
"output": "55041796"
},
{
"input": "1 293548 5",
"output": "12"
},
{
"input": "12059 259855 5874875",
"output": "543828"
},
{
"input": "46981 105809 585858",
"output": "305580"
},
{
"input": "9889 1221 2442",
"output": "7326"
},
{
"input": "100500 200600 300700",
"output": "601800"
},
{
"input": "318476 318476 318476",
"output": "955428"
},
{
"input": "23985 3353 75633",
"output": "54676"
},
{
"input": "120 1298 2222",
"output": "2836"
},
{
"input": "98437 23487 666672",
"output": "243848"
},
{
"input": "100000000 100000000 100000000",
"output": "300000000"
},
{
"input": "2 5 2",
"output": "8"
},
{
"input": "1 1000 1",
"output": "4"
},
{
"input": "1 100000000 1",
"output": "4"
}
] | 1,658,994,281 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 21 | 62 | 0 | a, b, c = map(int, input(). split())
q = a + b + c
w = (a * 2) + (b * 2)
e = (a * 2) + (c * 2)
r = (b * 2) + (c * 2)
print(min(q, w, e, r))
| Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. | ```python
a, b, c = map(int, input(). split())
q = a + b + c
w = (a * 2) + (b * 2)
e = (a * 2) + (c * 2)
r = (b * 2) + (c * 2)
print(min(q, w, e, r))
``` | 3 | |
508 | A | Pasha and Pixels | PROGRAMMING | 1,100 | [
"brute force"
] | null | null | Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.
Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed.
Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move.
Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed. | The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.
The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move. | If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed.
If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0. | [
"2 2 4\n1 1\n1 2\n2 1\n2 2\n",
"2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n",
"5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n"
] | [
"4\n",
"5\n",
"0\n"
] | none | 500 | [
{
"input": "2 2 4\n1 1\n1 2\n2 1\n2 2",
"output": "4"
},
{
"input": "2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1",
"output": "5"
},
{
"input": "5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2",
"output": "0"
},
{
"input": "3 3 11\n2 1\n3 1\n1 1\n1 3\n1 2\n2 3\n3 3\n3 2\n2 2\n1 3\n3 3",
"output": "9"
},
{
"input": "2 2 5\n1 1\n2 1\n2 1\n1 2\n2 2",
"output": "5"
},
{
"input": "518 518 10\n37 97\n47 278\n17 467\n158 66\n483 351\n83 123\n285 219\n513 187\n380 75\n304 352",
"output": "0"
},
{
"input": "1 1 5\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "0"
},
{
"input": "1 5 5\n1 1\n1 2\n1 3\n1 4\n1 5",
"output": "0"
},
{
"input": "5 1 5\n1 1\n2 1\n3 1\n4 1\n5 1",
"output": "0"
},
{
"input": "1 1 1\n1 1",
"output": "0"
},
{
"input": "10 10 4\n5 9\n6 9\n6 10\n5 10",
"output": "4"
},
{
"input": "1000 1000 4\n999 999\n999 1000\n1000 999\n1000 1000",
"output": "4"
},
{
"input": "2 3 5\n2 3\n1 3\n1 2\n1 1\n2 2",
"output": "5"
},
{
"input": "1000 1000 4\n1000 1000\n999 999\n1000 999\n999 1000",
"output": "4"
}
] | 1,663,827,778 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | n,m,k=map(int,input().split())
a=[[0]*(m+1) for i in range(n+1)]
for x in range(k):
i,j=map(int,input().split())
a[i][j]=1
if a[i][j+1]==1 and a[i+1][j]==1 and a[i+1][j+1]==1:
print(x+1)
break
elif a[i][j-1]==1 and a[i-1][j]==1 and a[i-1][j-1]==1:
print(x+1)
break
elif a[i][j-1]==1 and a[i+1][j]==1 and a[i+1][j-1]==1:
print(x+1)
break
elif a[i][j+1]==1 and a[i-1][j]==1 and a[i-1][j+1]==1:
print(x+1)
break
else:
print(0)
| Title: Pasha and Pixels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.
Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed.
Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move.
Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed.
Input Specification:
The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.
The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move.
Output Specification:
If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed.
If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0.
Demo Input:
['2 2 4\n1 1\n1 2\n2 1\n2 2\n', '2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n', '5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n']
Demo Output:
['4\n', '5\n', '0\n']
Note:
none | ```python
n,m,k=map(int,input().split())
a=[[0]*(m+1) for i in range(n+1)]
for x in range(k):
i,j=map(int,input().split())
a[i][j]=1
if a[i][j+1]==1 and a[i+1][j]==1 and a[i+1][j+1]==1:
print(x+1)
break
elif a[i][j-1]==1 and a[i-1][j]==1 and a[i-1][j-1]==1:
print(x+1)
break
elif a[i][j-1]==1 and a[i+1][j]==1 and a[i+1][j-1]==1:
print(x+1)
break
elif a[i][j+1]==1 and a[i-1][j]==1 and a[i-1][j+1]==1:
print(x+1)
break
else:
print(0)
``` | -1 | |
515 | C | Drazil and Factorial | PROGRAMMING | 1,400 | [
"greedy",
"math",
"sortings"
] | null | null | Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes. | Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. | [
"4\n1234\n",
"3\n555\n"
] | [
"33222\n",
"555\n"
] | In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 1,000 | [
{
"input": "4\n1234",
"output": "33222"
},
{
"input": "3\n555",
"output": "555"
},
{
"input": "15\n012345781234578",
"output": "7777553333222222222222"
},
{
"input": "1\n8",
"output": "7222"
},
{
"input": "10\n1413472614",
"output": "75333332222222"
},
{
"input": "8\n68931246",
"output": "77553333332222222"
},
{
"input": "7\n4424368",
"output": "75333332222222222"
},
{
"input": "6\n576825",
"output": "7755532222"
},
{
"input": "5\n97715",
"output": "7775332"
},
{
"input": "3\n915",
"output": "75332"
},
{
"input": "2\n26",
"output": "532"
},
{
"input": "1\n4",
"output": "322"
},
{
"input": "15\n028745260720699",
"output": "7777755533333332222222222"
},
{
"input": "13\n5761790121605",
"output": "7775555333322"
},
{
"input": "10\n3312667105",
"output": "755533332"
},
{
"input": "1\n7",
"output": "7"
},
{
"input": "15\n989898989898989",
"output": "777777777777777333333333333333322222222222222222222222222222"
},
{
"input": "15\n000000000000007",
"output": "7"
},
{
"input": "15\n999999999999990",
"output": "77777777777777333333333333333333333333333322222222222222"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "1\n3",
"output": "3"
},
{
"input": "1\n4",
"output": "322"
},
{
"input": "1\n5",
"output": "5"
},
{
"input": "1\n6",
"output": "53"
},
{
"input": "1\n7",
"output": "7"
},
{
"input": "1\n9",
"output": "7332"
},
{
"input": "2\n09",
"output": "7332"
},
{
"input": "13\n1337251172966",
"output": "777555333333222"
},
{
"input": "15\n987654329876543",
"output": "777777555533333333332222222222222"
},
{
"input": "9\n234567899",
"output": "777755333333322222222"
},
{
"input": "2\n99",
"output": "77333322"
},
{
"input": "2\n66",
"output": "5533"
},
{
"input": "3\n999",
"output": "777333333222"
},
{
"input": "5\n99999",
"output": "77777333333333322222"
},
{
"input": "9\n123456789",
"output": "77755333332222222"
},
{
"input": "9\n987654321",
"output": "77755333332222222"
},
{
"input": "3\n666",
"output": "555333"
},
{
"input": "2\n26",
"output": "532"
},
{
"input": "6\n555777",
"output": "777555"
},
{
"input": "10\n1234567899",
"output": "777755333333322222222"
},
{
"input": "4\n6666",
"output": "55553333"
},
{
"input": "4\n9754",
"output": "775333222"
},
{
"input": "2\n95",
"output": "75332"
},
{
"input": "14\n11122233344455",
"output": "55333333222222222"
},
{
"input": "12\n836544897832",
"output": "77777553333333222222222222222"
}
] | 1,656,827,635 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | n = int(input())
s=input()
def prime_check(x):
for i in range(2,x//2+1):
if x%i==0:
return False
return True
def fact(x):
l=[]
i = 2
while True:
if not prime_check(x):
if x%i==0:
x//=i
l.append(i)
else:
i+=1
else:
l.append(x)
break
return l
l=[]
n_p=[]
for i in s:
i=int(i)
if prime_check(i):
l.append(i)
else:
while i>1:
n_p.extend(fact(i))
i-=1
n_p.sort(reverse=True)
for i in n_p:
if i==3:
n_p.remove(2)
l.extend(n_p)
l.sort(reverse=True)
ans = ''
for i in l:
ans+=str(i)
print(int(ans))
| Title: Drazil and Factorial
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes.
Output Specification:
Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.
Demo Input:
['4\n1234\n', '3\n555\n']
Demo Output:
['33222\n', '555\n']
Note:
In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
n = int(input())
s=input()
def prime_check(x):
for i in range(2,x//2+1):
if x%i==0:
return False
return True
def fact(x):
l=[]
i = 2
while True:
if not prime_check(x):
if x%i==0:
x//=i
l.append(i)
else:
i+=1
else:
l.append(x)
break
return l
l=[]
n_p=[]
for i in s:
i=int(i)
if prime_check(i):
l.append(i)
else:
while i>1:
n_p.extend(fact(i))
i-=1
n_p.sort(reverse=True)
for i in n_p:
if i==3:
n_p.remove(2)
l.extend(n_p)
l.sort(reverse=True)
ans = ''
for i in l:
ans+=str(i)
print(int(ans))
``` | 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"
}
] | 1,605,897,076 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 5 | 140 | 0 | def add_vertex(graph,v,incidence):
if v not in graph:
incidence[v]=0
graph[v] = []
def add_edge(graph,v1, v2):
graph[v1].append(v2)
su=0
m=0
def dfs(visited, graph, node):
global su,m
if node not in visited:
visited.add(node)
su+=1
m=max(su,m)
for neighbour in graph[node]:
dfs(visited, graph, neighbour)
su-=1
graph={}
incidence={}
n=int(input())
for j in range(1,n+1):
s=list(map(str,input().split(" reposted ")))
add_vertex(graph,s[0].lower(),incidence)
add_vertex(graph,s[1].lower(),incidence)
add_edge(graph,s[0].lower(),s[1].lower())
try:
incidence[s[1].lower()]+=1
except(Exception):
incidence[s[1].lower()]=1
visited=set()
ma=0
for key,val in incidence.items():
if(val==0):
m,su=0,0
dfs(visited,graph,key)
ma=max(ma,m)
print(ma) | 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
def add_vertex(graph,v,incidence):
if v not in graph:
incidence[v]=0
graph[v] = []
def add_edge(graph,v1, v2):
graph[v1].append(v2)
su=0
m=0
def dfs(visited, graph, node):
global su,m
if node not in visited:
visited.add(node)
su+=1
m=max(su,m)
for neighbour in graph[node]:
dfs(visited, graph, neighbour)
su-=1
graph={}
incidence={}
n=int(input())
for j in range(1,n+1):
s=list(map(str,input().split(" reposted ")))
add_vertex(graph,s[0].lower(),incidence)
add_vertex(graph,s[1].lower(),incidence)
add_edge(graph,s[0].lower(),s[1].lower())
try:
incidence[s[1].lower()]+=1
except(Exception):
incidence[s[1].lower()]=1
visited=set()
ma=0
for key,val in incidence.items():
if(val==0):
m,su=0,0
dfs(visited,graph,key)
ma=max(ma,m)
print(ma)
``` | 0 | |
708 | A | Letters Cyclic Shift | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy",
"implementation",
"strings"
] | null | null | You are given a non-empty string *s* consisting of lowercase English letters. You have to pick exactly one non-empty substring of *s* and shift all its letters 'z' 'y' 'x' 'b' 'a' 'z'. In other words, each character is replaced with the previous character of English alphabet and 'a' is replaced with 'z'.
What is the lexicographically minimum string that can be obtained from *s* by performing this shift exactly once? | The only line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) consisting of lowercase English letters. | Print the lexicographically minimum string that can be obtained from *s* by shifting letters of exactly one non-empty substring. | [
"codeforces\n",
"abacaba\n"
] | [
"bncdenqbdr\n",
"aaacaba\n"
] | String *s* is lexicographically smaller than some other string *t* of the same length if there exists some 1 ≤ *i* ≤ |*s*|, such that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ..., *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, and *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>. | 500 | [
{
"input": "codeforces",
"output": "bncdenqbdr"
},
{
"input": "abacaba",
"output": "aaacaba"
},
{
"input": "babbbabaababbaa",
"output": "aabbbabaababbaa"
},
{
"input": "bcbacaabcababaccccaaaabacbbcbbaa",
"output": "abaacaabcababaccccaaaabacbbcbbaa"
},
{
"input": "cabaccaacccabaacdbdcbcdbccbccbabbdadbdcdcdbdbcdcdbdadcbcda",
"output": "babaccaacccabaacdbdcbcdbccbccbabbdadbdcdcdbdbcdcdbdadcbcda"
},
{
"input": "a",
"output": "z"
},
{
"input": "eeeedddccbceaabdaecaebaeaecccbdeeeaadcecdbeacecdcdcceabaadbcbbadcdaeddbcccaaeebccecaeeeaebcaaccbdaccbdcadadaaeacbbdcbaeeaecedeeeedadec",
"output": "ddddcccbbabdaabdaecaebaeaecccbdeeeaadcecdbeacecdcdcceabaadbcbbadcdaeddbcccaaeebccecaeeeaebcaaccbdaccbdcadadaaeacbbdcbaeeaecedeeeedadec"
},
{
"input": "fddfbabadaadaddfbfecadfaefaefefabcccdbbeeabcbbddefbafdcafdfcbdffeeaffcaebbbedabddeaecdddffcbeaafffcddccccfffdbcddcfccefafdbeaacbdeeebdeaaacdfdecadfeafaeaefbfdfffeeaefebdceebcebbfeaccfafdccdcecedeedadcadbfefccfdedfaaefabbaeebdebeecaadbebcfeafbfeeefcfaecadfe",
"output": "ecceaabadaadaddfbfecadfaefaefefabcccdbbeeabcbbddefbafdcafdfcbdffeeaffcaebbbedabddeaecdddffcbeaafffcddccccfffdbcddcfccefafdbeaacbdeeebdeaaacdfdecadfeafaeaefbfdfffeeaefebdceebcebbfeaccfafdccdcecedeedadcadbfefccfdedfaaefabbaeebdebeecaadbebcfeafbfeeefcfaecadfe"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaaaz"
},
{
"input": "abbabaaaaa",
"output": "aaaabaaaaa"
},
{
"input": "bbbbbbbbbbbb",
"output": "aaaaaaaaaaaa"
},
{
"input": "aabaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaz"
},
{
"input": "abaabaaaaaabbaaaaaaabaaaaaaaaabaaaabaaaaaaabaaaaaaaaaabaaaaaaaaaaaaaaabaaaabbaaaaabaaaaaaaabaaaaaaaa",
"output": "aaaabaaaaaabbaaaaaaabaaaaaaaaabaaaabaaaaaaabaaaaaaaaaabaaaaaaaaaaaaaaabaaaabbaaaaabaaaaaaaabaaaaaaaa"
},
{
"input": "abbbbbbbabbbbbbbbbbbbbbbbbbbbbbbabbabbbbbabbbbbbbbbbbabbbbbbbbabbabbbbbbbbbbbbbbabbabbbaababbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbabbbbbbbbbbbbbbbbabbbbabbbbbbbbbbbbbbbabbbbbbbbbaababbbbbbbbabbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbabbabbbbbbbbbbbbbbbbabbbabbbbbaabbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbbaabbbbbbbbbbbbababbabbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbabbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbbbabbbbbbbabbbbbbb",
"output": "aaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbabbabbbbbabbbbbbbbbbbabbbbbbbbabbabbbbbbbbbbbbbbabbabbbaababbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbabbbbbbbbbbbbbbbbabbbbabbbbbbbbbbbbbbbabbbbbbbbbaababbbbbbbbabbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbabbabbbbbbbbbbbbbbbbabbbabbbbbaabbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbbaabbbbbbbbbbbbababbabbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbabbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbbbabbbbbbbabbbbbbb"
},
{
"input": "aaaaa",
"output": "aaaaz"
},
{
"input": "aaa",
"output": "aaz"
},
{
"input": "aa",
"output": "az"
}
] | 1,655,259,905 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 46 | 204,800 | x=input()
a1=x.find('a')
if a1==-1:
for t in x:
print(chr(ord(t)-1),end='')
else:
if a1!=0:
for i in range(a1):
print(chr(ord(x[i])-1),end='')
print(x[a1:]);
else:
if len(x)==1:
print('z')
else:
a2=-1
for i in range(a1+1,len(x)):
if x[i]!='a':
a2=i
break
if a2==-1:
print(x[:-1],end='')
print('z')
else:
a3=x.find('a',a2+1)
print('a',end='')
for i in range(a2,a3):
print(chr(ord(x[i])-1),end='')
print(x[a3:])
| Title: Letters Cyclic Shift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a non-empty string *s* consisting of lowercase English letters. You have to pick exactly one non-empty substring of *s* and shift all its letters 'z' 'y' 'x' 'b' 'a' 'z'. In other words, each character is replaced with the previous character of English alphabet and 'a' is replaced with 'z'.
What is the lexicographically minimum string that can be obtained from *s* by performing this shift exactly once?
Input Specification:
The only line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) consisting of lowercase English letters.
Output Specification:
Print the lexicographically minimum string that can be obtained from *s* by shifting letters of exactly one non-empty substring.
Demo Input:
['codeforces\n', 'abacaba\n']
Demo Output:
['bncdenqbdr\n', 'aaacaba\n']
Note:
String *s* is lexicographically smaller than some other string *t* of the same length if there exists some 1 ≤ *i* ≤ |*s*|, such that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ..., *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, and *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>. | ```python
x=input()
a1=x.find('a')
if a1==-1:
for t in x:
print(chr(ord(t)-1),end='')
else:
if a1!=0:
for i in range(a1):
print(chr(ord(x[i])-1),end='')
print(x[a1:]);
else:
if len(x)==1:
print('z')
else:
a2=-1
for i in range(a1+1,len(x)):
if x[i]!='a':
a2=i
break
if a2==-1:
print(x[:-1],end='')
print('z')
else:
a3=x.find('a',a2+1)
print('a',end='')
for i in range(a2,a3):
print(chr(ord(x[i])-1),end='')
print(x[a3:])
``` | 0 | |
479 | B | Towers | PROGRAMMING | 1,400 | [
"brute force",
"constructive algorithms",
"greedy",
"implementation",
"sortings"
] | null | null | As you know, all the kids in Berland love playing with cubes. Little Petya has *n* towers consisting of cubes of the same size. Tower with number *i* consists of *a**i* cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than *k* such operations. Petya does not want to be late for class, so you have to help him accomplish this task. | The first line contains two space-separated positive integers *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains *n* space-separated positive integers *a**i* (1<=≤<=*a**i*<=≤<=104) — the towers' initial heights. | In the first line print two space-separated non-negative integers *s* and *m* (*m*<=≤<=*k*). The first number is the value of the minimum possible instability that can be obtained after performing at most *k* operations, the second number is the number of operations needed for that.
In the next *m* lines print the description of each operation as two positive integers *i* and *j*, each of them lies within limits from 1 to *n*. They represent that Petya took the top cube from the *i*-th tower and put in on the *j*-th one (*i*<=≠<=*j*). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them. | [
"3 2\n5 8 5\n",
"3 4\n2 2 4\n",
"5 3\n8 3 2 6 3\n"
] | [
"0 2\n2 1\n2 3\n",
"1 1\n3 2\n",
"3 3\n1 3\n1 2\n1 3\n"
] | In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6. | 1,000 | [
{
"input": "3 2\n5 8 5",
"output": "0 2\n2 1\n2 3"
},
{
"input": "3 4\n2 2 4",
"output": "1 4\n3 1\n1 2\n2 1\n1 2"
},
{
"input": "5 3\n8 3 2 6 3",
"output": "3 3\n1 3\n1 2\n1 3"
},
{
"input": "4 6\n1 10 8 2",
"output": "2 6\n2 1\n2 1\n2 4\n3 1\n2 4\n3 1"
},
{
"input": "5 3\n2 2 2 2 2",
"output": "0 0"
},
{
"input": "95 720\n9119 7480 7245 2431 7986 4088 3498 7724 9291 3853 9733 4018 1109 6476 6915 6456 8956 2877 8090 7557 1886 5944 9952 6275 2877 6488 2688 319 4635 8618 2724 7047 3261 8750 971 9114 6552 7749 4356 6059 1495 5713 2625 7230 7330 3879 8965 7337 5975 3263 646 9124 692 610 2091 2681 5354 1323 1933 4356 3164 8150 2234 4627 5773 1092 445 598 4949 8916 1335 1135 9004 395 665 9859 7949 4813 7543 6023 836 2465 4598 4516 5072 608 733 9299 5073 5333 7847 7921 3482 5422 8527",
"output": "8993 720\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 28\n23 2..."
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"input": "1 1\n1",
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{
"input": "1 1\n10000",
"output": "0 0"
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{
"input": "1 1\n1238",
"output": "0 0"
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{
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"output": "0 0"
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{
"input": "1 1000\n2153",
"output": "0 0"
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{
"input": "2 1\n13 13",
"output": "0 0"
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{
"input": "2 2\n27 29",
"output": "0 1\n2 1"
},
{
"input": "2 1\n27 29",
"output": "0 1\n2 1"
},
{
"input": "2 1\n37 41",
"output": "2 1\n2 1"
},
{
"input": "2 2\n37 41",
"output": "0 2\n2 1\n2 1"
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{
"input": "5 1000\n10000 10000 10000 10000 10000",
"output": "0 0"
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{
"input": "6 1000\n1 2 3 4 5 6",
"output": "1 1000\n6 1\n5 1\n6 2\n4 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2..."
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{
"input": "5 17\n10000 1 10000 1 10000",
"output": "9986 17\n1 2\n3 4\n5 2\n1 4\n3 2\n5 4\n1 2\n3 4\n5 2\n1 4\n3 2\n5 4\n1 2\n3 4\n5 2\n1 4\n3 2"
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{
"input": "9 325\n100 200 300 200 100 200 300 200 100",
"output": "1 325\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n7 9\n3 1\n7 5\n3 9\n7 1\n3 5\n..."
},
{
"input": "8 4\n9 8 7 5 6 8 10 12",
"output": "2 4\n8 4\n8 4\n7 5\n8 3"
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{
"input": "7 23\n1 1 1 1 1 1 1",
"output": "0 0"
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{
"input": "11 20\n10 12 14 7 9 11 13 15 5 7 88",
"output": "57 20\n11 9\n11 9\n11 4\n11 9\n11 10\n11 4\n11 9\n11 10\n11 4\n11 5\n11 9\n11 10\n11 1\n11 4\n11 5\n11 9\n11 10\n11 1\n11 4\n11 5"
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{
"input": "12 159\n987 654 321 123 456 789 963 852 147 741 852 963",
"output": "704 159\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9\n1 4\n7 9\n12 4\n1 9\n7 4\n12 9..."
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{
"input": "8 1000\n1 9999 1 9999 1 9999 1 9999",
"output": "9498 1000\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n2 1\n4 3\n6 5\n8 7\n..."
},
{
"input": "10 14\n18 26 19 37 37 37 12 2 11 39",
"output": "21 14\n10 8\n10 8\n4 8\n5 8\n6 8\n10 8\n4 8\n5 8\n6 8\n10 8\n4 9\n5 7\n6 8\n10 9"
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{
"input": "7 859\n12 10 11 12 10 15 18",
"output": "1 859\n7 2\n7 5\n7 2\n6 3\n7 5\n6 1\n7 2\n1 3\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n..."
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{
"input": "8 789\n80 88 86 88 89 80 85 86",
"output": "1 789\n5 1\n2 6\n4 1\n5 6\n2 1\n4 6\n5 1\n2 6\n3 1\n4 6\n5 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n..."
},
{
"input": "7 987\n670 658 658 658 679 690 684",
"output": "0 40\n6 2\n6 3\n6 4\n6 2\n6 3\n6 4\n6 2\n7 3\n6 4\n7 2\n6 3\n7 4\n6 2\n7 3\n6 4\n7 2\n5 3\n6 4\n7 2\n5 3\n6 4\n7 2\n5 3\n6 4\n7 2\n5 3\n6 4\n7 2\n5 3\n6 4\n7 2\n5 3\n6 4\n7 2\n5 3\n6 4\n7 1\n5 2\n6 3\n7 4"
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{
"input": "7 750\n96 101 102 103 104 105 103",
"output": "0 7\n6 1\n5 1\n6 1\n4 1\n5 1\n6 1\n7 2"
},
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"input": "5 1000\n300 300 2 3 5",
"output": "0 356\n1 3\n2 3\n1 4\n2 3\n1 4\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n2 3\n1 4\n2 5\n1 3\n2 4\n1 5\n..."
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"input": "55 534\n3524 154 2688 7004 4266 6818 7391 423 7879 8045 1874 5427 6158 2003 7257 9003 8493 5829 316 9797 6297 8719 5457 7939 5156 5078 1300 2974 3417 9610 494 5979 4779 6098 5504 4344 9563 5540 2247 4508 545 4620 1191 4460 704 6225 2066 82 756 9589 1136 294 468 2003 8982",
"output": "9162 534\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 48\n20 4..."
},
{
"input": "71 596\n8473 9181 6643 3064 9601 3196 3989 2178 115 4941 3471 3587 264 7530 1790 1006 4382 270 8350 6230 8004 951 5153 5411 1083 7861 7752 1981 4358 2761 7719 2207 3194 6694 9093 3221 9870 1171 4329 5662 5947 3527 302 4394 7325 8571 2063 4316 8240 3755 7979 4712 244 7045 8033 2246 6921 279 7189 9501 7075 6228 2137 3880 3510 5280 8692 6724 7386 4223 1854",
"output": "9114 596\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 9\n37 ..."
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{
"input": "83 307\n2877 1856 4790 3989 2777 3222 7883 1230 8702 1293 9261 8644 9505 7248 2131 3553 3919 3222 2737 2663 8222 9535 658 7075 3361 99 6364 989 6243 2808 9137 1139 8360 4042 6330 5346 177 3155 8571 9919 4454 5138 3060 5272 4891 1861 9356 3413 2477 81 4277 74 9076 8438 7085 676 5073 5501 5560 9977 9502 2091 9929 6411 754 5006 5719 8435 4348 4444 1034 5047 1509 6488 2205 8998 8417 287 522 9192 4140 3700 1092",
"output": "9656 307\n60 52\n60 52\n60 52\n60 52\n60 52\n60 52\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 50\n60 52\n60 26\n60 50\n60 52\n60 26\n60 50\n60 52\n63 26\n60 50\n63 52\n60 26\n63 50\n60 52\n63 26\n60 50\n63 52\n60 26\n63 50\n60 52\n63 26\n60 50\n63 52\n60 26\n63 50\n60 52\n63 26\n40 50\n60 52\n63 26\n40 5..."
},
{
"input": "99 313\n4178 5075 2937 8561 5408 9600 1777 3929 938 8188 8698 997 3611 2775 5720 5556 9808 7663 771 7608 2633 8118 2514 1444 5640 2882 5520 7292 5024 9607 9610 7367 2582 4639 7215 4224 6836 7298 654 1072 3504 4045 1626 2501 1513 4207 9353 7646 7258 2759 1120 300 8852 9287 9784 9107 9032 6531 9740 6806 4634 658 6232 9485 703 1627 6939 146 9823 7369 6565 5679 6851 5816 3753 5377 7175 1077 8588 4616 8889 1184 1363 2996 2823 4397 2729 6297 1205 1966 5030 4811 2580 2426 8136 6066 8139 2073 45",
"output": "9459 313\n69 99\n69 99\n69 99\n69 99\n69 99\n69 99\n69 99\n69 99\n69 99\n69 99\n69 99\n69 99\n69 99\n69 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n69 99\n17 99\n55 99\n69 99\n17 99\n55 99\n69 99\n17 99\n55 99\n69 9..."
},
{
"input": "100 742\n4447 7727 1809 6921 15 6873 3861 1648 4239 4725 3590 1680 8691 8691 6084 3747 9116 8916 9092 8292 4031 7149 6668 4222 1828 9699 5971 6464 5268 2621 2181 6171 9418 9429 3677 3266 1155 9382 1513 2792 8333 4670 8682 7925 1003 4717 5600 4273 6718 5444 2769 8199 8109 1927 4530 9283 9182 1115 4636 6404 5162 5416 3592 9178 3244 4567 2595 3864 7970 8180 7114 6498 3434 3676 4798 4000 2342 5696 5489 3129 1320 8400 9511 6616 2374 2897 2870 1994 3446 7191 5141 1294 3362 1899 6592 147 4485 5477 4098 5580",
"output": "8888 742\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 5\n26 ..."
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{
"input": "100 805\n1556 4594 5764 9334 9542 6899 459 699 6475 8916 9379 9841 6988 4217 6425 7239 8653 3357 9830 4725 8442 9381 2172 8590 7755 1937 4584 5471 9857 2668 9951 5103 3640 25 8210 6335 5109 7717 6300 1241 9544 6282 1440 7859 4920 711 5597 8506 1499 1770 1771 8425 589 9128 3582 7713 437 6337 3007 4720 293 3983 7192 8060 4681 4837 9622 5575 4932 8401 6294 4426 4584 5708 3642 6731 1100 9590 3555 4905 5125 76 1270 2012 5088 4180 56 2088 2243 909 1669 3352 5905 6690 850 9124 2100 581 544 6797",
"output": "9356 805\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 34\n31 87\n31 3..."
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"input": "100 811\n6505 7269 262 3906 5822 7469 4353 6102 4519 5812 8816 4898 2038 9744 3662 5594 4542 6309 7864 6965 2852 7965 7677 254 6930 4720 1036 4478 4446 5819 1368 1331 1511 7373 1799 1564 5416 9701 7838 8202 4946 5189 6 2929 5190 353 8698 7604 8984 8640 8614 3787 2525 4169 2633 6144 8589 4263 9891 1549 5425 5654 3495 1135 8277 8755 842 1478 407 3486 5473 866 5733 4091 1543 3111 4051 380 3108 3033 9874 4856 5734 5248 5641 4519 4139 4885 1040 979 1301 2307 2640 9321 9300 4454 8227 8789 3343 8958",
"output": "9188 811\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 43\n81 43\n59 4..."
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{
"input": "100 521\n910 6295 5705 5775 5349 3847 4599 5153 3107 5812 4605 3058 7631 5270 356 1789 4079 750 91 6102 3071 196 9533 7726 9208 3310 9648 9293 3228 6811 2786 3912 7221 1073 6332 442 4235 3844 9920 9356 3997 4096 2765 2862 2756 2699 8695 1837 7413 2262 4913 8205 2301 2859 5332 7278 2548 9485 8263 6217 7853 7868 1287 18 5522 8480 7869 3189 7369 3707 1005 4602 1075 3418 3091 5842 9161 1170 1173 6297 4623 9636 6005 3348 8355 5803 1326 4034 9837 1594 3636 7069 9374 464 3558 6136 2194 3893 9245 8015",
"output": "9355 521\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 64\n39 6..."
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{
"input": "100 527\n9506 8970 9660 4539 4333 3873 8492 4204 5343 9060 394 8115 5385 4989 4889 144 6320 4246 4477 2535 7482 8780 7742 9390 5135 8253 5156 5596 4169 9962 9068 140 1444 1669 9921 8375 4542 3123 515 7805 6695 5707 5523 92 5186 8693 5988 935 4898 4940 1755 9376 2077 7900 4384 5709 3804 515 2442 3046 6632 6435 4886 2548 5470 8750 6385 7604 7035 6633 7480 8338 8577 7609 5183 8574 7920 7768 6534 4425 5725 7120 7765 6584 1069 6142 8513 2640 9578 5312 2324 5480 3405 9447 2008 7817 5617 4805 2044 2880",
"output": "9377 527\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n30 44\n35 44\n30 44\n35 44\n30 44\n35 44\n30 44\n35 32\n30 44\n35 32\n30 44\n35 32\n30 44\n35 32\n30 44\n35 16\n30 32\n35 44\n30 16\n35 32\n30 44\n35 16\n30 32\n35 44\n30 16\n35 32\n30 44\n35 16\n30 32\n35 44\n30 1..."
},
{
"input": "100 324\n372 362 365 379 370 377 371 377 367 366 363 378 365 368 378 359 369 380 377 374 363 366 372 365 365 378 371 363 369 367 361 366 375 366 380 379 367 374 366 365 373 372 363 378 365 377 379 377 362 380 366 376 380 378 361 362 367 359 366 359 370 361 361 374 377 368 366 359 365 359 380 360 375 375 367 372 372 375 377 377 363 367 379 380 373 364 374 377 360 380 380 379 365 374 375 368 376 365 369 360",
"output": "1 324\n18 16\n35 58\n50 60\n53 68\n71 70\n84 16\n90 58\n91 60\n4 68\n18 70\n35 72\n36 89\n47 100\n50 16\n53 31\n71 55\n83 58\n84 60\n90 62\n91 63\n92 68\n4 70\n12 72\n15 89\n18 100\n26 2\n35 16\n36 31\n44 49\n47 55\n50 56\n53 58\n54 60\n71 62\n83 63\n84 68\n90 70\n91 72\n92 89\n4 100\n6 2\n8 11\n12 16\n15 21\n18 28\n19 31\n26 43\n35 49\n36 55\n44 56\n46 58\n47 60\n48 62\n50 63\n53 68\n54 70\n65 72\n71 81\n79 89\n80 100\n83 2\n84 11\n88 16\n90 21\n91 28\n92 31\n4 43\n6 49\n8 55\n12 56\n15 58\n18 60\n19 62\n..."
},
{
"input": "64 35\n369 370 362 370 380 377 379 360 369 380 366 369 360 379 359 374 360 373 370 375 363 367 380 361 364 367 369 370 368 370 375 362 359 364 367 371 362 377 372 367 377 373 375 364 359 359 370 368 377 372 374 378 376 365 368 370 369 373 370 365 365 372 365 367",
"output": "12 35\n5 15\n10 33\n23 45\n5 46\n7 8\n10 13\n14 15\n23 17\n5 33\n7 45\n10 46\n14 8\n23 13\n52 15\n5 17\n6 24\n7 33\n10 45\n14 46\n23 3\n38 8\n41 13\n49 15\n52 17\n5 24\n6 32\n7 33\n10 37\n14 45\n23 46\n38 3\n41 8\n49 13\n52 15\n53 17"
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{
"input": "76 41\n359 359 366 370 376 375 368 361 369 373 367 363 364 377 363 378 373 371 370 370 360 367 369 359 365 373 379 379 375 361 372 377 369 368 373 362 375 379 365 378 370 370 377 364 373 379 365 362 360 366 375 364 372 376 375 379 362 377 369 370 359 379 365 359 372 376 373 371 364 361 377 366 373 380 378 369",
"output": "13 41\n74 1\n27 2\n28 24\n38 61\n46 64\n56 1\n62 2\n74 21\n16 24\n27 49\n28 61\n38 64\n40 1\n46 2\n56 8\n62 21\n74 24\n75 30\n14 49\n16 61\n27 64\n28 70\n32 1\n38 2\n40 8\n43 21\n46 24\n56 30\n58 36\n62 48\n71 49\n74 57\n75 61\n5 64\n14 70\n16 1\n27 2\n28 8\n32 12\n38 15\n40 21"
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{
"input": "92 751\n380 368 363 379 359 379 380 368 379 365 370 376 359 366 378 371 362 374 370 376 375 367 379 367 363 360 373 362 360 360 370 375 376 362 370 378 368 364 367 371 374 371 372 368 380 377 364 375 377 362 362 366 366 365 363 365 364 374 377 376 372 366 360 366 380 370 362 368 369 376 371 376 364 371 360 375 360 373 359 367 370 365 374 374 379 379 359 371 367 377 359 360",
"output": "1 751\n1 5\n7 13\n45 79\n65 87\n1 91\n4 5\n6 13\n7 26\n9 29\n23 30\n45 63\n65 75\n85 77\n86 79\n1 87\n4 91\n6 92\n7 5\n9 13\n15 26\n23 29\n36 30\n45 63\n65 75\n85 77\n86 79\n1 87\n4 91\n6 92\n7 5\n9 13\n15 17\n23 26\n36 28\n45 29\n46 30\n49 34\n59 50\n65 51\n85 63\n86 67\n90 75\n1 77\n4 79\n6 87\n7 91\n9 92\n12 3\n15 5\n20 13\n23 17\n33 25\n36 26\n45 28\n46 29\n49 30\n59 34\n60 50\n65 51\n70 55\n72 63\n85 67\n86 75\n90 77\n1 79\n4 87\n6 91\n7 92\n9 3\n12 5\n15 13\n20 17\n21 25\n23 26\n32 28\n33 29\n36 30\n..."
},
{
"input": "4 917\n371 379 368 369",
"output": "1 917\n2 3\n2 3\n2 4\n2 3\n2 4\n2 1\n2 3\n1 4\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n..."
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{
"input": "40 543\n817 818 818 817 817 818 818 818 818 817 818 818 818 818 818 818 818 818 817 818 817 817 818 817 818 817 818 817 818 818 817 818 818 818 817 818 818 818 818 818",
"output": "1 543\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n..."
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{
"input": "56 253\n818 817 817 818 817 818 817 817 818 818 817 817 818 817 817 817 817 817 818 817 818 818 818 817 817 818 818 818 817 817 818 817 817 818 818 817 818 818 817 817 817 817 818 817 818 818 818 817 817 818 817 817 818 818 818 817",
"output": "1 253\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n..."
},
{
"input": "16 611\n817 817 818 818 818 818 817 817 818 818 818 817 817 817 818 818",
"output": "1 611\n3 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n1 2\n2 1\n..."
},
{
"input": "87 161\n1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024",
"output": "0 0"
},
{
"input": "100 1000\n18 12 14 17 19 20 18 16 15 12 10 20 20 19 18 12 13 14 19 11 17 10 18 13 11 11 14 15 13 20 16 13 17 12 12 19 18 15 16 11 12 13 15 20 12 16 20 10 13 20 13 15 13 20 17 16 19 20 19 11 20 19 12 15 12 20 16 10 19 19 11 19 13 11 18 12 19 10 14 15 13 19 20 20 11 10 11 19 19 12 20 20 10 13 10 18 16 15 20 16",
"output": "1 1000\n6 11\n12 22\n13 48\n30 68\n44 78\n47 86\n50 93\n54 95\n58 11\n61 20\n66 22\n83 25\n84 26\n91 40\n92 48\n99 60\n5 68\n6 71\n12 74\n13 78\n14 85\n19 86\n30 87\n36 93\n44 95\n47 2\n50 10\n54 11\n57 16\n58 20\n59 22\n61 25\n62 26\n66 34\n69 35\n70 40\n72 41\n77 45\n82 48\n83 60\n84 63\n88 65\n89 68\n91 71\n92 74\n99 76\n1 78\n5 85\n6 86\n7 87\n12 90\n13 93\n14 95\n15 2\n19 10\n23 11\n30 16\n36 17\n37 20\n44 22\n47 24\n50 25\n54 26\n57 29\n58 32\n59 34\n61 35\n62 40\n66 41\n69 42\n70 45\n72 48\n75 49\n7..."
},
{
"input": "100 1000\n22 12 16 15 19 16 15 21 13 20 12 16 14 13 21 14 16 12 12 12 18 16 19 18 15 14 21 11 12 22 20 12 20 22 13 22 11 19 11 17 21 16 13 13 22 13 14 21 13 21 21 20 19 17 13 18 15 14 15 13 13 16 16 22 17 11 17 22 22 22 19 19 21 16 19 11 12 12 13 18 13 18 22 21 11 18 19 19 18 14 21 20 16 20 13 17 12 15 19 17",
"output": "1 1000\n1 28\n30 37\n34 39\n36 66\n45 76\n64 85\n68 2\n69 11\n70 18\n83 19\n1 20\n8 28\n15 29\n27 32\n30 37\n34 39\n36 66\n41 76\n45 77\n48 78\n50 85\n51 97\n64 2\n68 9\n69 11\n70 14\n73 18\n83 19\n84 20\n91 28\n1 29\n8 32\n10 35\n15 37\n27 39\n30 43\n31 44\n33 46\n34 49\n36 55\n41 60\n45 61\n48 66\n50 76\n51 77\n52 78\n64 79\n68 81\n69 85\n70 95\n73 97\n83 2\n84 9\n91 11\n92 13\n94 14\n1 16\n5 18\n8 19\n10 20\n15 26\n23 28\n27 29\n30 32\n31 35\n33 37\n34 39\n36 43\n38 44\n41 46\n45 47\n48 49\n50 55\n51 58..."
},
{
"input": "100 1000\n18 19 23 15 19 23 22 19 14 14 14 17 17 18 13 18 18 16 22 12 21 20 13 13 14 22 24 13 23 24 19 23 23 21 16 17 24 19 24 20 17 15 16 16 19 22 16 14 19 20 24 22 20 15 16 15 17 12 12 17 13 13 24 15 24 18 14 16 23 22 18 15 24 20 13 17 14 22 16 13 14 21 18 18 21 16 18 12 19 23 14 23 22 16 20 17 12 22 13 22",
"output": "1 1000\n27 20\n30 58\n37 59\n39 88\n51 97\n63 15\n65 20\n73 23\n3 24\n6 28\n27 58\n29 59\n30 61\n32 62\n33 75\n37 80\n39 88\n51 97\n63 99\n65 9\n69 10\n73 11\n90 15\n92 20\n3 23\n6 24\n7 25\n19 28\n26 48\n27 58\n29 59\n30 61\n32 62\n33 67\n37 75\n39 77\n46 80\n51 81\n52 88\n63 91\n65 97\n69 99\n70 4\n73 9\n78 10\n90 11\n92 15\n93 20\n98 23\n100 24\n3 25\n6 28\n7 42\n19 48\n21 54\n26 56\n27 58\n29 59\n30 61\n32 62\n33 64\n34 67\n37 72\n39 75\n46 77\n51 80\n52 81\n63 88\n65 91\n69 97\n70 99\n73 4\n78 9\n82 1..."
},
{
"input": "100 1000\n19 21 14 23 21 13 17 19 25 20 21 20 17 22 22 23 13 18 15 25 17 19 19 20 16 25 20 22 20 18 22 23 24 16 21 25 21 13 13 17 14 14 22 18 25 23 23 20 23 20 24 23 21 13 13 19 15 26 19 18 25 20 26 25 16 23 15 13 20 24 21 24 14 23 16 14 14 14 22 18 26 18 22 13 15 17 15 18 20 24 15 25 19 19 18 24 16 23 21 18",
"output": "1 1000\n58 6\n63 17\n81 38\n9 39\n20 54\n26 55\n36 68\n45 84\n58 3\n61 6\n63 17\n64 38\n81 39\n92 41\n9 42\n20 54\n26 55\n33 68\n36 73\n45 76\n51 77\n58 78\n61 84\n63 3\n64 6\n70 17\n72 19\n81 38\n90 39\n92 41\n96 42\n4 54\n9 55\n16 57\n20 67\n26 68\n32 73\n33 76\n36 77\n45 78\n46 84\n47 85\n49 87\n51 91\n52 3\n58 6\n61 17\n63 19\n64 25\n66 34\n70 38\n72 39\n74 41\n81 42\n90 54\n92 55\n96 57\n98 65\n4 67\n9 68\n14 73\n15 75\n16 76\n20 77\n26 78\n28 84\n31 85\n32 87\n33 91\n36 97\n43 3\n45 6\n46 7\n47 13\n4..."
},
{
"input": "100 1000\n28 27 21 24 21 19 24 22 18 21 19 22 19 18 19 17 14 16 27 22 17 25 22 14 14 18 26 27 23 24 25 16 27 26 15 16 24 23 25 28 20 23 19 23 19 16 23 26 22 18 23 24 24 15 25 15 23 15 20 22 21 20 24 15 28 17 28 28 14 25 17 18 27 20 20 24 27 20 14 22 22 26 26 18 21 25 15 24 27 28 16 25 19 26 16 21 26 25 18 25",
"output": "1 1000\n1 17\n40 24\n65 25\n67 69\n68 79\n90 17\n1 24\n2 25\n19 35\n28 54\n33 56\n40 58\n65 64\n67 69\n68 79\n73 87\n77 17\n89 18\n90 24\n1 25\n2 32\n19 35\n27 36\n28 46\n33 54\n34 56\n40 58\n48 64\n65 69\n67 79\n68 87\n73 91\n77 95\n82 16\n83 17\n89 18\n90 21\n94 24\n97 25\n1 32\n2 35\n19 36\n22 46\n27 54\n28 56\n31 58\n33 64\n34 66\n39 69\n40 71\n48 79\n55 87\n65 91\n67 95\n68 9\n70 14\n73 16\n77 17\n82 18\n83 21\n86 24\n89 25\n90 26\n92 32\n94 35\n97 36\n98 46\n100 50\n1 54\n2 56\n4 58\n7 64\n19 66\n22 ..."
},
{
"input": "100 1000\n21 30 21 17 25 18 25 29 30 23 17 24 18 22 21 28 22 25 29 21 29 15 22 17 26 21 29 17 21 29 17 21 22 26 20 19 24 18 27 17 23 26 25 22 18 29 23 28 25 21 15 16 15 16 27 16 24 27 26 27 28 26 28 20 18 30 16 15 21 28 28 18 18 28 21 21 23 23 18 18 21 24 29 21 15 28 17 28 24 17 28 23 24 26 30 25 23 24 24 20",
"output": "1 1000\n2 22\n9 51\n66 53\n95 68\n2 85\n8 22\n9 51\n19 52\n21 53\n27 54\n30 56\n46 67\n66 68\n83 85\n95 4\n2 11\n8 22\n9 24\n16 28\n19 31\n21 40\n27 51\n30 52\n46 53\n48 54\n61 56\n63 67\n66 68\n70 85\n71 87\n74 90\n83 4\n86 6\n88 11\n91 13\n95 22\n2 24\n8 28\n9 31\n16 38\n19 40\n21 45\n27 51\n30 52\n39 53\n46 54\n48 56\n55 65\n58 67\n60 68\n61 72\n63 73\n66 79\n70 80\n71 85\n74 87\n83 90\n86 4\n88 6\n91 11\n95 13\n2 22\n8 24\n9 28\n16 31\n19 36\n21 38\n25 40\n27 45\n30 51\n34 52\n39 53\n42 54\n46 56\n48 6..."
},
{
"input": "25 166\n7516 5616 98 9956 8158 983 3595 4018 2527 8817 2733 7227 9335 2777 611 9941 7640 9697 3350 3056 5024 4597 4279 1190 4580",
"output": "9602 166\n4 3\n4 3\n4 3\n4 3\n4 3\n4 3\n4 3\n4 3\n4 3\n4 3\n4 3\n4 3\n4 3\n4 3\n4 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 3\n4 3\n16 ..."
},
{
"input": "37 228\n1921 4643 8245 4529 7141 3713 193 3069 2058 3009 4874 2284 7088 5200 5144 1944 9882 9945 1385 5297 5243 3180 6136 2854 9563 8091 2437 5001 7489 8247 6627 5247 9162 7414 9188 3037 9441",
"output": "9379 228\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n18 7\n17 7\n18 7\n17 7\n18 7\n17 7\n18 7\n17 7\n18 7\n17 7\n18 7\n17 7\n18 7\n17 7\n18 7\n17 7\n18 7\n17 7\n18 7\n17 7\n18 7\n17 ..."
},
{
"input": "53 234\n6870 7317 2200 589 6125 4283 4087 8472 6454 9905 663 444 2138 726 5486 4491 5771 7089 3611 5378 9653 9604 1640 4518 1841 6681 3753 7657 2078 9238 205 1475 4872 1114 2777 1914 3395 7263 5105 7773 3567 4552 2673 1778 5603 4210 7133 6077 5475 8420 919 4794 8123",
"output": "9232 234\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 31\n10 3..."
},
{
"input": "17 944\n1274 536 346 5161 2949 4309 7981 7523 8690 9905 3748 5501 435 6253 19 6495 5308",
"output": "8391 944\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 15\n10 1..."
},
{
"input": "29 462\n9871 507 4301 7030 1932 687 4578 6574 7278 3153 1697 7854 5485 1779 6712 9042 4845 4482 6031 6756 2123 419 1705 6358 46 1702 8818 2967 2745",
"output": "8946 462\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 25\n1 2..."
},
{
"input": "4 1\n10000 3 3 3",
"output": "9996 1\n1 2"
},
{
"input": "4 4\n2 2 2 3",
"output": "1 4\n4 1\n1 2\n2 1\n1 2"
},
{
"input": "2 1\n10000 1",
"output": "9997 1\n1 2"
},
{
"input": "4 3\n10000 3 3 3",
"output": "9993 3\n1 2\n1 3\n1 4"
},
{
"input": "2 2\n1 3",
"output": "0 1\n2 1"
}
] | 1,687,668,828 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 45 | 93 | 3,276,800 | import sys
def main():
read = sys.stdin.readline
n, moves = (int(i) for i in read().split())
values = [(int(val), i) for i, val in enumerate(read().split())]
diff = float('inf')
last_move = (-1, -1)
num_moves = 0
history = []
if n == 1:
print(0, 0)
return
while moves > 0:
max_val, i = max(values, key=lambda x: x[0])
min_val, j = min(values, key=lambda x: x[0])
# Check if making a transfer from max to min changes the diff
if last_move[1] == i and last_move[0] == j or max_val - 1 < min_val + 1:
break
else:
values[i] = (values[i][0]-1, values[i][1])
values[j] = (values[j][0]+1, values[j][1])
max_val, _ = max(values, key=lambda x: x[0])
min_val, _ = min(values, key=lambda x: x[0])
if max_val - min_val <= diff:
diff = max_val - min_val
history.append((str(i + 1), str(j +1)))
last_move = (i, j)
else:
# It is not possible to decr further so break
break
moves -= 1
if len(history) == 0:
diff = 0
print(diff, len(history))
for r in history:
print(' '.join(r))
if __name__ == '__main__':
main() | Title: Towers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you know, all the kids in Berland love playing with cubes. Little Petya has *n* towers consisting of cubes of the same size. Tower with number *i* consists of *a**i* cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it's a waste of time.
Before going to school, the boy will have time to perform no more than *k* such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
Input Specification:
The first line contains two space-separated positive integers *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains *n* space-separated positive integers *a**i* (1<=≤<=*a**i*<=≤<=104) — the towers' initial heights.
Output Specification:
In the first line print two space-separated non-negative integers *s* and *m* (*m*<=≤<=*k*). The first number is the value of the minimum possible instability that can be obtained after performing at most *k* operations, the second number is the number of operations needed for that.
In the next *m* lines print the description of each operation as two positive integers *i* and *j*, each of them lies within limits from 1 to *n*. They represent that Petya took the top cube from the *i*-th tower and put in on the *j*-th one (*i*<=≠<=*j*). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
Demo Input:
['3 2\n5 8 5\n', '3 4\n2 2 4\n', '5 3\n8 3 2 6 3\n']
Demo Output:
['0 2\n2 1\n2 3\n', '1 1\n3 2\n', '3 3\n1 3\n1 2\n1 3\n']
Note:
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6. | ```python
import sys
def main():
read = sys.stdin.readline
n, moves = (int(i) for i in read().split())
values = [(int(val), i) for i, val in enumerate(read().split())]
diff = float('inf')
last_move = (-1, -1)
num_moves = 0
history = []
if n == 1:
print(0, 0)
return
while moves > 0:
max_val, i = max(values, key=lambda x: x[0])
min_val, j = min(values, key=lambda x: x[0])
# Check if making a transfer from max to min changes the diff
if last_move[1] == i and last_move[0] == j or max_val - 1 < min_val + 1:
break
else:
values[i] = (values[i][0]-1, values[i][1])
values[j] = (values[j][0]+1, values[j][1])
max_val, _ = max(values, key=lambda x: x[0])
min_val, _ = min(values, key=lambda x: x[0])
if max_val - min_val <= diff:
diff = max_val - min_val
history.append((str(i + 1), str(j +1)))
last_move = (i, j)
else:
# It is not possible to decr further so break
break
moves -= 1
if len(history) == 0:
diff = 0
print(diff, len(history))
for r in history:
print(' '.join(r))
if __name__ == '__main__':
main()
``` | 0 | |
479 | A | Expression | PROGRAMMING | 1,000 | [
"brute force",
"math"
] | null | null | Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get. | The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10). | Print the maximum value of the expression that you can obtain. | [
"1\n2\n3\n",
"2\n10\n3\n"
] | [
"9\n",
"60\n"
] | none | 500 | [
{
"input": "1\n2\n3",
"output": "9"
},
{
"input": "2\n10\n3",
"output": "60"
},
{
"input": "1\n1\n1",
"output": "3"
},
{
"input": "1\n2\n1",
"output": "4"
},
{
"input": "10\n10\n10",
"output": "1000"
},
{
"input": "5\n1\n3",
"output": "20"
},
{
"input": "3\n1\n5",
"output": "20"
},
{
"input": "6\n7\n1",
"output": "48"
},
{
"input": "1\n8\n3",
"output": "27"
},
{
"input": "9\n7\n2",
"output": "126"
},
{
"input": "1\n1\n10",
"output": "20"
},
{
"input": "9\n1\n1",
"output": "18"
},
{
"input": "10\n5\n6",
"output": "300"
},
{
"input": "8\n9\n7",
"output": "504"
},
{
"input": "4\n2\n10",
"output": "80"
},
{
"input": "2\n8\n3",
"output": "48"
},
{
"input": "3\n5\n7",
"output": "105"
},
{
"input": "1\n10\n1",
"output": "12"
},
{
"input": "2\n2\n2",
"output": "8"
},
{
"input": "5\n6\n1",
"output": "35"
},
{
"input": "10\n1\n1",
"output": "20"
},
{
"input": "1\n6\n1",
"output": "8"
},
{
"input": "1\n9\n1",
"output": "11"
},
{
"input": "2\n1\n2",
"output": "6"
},
{
"input": "2\n6\n1",
"output": "14"
},
{
"input": "9\n2\n1",
"output": "27"
},
{
"input": "1\n9\n2",
"output": "20"
},
{
"input": "1\n3\n1",
"output": "5"
},
{
"input": "2\n1\n1",
"output": "4"
}
] | 1,693,598,990 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 31 | 0 | a=int(input())
b=int(input())
c=int(input())
d=[]
d.append(a*b*c)
d.append(a*(b+c))
d.append(b*(a+c))
d.append(c*(a+b))
print(max(d)) | Title: Expression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
Input Specification:
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
Output Specification:
Print the maximum value of the expression that you can obtain.
Demo Input:
['1\n2\n3\n', '2\n10\n3\n']
Demo Output:
['9\n', '60\n']
Note:
none | ```python
a=int(input())
b=int(input())
c=int(input())
d=[]
d.append(a*b*c)
d.append(a*(b+c))
d.append(b*(a+c))
d.append(c*(a+b))
print(max(d))
``` | 0 | |
758 | A | Holiday Of Equality | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them. | The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen. | In the only line print the integer *S* — the minimum number of burles which are had to spend. | [
"5\n0 1 2 3 4\n",
"5\n1 1 0 1 1\n",
"3\n1 3 1\n",
"1\n12\n"
] | [
"10",
"1",
"4",
"0"
] | In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | 500 | [
{
"input": "5\n0 1 2 3 4",
"output": "10"
},
{
"input": "5\n1 1 0 1 1",
"output": "1"
},
{
"input": "3\n1 3 1",
"output": "4"
},
{
"input": "1\n12",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "14\n52518 718438 358883 462189 853171 592966 225788 46977 814826 295697 676256 561479 56545 764281",
"output": "5464380"
},
{
"input": "21\n842556 216391 427181 626688 775504 168309 851038 448402 880826 73697 593338 519033 135115 20128 424606 939484 846242 756907 377058 241543 29353",
"output": "9535765"
},
{
"input": "3\n1 3 2",
"output": "3"
},
{
"input": "3\n2 1 3",
"output": "3"
},
{
"input": "3\n2 3 1",
"output": "3"
},
{
"input": "3\n3 1 2",
"output": "3"
},
{
"input": "3\n3 2 1",
"output": "3"
},
{
"input": "1\n228503",
"output": "0"
},
{
"input": "2\n32576 550340",
"output": "517764"
},
{
"input": "3\n910648 542843 537125",
"output": "741328"
},
{
"input": "4\n751720 572344 569387 893618",
"output": "787403"
},
{
"input": "6\n433864 631347 597596 794426 713555 231193",
"output": "1364575"
},
{
"input": "9\n31078 645168 695751 126111 375934 150495 838412 434477 993107",
"output": "4647430"
},
{
"input": "30\n315421 772664 560686 654312 151528 356749 351486 707462 820089 226682 546700 136028 824236 842130 578079 337807 665903 764100 617900 822937 992759 591749 651310 742085 767695 695442 17967 515106 81059 186025",
"output": "13488674"
},
{
"input": "45\n908719 394261 815134 419990 926993 383792 772842 277695 527137 655356 684956 695716 273062 550324 106247 399133 442382 33076 462920 294674 846052 817752 421365 474141 290471 358990 109812 74492 543281 169434 919692 786809 24028 197184 310029 801476 699355 429672 51343 374128 776726 850380 293868 981569 550763",
"output": "21993384"
},
{
"input": "56\n100728 972537 13846 385421 756708 184642 259487 319707 376662 221694 675284 972837 499419 13846 38267 289898 901299 831197 954715 197515 514102 910423 127555 883934 362472 870788 538802 741008 973434 448124 391526 363321 947321 544618 68006 782313 955075 741981 815027 723297 585059 718114 700739 413489 454091 736144 308999 98065 3716 347323 9635 289003 986510 607065 60236 273351",
"output": "26984185"
},
{
"input": "70\n644488 5444 150441 714420 602059 335330 510670 196555 546346 740011 509449 850947 692874 524857 750434 952985 223744 374727 896124 753037 367352 679050 560202 172728 569291 778616 332116 286927 843598 372698 244906 498046 900681 709791 420904 724593 864493 813094 791377 39998 296710 625656 403891 579231 706693 984045 16901 574259 562265 761104 930361 256045 124461 538980 573508 372148 988722 108592 784354 55302 232524 277205 782251 299943 436488 743389 324618 742543 266915 99642",
"output": "32816391"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n1000000",
"output": "0"
},
{
"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": "0"
},
{
"input": "100\n1000000 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": "99000000"
},
{
"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 1000000",
"output": "99000000"
},
{
"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 1000000 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": "99000000"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "50\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",
"output": "0"
},
{
"input": "3\n5 0 0",
"output": "10"
},
{
"input": "5\n2 10 0 0 0",
"output": "38"
}
] | 1,682,848,212 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 46 | 0 | n = int(input())
lst = [int(i) for i in input().split()]
m = lst[0]
s = 0
for i in range(1, len(lst)):
if lst[i] >= m:
m = lst[i]
for i in range(len(lst)):
s += m - lst[i]
lst[i]+= m - lst[i]
print(s) | Title: Holiday Of Equality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
Output Specification:
In the only line print the integer *S* — the minimum number of burles which are had to spend.
Demo Input:
['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n']
Demo Output:
['10', '1', '4', '0']
Note:
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | ```python
n = int(input())
lst = [int(i) for i in input().split()]
m = lst[0]
s = 0
for i in range(1, len(lst)):
if lst[i] >= m:
m = lst[i]
for i in range(len(lst)):
s += m - lst[i]
lst[i]+= m - lst[i]
print(s)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.
An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself. | The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries.
*q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query. | For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings. | [
"1\n12\n",
"2\n6\n8\n",
"3\n1\n2\n3\n"
] | [
"3\n",
"1\n2\n",
"-1\n-1\n-1\n"
] | 12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.
8 = 4 + 4, 6 can't be split into several composite summands.
1, 2, 3 are less than any composite number, so they do not have valid splittings. | 0 | [
{
"input": "1\n12",
"output": "3"
},
{
"input": "2\n6\n8",
"output": "1\n2"
},
{
"input": "3\n1\n2\n3",
"output": "-1\n-1\n-1"
},
{
"input": "6\n1\n2\n3\n5\n7\n11",
"output": "-1\n-1\n-1\n-1\n-1\n-1"
},
{
"input": "3\n4\n6\n9",
"output": "1\n1\n1"
},
{
"input": "20\n8\n13\n20\n12\n9\n16\n4\n19\n7\n15\n10\n6\n14\n11\n3\n2\n5\n17\n18\n1",
"output": "2\n2\n5\n3\n1\n4\n1\n3\n-1\n2\n2\n1\n3\n-1\n-1\n-1\n-1\n3\n4\n-1"
},
{
"input": "100\n611\n513\n544\n463\n38\n778\n347\n317\n848\n664\n382\n108\n718\n33\n334\n876\n234\n22\n944\n305\n159\n245\n513\n691\n639\n135\n308\n324\n813\n459\n304\n116\n331\n993\n184\n224\n853\n769\n121\n687\n93\n930\n751\n308\n485\n914\n400\n695\n95\n981\n175\n972\n121\n654\n242\n610\n617\n999\n237\n548\n742\n767\n613\n172\n223\n391\n102\n907\n673\n116\n230\n355\n189\n552\n399\n493\n903\n201\n985\n459\n776\n641\n693\n919\n253\n540\n427\n394\n655\n101\n461\n854\n417\n249\n66\n380\n213\n906\n212\n528",
"output": "151\n127\n136\n114\n9\n194\n85\n78\n212\n166\n95\n27\n179\n7\n83\n219\n58\n5\n236\n75\n38\n60\n127\n171\n158\n32\n77\n81\n202\n113\n76\n29\n81\n247\n46\n56\n212\n191\n29\n170\n22\n232\n186\n77\n120\n228\n100\n172\n22\n244\n42\n243\n29\n163\n60\n152\n153\n248\n58\n137\n185\n190\n152\n43\n54\n96\n25\n225\n167\n29\n57\n87\n46\n138\n98\n122\n224\n49\n245\n113\n194\n159\n172\n228\n62\n135\n105\n98\n162\n24\n114\n213\n103\n61\n16\n95\n52\n226\n53\n132"
},
{
"input": "1\n10000001",
"output": "2499999"
}
] | 1,524,503,926 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 77 | 7,065,600 | t = int(input())
while t>0:
a = int(input())
if a == 1 or a ==2 or a == 3 or a == 5 or a == 7 or a == 9 or a == 11:
print(-1)
else:
temp = a%4
if temp == 0:
print(a//4)
elif temp == 1:
print(a//4-2+1)
elif temp == 2:
print(a//4-1+1)
else:
print(a//4-2+1)
t-=1 | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.
An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.
Input Specification:
The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries.
*q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query.
Output Specification:
For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.
Demo Input:
['1\n12\n', '2\n6\n8\n', '3\n1\n2\n3\n']
Demo Output:
['3\n', '1\n2\n', '-1\n-1\n-1\n']
Note:
12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.
8 = 4 + 4, 6 can't be split into several composite summands.
1, 2, 3 are less than any composite number, so they do not have valid splittings. | ```python
t = int(input())
while t>0:
a = int(input())
if a == 1 or a ==2 or a == 3 or a == 5 or a == 7 or a == 9 or a == 11:
print(-1)
else:
temp = a%4
if temp == 0:
print(a//4)
elif temp == 1:
print(a//4-2+1)
elif temp == 2:
print(a//4-1+1)
else:
print(a//4-2+1)
t-=1
``` | 0 | |
834 | B | The Festive Evening | PROGRAMMING | 1,100 | [
"data structures",
"implementation"
] | null | null | It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in.
There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously.
For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed.
Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened. | Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26).
In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest. | Output «YES» if at least one door was unguarded during some time, and «NO» otherwise.
You can output each letter in arbitrary case (upper or lower). | [
"5 1\nAABBB\n",
"5 1\nABABB\n"
] | [
"NO\n",
"YES\n"
] | In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened.
In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door. | 1,000 | [
{
"input": "5 1\nAABBB",
"output": "NO"
},
{
"input": "5 1\nABABB",
"output": "YES"
},
{
"input": "26 1\nABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "NO"
},
{
"input": "27 1\nABCDEFGHIJKLMNOPQRSTUVWXYZA",
"output": "YES"
},
{
"input": "5 2\nABACA",
"output": "NO"
},
{
"input": "6 2\nABCABC",
"output": "YES"
},
{
"input": "8 3\nABCBCDCA",
"output": "NO"
},
{
"input": "73 2\nDEBECECBBADAADEAABEAEEEAEBEAEBCDDBABBAEBACCBEEBBAEADEECACEDEEDABACDCDBBBD",
"output": "YES"
},
{
"input": "44 15\nHGJIFCGGCDGIJDHBIBGAEABCIABIGBDEADBBBAGDFDHA",
"output": "NO"
},
{
"input": "41 19\nTMEYYIIELFDCMBDKWWKYNRNDUPRONYROXQCLVQALP",
"output": "NO"
},
{
"input": "377 3\nEADADBBBBDEAABBAEBABACDBDBBCACAADBEAEACDEAABACADEEDEACACDADABBBBDDEECBDABACACBAECBADAEBDEEBDBCDAEADBCDDACACDCCEEDBCCBBCEDBECBABCDDBBDEADEDAEACDECECBEBACBCCDCDBDAECDECADBCBEDBBDAAEBCAAECCDCCDBDDEBADEEBDCAEABBDEDBBDDEAECCBDDCDEACDAECCBDDABABEAEDCDEDBAECBDEACEBCECEACDCBABCBAAEAADACADBBBBABEADBCADEBCBECCABBDDDEEBCDEBADEBDAAABBEABADEDEAEABCEEBEEDEAEBEABCEDDBACBCCADEBAAAAAEABABBCE",
"output": "YES"
},
{
"input": "433 3\nFZDDHMJGBZCHFUXBBPIEBBEFDWOMXXEPOMDGSMPIUZOMRZQNSJAVNATGIWPDFISKFQXJNVFXPHOZDAEZFDAHDXXQKZMGNSGKQNWGNGJGJZVVITKNFLVCPMZSDMCHBTVAWYVZLIXXIADXNYILEYNIQHKMOGMVOCWGHCWIYMPEPADSJAAKEGTUSEDWAHMNYJDIHBKHVUHLYGNGZDBULRXLSAJHPCMNWCEAAPYMHDTYWPADOTJTXTXUKLCHWKUSZRHEKQEFPVJEJJHRWCKYOIWALRTIBUMNOCRXLSIKQCJVQXEPGOHRUDJDKMUUUDORURWXJNVRVMNOUNRFKSVMTMZGOIJLXEPAMVGESOADYIGZXRBJDIWKNOWTCSROAQTBECHTOZVSQUOOJRZIBAUHMKAXDCIMDZJFMABGRNTGPUJAUNFPFWCJG",
"output": "YES"
},
{
"input": "5 2\nABCAB",
"output": "YES"
},
{
"input": "5 1\nAZAZA",
"output": "YES"
},
{
"input": "7 2\nABCDBCD",
"output": "YES"
},
{
"input": "3 26\nAAB",
"output": "NO"
}
] | 1,598,051,060 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 77 | 6,656,000 | from sys import stdin
def unguarded_door(n, k, entrances):
open_entrances = set()
for entrance in entrances:
open_entrances.add(entrance)
if len(open_entrances) > k:
return "YES"
return "NO"
if __name__ == "__main__":
n, k = [int(c) for c in stdin.readline().split(" ")]
entrances = stdin.readline()
print(unguarded_door(n, k, entrances))
| Title: The Festive Evening
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in.
There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously.
For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed.
Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened.
Input Specification:
Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26).
In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest.
Output Specification:
Output «YES» if at least one door was unguarded during some time, and «NO» otherwise.
You can output each letter in arbitrary case (upper or lower).
Demo Input:
['5 1\nAABBB\n', '5 1\nABABB\n']
Demo Output:
['NO\n', 'YES\n']
Note:
In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened.
In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door. | ```python
from sys import stdin
def unguarded_door(n, k, entrances):
open_entrances = set()
for entrance in entrances:
open_entrances.add(entrance)
if len(open_entrances) > k:
return "YES"
return "NO"
if __name__ == "__main__":
n, k = [int(c) for c in stdin.readline().split(" ")]
entrances = stdin.readline()
print(unguarded_door(n, k, entrances))
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Natasha is going to fly on a rocket to Mars and return to Earth. Also, on the way to Mars, she will land on $n - 2$ intermediate planets. Formally: we number all the planets from $1$ to $n$. $1$ is Earth, $n$ is Mars. Natasha will make exactly $n$ flights: $1 \to 2 \to \ldots n \to 1$.
Flight from $x$ to $y$ consists of two phases: take-off from planet $x$ and landing to planet $y$. This way, the overall itinerary of the trip will be: the $1$-st planet $\to$ take-off from the $1$-st planet $\to$ landing to the $2$-nd planet $\to$ $2$-nd planet $\to$ take-off from the $2$-nd planet $\to$ $\ldots$ $\to$ landing to the $n$-th planet $\to$ the $n$-th planet $\to$ take-off from the $n$-th planet $\to$ landing to the $1$-st planet $\to$ the $1$-st planet.
The mass of the rocket together with all the useful cargo (but without fuel) is $m$ tons. However, Natasha does not know how much fuel to load into the rocket. Unfortunately, fuel can only be loaded on Earth, so if the rocket runs out of fuel on some other planet, Natasha will not be able to return home. Fuel is needed to take-off from each planet and to land to each planet. It is known that $1$ ton of fuel can lift off $a_i$ tons of rocket from the $i$-th planet or to land $b_i$ tons of rocket onto the $i$-th planet.
For example, if the weight of rocket is $9$ tons, weight of fuel is $3$ tons and take-off coefficient is $8$ ($a_i = 8$), then $1.5$ tons of fuel will be burnt (since $1.5 \cdot 8 = 9 + 3$). The new weight of fuel after take-off will be $1.5$ tons.
Please note, that it is allowed to burn non-integral amount of fuel during take-off or landing, and the amount of initial fuel can be non-integral as well.
Help Natasha to calculate the minimum mass of fuel to load into the rocket. Note, that the rocket must spend fuel to carry both useful cargo and the fuel itself. However, it doesn't need to carry the fuel which has already been burnt. Assume, that the rocket takes off and lands instantly. | The first line contains a single integer $n$ ($2 \le n \le 1000$) — number of planets.
The second line contains the only integer $m$ ($1 \le m \le 1000$) — weight of the payload.
The third line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 1000$), where $a_i$ is the number of tons, which can be lifted off by one ton of fuel.
The fourth line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 1000$), where $b_i$ is the number of tons, which can be landed by one ton of fuel.
It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel. | If Natasha can fly to Mars through $(n - 2)$ planets and return to Earth, print the minimum mass of fuel (in tons) that Natasha should take. Otherwise, print a single number $-1$.
It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel.
The answer will be considered correct if its absolute or relative error doesn't exceed $10^{-6}$. Formally, let your answer be $p$, and the jury's answer be $q$. Your answer is considered correct if $\frac{|p - q|}{\max{(1, |q|)}} \le 10^{-6}$. | [
"2\n12\n11 8\n7 5\n",
"3\n1\n1 4 1\n2 5 3\n",
"6\n2\n4 6 3 3 5 6\n2 6 3 6 5 3\n"
] | [
"10.0000000000\n",
"-1\n",
"85.4800000000\n"
] | Let's consider the first example.
Initially, the mass of a rocket with fuel is $22$ tons.
- At take-off from Earth one ton of fuel can lift off $11$ tons of cargo, so to lift off $22$ tons you need to burn $2$ tons of fuel. Remaining weight of the rocket with fuel is $20$ tons.- During landing on Mars, one ton of fuel can land $5$ tons of cargo, so for landing $20$ tons you will need to burn $4$ tons of fuel. There will be $16$ tons of the rocket with fuel remaining.- While taking off from Mars, one ton of fuel can raise $8$ tons of cargo, so to lift off $16$ tons you will need to burn $2$ tons of fuel. There will be $14$ tons of rocket with fuel after that.- During landing on Earth, one ton of fuel can land $7$ tons of cargo, so for landing $14$ tons you will need to burn $2$ tons of fuel. Remaining weight is $12$ tons, that is, a rocket without any fuel.
In the second case, the rocket will not be able even to take off from Earth. | 0 | [
{
"input": "2\n12\n11 8\n7 5",
"output": "10.0000000000"
},
{
"input": "3\n1\n1 4 1\n2 5 3",
"output": "-1"
},
{
"input": "6\n2\n4 6 3 3 5 6\n2 6 3 6 5 3",
"output": "85.4800000000"
},
{
"input": "3\n3\n1 2 1\n2 2 2",
"output": "-1"
},
{
"input": "4\n4\n2 3 2 2\n2 3 4 3",
"output": "284.0000000000"
},
{
"input": "5\n2\n1 2 2 1 2\n4 5 1 4 1",
"output": "-1"
},
{
"input": "7\n7\n3 2 6 2 2 2 5\n4 7 5 6 2 2 2",
"output": "4697.0000000000"
},
{
"input": "2\n1000\n12 34\n56 78",
"output": "159.2650775220"
},
{
"input": "8\n4\n1 1 4 1 3 1 8 1\n1 1 1 1 1 3 1 2",
"output": "-1"
},
{
"input": "9\n2\n8 7 1 1 3 7 1 2 4\n4 1 1 8 7 7 1 1 5",
"output": "-1"
},
{
"input": "10\n10\n9 8 8 7 2 10 2 9 2 4\n3 10 6 2 6 6 5 9 4 5",
"output": "3075.7142857143"
},
{
"input": "20\n12\n3 9 12 13 16 18 9 9 19 7 2 5 17 14 7 7 15 16 5 7\n16 9 13 5 14 10 4 3 16 16 12 20 17 11 4 5 5 14 6 15",
"output": "4670.8944493007"
},
{
"input": "30\n5\n25 1 28 1 27 25 24 1 28 1 12 1 29 16 1 1 1 1 27 1 24 1 1 1 1 1 1 1 30 3\n1 22 1 1 24 2 13 1 16 21 1 27 14 16 1 1 7 1 1 18 1 23 10 1 15 16 16 15 10 1",
"output": "-1"
},
{
"input": "40\n13\n1 1 1 23 21 1 1 1 1 1 40 32 1 21 1 8 1 1 36 15 33 1 30 1 1 37 22 1 4 39 7 1 9 37 1 1 1 28 1 1\n1 34 17 1 38 20 8 14 1 18 29 3 21 21 18 14 1 11 1 1 23 1 25 1 14 1 7 31 9 20 25 1 1 1 1 8 26 12 1 1",
"output": "-1"
},
{
"input": "50\n19\n17 7 13 42 19 25 10 25 2 36 17 40 30 48 34 43 34 20 5 15 8 7 43 35 21 40 40 19 30 11 49 7 24 23 43 30 38 49 10 8 30 11 28 50 48 25 25 20 48 24\n49 35 10 22 24 50 50 7 6 13 16 35 12 43 50 44 35 33 38 49 26 18 23 37 7 38 23 20 28 48 41 16 6 32 32 34 11 39 38 9 38 23 16 31 37 47 33 20 46 30",
"output": "7832.1821424977"
},
{
"input": "60\n21\n11 35 1 28 39 13 19 56 13 13 21 25 1 1 23 1 52 26 53 1 1 1 30 39 1 7 1 1 3 1 1 10 1 1 37 1 1 25 1 1 1 53 1 3 48 1 6 5 4 15 1 14 25 53 25 38 27 1 1 1\n1 1 1 35 40 58 10 22 1 56 1 59 1 6 33 1 1 1 1 18 14 1 1 40 25 47 1 34 1 1 53 1 1 25 1 45 1 1 25 34 3 1 1 1 53 27 11 58 1 1 1 10 12 1 1 1 31 52 1 1",
"output": "-1"
},
{
"input": "70\n69\n70 66 57 58 24 60 39 2 48 61 65 22 10 26 68 62 48 25 12 14 45 57 6 30 48 15 46 33 42 28 69 42 64 25 24 8 62 12 68 53 55 20 32 70 3 5 41 49 16 26 2 34 34 20 39 65 18 47 62 31 39 28 61 67 7 14 31 31 53 54\n40 33 24 20 68 20 22 39 53 56 48 38 59 45 47 46 7 69 11 58 61 40 35 38 62 66 18 36 44 48 67 24 14 27 67 63 68 30 50 6 58 7 6 35 20 58 6 12 12 23 14 2 63 27 29 22 49 16 55 40 70 27 27 70 42 38 66 55 69 47",
"output": "217989.4794743629"
},
{
"input": "80\n21\n65 4 26 25 1 1 1 1 1 1 60 1 29 43 48 6 48 13 29 1 1 62 1 1 1 1 1 1 1 26 9 1 22 1 35 13 66 36 1 1 1 38 55 21 70 1 58 70 1 1 38 1 1 20 1 1 51 1 1 28 1 23 11 1 39 47 1 52 41 1 63 1 1 52 1 45 11 10 80 1\n1 1 25 30 1 1 55 54 1 48 10 37 22 1 74 1 78 13 1 65 32 1 1 1 1 69 5 59 1 1 65 1 40 1 31 1 1 75 54 1 60 1 1 1 1 1 1 1 11 29 36 1 72 71 52 1 1 1 37 1 1 75 43 9 53 1 62 1 29 1 40 27 59 74 41 53 19 30 1 73",
"output": "-1"
},
{
"input": "90\n35\n1 68 16 30 24 1 1 1 35 1 1 67 1 1 1 1 33 16 37 77 83 1 77 26 1 1 68 67 70 62 1 47 1 1 1 84 1 65 1 32 83 1 1 1 28 1 71 76 84 1 1 5 1 74 10 1 1 1 38 87 13 1 7 66 81 49 1 9 1 11 1 25 1 1 1 1 7 1 1 36 61 47 51 1 1 69 40 1 37 1\n40 1 21 1 19 51 37 52 64 1 86 1 5 24 1 1 1 19 36 1 1 77 24 4 1 18 89 1 1 1 1 1 29 22 1 80 32 36 6 1 63 1 30 1 1 1 86 79 73 52 9 1 1 11 7 1 25 20 1 20 1 49 1 37 1 41 1 1 1 1 54 55 1 10 1 1 1 1 1 1 66 1 68 1 1 1 1 53 1 1",
"output": "-1"
},
{
"input": "2\n1\n1 1\n1 1",
"output": "-1"
},
{
"input": "2\n1\n1 1\n2 2",
"output": "-1"
},
{
"input": "2\n1\n2 2\n1 1",
"output": "-1"
},
{
"input": "2\n1\n2 2\n2 2",
"output": "15.0000000000"
},
{
"input": "2\n2\n1 1\n1 1",
"output": "-1"
},
{
"input": "2\n2\n1 1\n2 2",
"output": "-1"
},
{
"input": "2\n2\n2 2\n1 1",
"output": "-1"
},
{
"input": "2\n2\n2 2\n2 2",
"output": "30.0000000000"
},
{
"input": "40\n55\n1 382 1 1 1 629 111 689 396 614 1 1 995 148 7 820 913 1 1 169 157 1 702 1 159 1 1 226 1 253 1 319 1 130 1 1 1 466 1 756\n1 23 555 1 412 1 1 373 316 234 888 1 112 818 33 443 313 1 235 1 1 610 110 535 1 445 1 386 1 1 758 1 292 1 862 1 244 428 530 1",
"output": "-1"
},
{
"input": "49\n1\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\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",
"output": "695580114.6380882263"
},
{
"input": "2\n12\n11 8\n1 1",
"output": "-1"
},
{
"input": "3\n3\n7 11 17\n19 31 33",
"output": "1.6012429470"
}
] | 1,561,221,802 | 2,842 | PyPy 3 | OK | TESTS | 76 | 140 | 1,331,200 |
n = int(input())
m = float(input())
a = list(map(int, input().strip().split()))
b = list(map(int, input().strip().split()))
if min(a) == 1 or min(b) == 1:
print(-1)
else:
C = 1.0
for i in a:
C *= i
C /= (i-1)
for i in b:
C *= i
C /= (i-1)
print(m * (C - 1.0)) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is going to fly on a rocket to Mars and return to Earth. Also, on the way to Mars, she will land on $n - 2$ intermediate planets. Formally: we number all the planets from $1$ to $n$. $1$ is Earth, $n$ is Mars. Natasha will make exactly $n$ flights: $1 \to 2 \to \ldots n \to 1$.
Flight from $x$ to $y$ consists of two phases: take-off from planet $x$ and landing to planet $y$. This way, the overall itinerary of the trip will be: the $1$-st planet $\to$ take-off from the $1$-st planet $\to$ landing to the $2$-nd planet $\to$ $2$-nd planet $\to$ take-off from the $2$-nd planet $\to$ $\ldots$ $\to$ landing to the $n$-th planet $\to$ the $n$-th planet $\to$ take-off from the $n$-th planet $\to$ landing to the $1$-st planet $\to$ the $1$-st planet.
The mass of the rocket together with all the useful cargo (but without fuel) is $m$ tons. However, Natasha does not know how much fuel to load into the rocket. Unfortunately, fuel can only be loaded on Earth, so if the rocket runs out of fuel on some other planet, Natasha will not be able to return home. Fuel is needed to take-off from each planet and to land to each planet. It is known that $1$ ton of fuel can lift off $a_i$ tons of rocket from the $i$-th planet or to land $b_i$ tons of rocket onto the $i$-th planet.
For example, if the weight of rocket is $9$ tons, weight of fuel is $3$ tons and take-off coefficient is $8$ ($a_i = 8$), then $1.5$ tons of fuel will be burnt (since $1.5 \cdot 8 = 9 + 3$). The new weight of fuel after take-off will be $1.5$ tons.
Please note, that it is allowed to burn non-integral amount of fuel during take-off or landing, and the amount of initial fuel can be non-integral as well.
Help Natasha to calculate the minimum mass of fuel to load into the rocket. Note, that the rocket must spend fuel to carry both useful cargo and the fuel itself. However, it doesn't need to carry the fuel which has already been burnt. Assume, that the rocket takes off and lands instantly.
Input Specification:
The first line contains a single integer $n$ ($2 \le n \le 1000$) — number of planets.
The second line contains the only integer $m$ ($1 \le m \le 1000$) — weight of the payload.
The third line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 1000$), where $a_i$ is the number of tons, which can be lifted off by one ton of fuel.
The fourth line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 1000$), where $b_i$ is the number of tons, which can be landed by one ton of fuel.
It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel.
Output Specification:
If Natasha can fly to Mars through $(n - 2)$ planets and return to Earth, print the minimum mass of fuel (in tons) that Natasha should take. Otherwise, print a single number $-1$.
It is guaranteed, that if Natasha can make a flight, then it takes no more than $10^9$ tons of fuel.
The answer will be considered correct if its absolute or relative error doesn't exceed $10^{-6}$. Formally, let your answer be $p$, and the jury's answer be $q$. Your answer is considered correct if $\frac{|p - q|}{\max{(1, |q|)}} \le 10^{-6}$.
Demo Input:
['2\n12\n11 8\n7 5\n', '3\n1\n1 4 1\n2 5 3\n', '6\n2\n4 6 3 3 5 6\n2 6 3 6 5 3\n']
Demo Output:
['10.0000000000\n', '-1\n', '85.4800000000\n']
Note:
Let's consider the first example.
Initially, the mass of a rocket with fuel is $22$ tons.
- At take-off from Earth one ton of fuel can lift off $11$ tons of cargo, so to lift off $22$ tons you need to burn $2$ tons of fuel. Remaining weight of the rocket with fuel is $20$ tons.- During landing on Mars, one ton of fuel can land $5$ tons of cargo, so for landing $20$ tons you will need to burn $4$ tons of fuel. There will be $16$ tons of the rocket with fuel remaining.- While taking off from Mars, one ton of fuel can raise $8$ tons of cargo, so to lift off $16$ tons you will need to burn $2$ tons of fuel. There will be $14$ tons of rocket with fuel after that.- During landing on Earth, one ton of fuel can land $7$ tons of cargo, so for landing $14$ tons you will need to burn $2$ tons of fuel. Remaining weight is $12$ tons, that is, a rocket without any fuel.
In the second case, the rocket will not be able even to take off from Earth. | ```python
n = int(input())
m = float(input())
a = list(map(int, input().strip().split()))
b = list(map(int, input().strip().split()))
if min(a) == 1 or min(b) == 1:
print(-1)
else:
C = 1.0
for i in a:
C *= i
C /= (i-1)
for i in b:
C *= i
C /= (i-1)
print(m * (C - 1.0))
``` | 3 | |
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"
}
] | 1,624,247,740 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 3 | 46 | 0 | n=int(input())
a=list(map(int,input().split()))
if n%2==0:
for i in range(0, n//2):
a.remove(max(a))
a.remove(min(a))
else:
for k in range(0, (n-1)//2):
a.remove(max(a))
a.remove(min(a))
print(a[0])
| 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()))
if n%2==0:
for i in range(0, n//2):
a.remove(max(a))
a.remove(min(a))
else:
for k in range(0, (n-1)//2):
a.remove(max(a))
a.remove(min(a))
print(a[0])
``` | -1 | |
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"
}
] | 1,539,565,560 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 109 | 0 | D = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
n = int(input())
if n < 6:
print(D[n-1])
exit()
i = 0
while (1 << i) * 5 < n:
n -= (1 << i) * 5
i += 1
print(D[n // (1 << i)]) | 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
D = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
n = int(input())
if n < 6:
print(D[n-1])
exit()
i = 0
while (1 << i) * 5 < n:
n -= (1 << i) * 5
i += 1
print(D[n // (1 << i)])
``` | 3.9455 |
2 | B | The least round way | PROGRAMMING | 2,000 | [
"dp",
"math"
] | B. The least round way | 2 | 64 | There is a square matrix *n*<=×<=*n*, consisting of non-negative integer numbers. You should find such a way on it that
- starts in the upper left cell of the matrix; - each following cell is to the right or down from the current cell; - the way ends in the bottom right cell.
Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros. | The first line contains an integer number *n* (2<=≤<=*n*<=≤<=1000), *n* is the size of the matrix. Then follow *n* lines containing the matrix elements (non-negative integer numbers not exceeding 109). | In the first line print the least number of trailing zeros. In the second line print the correspondent way itself. | [
"3\n1 2 3\n4 5 6\n7 8 9\n"
] | [
"0\nDDRR\n"
] | none | 0 | [
{
"input": "3\n1 2 3\n4 5 6\n7 8 9",
"output": "0\nDDRR"
},
{
"input": "2\n7 6\n3 8",
"output": "0\nDR"
},
{
"input": "3\n4 10 5\n10 9 4\n6 5 3",
"output": "1\nDRRD"
},
{
"input": "4\n1 1 9 9\n3 4 7 3\n7 9 1 7\n1 7 1 5",
"output": "0\nDDDRRR"
},
{
"input": "5\n8 3 2 1 4\n3 7 2 4 8\n9 2 8 9 10\n2 3 6 10 1\n8 2 2 8 4",
"output": "0\nDDDDRRRR"
},
{
"input": "6\n5 5 4 10 5 5\n7 10 8 7 6 6\n7 1 7 9 7 8\n5 5 3 3 10 9\n5 8 10 6 3 8\n3 10 5 4 3 4",
"output": "1\nDDRRDRDDRR"
},
{
"input": "7\n2 9 8 2 7 4 8\n9 5 4 4 8 5 3\n5 7 2 10 8 1 8\n2 7 10 7 5 7 7\n9 2 7 6 4 8 4\n7 2 4 7 4 1 8\n9 5 3 10 1 6 2",
"output": "0\nRRDRRDRDDDDR"
},
{
"input": "8\n1 1 10 1 8 4 8 7\n9 3 3 2 2 6 2 4\n7 4 3 5 10 3 5 1\n8 4 4 10 4 5 9 4\n5 5 5 2 6 7 1 8\n4 10 1 3 2 4 8 3\n8 1 10 2 8 2 2 4\n2 10 6 8 10 2 8 4",
"output": "0\nDRRRRRRRDDDDDD"
},
{
"input": "9\n8 3 3 3 10 3 10 5 6\n2 1 6 1 8 1 9 1 6\n6 1 5 4 2 2 10 4 9\n1 9 1 3 10 6 10 5 5\n1 10 5 4 7 2 5 9 10\n6 6 1 3 1 9 4 9 9\n5 3 7 6 4 6 2 10 2\n9 3 3 10 5 6 7 6 4\n4 9 6 7 4 3 7 6 5",
"output": "1\nDDDDDRDDDRRRRRRR"
},
{
"input": "10\n10 8 6 5 9 8 2 5 3 2\n3 1 8 6 8 10 5 5 7 8\n5 9 7 7 4 9 7 2 5 2\n5 9 9 5 4 2 6 6 8 1\n10 6 9 9 10 5 6 3 5 9\n6 7 10 3 1 4 3 6 7 6\n10 10 4 7 4 4 8 6 7 4\n10 5 8 2 2 7 4 4 1 4\n8 4 6 10 10 6 1 3 3 1\n9 9 7 2 9 5 1 8 6 3",
"output": "1\nDRDDDRRDDDRRDRDRRR"
}
] | 1,467,908,414 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 61 | 204,800 | #!/usr/bin/python
def get25(n):
t = n
c2 = 0
while t%2 == 0:
c2 += 1
t /= 2
t = n
c5 = 0
while t%5 == 0:
c5 += 1
t /= 5
return [c2, c5]
def main():
n = int(input())
a = [None] * n
f = [None] * n
for i in range(n):
b = tuple(map(int, input().split()))
a[i] = []
f[i] = []
for j in range(n):
a[i].append(get25(b[j]))
f[i].append(None)
f[0][0] = 0
for i in range(n):
for j in range(n):
p = None
q = None
if i-1 >= 0:
p = (a[i-1][j][0] + a[i][j][0], a[i-1][j][1] + a[i][j][1])
if j-1 >= 0:
q = (a[i][j-1][0] + a[i][j][0], a[i][j-1][1] + a[i][j][1])
if p != None and q != None:
if min(p) < min(q):
a[i][j] = p
f[i][j] = f[i-1][j] + 1
else:
a[i][j] = q
f[i][j] = f[i][j-1] + 1
elif p == None and q != None:
a[i][j] = q
f[i][j] = f[i][j-1] + 1
elif p != None and q == None:
a[i][j] = p
f[i][j] = f[i-1][j] + 1
i = 0
j = 0
path = ''
while i != n-1 or j != n-1:
if i+1 < n and f[i][j] + 1 == f[i+1][j]:
i += 1
path += 'D'
elif j+1 < n and f[i][j] + 1 == f[i][j+1]:
j += 1
path += 'R'
print(min(a[n-1][n-1]))
print(path)
if __name__ == '__main__':
main()
| Title: The least round way
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
There is a square matrix *n*<=×<=*n*, consisting of non-negative integer numbers. You should find such a way on it that
- starts in the upper left cell of the matrix; - each following cell is to the right or down from the current cell; - the way ends in the bottom right cell.
Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.
Input Specification:
The first line contains an integer number *n* (2<=≤<=*n*<=≤<=1000), *n* is the size of the matrix. Then follow *n* lines containing the matrix elements (non-negative integer numbers not exceeding 109).
Output Specification:
In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.
Demo Input:
['3\n1 2 3\n4 5 6\n7 8 9\n']
Demo Output:
['0\nDDRR\n']
Note:
none | ```python
#!/usr/bin/python
def get25(n):
t = n
c2 = 0
while t%2 == 0:
c2 += 1
t /= 2
t = n
c5 = 0
while t%5 == 0:
c5 += 1
t /= 5
return [c2, c5]
def main():
n = int(input())
a = [None] * n
f = [None] * n
for i in range(n):
b = tuple(map(int, input().split()))
a[i] = []
f[i] = []
for j in range(n):
a[i].append(get25(b[j]))
f[i].append(None)
f[0][0] = 0
for i in range(n):
for j in range(n):
p = None
q = None
if i-1 >= 0:
p = (a[i-1][j][0] + a[i][j][0], a[i-1][j][1] + a[i][j][1])
if j-1 >= 0:
q = (a[i][j-1][0] + a[i][j][0], a[i][j-1][1] + a[i][j][1])
if p != None and q != None:
if min(p) < min(q):
a[i][j] = p
f[i][j] = f[i-1][j] + 1
else:
a[i][j] = q
f[i][j] = f[i][j-1] + 1
elif p == None and q != None:
a[i][j] = q
f[i][j] = f[i][j-1] + 1
elif p != None and q == None:
a[i][j] = p
f[i][j] = f[i-1][j] + 1
i = 0
j = 0
path = ''
while i != n-1 or j != n-1:
if i+1 < n and f[i][j] + 1 == f[i+1][j]:
i += 1
path += 'D'
elif j+1 < n and f[i][j] + 1 == f[i][j+1]:
j += 1
path += 'R'
print(min(a[n-1][n-1]))
print(path)
if __name__ == '__main__':
main()
``` | 0 |
120 | A | Elevator | PROGRAMMING | 1,000 | [
"brute force",
"implementation",
"math"
] | null | null | A sky scraper with 1000 floors has been built in the city of N. It has modern superfast elevators to help to travel from one floor to another. Each elevator has two doors, the front one and the back one. If one goes in through the front door, he goes out through the back one and vice versa. The elevator has two rails numbered with numbers 1 and 2. Rail 1 is located to the left of the entrance to the front door (or correspondingly, to the right of the entrance to the back door). Rail 2 is located opposite it, to the right of the entrance to the front door and to the left of the entrance to the back door. We know that each person in the city of N holds at a rail with the strongest hand.
One day a VIP person visited the city and of course, he took a look at the skyscraper and took a ride in the elevator. We know the door through which he entered and the rail he was holding at. Now we need to determine as soon as possible whether he is left-handed or right-handed. | The first line indicates the door through which the very important person entered the elevator. It contains "front" if the person enters the elevator through the front door and "back" if he entered the elevator through the back door. The second line contains integer *a* (1<=≤<=*a*<=≤<=2) which denotes the number of the rail at which the person was holding. | Print character "R" if the VIP is right-handed or "L" if he is left-handed. | [
"front\n1\n"
] | [
"L\n"
] | none | 0 | [
{
"input": "front\n1",
"output": "L"
},
{
"input": "back\n1",
"output": "R"
},
{
"input": "front\n2",
"output": "R"
},
{
"input": "back\n2",
"output": "L"
}
] | 1,554,385,733 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 374 | 1,024,000 | """""""""""""""""""""""""""""""""""""""""""""
| author: mr.math - Hakimov Rahimjon |
| e-mail: [email protected] |
"""""""""""""""""""""""""""""""""""""""""""""
#inp = open("lepus.in", "r"); input = inp.readline; out = open("lepus.out", "w"); print = out.write
TN = 1
# ===========================================
def solution():
s = input()
a = int(input())
if (s == "front" and a == 1) or (s == "back" and a == 2): print("L")
else: print("R")
# ===========================================
while TN != 0:
solution()
TN -= 1
# ===========================================
#inp.close()
#out.close() | Title: Elevator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A sky scraper with 1000 floors has been built in the city of N. It has modern superfast elevators to help to travel from one floor to another. Each elevator has two doors, the front one and the back one. If one goes in through the front door, he goes out through the back one and vice versa. The elevator has two rails numbered with numbers 1 and 2. Rail 1 is located to the left of the entrance to the front door (or correspondingly, to the right of the entrance to the back door). Rail 2 is located opposite it, to the right of the entrance to the front door and to the left of the entrance to the back door. We know that each person in the city of N holds at a rail with the strongest hand.
One day a VIP person visited the city and of course, he took a look at the skyscraper and took a ride in the elevator. We know the door through which he entered and the rail he was holding at. Now we need to determine as soon as possible whether he is left-handed or right-handed.
Input Specification:
The first line indicates the door through which the very important person entered the elevator. It contains "front" if the person enters the elevator through the front door and "back" if he entered the elevator through the back door. The second line contains integer *a* (1<=≤<=*a*<=≤<=2) which denotes the number of the rail at which the person was holding.
Output Specification:
Print character "R" if the VIP is right-handed or "L" if he is left-handed.
Demo Input:
['front\n1\n']
Demo Output:
['L\n']
Note:
none | ```python
"""""""""""""""""""""""""""""""""""""""""""""
| author: mr.math - Hakimov Rahimjon |
| e-mail: [email protected] |
"""""""""""""""""""""""""""""""""""""""""""""
#inp = open("lepus.in", "r"); input = inp.readline; out = open("lepus.out", "w"); print = out.write
TN = 1
# ===========================================
def solution():
s = input()
a = int(input())
if (s == "front" and a == 1) or (s == "back" and a == 2): print("L")
else: print("R")
# ===========================================
while TN != 0:
solution()
TN -= 1
# ===========================================
#inp.close()
#out.close()
``` | -1 | |
837 | B | Flag of Berland | PROGRAMMING | 1,600 | [
"brute force",
"implementation"
] | null | null | The flag of Berland is such rectangular field *n*<=×<=*m* that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'. - Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color. - Each color should be used in exactly one stripe.
You are given a field *n*<=×<=*m*, consisting of characters 'R', 'G' and 'B'. Output "YES" (without quotes) if this field corresponds to correct flag of Berland. Otherwise, print "NO" (without quotes). | The first line contains two integer numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the sizes of the field.
Each of the following *n* lines consisting of *m* characters 'R', 'G' and 'B' — the description of the field. | Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes). | [
"6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG\n",
"4 3\nBRG\nBRG\nBRG\nBRG\n",
"6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB\n",
"4 4\nRRRR\nRRRR\nBBBB\nGGGG\n"
] | [
"YES\n",
"YES\n",
"NO\n",
"NO\n"
] | The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | 0 | [
{
"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG",
"output": "YES"
},
{
"input": "4 3\nBRG\nBRG\nBRG\nBRG",
"output": "YES"
},
{
"input": "6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB",
"output": "NO"
},
{
"input": "4 4\nRRRR\nRRRR\nBBBB\nGGGG",
"output": "NO"
},
{
"input": "1 3\nGRB",
"output": "YES"
},
{
"input": "3 1\nR\nG\nB",
"output": "YES"
},
{
"input": "4 3\nRGB\nGRB\nGRB\nGRB",
"output": "NO"
},
{
"input": "4 6\nGGRRBB\nGGRRBB\nGGRRBB\nRRGGBB",
"output": "NO"
},
{
"input": "100 3\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nRGB\nGRB",
"output": "NO"
},
{
"input": "3 100\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRG",
"output": "NO"
},
{
"input": "3 1\nR\nR\nB",
"output": "NO"
},
{
"input": "3 2\nRR\nBB\nRR",
"output": "NO"
},
{
"input": "3 2\nRR\nBG\nBG",
"output": "NO"
},
{
"input": "3 2\nBB\nRR\nBB",
"output": "NO"
},
{
"input": "3 3\nRRR\nRRR\nRRR",
"output": "NO"
},
{
"input": "3 3\nGGG\nGGG\nGGG",
"output": "NO"
},
{
"input": "1 3\nRGG",
"output": "NO"
},
{
"input": "4 3\nRGR\nRGR\nRGR\nRGR",
"output": "NO"
},
{
"input": "3 4\nRRGG\nRRGG\nBBBB",
"output": "NO"
},
{
"input": "3 3\nBRG\nBRG\nBRG",
"output": "YES"
},
{
"input": "3 1\nR\nG\nR",
"output": "NO"
},
{
"input": "5 3\nBBG\nBBG\nBBG\nBBG\nBBG",
"output": "NO"
},
{
"input": "3 3\nRRR\nGGG\nRRR",
"output": "NO"
},
{
"input": "1 3\nRGR",
"output": "NO"
},
{
"input": "3 6\nRRBBGG\nRRBBGG\nRRBBGG",
"output": "YES"
},
{
"input": "6 6\nRRBBGG\nRRBBGG\nRRBBGG\nRRBBGG\nRRBBGG\nRRBBGG",
"output": "YES"
},
{
"input": "4 3\nRRR\nGGG\nBBB\nBBB",
"output": "NO"
},
{
"input": "3 3\nRRR\nBBB\nRRR",
"output": "NO"
},
{
"input": "3 1\nB\nR\nB",
"output": "NO"
},
{
"input": "1 3\nBGB",
"output": "NO"
},
{
"input": "3 1\nB\nB\nB",
"output": "NO"
},
{
"input": "3 4\nRRRR\nBBBB\nRRRR",
"output": "NO"
},
{
"input": "1 6\nRGGGBB",
"output": "NO"
},
{
"input": "9 3\nBBB\nBBB\nBBB\nGGG\nGGG\nGRG\nRGR\nRRR\nRRR",
"output": "NO"
},
{
"input": "4 4\nRGBB\nRGBB\nRGBB\nRGBB",
"output": "NO"
},
{
"input": "3 3\nRBR\nRBR\nRBR",
"output": "NO"
},
{
"input": "1 6\nRRRRBB",
"output": "NO"
},
{
"input": "1 6\nRRRRRR",
"output": "NO"
},
{
"input": "1 6\nRRGGGG",
"output": "NO"
},
{
"input": "4 4\nRRRR\nRRRR\nRRRR\nRRRR",
"output": "NO"
},
{
"input": "3 1\nB\nG\nB",
"output": "NO"
},
{
"input": "3 1\nR\nR\nR",
"output": "NO"
},
{
"input": "1 9\nRRRGGGBBB",
"output": "YES"
},
{
"input": "1 3\nRRR",
"output": "NO"
},
{
"input": "3 5\nRRRRR\nBBBBB\nBBBBB",
"output": "NO"
},
{
"input": "3 3\nRRR\nGGG\nGGG",
"output": "NO"
},
{
"input": "1 1\nR",
"output": "NO"
},
{
"input": "3 3\nRGR\nRGR\nRGR",
"output": "NO"
},
{
"input": "1 3\nGGG",
"output": "NO"
},
{
"input": "3 3\nRBG\nGBR\nRGB",
"output": "NO"
},
{
"input": "3 3\nRGB\nRGB\nRGB",
"output": "YES"
},
{
"input": "1 3\nBRB",
"output": "NO"
},
{
"input": "2 1\nR\nB",
"output": "NO"
},
{
"input": "1 3\nRBR",
"output": "NO"
},
{
"input": "3 5\nRRGBB\nRRGBB\nRRGBB",
"output": "NO"
},
{
"input": "5 3\nBBR\nBBR\nBBR\nBBR\nBBR",
"output": "NO"
},
{
"input": "3 3\nRGB\nRBG\nRGB",
"output": "NO"
},
{
"input": "1 2\nRB",
"output": "NO"
},
{
"input": "4 3\nBBB\nBBB\nBBB\nBBB",
"output": "NO"
},
{
"input": "36 6\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR\nBBRRRR",
"output": "NO"
},
{
"input": "4 1\nR\nB\nG\nR",
"output": "NO"
},
{
"input": "13 12\nRRRRGGGGRRRR\nRRRRGGGGRRRR\nRRRRGGGGRRRR\nRRRRGGGGRRRR\nRRRRGGGGRRRR\nRRRRGGGGRRRR\nRRRRGGGGRRRR\nRRRRGGGGRRRR\nRRRRGGGGRRRR\nRRRRGGGGRRRR\nRRRRGGGGRRRR\nRRRRGGGGRRRR\nRRRRGGGGRRRR",
"output": "NO"
},
{
"input": "2 2\nRR\nRR",
"output": "NO"
},
{
"input": "6 6\nRRGGBB\nGRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB",
"output": "NO"
},
{
"input": "70 3\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG\nBGG",
"output": "NO"
},
{
"input": "4 3\nBBG\nBBG\nBBG\nBBG",
"output": "NO"
},
{
"input": "6 3\nBBB\nGGG\nRRR\nBRG\nBRG\nBRG",
"output": "NO"
},
{
"input": "3 6\nRRBBGG\nRBBBGG\nRBBBGG",
"output": "NO"
},
{
"input": "6 6\nGGGGGG\nGGGGGG\nBBBBBB\nBBBBBB\nGGGGGG\nGGGGGG",
"output": "NO"
},
{
"input": "6 1\nR\nB\nG\nR\nB\nG",
"output": "NO"
},
{
"input": "6 5\nRRRRR\nBBBBB\nGGGGG\nRRRRR\nBBBBB\nGGGGG",
"output": "NO"
},
{
"input": "6 3\nRRR\nGGG\nBBB\nRRR\nGGG\nBBB",
"output": "NO"
},
{
"input": "6 5\nRRRRR\nRRRRR\nRRRRR\nGGGGG\nGGGGG\nGGGGG",
"output": "NO"
},
{
"input": "15 28\nBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nGGGGGGGGGGGGGGGGGGGGGGGGGGGG\nGGGGGGGGGGGGGGGGGGGGGGGGGGGG\nGGGGGGGGGGGGGGGGGGGGGGGGGGGG\nGGGGGGGGGGGGGGGGGGGGGGGGGGGG\nGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "NO"
},
{
"input": "21 10\nRRRRRRRRRR\nRRRRRRRRRR\nRRRRRRRRRR\nRRRRRRRRRR\nRRRRRRRRRR\nRRRRRRRRRR\nRRRRRRRRRR\nBBBBBBBBBB\nBBBBBBBBBB\nBBBBBGBBBB\nBBBBBBBBBB\nBBBBBBBBBB\nBBBBBBBBBB\nBBBBBBBBBB\nGGGGGGGGGG\nGGGGGGGGGG\nGGGGGGGGGG\nGGGGGGGGGG\nGGGGGGGGGG\nGGGGGGGGGG\nGGGGGGGGGG",
"output": "NO"
},
{
"input": "3 2\nRR\nGB\nGB",
"output": "NO"
},
{
"input": "3 2\nRG\nRG\nBB",
"output": "NO"
},
{
"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nRRRRR\nRRRRR",
"output": "NO"
},
{
"input": "3 3\nRGB\nGBR\nBRG",
"output": "NO"
},
{
"input": "1 3\nRBB",
"output": "NO"
},
{
"input": "3 3\nBGR\nBGR\nBGR",
"output": "YES"
},
{
"input": "6 6\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB",
"output": "YES"
},
{
"input": "4 2\nRR\nGG\nRR\nBB",
"output": "NO"
},
{
"input": "3 3\nRRR\nRRR\nGGG",
"output": "NO"
},
{
"input": "8 6\nRRRRRR\nRRRRRR\nRRRRRR\nRRRRRR\nRRRRRR\nRRRRRR\nRRRRRR\nRRRRRR",
"output": "NO"
},
{
"input": "3 4\nRRRR\nRRRR\nGGGG",
"output": "NO"
},
{
"input": "3 4\nRRRR\nRRRR\nRRRR",
"output": "NO"
},
{
"input": "6 1\nR\nR\nR\nR\nR\nR",
"output": "NO"
},
{
"input": "1 6\nRRBBGG",
"output": "YES"
},
{
"input": "1 6\nRGBRGB",
"output": "NO"
},
{
"input": "3 4\nRRRR\nGGGG\nRRRR",
"output": "NO"
},
{
"input": "3 3\nRRB\nGRG\nGBB",
"output": "NO"
},
{
"input": "3 7\nRRGGBBB\nRRGGBBB\nRRGGBBB",
"output": "NO"
},
{
"input": "3 1\nG\nR\nR",
"output": "NO"
},
{
"input": "2 3\nRGG\nRBB",
"output": "NO"
},
{
"input": "3 3\nRRG\nGGG\nBBB",
"output": "NO"
},
{
"input": "3 3\nRGB\nRBB\nRGB",
"output": "NO"
},
{
"input": "3 3\nRGR\nRGB\nRGB",
"output": "NO"
},
{
"input": "3 1\nB\nR\nR",
"output": "NO"
},
{
"input": "1 3\nGRR",
"output": "NO"
},
{
"input": "4 4\nRRRR\nGGGG\nBBBB\nBBBB",
"output": "NO"
},
{
"input": "1 3\nGGR",
"output": "NO"
},
{
"input": "3 3\nRGB\nGGB\nRGB",
"output": "NO"
},
{
"input": "3 3\nRGR\nGGG\nBBB",
"output": "NO"
},
{
"input": "6 6\nRRRRRR\nGGGGGG\nGGGGGG\nGGGGGG\nBBBBBB\nBBBBBB",
"output": "NO"
},
{
"input": "6 6\nRRRRRR\nRRRRRR\nGGGGGG\nBBBBBB\nBBBBBB\nBBBBBB",
"output": "NO"
},
{
"input": "3 1\nG\nB\nR",
"output": "YES"
},
{
"input": "3 3\nGGB\nRGB\nRGB",
"output": "NO"
},
{
"input": "3 3\nGRR\nGGG\nBBB",
"output": "NO"
},
{
"input": "6 6\nRRRRRR\nRRRRRR\nGGGGGG\nGGGGGG\nBBBBBB\nRRRRRR",
"output": "NO"
},
{
"input": "3 3\nRRR\nGBG\nBBB",
"output": "NO"
},
{
"input": "3 8\nRRGGBBBB\nRRGGBBBB\nRRGGBBBB",
"output": "NO"
},
{
"input": "2 2\nRR\nGG",
"output": "NO"
},
{
"input": "3 3\nRGB\nRGR\nRGB",
"output": "NO"
},
{
"input": "1 3\nRBG",
"output": "YES"
},
{
"input": "2 6\nRRGGBB\nGGRRBB",
"output": "NO"
},
{
"input": "6 2\nRR\nGG\nBB\nRR\nGG\nBB",
"output": "NO"
},
{
"input": "1 5\nRRGGB",
"output": "NO"
},
{
"input": "1 2\nRG",
"output": "NO"
},
{
"input": "1 6\nRGBRBG",
"output": "NO"
},
{
"input": "1 6\nRRRGGB",
"output": "NO"
},
{
"input": "1 3\nRGB",
"output": "YES"
},
{
"input": "4 3\nRRR\nBBR\nGBB\nGGG",
"output": "NO"
},
{
"input": "6 3\nRRR\nBBB\nBBB\nBBB\nGGG\nGGG",
"output": "NO"
},
{
"input": "3 3\nRBG\nRBG\nRBG",
"output": "YES"
},
{
"input": "6 3\nRRR\nBBB\nGGG\nRRR\nBBB\nGGG",
"output": "NO"
},
{
"input": "1 4\nRGBB",
"output": "NO"
},
{
"input": "6 6\nRRRRRR\nRRRRRR\nRRRRRR\nRRRRRR\nRRRRRR\nRRRRRR",
"output": "NO"
},
{
"input": "6 5\nRRRRR\nRRRRR\nGGGGG\nGGGGG\nRRRRR\nRRRRR",
"output": "NO"
},
{
"input": "3 3\nRGB\nBRG\nGBR",
"output": "NO"
},
{
"input": "6 10\nRRRRRRRRRR\nGGGGGGGGGG\nBBBBBBBBBB\nRRRRRRRRRR\nGGGGGGGGGG\nBBBBBBBBBB",
"output": "NO"
},
{
"input": "20 6\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB\nRRGGBB",
"output": "YES"
},
{
"input": "4 1\nR\nG\nB\nR",
"output": "NO"
},
{
"input": "1 4\nRGBR",
"output": "NO"
},
{
"input": "2 4\nRGBB\nRRGB",
"output": "NO"
}
] | 1,600,576,587 | 687 | PyPy 3 | WRONG_ANSWER | TESTS | 66 | 155 | 1,638,400 | import sys
n, m = map(int, input().split())
flag = [list(input()) for _ in range(n)]
if len(set(flag[0])) > 1:
_flag = [['']*n for _ in range(m)]
for i in range(n):
for j in range(m):
_flag[j][i] = flag[i][j]
flag = _flag
n, m = m, n
if not all(len(set(row)) == 1 for row in flag):
print('NO')
exit()
color = {flag[0][0]}
length = [1]
prev = flag[0][0]
for i in range(1, n):
if prev == flag[i][0]:
length[-1] += 1
else:
prev = flag[i][0]
color.add(flag[i][0])
length.append(1)
if len(color) == 3 and len(set(length)) == 1:
print('YES')
else:
print('NO')
| Title: Flag of Berland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The flag of Berland is such rectangular field *n*<=×<=*m* that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'. - Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color. - Each color should be used in exactly one stripe.
You are given a field *n*<=×<=*m*, consisting of characters 'R', 'G' and 'B'. Output "YES" (without quotes) if this field corresponds to correct flag of Berland. Otherwise, print "NO" (without quotes).
Input Specification:
The first line contains two integer numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the sizes of the field.
Each of the following *n* lines consisting of *m* characters 'R', 'G' and 'B' — the description of the field.
Output Specification:
Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes).
Demo Input:
['6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG\n', '4 3\nBRG\nBRG\nBRG\nBRG\n', '6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB\n', '4 4\nRRRR\nRRRR\nBBBB\nGGGG\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n', 'NO\n']
Note:
The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | ```python
import sys
n, m = map(int, input().split())
flag = [list(input()) for _ in range(n)]
if len(set(flag[0])) > 1:
_flag = [['']*n for _ in range(m)]
for i in range(n):
for j in range(m):
_flag[j][i] = flag[i][j]
flag = _flag
n, m = m, n
if not all(len(set(row)) == 1 for row in flag):
print('NO')
exit()
color = {flag[0][0]}
length = [1]
prev = flag[0][0]
for i in range(1, n):
if prev == flag[i][0]:
length[-1] += 1
else:
prev = flag[i][0]
color.add(flag[i][0])
length.append(1)
if len(color) == 3 and len(set(length)) == 1:
print('YES')
else:
print('NO')
``` | 0 | |
621 | C | Wet Shark and Flowers | PROGRAMMING | 1,700 | [
"combinatorics",
"math",
"number theory",
"probabilities"
] | null | null | There are *n* sharks who grow flowers for Wet Shark. They are all sitting around the table, such that sharks *i* and *i*<=+<=1 are neighbours for all *i* from 1 to *n*<=-<=1. Sharks *n* and 1 are neighbours too.
Each shark will grow some number of flowers *s**i*. For *i*-th shark value *s**i* is random integer equiprobably chosen in range from *l**i* to *r**i*. Wet Shark has it's favourite prime number *p*, and he really likes it! If for any pair of neighbouring sharks *i* and *j* the product *s**i*·*s**j* is divisible by *p*, then Wet Shark becomes happy and gives 1000 dollars to each of these sharks.
At the end of the day sharks sum all the money Wet Shark granted to them. Find the expectation of this value. | The first line of the input contains two space-separated integers *n* and *p* (3<=≤<=*n*<=≤<=100<=000,<=2<=≤<=*p*<=≤<=109) — the number of sharks and Wet Shark's favourite prime number. It is guaranteed that *p* is prime.
The *i*-th of the following *n* lines contains information about *i*-th shark — two space-separated integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109), the range of flowers shark *i* can produce. Remember that *s**i* is chosen equiprobably among all integers from *l**i* to *r**i*, inclusive. | Print a single real number — the expected number of dollars that the sharks receive in total. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if . | [
"3 2\n1 2\n420 421\n420420 420421\n",
"3 5\n1 4\n2 3\n11 14\n"
] | [
"4500.0\n",
"0.0\n"
] | A prime number is a positive integer number that is divisible only by 1 and itself. 1 is not considered to be prime.
Consider the first sample. First shark grows some number of flowers from 1 to 2, second sharks grows from 420 to 421 flowers and third from 420420 to 420421. There are eight cases for the quantities of flowers (*s*<sub class="lower-index">0</sub>, *s*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub>) each shark grows:
1. (1, 420, 420420): note that *s*<sub class="lower-index">0</sub>·*s*<sub class="lower-index">1</sub> = 420, *s*<sub class="lower-index">1</sub>·*s*<sub class="lower-index">2</sub> = 176576400, and *s*<sub class="lower-index">2</sub>·*s*<sub class="lower-index">0</sub> = 420420. For each pair, 1000 dollars will be awarded to each shark. Therefore, each shark will be awarded 2000 dollars, for a total of 6000 dollars.1. (1, 420, 420421): now, the product *s*<sub class="lower-index">2</sub>·*s*<sub class="lower-index">0</sub> is not divisible by 2. Therefore, sharks *s*<sub class="lower-index">0</sub> and *s*<sub class="lower-index">2</sub> will receive 1000 dollars, while shark *s*<sub class="lower-index">1</sub> will receive 2000. The total is 4000.1. (1, 421, 420420): total is 4000 1. (1, 421, 420421): total is 0. 1. (2, 420, 420420): total is 6000. 1. (2, 420, 420421): total is 6000. 1. (2, 421, 420420): total is 6000. 1. (2, 421, 420421): total is 4000.
The expected value is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dfe520d00a8615f7c270ccbccbebe182cc7db883.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample, no combination of quantities will garner the sharks any money. | 1,500 | [
{
"input": "3 2\n1 2\n420 421\n420420 420421",
"output": "4500.0"
},
{
"input": "3 5\n1 4\n2 3\n11 14",
"output": "0.0"
},
{
"input": "3 3\n3 3\n2 4\n1 1",
"output": "4666.666666666667"
},
{
"input": "5 5\n5 204\n420 469\n417 480\n442 443\n44 46",
"output": "3451.25"
},
{
"input": "3 2\n2 2\n3 3\n4 4",
"output": "6000.0"
},
{
"input": "6 7\n8 13\n14 14\n8 13\n14 14\n8 13\n14 14",
"output": "12000.0"
},
{
"input": "3 7\n7 14\n700000000 700000007\n420 4200",
"output": "2304.2515207617034"
},
{
"input": "5 999999937\n999999935 999999936\n999999937 999999938\n999999939 999999940\n999999941 999999942\n999999943 999999944",
"output": "2000.0"
},
{
"input": "5 999999937\n1 999999936\n1 999999936\n1 999999936\n1 999999936\n1 999999936",
"output": "0.0"
},
{
"input": "20 999999937\n999999936 999999937\n999999937 999999938\n999999936 999999937\n999999937 999999938\n999999936 999999937\n999999937 999999938\n999999936 999999937\n999999937 999999938\n999999936 999999937\n999999937 999999938\n999999936 999999937\n999999937 999999938\n999999936 999999937\n999999937 999999938\n999999936 999999937\n999999937 999999938\n999999936 999999937\n999999937 999999938\n999999936 999999937\n999999937 999999938",
"output": "30000.0"
},
{
"input": "9 41\n40 42\n42 44\n44 46\n82 84\n82 83\n80 83\n40 83\n40 82\n42 82",
"output": "5503.274377352654"
},
{
"input": "3 2\n1 1\n1 2\n1 1",
"output": "2000.0"
},
{
"input": "12 3\n697806 966852\n802746 974920\n579567 821770\n628655 642480\n649359 905832\n87506 178848\n605628 924780\n843338 925533\n953514 978612\n375312 997707\n367620 509906\n277106 866177",
"output": "13333.518289809368"
},
{
"input": "5 3\n67050 461313\n927808 989615\n169239 201720\n595515 756354\n392844 781910",
"output": "5555.597086312073"
},
{
"input": "6 7\n984774 984865\n720391 916269\n381290 388205\n628383 840455\n747138 853964\n759705 959629",
"output": "3215.6233297395006"
},
{
"input": "3 5\n99535 124440\n24114 662840\n529335 875935",
"output": "2160.11317825774"
},
{
"input": "4 3\n561495 819666\n718673 973130\n830124 854655\n430685 963699",
"output": "4444.521972611004"
},
{
"input": "10 3\n311664 694971\n364840 366487\n560148 821101\n896470 923613\n770019 828958\n595743 827536\n341418 988218\n207847 366132\n517968 587855\n168695 878142",
"output": "11110.602699850484"
},
{
"input": "11 3\n66999 737907\n499872 598806\n560583 823299\n579017 838419\n214308 914576\n31820 579035\n373821 695652\n438988 889317\n181332 513682\n740575 769488\n597348 980891",
"output": "12222.259608784536"
},
{
"input": "12 3\n158757 341790\n130709 571435\n571161 926255\n851779 952236\n914910 941369\n774359 860799\n224067 618483\n411639 902888\n264423 830336\n33133 608526\n951696 976379\n923880 968563",
"output": "13333.377729413933"
},
{
"input": "9 2\n717582 964152\n268030 456147\n400022 466269\n132600 698200\n658890 807357\n196658 849497\n257020 380298\n267729 284534\n311978 917744",
"output": "13500.015586135814"
},
{
"input": "10 7\n978831 984305\n843967 844227\n454356 748444\n219513 623868\n472997 698189\n542337 813387\n867615 918554\n413076 997267\n79310 138855\n195703 296681",
"output": "5303.027968302269"
}
] | 1,680,587,089 | 2,089 | PyPy 3-64 | OK | TESTS | 94 | 655 | 11,980,800 | n,p=map(int,input().split())
num=[]
for _ in range(n):
l,r=map(int,input().split())
right=r//p
left=(l-1)//p
num.append((right-left,r-l+1))
ans=0
for i,(p,q) in enumerate(num):
p1=(p/q)
i+=1
i%=n
p,q=num[i]
p2=p/q
t=p1+p2-p1*p2
ans+=t*2000
print(ans) | Title: Wet Shark and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* sharks who grow flowers for Wet Shark. They are all sitting around the table, such that sharks *i* and *i*<=+<=1 are neighbours for all *i* from 1 to *n*<=-<=1. Sharks *n* and 1 are neighbours too.
Each shark will grow some number of flowers *s**i*. For *i*-th shark value *s**i* is random integer equiprobably chosen in range from *l**i* to *r**i*. Wet Shark has it's favourite prime number *p*, and he really likes it! If for any pair of neighbouring sharks *i* and *j* the product *s**i*·*s**j* is divisible by *p*, then Wet Shark becomes happy and gives 1000 dollars to each of these sharks.
At the end of the day sharks sum all the money Wet Shark granted to them. Find the expectation of this value.
Input Specification:
The first line of the input contains two space-separated integers *n* and *p* (3<=≤<=*n*<=≤<=100<=000,<=2<=≤<=*p*<=≤<=109) — the number of sharks and Wet Shark's favourite prime number. It is guaranteed that *p* is prime.
The *i*-th of the following *n* lines contains information about *i*-th shark — two space-separated integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109), the range of flowers shark *i* can produce. Remember that *s**i* is chosen equiprobably among all integers from *l**i* to *r**i*, inclusive.
Output Specification:
Print a single real number — the expected number of dollars that the sharks receive in total. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['3 2\n1 2\n420 421\n420420 420421\n', '3 5\n1 4\n2 3\n11 14\n']
Demo Output:
['4500.0\n', '0.0\n']
Note:
A prime number is a positive integer number that is divisible only by 1 and itself. 1 is not considered to be prime.
Consider the first sample. First shark grows some number of flowers from 1 to 2, second sharks grows from 420 to 421 flowers and third from 420420 to 420421. There are eight cases for the quantities of flowers (*s*<sub class="lower-index">0</sub>, *s*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub>) each shark grows:
1. (1, 420, 420420): note that *s*<sub class="lower-index">0</sub>·*s*<sub class="lower-index">1</sub> = 420, *s*<sub class="lower-index">1</sub>·*s*<sub class="lower-index">2</sub> = 176576400, and *s*<sub class="lower-index">2</sub>·*s*<sub class="lower-index">0</sub> = 420420. For each pair, 1000 dollars will be awarded to each shark. Therefore, each shark will be awarded 2000 dollars, for a total of 6000 dollars.1. (1, 420, 420421): now, the product *s*<sub class="lower-index">2</sub>·*s*<sub class="lower-index">0</sub> is not divisible by 2. Therefore, sharks *s*<sub class="lower-index">0</sub> and *s*<sub class="lower-index">2</sub> will receive 1000 dollars, while shark *s*<sub class="lower-index">1</sub> will receive 2000. The total is 4000.1. (1, 421, 420420): total is 4000 1. (1, 421, 420421): total is 0. 1. (2, 420, 420420): total is 6000. 1. (2, 420, 420421): total is 6000. 1. (2, 421, 420420): total is 6000. 1. (2, 421, 420421): total is 4000.
The expected value is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dfe520d00a8615f7c270ccbccbebe182cc7db883.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample, no combination of quantities will garner the sharks any money. | ```python
n,p=map(int,input().split())
num=[]
for _ in range(n):
l,r=map(int,input().split())
right=r//p
left=(l-1)//p
num.append((right-left,r-l+1))
ans=0
for i,(p,q) in enumerate(num):
p1=(p/q)
i+=1
i%=n
p,q=num[i]
p2=p/q
t=p1+p2-p1*p2
ans+=t*2000
print(ans)
``` | 3 | |
4 | B | Before an Exam | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy"
] | B. Before an Exam | 0 | 64 | Tomorrow Peter has a Biology exam. He does not like this subject much, but *d* days ago he learnt that he would have to take this exam. Peter's strict parents made him prepare for the exam immediately, for this purpose he has to study not less than *minTime**i* and not more than *maxTime**i* hours per each *i*-th day. Moreover, they warned Peter that a day before the exam they would check how he has followed their instructions.
So, today is the day when Peter's parents ask him to show the timetable of his preparatory studies. But the boy has counted only the sum of hours *sumTime* spent him on preparation, and now he wants to know if he can show his parents a timetable *sсhedule* with *d* numbers, where each number *sсhedule**i* stands for the time in hours spent by Peter each *i*-th day on biology studies, and satisfying the limitations imposed by his parents, and at the same time the sum total of all *schedule**i* should equal to *sumTime*. | The first input line contains two integer numbers *d*,<=*sumTime* (1<=≤<=*d*<=≤<=30,<=0<=≤<=*sumTime*<=≤<=240) — the amount of days, during which Peter studied, and the total amount of hours, spent on preparation. Each of the following *d* lines contains two integer numbers *minTime**i*,<=*maxTime**i* (0<=≤<=*minTime**i*<=≤<=*maxTime**i*<=≤<=8), separated by a space — minimum and maximum amount of hours that Peter could spent in the *i*-th day. | In the first line print YES, and in the second line print *d* numbers (separated by a space), each of the numbers — amount of hours, spent by Peter on preparation in the corresponding day, if he followed his parents' instructions; or print NO in the unique line. If there are many solutions, print any of them. | [
"1 48\n5 7\n",
"2 5\n0 1\n3 5\n"
] | [
"NO\n",
"YES\n1 4 "
] | none | 0 | [
{
"input": "1 48\n5 7",
"output": "NO"
},
{
"input": "2 5\n0 1\n3 5",
"output": "YES\n1 4 "
},
{
"input": "1 1\n5 6",
"output": "NO"
},
{
"input": "1 4\n2 4",
"output": "YES\n4 "
},
{
"input": "2 5\n4 6\n0 0",
"output": "YES\n5 0 "
},
{
"input": "27 97\n2 8\n0 5\n5 6\n3 6\n5 5\n1 2\n3 5\n1 8\n0 4\n3 3\n0 2\n0 0\n4 8\n5 6\n5 8\n0 7\n1 4\n0 4\n1 5\n3 7\n2 5\n5 6\n4 7\n3 8\n0 1\n3 4\n5 7",
"output": "YES\n8 5 6 6 5 2 5 8 4 3 2 0 6 5 5 0 1 0 1 3 2 5 4 3 0 3 5 "
},
{
"input": "30 92\n4 5\n4 7\n2 6\n8 8\n7 8\n4 5\n1 5\n7 8\n1 2\n6 8\n2 7\n2 4\n0 0\n1 3\n4 5\n1 1\n0 7\n2 5\n2 5\n3 3\n1 2\n1 7\n5 5\n5 8\n6 7\n0 3\n2 6\n0 7\n5 6\n2 5",
"output": "YES\n5 7 2 8 7 4 1 7 1 6 2 2 0 1 4 1 0 2 2 3 1 1 5 5 6 0 2 0 5 2 "
},
{
"input": "30 178\n1 6\n2 7\n2 5\n2 8\n1 6\n2 8\n3 4\n2 7\n0 2\n0 8\n0 3\n0 2\n2 4\n4 8\n6 8\n0 8\n0 6\n1 8\n0 3\n6 7\n4 8\n2 7\n1 1\n3 7\n3 6\n2 5\n4 7\n2 2\n1 8\n5 6",
"output": "NO"
},
{
"input": "30 71\n1 3\n0 6\n3 5\n3 6\n2 4\n2 8\n2 4\n3 8\n3 5\n2 4\n2 3\n3 7\n0 0\n5 7\n0 2\n5 8\n0 8\n4 7\n0 3\n3 7\n2 3\n4 5\n7 8\n7 7\n2 7\n1 3\n0 1\n1 5\n6 7\n5 8",
"output": "NO"
},
{
"input": "30 119\n2 7\n1 3\n0 3\n3 4\n7 7\n7 7\n0 5\n2 3\n0 8\n0 8\n0 5\n5 7\n2 2\n2 6\n2 5\n3 7\n0 8\n0 2\n1 3\n2 3\n1 4\n0 1\n3 7\n7 8\n1 2\n0 6\n1 8\n1 7\n4 8\n1 4",
"output": "YES\n7 3 3 4 7 7 5 3 8 8 5 7 2 6 5 7 8 2 1 2 1 0 3 7 1 0 1 1 4 1 "
},
{
"input": "1 0\n0 0",
"output": "YES\n0 "
},
{
"input": "30 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",
"output": "YES\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 "
},
{
"input": "30 240\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8",
"output": "YES\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 "
},
{
"input": "30 64\n7 7\n3 4\n0 4\n5 8\n1 3\n1 1\n0 2\n0 4\n7 7\n3 5\n4 4\n0 4\n4 5\n1 4\n4 6\n5 7\n1 4\n1 2\n3 3\n0 2\n3 7\n3 4\n0 3\n3 8\n0 8\n0 1\n0 3\n4 5\n1 5\n0 3",
"output": "YES\n7 3 0 5 1 1 0 0 7 3 4 0 4 1 4 5 1 1 3 0 3 3 0 3 0 0 0 4 1 0 "
},
{
"input": "30 189\n0 8\n3 8\n5 8\n2 2\n7 7\n2 8\n7 8\n2 5\n1 7\n0 1\n3 4\n1 7\n1 6\n5 6\n3 7\n4 7\n4 7\n4 7\n5 5\n0 5\n4 7\n3 5\n0 6\n4 8\n2 5\n5 6\n5 6\n2 8\n0 7\n7 8",
"output": "YES\n8 8 8 2 7 8 8 5 7 1 4 7 6 6 7 7 7 7 5 5 7 5 6 8 5 6 6 8 7 8 "
}
] | 1,696,377,069 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 15 | 77 | 0 | d, time = map(int, input().split())
minTime = []
maxTime = []
for i in range(d):
mn, mx = map(int, input().split())
minTime.append(mn)
maxTime.append(mx)
minSum = sum(minTime)
maxSum = sum(maxTime)
if minSum <= time <= maxSum:
print("YES")
for i in range(d):
temp = min(minTime[i] + time - minSum, maxTime[i])
print(temp, end = " " if i < d - 1 else "\n")
time -= (temp - minTime[i])
else:
print("NO") | Title: Before an Exam
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
Tomorrow Peter has a Biology exam. He does not like this subject much, but *d* days ago he learnt that he would have to take this exam. Peter's strict parents made him prepare for the exam immediately, for this purpose he has to study not less than *minTime**i* and not more than *maxTime**i* hours per each *i*-th day. Moreover, they warned Peter that a day before the exam they would check how he has followed their instructions.
So, today is the day when Peter's parents ask him to show the timetable of his preparatory studies. But the boy has counted only the sum of hours *sumTime* spent him on preparation, and now he wants to know if he can show his parents a timetable *sсhedule* with *d* numbers, where each number *sсhedule**i* stands for the time in hours spent by Peter each *i*-th day on biology studies, and satisfying the limitations imposed by his parents, and at the same time the sum total of all *schedule**i* should equal to *sumTime*.
Input Specification:
The first input line contains two integer numbers *d*,<=*sumTime* (1<=≤<=*d*<=≤<=30,<=0<=≤<=*sumTime*<=≤<=240) — the amount of days, during which Peter studied, and the total amount of hours, spent on preparation. Each of the following *d* lines contains two integer numbers *minTime**i*,<=*maxTime**i* (0<=≤<=*minTime**i*<=≤<=*maxTime**i*<=≤<=8), separated by a space — minimum and maximum amount of hours that Peter could spent in the *i*-th day.
Output Specification:
In the first line print YES, and in the second line print *d* numbers (separated by a space), each of the numbers — amount of hours, spent by Peter on preparation in the corresponding day, if he followed his parents' instructions; or print NO in the unique line. If there are many solutions, print any of them.
Demo Input:
['1 48\n5 7\n', '2 5\n0 1\n3 5\n']
Demo Output:
['NO\n', 'YES\n1 4 ']
Note:
none | ```python
d, time = map(int, input().split())
minTime = []
maxTime = []
for i in range(d):
mn, mx = map(int, input().split())
minTime.append(mn)
maxTime.append(mx)
minSum = sum(minTime)
maxSum = sum(maxTime)
if minSum <= time <= maxSum:
print("YES")
for i in range(d):
temp = min(minTime[i] + time - minSum, maxTime[i])
print(temp, end = " " if i < d - 1 else "\n")
time -= (temp - minTime[i])
else:
print("NO")
``` | 3 |
109 | A | Lucky Sum of Digits | PROGRAMMING | 1,000 | [
"brute force",
"implementation"
] | A. Lucky Sum of Digits | 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.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task. | The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number. | Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1. | [
"11\n",
"10\n"
] | [
"47\n",
"-1\n"
] | none | 500 | [
{
"input": "11",
"output": "47"
},
{
"input": "10",
"output": "-1"
},
{
"input": "64",
"output": "4477777777"
},
{
"input": "1",
"output": "-1"
},
{
"input": "4",
"output": "4"
},
{
"input": "7",
"output": "7"
},
{
"input": "12",
"output": "444"
},
{
"input": "1000000",
"output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "47",
"output": "44477777"
},
{
"input": "100",
"output": "4444777777777777"
},
{
"input": "700",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "485",
"output": "44447777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "111",
"output": "444447777777777777"
},
{
"input": "85",
"output": "4477777777777"
},
{
"input": "114",
"output": "444477777777777777"
},
{
"input": "474",
"output": "444777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "74",
"output": "47777777777"
},
{
"input": "1000",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "1024",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "4444",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "45784",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10000",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "9854",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
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"output": "477777777777777777777777777"
},
{
"input": "10416",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10417",
"output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "3840",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "100000",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "9876",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "99999",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "777777",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
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},
{
"input": "11000",
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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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},
{
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},
{
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},
{
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},
{
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},
{
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"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999980",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "800000",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "980000",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
}
] | 1,540,615,420 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 216 | 0 | n = int(input())
for x in range(n//7,0,-1):
if (n-x*7)%4 == 0:
print ('4'*((n-x*7)//4 )+ '7'*x)
break
else:
print (-1) | Title: Lucky Sum of Digits
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.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.
Output Specification:
Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.
Demo Input:
['11\n', '10\n']
Demo Output:
['47\n', '-1\n']
Note:
none | ```python
n = int(input())
for x in range(n//7,0,-1):
if (n-x*7)%4 == 0:
print ('4'*((n-x*7)//4 )+ '7'*x)
break
else:
print (-1)
``` | 0 |
893 | A | Chess For Three | PROGRAMMING | 900 | [
"implementation"
] | null | null | Alex, Bob and Carl will soon participate in a team chess tournament. Since they are all in the same team, they have decided to practise really hard before the tournament. But it's a bit difficult for them because chess is a game for two players, not three.
So they play with each other according to following rules:
- Alex and Bob play the first game, and Carl is spectating; - When the game ends, the one who lost the game becomes the spectator in the next game, and the one who was spectating plays against the winner.
Alex, Bob and Carl play in such a way that there are no draws.
Today they have played *n* games, and for each of these games they remember who was the winner. They decided to make up a log of games describing who won each game. But now they doubt if the information in the log is correct, and they want to know if the situation described in the log they made up was possible (that is, no game is won by someone who is spectating if Alex, Bob and Carl play according to the rules). Help them to check it! | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of games Alex, Bob and Carl played.
Then *n* lines follow, describing the game log. *i*-th line contains one integer *a**i* (1<=≤<=*a**i*<=≤<=3) which is equal to 1 if Alex won *i*-th game, to 2 if Bob won *i*-th game and 3 if Carl won *i*-th game. | Print YES if the situation described in the log was possible. Otherwise print NO. | [
"3\n1\n1\n2\n",
"2\n1\n2\n"
] | [
"YES\n",
"NO\n"
] | In the first example the possible situation is:
1. Alex wins, Carl starts playing instead of Bob; 1. Alex wins, Bob replaces Carl; 1. Bob wins.
The situation in the second example is impossible because Bob loses the first game, so he cannot win the second one. | 0 | [
{
"input": "3\n1\n1\n2",
"output": "YES"
},
{
"input": "2\n1\n2",
"output": "NO"
},
{
"input": "100\n2\n3\n1\n2\n3\n3\n3\n1\n1\n1\n1\n3\n3\n3\n3\n1\n2\n3\n3\n3\n3\n3\n3\n3\n1\n2\n2\n2\n3\n1\n1\n3\n3\n3\n3\n3\n3\n3\n3\n1\n2\n3\n3\n3\n1\n1\n1\n1\n3\n3\n3\n3\n1\n2\n3\n1\n2\n2\n2\n3\n3\n2\n1\n3\n3\n1\n2\n3\n1\n1\n1\n2\n2\n2\n3\n1\n1\n1\n1\n1\n1\n3\n2\n2\n2\n2\n2\n2\n3\n1\n2\n2\n2\n2\n2\n3\n3\n2\n1\n1",
"output": "YES"
},
{
"input": "99\n1\n3\n2\n2\n3\n1\n1\n3\n3\n3\n3\n3\n3\n1\n1\n3\n3\n3\n3\n1\n1\n3\n2\n1\n1\n1\n1\n1\n1\n1\n3\n2\n2\n2\n1\n3\n3\n1\n1\n3\n2\n1\n3\n3\n1\n2\n3\n3\n3\n1\n2\n2\n2\n3\n3\n3\n3\n3\n3\n2\n2\n2\n2\n3\n3\n3\n1\n1\n3\n2\n1\n1\n2\n2\n2\n3\n3\n2\n1\n1\n2\n2\n1\n3\n2\n1\n1\n2\n3\n3\n3\n3\n2\n2\n2\n2\n2\n1\n3",
"output": "YES"
},
{
"input": "100\n2\n2\n1\n3\n1\n3\n3\n1\n1\n3\n1\n1\n3\n2\n1\n3\n1\n1\n3\n3\n2\n2\n3\n1\n1\n2\n3\n2\n2\n3\n1\n1\n2\n3\n2\n1\n2\n2\n3\n3\n1\n1\n3\n1\n2\n1\n3\n1\n1\n3\n2\n2\n2\n1\n1\n1\n3\n1\n3\n2\n1\n2\n2\n2\n3\n3\n2\n1\n1\n3\n3\n2\n1\n2\n1\n1\n3\n1\n2\n3\n2\n3\n3\n3\n2\n2\n1\n3\n1\n2\n3\n1\n2\n3\n3\n1\n2\n1\n3\n1",
"output": "NO"
},
{
"input": "10\n2\n3\n3\n3\n3\n2\n2\n2\n3\n2",
"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": "YES"
},
{
"input": "1\n3",
"output": "NO"
},
{
"input": "1\n2",
"output": "YES"
},
{
"input": "42\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": "YES"
},
{
"input": "4\n2\n3\n3\n3",
"output": "YES"
},
{
"input": "3\n1\n2\n3",
"output": "NO"
},
{
"input": "5\n1\n1\n1\n1\n3",
"output": "NO"
},
{
"input": "5\n2\n3\n3\n3\n3",
"output": "YES"
},
{
"input": "2\n3\n3",
"output": "NO"
},
{
"input": "4\n1\n2\n2\n1",
"output": "NO"
},
{
"input": "3\n2\n2\n3",
"output": "NO"
},
{
"input": "5\n2\n3\n3\n1\n1",
"output": "NO"
},
{
"input": "3\n3\n1\n3",
"output": "NO"
},
{
"input": "3\n3\n3\n1",
"output": "NO"
},
{
"input": "2\n2\n1",
"output": "NO"
},
{
"input": "3\n1\n1\n3",
"output": "NO"
},
{
"input": "6\n2\n2\n2\n3\n1\n3",
"output": "NO"
},
{
"input": "2\n3\n1",
"output": "NO"
},
{
"input": "2\n3\n2",
"output": "NO"
},
{
"input": "2\n1\n3",
"output": "YES"
},
{
"input": "3\n1\n3\n1",
"output": "NO"
},
{
"input": "5\n1\n1\n2\n2\n3",
"output": "NO"
},
{
"input": "3\n2\n1\n1",
"output": "NO"
},
{
"input": "2\n2\n2",
"output": "YES"
},
{
"input": "3\n2\n2\n1",
"output": "YES"
},
{
"input": "5\n2\n2\n2\n2\n2",
"output": "YES"
},
{
"input": "8\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "YES"
},
{
"input": "3\n3\n2\n2",
"output": "NO"
},
{
"input": "3\n3\n2\n3",
"output": "NO"
},
{
"input": "7\n2\n2\n2\n2\n2\n2\n2",
"output": "YES"
},
{
"input": "3\n2\n2\n2",
"output": "YES"
}
] | 1,673,407,867 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 46 | 0 | import sys
L=[]
b=int(input())
for i in range(b):
c=int(input())
L.append(c)
A=1
B=1
C=0
if L[0]==1:
A=1
B=0
C=1
else:
A=0
B=1
C=1
for i in range(1,b):
if L[i]==1 and A==0:
print("NO")
sys.exit()
elif L[i]==2 and B==0:
print("NO")
sys.exit()
elif L[i]==3 and C==0:
print("NO")
sys.exit()
if L[i]==1 and B==0:
B=1
C=0
elif L[i]==1 and C==0:
C=1
B=0
if L[i]==2 and A==0:
A=1
C=0
elif L[i]==2 and C==0:
C=1
A=0
if L[i]==3 and A==0:
A=1
B=0
elif L[i]==3 and B==0:
B=1
A=0
print("YES")
| Title: Chess For Three
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alex, Bob and Carl will soon participate in a team chess tournament. Since they are all in the same team, they have decided to practise really hard before the tournament. But it's a bit difficult for them because chess is a game for two players, not three.
So they play with each other according to following rules:
- Alex and Bob play the first game, and Carl is spectating; - When the game ends, the one who lost the game becomes the spectator in the next game, and the one who was spectating plays against the winner.
Alex, Bob and Carl play in such a way that there are no draws.
Today they have played *n* games, and for each of these games they remember who was the winner. They decided to make up a log of games describing who won each game. But now they doubt if the information in the log is correct, and they want to know if the situation described in the log they made up was possible (that is, no game is won by someone who is spectating if Alex, Bob and Carl play according to the rules). Help them to check it!
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of games Alex, Bob and Carl played.
Then *n* lines follow, describing the game log. *i*-th line contains one integer *a**i* (1<=≤<=*a**i*<=≤<=3) which is equal to 1 if Alex won *i*-th game, to 2 if Bob won *i*-th game and 3 if Carl won *i*-th game.
Output Specification:
Print YES if the situation described in the log was possible. Otherwise print NO.
Demo Input:
['3\n1\n1\n2\n', '2\n1\n2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example the possible situation is:
1. Alex wins, Carl starts playing instead of Bob; 1. Alex wins, Bob replaces Carl; 1. Bob wins.
The situation in the second example is impossible because Bob loses the first game, so he cannot win the second one. | ```python
import sys
L=[]
b=int(input())
for i in range(b):
c=int(input())
L.append(c)
A=1
B=1
C=0
if L[0]==1:
A=1
B=0
C=1
else:
A=0
B=1
C=1
for i in range(1,b):
if L[i]==1 and A==0:
print("NO")
sys.exit()
elif L[i]==2 and B==0:
print("NO")
sys.exit()
elif L[i]==3 and C==0:
print("NO")
sys.exit()
if L[i]==1 and B==0:
B=1
C=0
elif L[i]==1 and C==0:
C=1
B=0
if L[i]==2 and A==0:
A=1
C=0
elif L[i]==2 and C==0:
C=1
A=0
if L[i]==3 and A==0:
A=1
B=0
elif L[i]==3 and B==0:
B=1
A=0
print("YES")
``` | 0 | |
588 | B | Duff in Love | PROGRAMMING | 1,300 | [
"math"
] | null | null | Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*.
Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.
Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store. | The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012). | Print the answer in one line. | [
"10\n",
"12\n"
] | [
"10\n",
"6\n"
] | In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.
In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 2<sup class="upper-index">2</sup>, so 12 is not lovely, while 6 is indeed lovely. | 1,000 | [
{
"input": "10",
"output": "10"
},
{
"input": "12",
"output": "6"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "2"
},
{
"input": "4",
"output": "2"
},
{
"input": "8",
"output": "2"
},
{
"input": "3",
"output": "3"
},
{
"input": "31",
"output": "31"
},
{
"input": "97",
"output": "97"
},
{
"input": "1000000000000",
"output": "10"
},
{
"input": "15",
"output": "15"
},
{
"input": "894",
"output": "894"
},
{
"input": "271",
"output": "271"
},
{
"input": "2457",
"output": "273"
},
{
"input": "2829",
"output": "2829"
},
{
"input": "5000",
"output": "10"
},
{
"input": "20",
"output": "10"
},
{
"input": "68",
"output": "34"
},
{
"input": "3096",
"output": "258"
},
{
"input": "1024",
"output": "2"
},
{
"input": "1048576",
"output": "2"
},
{
"input": "413933789280",
"output": "25870861830"
},
{
"input": "817634153013",
"output": "817634153013"
},
{
"input": "56517269141",
"output": "56517269141"
},
{
"input": "30707328551",
"output": "30707328551"
},
{
"input": "279564127218",
"output": "10354226934"
},
{
"input": "491159577042",
"output": "18191095446"
},
{
"input": "734337660466",
"output": "734337660466"
},
{
"input": "808453785117",
"output": "808453785117"
},
{
"input": "55926835837",
"output": "55926835837"
},
{
"input": "294809951965",
"output": "294809951965"
},
{
"input": "537988035389",
"output": "76855433627"
},
{
"input": "822722434952",
"output": "205680608738"
},
{
"input": "699511759613",
"output": "699511759613"
},
{
"input": "942689843037",
"output": "104743315893"
},
{
"input": "663634158717",
"output": "663634158717"
},
{
"input": "213612977250",
"output": "11730"
},
{
"input": "999999999989",
"output": "999999999989"
},
{
"input": "999999999988",
"output": "499999999994"
},
{
"input": "87178291200",
"output": "30030"
},
{
"input": "927668721948",
"output": "463834360974"
},
{
"input": "562436815639",
"output": "37927"
},
{
"input": "302981118597",
"output": "35853"
},
{
"input": "5",
"output": "5"
},
{
"input": "9",
"output": "3"
},
{
"input": "36",
"output": "6"
},
{
"input": "2231",
"output": "2231"
},
{
"input": "27648",
"output": "6"
},
{
"input": "40320",
"output": "210"
},
{
"input": "648000",
"output": "30"
},
{
"input": "999966000289",
"output": "999983"
},
{
"input": "999985999949",
"output": "999985999949"
},
{
"input": "991921850317",
"output": "9973"
}
] | 1,667,364,818 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 514 | 0 | n = int(input())
for i in range(2,10**6):
while(n%i**2 == 0):
n //= i
print(n) | Title: Duff in Love
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*.
Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.
Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.
Input Specification:
The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012).
Output Specification:
Print the answer in one line.
Demo Input:
['10\n', '12\n']
Demo Output:
['10\n', '6\n']
Note:
In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.
In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 2<sup class="upper-index">2</sup>, so 12 is not lovely, while 6 is indeed lovely. | ```python
n = int(input())
for i in range(2,10**6):
while(n%i**2 == 0):
n //= i
print(n)
``` | 3 | |
572 | A | Arrays | PROGRAMMING | 900 | [
"sortings"
] | null | null | You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array. | The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly.
The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space.
The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*.
The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*. | Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes). | [
"3 3\n2 1\n1 2 3\n3 4 5\n",
"3 3\n3 3\n1 2 3\n3 4 5\n",
"5 2\n3 1\n1 1 1 1 1\n2 2\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3).
In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 500 | [
{
"input": "3 3\n2 1\n1 2 3\n3 4 5",
"output": "YES"
},
{
"input": "3 3\n3 3\n1 2 3\n3 4 5",
"output": "NO"
},
{
"input": "5 2\n3 1\n1 1 1 1 1\n2 2",
"output": "YES"
},
{
"input": "3 5\n1 1\n5 5 5\n5 5 5 5 5",
"output": "NO"
},
{
"input": "1 1\n1 1\n1\n1",
"output": "NO"
},
{
"input": "3 3\n1 1\n1 2 3\n1 2 3",
"output": "YES"
},
{
"input": "3 3\n1 2\n1 2 3\n1 2 3",
"output": "YES"
},
{
"input": "3 3\n2 2\n1 2 3\n1 2 3",
"output": "NO"
},
{
"input": "10 15\n10 1\n1 1 5 17 22 29 32 36 39 48\n9 10 20 23 26 26 32 32 33 39 43 45 47 49 49",
"output": "YES"
},
{
"input": "10 15\n1 15\n91 91 91 92 92 94 94 95 98 100\n92 92 93 93 93 94 95 96 97 98 98 99 99 100 100",
"output": "YES"
},
{
"input": "15 10\n12 5\n9 25 25 32 32 38 40 41 46 46 48 51 64 64 73\n5 14 30 35 50 52 67 79 89 99",
"output": "YES"
},
{
"input": "15 10\n4 10\n22 32 35 45 45 50 51 55 79 80 83 88 90 92 93\n46 48 52 55 60 60 68 75 80 81",
"output": "YES"
},
{
"input": "20 30\n2 8\n6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 10\n1 1 2 2 2 2 2 2 2 3 3 4 5 5 5 5 6 6 6 6 6 6 7 7 7 8 8 9 10 10",
"output": "NO"
},
{
"input": "20 30\n19 29\n1 1 2 2 2 3 4 4 7 7 7 8 8 8 8 8 9 9 9 9\n6 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "1 1\n1 1\n-1000000000\n30",
"output": "YES"
},
{
"input": "3 3\n1 3\n1 3 3\n3 3 3",
"output": "YES"
},
{
"input": "3 3\n1 1\n3 3 3\n2 2 2",
"output": "NO"
},
{
"input": "5 5\n3 3\n1 5 6 7 8\n1 2 5 6 7",
"output": "NO"
},
{
"input": "3 4\n2 2\n5 6 7\n1 2 3 4",
"output": "NO"
},
{
"input": "3 3\n3 3\n1 2 3\n4 5 6",
"output": "YES"
},
{
"input": "5 5\n4 5\n2 2 3 4 5\n5 6 7 8 9",
"output": "YES"
}
] | 1,621,020,225 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 77 | 0 | na, nb = map(int, input().split())
k, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
m_min = min(b)
m_max = max(b)
count = 0
if k == na and m == nb :
for i in range(k):
if a[i] < m_min:
count += 1
if count >= k:
print("YES")
else:
print("NO")
else:
for i in range(k):
if a[i] < m_max:
count += 1
if count >= k:
print("YES")
else:
print("NO")
| Title: Arrays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array.
Input Specification:
The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly.
The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space.
The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*.
The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*.
Output Specification:
Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes).
Demo Input:
['3 3\n2 1\n1 2 3\n3 4 5\n', '3 3\n3 3\n1 2 3\n3 4 5\n', '5 2\n3 1\n1 1 1 1 1\n2 2\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3).
In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
na, nb = map(int, input().split())
k, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
m_min = min(b)
m_max = max(b)
count = 0
if k == na and m == nb :
for i in range(k):
if a[i] < m_min:
count += 1
if count >= k:
print("YES")
else:
print("NO")
else:
for i in range(k):
if a[i] < m_max:
count += 1
if count >= k:
print("YES")
else:
print("NO")
``` | 0 | |
479 | A | Expression | PROGRAMMING | 1,000 | [
"brute force",
"math"
] | null | null | Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get. | The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10). | Print the maximum value of the expression that you can obtain. | [
"1\n2\n3\n",
"2\n10\n3\n"
] | [
"9\n",
"60\n"
] | none | 500 | [
{
"input": "1\n2\n3",
"output": "9"
},
{
"input": "2\n10\n3",
"output": "60"
},
{
"input": "1\n1\n1",
"output": "3"
},
{
"input": "1\n2\n1",
"output": "4"
},
{
"input": "10\n10\n10",
"output": "1000"
},
{
"input": "5\n1\n3",
"output": "20"
},
{
"input": "3\n1\n5",
"output": "20"
},
{
"input": "6\n7\n1",
"output": "48"
},
{
"input": "1\n8\n3",
"output": "27"
},
{
"input": "9\n7\n2",
"output": "126"
},
{
"input": "1\n1\n10",
"output": "20"
},
{
"input": "9\n1\n1",
"output": "18"
},
{
"input": "10\n5\n6",
"output": "300"
},
{
"input": "8\n9\n7",
"output": "504"
},
{
"input": "4\n2\n10",
"output": "80"
},
{
"input": "2\n8\n3",
"output": "48"
},
{
"input": "3\n5\n7",
"output": "105"
},
{
"input": "1\n10\n1",
"output": "12"
},
{
"input": "2\n2\n2",
"output": "8"
},
{
"input": "5\n6\n1",
"output": "35"
},
{
"input": "10\n1\n1",
"output": "20"
},
{
"input": "1\n6\n1",
"output": "8"
},
{
"input": "1\n9\n1",
"output": "11"
},
{
"input": "2\n1\n2",
"output": "6"
},
{
"input": "2\n6\n1",
"output": "14"
},
{
"input": "9\n2\n1",
"output": "27"
},
{
"input": "1\n9\n2",
"output": "20"
},
{
"input": "1\n3\n1",
"output": "5"
},
{
"input": "2\n1\n1",
"output": "4"
}
] | 1,694,379,178 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 31 | 0 | max=0
min=0
mid=0
count=0
number_index=0
mid_index=0
max_index=0
min_index=0
for i in range(3):
number=int(input())
number_index=i+1
if(number==1):
count+=1
if(i==0):
mid=number
mid_index=number_index
max=number
max_index=number_index
min=number
min_index=number_index
continue
if (number>max):
mid=max
mid_index=max_index
max=number
max_index=number_index
elif(number>min):
mid=number
mid_index=number_index
if(number<min):
min=number
min_index=number_index
if(count):
if (count==3):
print(3)
elif(count==2):
if(abs(min_index-mid_index)==1):
print((mid+min)*max)
else:
print((mid+max)*min)
else:
if(abs(mid_index-min_index)==1):
print((mid+min)*max)
else:
print((max+min)*mid)
else:
print(mid*min*max) | Title: Expression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
Input Specification:
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
Output Specification:
Print the maximum value of the expression that you can obtain.
Demo Input:
['1\n2\n3\n', '2\n10\n3\n']
Demo Output:
['9\n', '60\n']
Note:
none | ```python
max=0
min=0
mid=0
count=0
number_index=0
mid_index=0
max_index=0
min_index=0
for i in range(3):
number=int(input())
number_index=i+1
if(number==1):
count+=1
if(i==0):
mid=number
mid_index=number_index
max=number
max_index=number_index
min=number
min_index=number_index
continue
if (number>max):
mid=max
mid_index=max_index
max=number
max_index=number_index
elif(number>min):
mid=number
mid_index=number_index
if(number<min):
min=number
min_index=number_index
if(count):
if (count==3):
print(3)
elif(count==2):
if(abs(min_index-mid_index)==1):
print((mid+min)*max)
else:
print((mid+max)*min)
else:
if(abs(mid_index-min_index)==1):
print((mid+min)*max)
else:
print((max+min)*mid)
else:
print(mid*min*max)
``` | 0 | |
678 | A | Johny Likes Numbers | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*. | The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109). | Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*. | [
"5 3\n",
"25 13\n",
"26 13\n"
] | [
"6\n",
"26\n",
"39\n"
] | none | 0 | [
{
"input": "5 3",
"output": "6"
},
{
"input": "25 13",
"output": "26"
},
{
"input": "26 13",
"output": "39"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "8 8",
"output": "16"
},
{
"input": "14 15",
"output": "15"
},
{
"input": "197 894",
"output": "894"
},
{
"input": "6058 8581",
"output": "8581"
},
{
"input": "97259 41764",
"output": "125292"
},
{
"input": "453145 333625",
"output": "667250"
},
{
"input": "2233224 4394826",
"output": "4394826"
},
{
"input": "76770926 13350712",
"output": "80104272"
},
{
"input": "687355301 142098087",
"output": "710490435"
},
{
"input": "1000000000 999999999",
"output": "1999999998"
},
{
"input": "1000000000 1000000000",
"output": "2000000000"
},
{
"input": "999999999 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 1",
"output": "1000000001"
},
{
"input": "1000000000 2",
"output": "1000000002"
},
{
"input": "999999999 1",
"output": "1000000000"
},
{
"input": "100000000 1",
"output": "100000001"
},
{
"input": "999999999 500000000",
"output": "1000000000"
},
{
"input": "999999990 10",
"output": "1000000000"
},
{
"input": "1000000000 999999997",
"output": "1999999994"
},
{
"input": "999999999 2",
"output": "1000000000"
},
{
"input": "999999984 1",
"output": "999999985"
},
{
"input": "999999983 1",
"output": "999999984"
},
{
"input": "666666666 1",
"output": "666666667"
},
{
"input": "1000000000 990000000",
"output": "1980000000"
},
{
"input": "41 48",
"output": "48"
},
{
"input": "123456 2",
"output": "123458"
},
{
"input": "111 111",
"output": "222"
},
{
"input": "878787 1",
"output": "878788"
},
{
"input": "121 1",
"output": "122"
},
{
"input": "114514 114514",
"output": "229028"
},
{
"input": "500000001 1000000000",
"output": "1000000000"
},
{
"input": "999999997 1",
"output": "999999998"
},
{
"input": "100000000 10",
"output": "100000010"
}
] | 1,681,317,133 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 77 | 0 | import math
n,k=input().split()
n=int(n)
k=int(k)
print(n+k) | Title: Johny Likes Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
Output Specification:
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
Demo Input:
['5 3\n', '25 13\n', '26 13\n']
Demo Output:
['6\n', '26\n', '39\n']
Note:
none | ```python
import math
n,k=input().split()
n=int(n)
k=int(k)
print(n+k)
``` | 0 | |
339 | A | Helpful Maths | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings",
"strings"
] | null | null | Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum. | The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long. | Print the new sum that Xenia can count. | [
"3+2+1\n",
"1+1+3+1+3\n",
"2\n"
] | [
"1+2+3\n",
"1+1+1+3+3\n",
"2\n"
] | none | 500 | [
{
"input": "3+2+1",
"output": "1+2+3"
},
{
"input": "1+1+3+1+3",
"output": "1+1+1+3+3"
},
{
"input": "2",
"output": "2"
},
{
"input": "2+2+1+1+3",
"output": "1+1+2+2+3"
},
{
"input": "2+1+2+2+2+3+1+3+1+2",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "1+2+1+2+2+2+2+1+3+3",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "2+3+3+1+2+2+2+1+1+2+1+3+2+2+3+3+2+2+3+3+3+1+1+1+3+3+3+2+1+3+2+3+2+1+1+3+3+3+1+2+2+1+2+2+1+2+1+3+1+1",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "1",
"output": "1"
},
{
"input": "2+1+2+2+1+3+2+3+1+1+2+1+2+2+3+1+1+3+3+3+2+2+3+2+2+2+1+2+1+2+3+2+2+2+1+3+1+3+3+3+1+2+1+2+2+2+2+3+1+1",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "2+2+1+1+1+3+1+1+3+3+2+3+1+3+1+1+3+1+1+2+2+2+2+1+2+1+2+1+1+1+3+1+3+2+3+2+3+3+1+1+1+2+3+2+1+3+1+3+2+2",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3+2+3+3+2+2+1+2+1+2+3+1+2+3+2+3+2+1+2+2+1+1+2+2+3+2+1+3+1+1+3+2+2+2+2+3+3+2+2+3+3+1+1+2+3+3+2+3+3+3",
"output": "1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3",
"output": "3"
},
{
"input": "1+1",
"output": "1+1"
},
{
"input": "1+2",
"output": "1+2"
},
{
"input": "1+3",
"output": "1+3"
},
{
"input": "2+1",
"output": "1+2"
},
{
"input": "2+2",
"output": "2+2"
},
{
"input": "2+3",
"output": "2+3"
},
{
"input": "3+1",
"output": "1+3"
},
{
"input": "3+2",
"output": "2+3"
},
{
"input": "3+3",
"output": "3+3"
}
] | 1,698,065,860 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 92 | 0 | s = input()
numbers = s.split("+")
numbers.sort()
numbers="+".join(numbers)
print(numbers)
| Title: Helpful Maths
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
Input Specification:
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
Output Specification:
Print the new sum that Xenia can count.
Demo Input:
['3+2+1\n', '1+1+3+1+3\n', '2\n']
Demo Output:
['1+2+3\n', '1+1+1+3+3\n', '2\n']
Note:
none | ```python
s = input()
numbers = s.split("+")
numbers.sort()
numbers="+".join(numbers)
print(numbers)
``` | 3 | |
417 | C | Football | PROGRAMMING | 1,400 | [
"constructive algorithms",
"graphs",
"implementation"
] | null | null | One day, at the "Russian Code Cup" event it was decided to play football as an out of competition event. All participants was divided into *n* teams and played several matches, two teams could not play against each other more than once.
The appointed Judge was the most experienced member — Pavel. But since he was the wisest of all, he soon got bored of the game and fell asleep. Waking up, he discovered that the tournament is over and the teams want to know the results of all the matches.
Pavel didn't want anyone to discover about him sleeping and not keeping an eye on the results, so he decided to recover the results of all games. To do this, he asked all the teams and learned that the real winner was friendship, that is, each team beat the other teams exactly *k* times. Help Pavel come up with chronology of the tournir that meets all the conditions, or otherwise report that there is no such table. | The first line contains two integers — *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1000). | In the first line print an integer *m* — number of the played games. The following *m* lines should contain the information about all the matches, one match per line. The *i*-th line should contain two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*). The numbers *a**i* and *b**i* mean, that in the *i*-th match the team with number *a**i* won against the team with number *b**i*. You can assume, that the teams are numbered from 1 to *n*.
If a tournir that meets the conditions of the problem does not exist, then print -1. | [
"3 1\n"
] | [
"3\n1 2\n2 3\n3 1\n"
] | none | 1,500 | [
{
"input": "3 1",
"output": "3\n1 2\n2 3\n3 1"
},
{
"input": "7 3",
"output": "21\n1 2\n1 3\n1 4\n2 3\n2 4\n2 5\n3 4\n3 5\n3 6\n4 5\n4 6\n4 7\n5 6\n5 7\n5 1\n6 7\n6 1\n6 2\n7 1\n7 2\n7 3"
},
{
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"output": "4\n1 2\n2 3\n3 4\n4 1"
},
{
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"output": "10\n1 2\n1 3\n2 3\n2 4\n3 4\n3 5\n4 5\n4 1\n5 1\n5 2"
},
{
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"output": "10\n1 2\n1 3\n2 3\n2 4\n3 4\n3 5\n4 5\n4 1\n5 1\n5 2"
},
{
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"output": "-1"
},
{
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"output": "55\n1 2\n1 3\n1 4\n1 5\n1 6\n2 3\n2 4\n2 5\n2 6\n2 7\n3 4\n3 5\n3 6\n3 7\n3 8\n4 5\n4 6\n4 7\n4 8\n4 9\n5 6\n5 7\n5 8\n5 9\n5 10\n6 7\n6 8\n6 9\n6 10\n6 11\n7 8\n7 9\n7 10\n7 11\n7 1\n8 9\n8 10\n8 11\n8 1\n8 2\n9 10\n9 11\n9 1\n9 2\n9 3\n10 11\n10 1\n10 2\n10 3\n10 4\n11 1\n11 2\n11 3\n11 4\n11 5"
},
{
"input": "1 1",
"output": "-1"
},
{
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"output": "-1"
},
{
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},
{
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"output": "-1"
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{
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"output": "-1"
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"output": "-1"
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{
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},
{
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},
{
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{
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{
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},
{
"input": "137 68",
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},
{
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{
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},
{
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},
{
"input": "863 431",
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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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},
{
"input": "999 499",
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},
{
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{
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{
"input": "999 2",
"output": "1998\n1 2\n1 3\n2 3\n2 4\n3 4\n3 5\n4 5\n4 6\n5 6\n5 7\n6 7\n6 8\n7 8\n7 9\n8 9\n8 10\n9 10\n9 11\n10 11\n10 12\n11 12\n11 13\n12 13\n12 14\n13 14\n13 15\n14 15\n14 16\n15 16\n15 17\n16 17\n16 18\n17 18\n17 19\n18 19\n18 20\n19 20\n19 21\n20 21\n20 22\n21 22\n21 23\n22 23\n22 24\n23 24\n23 25\n24 25\n24 26\n25 26\n25 27\n26 27\n26 28\n27 28\n27 29\n28 29\n28 30\n29 30\n29 31\n30 31\n30 32\n31 32\n31 33\n32 33\n32 34\n33 34\n33 35\n34 35\n34 36\n35 36\n35 37\n36 37\n36 38\n37 38\n37 39\n38 39\n38 40\n39 40\n..."
},
{
"input": "1000 490",
"output": "490000\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1..."
}
] | 1,613,771,830 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 33 | 1,000 | 307,200 | # Collaborated with No One
input1 = list(map(int, input().split(" ")))
n = input1[0]
k = input1[1]
m = n-1
totalGamesPlayed = m+(m*(m-1))//2
if(k*n>totalGamesPlayed):
print("-1")
else:
print(n*k)
for i in range(1,n+1):
j=i+1
while(j<=i+k):
if(j>n):
print(str(i) + " " + str(j%n))
else:
print(str(i) + " " + str(j))
j+=1
| Title: Football
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, at the "Russian Code Cup" event it was decided to play football as an out of competition event. All participants was divided into *n* teams and played several matches, two teams could not play against each other more than once.
The appointed Judge was the most experienced member — Pavel. But since he was the wisest of all, he soon got bored of the game and fell asleep. Waking up, he discovered that the tournament is over and the teams want to know the results of all the matches.
Pavel didn't want anyone to discover about him sleeping and not keeping an eye on the results, so he decided to recover the results of all games. To do this, he asked all the teams and learned that the real winner was friendship, that is, each team beat the other teams exactly *k* times. Help Pavel come up with chronology of the tournir that meets all the conditions, or otherwise report that there is no such table.
Input Specification:
The first line contains two integers — *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1000).
Output Specification:
In the first line print an integer *m* — number of the played games. The following *m* lines should contain the information about all the matches, one match per line. The *i*-th line should contain two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*). The numbers *a**i* and *b**i* mean, that in the *i*-th match the team with number *a**i* won against the team with number *b**i*. You can assume, that the teams are numbered from 1 to *n*.
If a tournir that meets the conditions of the problem does not exist, then print -1.
Demo Input:
['3 1\n']
Demo Output:
['3\n1 2\n2 3\n3 1\n']
Note:
none | ```python
# Collaborated with No One
input1 = list(map(int, input().split(" ")))
n = input1[0]
k = input1[1]
m = n-1
totalGamesPlayed = m+(m*(m-1))//2
if(k*n>totalGamesPlayed):
print("-1")
else:
print(n*k)
for i in range(1,n+1):
j=i+1
while(j<=i+k):
if(j>n):
print(str(i) + " " + str(j%n))
else:
print(str(i) + " " + str(j))
j+=1
``` | 0 | |
80 | A | Panoramix's Prediction | PROGRAMMING | 800 | [
"brute force"
] | A. Panoramix's Prediction | 2 | 256 | A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix? | The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4. | Print YES, if *m* is the next prime number after *n*, or NO otherwise. | [
"3 5\n",
"7 11\n",
"7 9\n"
] | [
"YES",
"YES",
"NO"
] | none | 500 | [
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
"output": "YES"
},
{
"input": "5 7",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "11 13",
"output": "YES"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "17 19",
"output": "YES"
},
{
"input": "19 23",
"output": "YES"
},
{
"input": "23 29",
"output": "YES"
},
{
"input": "29 31",
"output": "YES"
},
{
"input": "31 37",
"output": "YES"
},
{
"input": "37 41",
"output": "YES"
},
{
"input": "41 43",
"output": "YES"
},
{
"input": "43 47",
"output": "YES"
},
{
"input": "2 5",
"output": "NO"
},
{
"input": "2 7",
"output": "NO"
},
{
"input": "2 6",
"output": "NO"
},
{
"input": "2 11",
"output": "NO"
},
{
"input": "3 6",
"output": "NO"
},
{
"input": "3 7",
"output": "NO"
},
{
"input": "3 9",
"output": "NO"
},
{
"input": "5 6",
"output": "NO"
},
{
"input": "5 9",
"output": "NO"
},
{
"input": "5 11",
"output": "NO"
},
{
"input": "5 13",
"output": "NO"
},
{
"input": "5 15",
"output": "NO"
},
{
"input": "7 8",
"output": "NO"
},
{
"input": "7 13",
"output": "NO"
},
{
"input": "13 15",
"output": "NO"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "19 21",
"output": "NO"
},
{
"input": "13 20",
"output": "NO"
},
{
"input": "41 49",
"output": "NO"
},
{
"input": "43 49",
"output": "NO"
},
{
"input": "47 50",
"output": "NO"
},
{
"input": "47 49",
"output": "NO"
},
{
"input": "47 48",
"output": "NO"
},
{
"input": "23 25",
"output": "NO"
},
{
"input": "2 50",
"output": "NO"
},
{
"input": "31 33",
"output": "NO"
}
] | 1,542,019,234 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 280 | 0 | n=input()
m=n.split()
a=int(m[0])
b=int(m[1])
def isPrime(n):
for i in range(2,n):
if n%i==0 : return False
return True
if isPrime(b)==False : print("NO")
else :
j=a+1
while isPrime(j)==False:
j+=1
if j==b : print("YES")
else : print("NO")
| Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none | ```python
n=input()
m=n.split()
a=int(m[0])
b=int(m[1])
def isPrime(n):
for i in range(2,n):
if n%i==0 : return False
return True
if isPrime(b)==False : print("NO")
else :
j=a+1
while isPrime(j)==False:
j+=1
if j==b : print("YES")
else : print("NO")
``` | 3.93 |
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