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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1,011 | B | Planning The Expedition | PROGRAMMING | 1,200 | [
"binary search",
"brute force",
"implementation"
] | null | null | Natasha is planning an expedition to Mars for $n$ people. One of the important tasks is to provide food for each participant.
The warehouse has $m$ daily food packages. Each package has some food type $a_i$.
Each participant must eat exactly one food package each day. Due to extreme loads, each participant must eat the same food type throughout the expedition. Different participants may eat different (or the same) types of food.
Formally, for each participant $j$ Natasha should select his food type $b_j$ and each day $j$-th participant will eat one food package of type $b_j$. The values $b_j$ for different participants may be different.
What is the maximum possible number of days the expedition can last, following the requirements above? | The first line contains two integers $n$ and $m$ ($1 \le n \le 100$, $1 \le m \le 100$) — the number of the expedition participants and the number of the daily food packages available.
The second line contains sequence of integers $a_1, a_2, \dots, a_m$ ($1 \le a_i \le 100$), where $a_i$ is the type of $i$-th food package. | Print the single integer — the number of days the expedition can last. If it is not possible to plan the expedition for even one day, print 0. | [
"4 10\n1 5 2 1 1 1 2 5 7 2\n",
"100 1\n1\n",
"2 5\n5 4 3 2 1\n",
"3 9\n42 42 42 42 42 42 42 42 42\n"
] | [
"2\n",
"0\n",
"1\n",
"3\n"
] | In the first example, Natasha can assign type $1$ food to the first participant, the same type $1$ to the second, type $5$ to the third and type $2$ to the fourth. In this case, the expedition can last for $2$ days, since each participant can get two food packages of his food type (there will be used $4$ packages of type $1$, two packages of type $2$ and two packages of type $5$).
In the second example, there are $100$ participants and only $1$ food package. In this case, the expedition can't last even $1$ day. | 1,000 | [
{
"input": "4 10\n1 5 2 1 1 1 2 5 7 2",
"output": "2"
},
{
"input": "100 1\n1",
"output": "0"
},
{
"input": "2 5\n5 4 3 2 1",
"output": "1"
},
{
"input": "3 9\n42 42 42 42 42 42 42 42 42",
"output": "3"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "4 100\n84 99 66 69 86 94 89 96 98 93 93 82 87 93 91 100 69 99 93 81 99 84 75 100 86 88 98 100 84 96 44 70 94 91 85 78 86 79 45 88 91 78 98 94 81 87 93 72 96 88 96 97 96 62 86 72 94 84 80 98 88 90 93 73 73 98 78 50 91 96 97 82 85 90 87 41 97 82 97 77 100 100 92 83 98 81 70 81 74 78 84 79 98 98 55 99 97 99 79 98",
"output": "5"
},
{
"input": "100 100\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": "1"
},
{
"input": "1 100\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": "100"
},
{
"input": "6 100\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 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 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "15"
},
{
"input": "1 1\n59",
"output": "1"
},
{
"input": "1 50\n39 1 46 21 23 28 100 32 63 63 18 15 40 29 34 49 56 74 47 42 96 97 59 62 76 62 69 61 36 21 66 18 92 58 63 85 5 6 77 75 91 66 38 10 66 43 20 74 37 83",
"output": "3"
},
{
"input": "1 100\n83 72 21 55 49 5 61 60 87 21 89 88 3 75 49 81 36 25 50 61 96 19 36 55 48 8 97 69 50 24 23 39 26 25 41 90 69 20 19 62 38 52 60 6 66 31 9 45 36 12 69 94 22 60 91 65 35 58 13 85 33 87 83 11 95 20 20 85 13 21 57 69 17 94 78 37 59 45 60 7 64 51 60 89 91 22 6 58 95 96 51 53 89 22 28 16 27 56 1 54",
"output": "5"
},
{
"input": "50 1\n75",
"output": "0"
},
{
"input": "50 50\n85 20 12 73 52 78 70 95 88 43 31 88 81 41 80 99 16 11 97 11 21 44 2 34 47 38 87 2 32 47 97 93 52 14 35 37 97 48 58 19 52 55 97 72 17 25 16 85 90 58",
"output": "1"
},
{
"input": "50 100\n2 37 74 32 99 75 73 86 67 33 62 30 15 21 51 41 73 75 67 39 90 10 56 74 72 26 38 65 75 55 46 99 34 49 92 82 11 100 15 71 75 12 22 56 47 74 20 98 59 65 14 76 1 40 89 36 43 93 83 73 75 100 50 95 27 10 72 51 25 69 15 3 57 60 84 99 31 44 12 61 69 95 51 31 28 36 57 35 31 52 44 19 79 12 27 27 7 81 68 1",
"output": "1"
},
{
"input": "100 1\n26",
"output": "0"
},
{
"input": "100 50\n8 82 62 11 85 57 5 32 99 92 77 2 61 86 8 88 10 28 83 4 68 79 8 64 56 98 4 88 22 54 30 60 62 79 72 38 17 28 32 16 62 26 56 44 72 33 22 84 77 45",
"output": "0"
},
{
"input": "100 100\n13 88 64 65 78 10 61 97 16 32 76 9 60 1 40 35 90 61 60 85 26 16 38 36 33 95 24 55 82 88 13 9 47 34 94 2 90 74 11 81 46 70 94 11 55 32 19 36 97 16 17 35 38 82 89 16 74 94 97 79 9 94 88 12 28 2 4 25 72 95 49 31 88 82 6 77 70 98 90 57 57 33 38 61 26 75 2 66 22 44 13 35 16 4 33 16 12 66 32 86",
"output": "1"
},
{
"input": "34 64\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",
"output": "1"
},
{
"input": "53 98\n1 1 2 2 2 2 2 1 2 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 1 2 1 1 1 2 1 2 1 1 1 2 2 1 2 1 1 1 2 2 1 2 1 1 2 1 2 2 1 2 2 2 2 2 2 2 2 2 1 1 2 2 1 2 1 2 1 2 1 1 2 2 2 1 1 2 1 2 1 1 1 1 2 2 2 2 2 1 1 2 2 2 1 1",
"output": "1"
},
{
"input": "17 8\n2 5 3 4 3 2 2 2",
"output": "0"
},
{
"input": "24 77\n8 6 10 4 6 6 4 10 9 7 7 5 5 4 6 7 10 6 3 4 6 6 4 9 4 6 2 5 3 4 4 1 4 6 6 8 1 1 6 4 6 2 5 7 7 2 4 4 10 1 10 9 2 3 8 1 10 4 3 9 3 8 3 5 6 3 4 9 5 3 4 1 1 6 1 2 1",
"output": "2"
},
{
"input": "65 74\n7 19 2 38 28 44 34 49 14 13 30 22 11 4 4 12 8 1 40 8 34 31 44 38 21 35 13 7 19 32 37 5 36 26 7 2 15 11 47 45 48 2 49 10 10 42 42 31 50 24 29 34 31 38 39 48 43 47 32 46 10 1 33 21 12 50 13 44 38 11 41 41 10 7",
"output": "1"
},
{
"input": "37 71\n50 93 15 80 82 23 35 90 70 73 55 23 23 6 86 63 38 70 38 52 88 34 25 75 32 19 6 98 31 38 21 8 66 8 59 71 7 80 69 23 17 70 6 40 72 5 48 59 18 1 48 91 17 41 11 27 53 95 87 31 62 94 94 60 38 99 70 50 81 86 44",
"output": "1"
},
{
"input": "35 4\n100 100 100 100",
"output": "0"
},
{
"input": "68 12\n100 100 100 99 99 100 100 100 99 99 99 99",
"output": "0"
},
{
"input": "91 33\n97 100 96 96 97 100 97 97 96 96 99 99 98 97 97 99 99 98 100 96 96 99 100 96 97 100 97 97 99 98 96 98 97",
"output": "0"
},
{
"input": "59 72\n99 96 96 93 94 97 93 96 100 99 93 99 96 92 92 97 93 100 92 100 91 96 99 100 97 98 93 93 96 100 97 99 100 100 91 93 96 91 93 100 95 96 92 97 91 91 99 100 91 95 95 100 92 93 98 93 95 92 95 100 91 95 99 95 92 91 100 93 98 98 96 99",
"output": "1"
},
{
"input": "9 35\n95 54 78 94 56 63 68 73 61 85 73 62 61 94 67 91 66 54 89 69 92 60 91 87 63 62 85 70 62 68 97 88 96 55 62",
"output": "2"
},
{
"input": "99 100\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": "1"
},
{
"input": "1 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": "50"
}
] | 1,612,335,224 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 62 | 307,200 | def solve(particpant,package,seq) :
maximumDays = package//particpant
seq.sort()
while maximumDays > 0 :
participantFed = 0
type = seq[0]
temp = 0
for x in seq :
if temp == maximumDays :
participantFed += 1
temp = 1
type = x
else :
if type == x :
temp += 1
else :
type = x
temp = 1
if temp == maximumDays :
participantFed += 1
if participantFed >= particpant :
return maximumDays
maximumDays -= 1
return maximumDays
p,pa = list(map(int,input().split()))
seq = list(map(int,input().split()))
print (solve(p,pa,seq))
| Title: Planning The Expedition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is planning an expedition to Mars for $n$ people. One of the important tasks is to provide food for each participant.
The warehouse has $m$ daily food packages. Each package has some food type $a_i$.
Each participant must eat exactly one food package each day. Due to extreme loads, each participant must eat the same food type throughout the expedition. Different participants may eat different (or the same) types of food.
Formally, for each participant $j$ Natasha should select his food type $b_j$ and each day $j$-th participant will eat one food package of type $b_j$. The values $b_j$ for different participants may be different.
What is the maximum possible number of days the expedition can last, following the requirements above?
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n \le 100$, $1 \le m \le 100$) — the number of the expedition participants and the number of the daily food packages available.
The second line contains sequence of integers $a_1, a_2, \dots, a_m$ ($1 \le a_i \le 100$), where $a_i$ is the type of $i$-th food package.
Output Specification:
Print the single integer — the number of days the expedition can last. If it is not possible to plan the expedition for even one day, print 0.
Demo Input:
['4 10\n1 5 2 1 1 1 2 5 7 2\n', '100 1\n1\n', '2 5\n5 4 3 2 1\n', '3 9\n42 42 42 42 42 42 42 42 42\n']
Demo Output:
['2\n', '0\n', '1\n', '3\n']
Note:
In the first example, Natasha can assign type $1$ food to the first participant, the same type $1$ to the second, type $5$ to the third and type $2$ to the fourth. In this case, the expedition can last for $2$ days, since each participant can get two food packages of his food type (there will be used $4$ packages of type $1$, two packages of type $2$ and two packages of type $5$).
In the second example, there are $100$ participants and only $1$ food package. In this case, the expedition can't last even $1$ day. | ```python
def solve(particpant,package,seq) :
maximumDays = package//particpant
seq.sort()
while maximumDays > 0 :
participantFed = 0
type = seq[0]
temp = 0
for x in seq :
if temp == maximumDays :
participantFed += 1
temp = 1
type = x
else :
if type == x :
temp += 1
else :
type = x
temp = 1
if temp == maximumDays :
participantFed += 1
if participantFed >= particpant :
return maximumDays
maximumDays -= 1
return maximumDays
p,pa = list(map(int,input().split()))
seq = list(map(int,input().split()))
print (solve(p,pa,seq))
``` | 3 | |
234 | B | Reading | PROGRAMMING | 1,000 | [
"sortings"
] | null | null | Vasya is going to the Olympics in the city Ntown by train. The boy wants to read the textbook to prepare for the Olympics. He counted that he needed *k* hours for this. He also found that the light in the train changes every hour. The light is measured on a scale from 0 to 100, where 0 is very dark, and 100 is very light.
Vasya has a train lighting schedule for all *n* hours of the trip — *n* numbers from 0 to 100 each (the light level in the first hour, the second hour and so on). During each of those hours he will either read the whole time, or not read at all. He wants to choose *k* hours to read a book, not necessarily consecutive, so that the minimum level of light among the selected hours were maximum. Vasya is very excited before the upcoming contest, help him choose reading hours. | The first input line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*k*<=≤<=*n*) — the number of hours on the train and the number of hours to read, correspondingly. The second line contains *n* space-separated integers *a**i* (0<=≤<=*a**i*<=≤<=100), *a**i* is the light level at the *i*-th hour. | In the first output line print the minimum light level Vasya will read at. In the second line print *k* distinct space-separated integers *b*1,<=*b*2,<=...,<=*b**k*, — the indexes of hours Vasya will read at (1<=≤<=*b**i*<=≤<=*n*). The hours are indexed starting from 1. If there are multiple optimal solutions, print any of them. Print the numbers *b**i* in an arbitrary order. | [
"5 3\n20 10 30 40 10\n",
"6 5\n90 20 35 40 60 100\n"
] | [
"20\n1 3 4 \n",
"35\n1 3 4 5 6 \n"
] | In the first sample Vasya should read at the first hour (light 20), third hour (light 30) and at the fourth hour (light 40). The minimum light Vasya will have to read at is 20. | 0 | [
{
"input": "5 3\n20 10 30 40 10",
"output": "20\n1 3 4 "
},
{
"input": "6 5\n90 20 35 40 60 100",
"output": "35\n1 3 4 5 6 "
},
{
"input": "100 7\n85 66 9 91 50 46 61 12 55 65 95 1 25 97 95 4 59 59 52 34 94 30 60 11 68 36 17 84 87 68 72 87 46 99 24 66 75 77 75 2 19 3 33 19 7 20 22 3 71 29 88 63 89 47 7 52 47 55 87 77 9 81 44 13 30 43 66 74 9 42 9 72 97 61 9 94 92 29 18 7 92 68 76 43 35 71 54 49 77 50 77 68 57 24 84 73 32 85 24 37",
"output": "94\n11 14 15 21 34 73 76 "
},
{
"input": "1 1\n10",
"output": "10\n1 "
},
{
"input": "1 1\n86",
"output": "86\n1 "
},
{
"input": "100 79\n83 83 83 83 83 94 94 83 83 83 83 90 83 99 83 91 83 83 83 83 83 83 83 83 83 83 83 91 83 83 83 83 83 96 83 83 83 91 83 83 92 83 83 83 83 83 83 83 83 83 83 83 83 83 83 83 83 83 83 98 83 83 91 97 83 83 83 83 83 83 83 92 83 83 83 83 83 83 83 93 83 83 91 83 83 83 83 83 83 83 83 83 83 83 96 83 83 83 83 83",
"output": "83\n6 7 12 14 16 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 "
},
{
"input": "20 3\n17 76 98 17 55 17 17 99 65 17 17 17 17 52 17 17 69 88 17 17",
"output": "88\n3 8 18 "
},
{
"input": "15 1\n0 78 24 24 61 60 0 65 52 57 97 51 56 13 10",
"output": "97\n11 "
},
{
"input": "50 50\n59 40 52 0 65 49 3 58 57 22 86 37 55 72 11 3 30 30 20 64 44 45 12 48 96 96 39 14 8 53 40 37 8 58 97 16 96 48 30 89 66 19 31 50 23 80 67 16 11 7",
"output": "0\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 "
},
{
"input": "60 8\n59 12 34 86 57 65 42 24 62 18 94 92 43 29 95 33 73 3 69 18 36 18 34 97 85 65 74 25 26 70 46 31 57 73 78 89 95 77 94 71 38 23 30 97 69 97 76 43 76 31 38 50 13 16 55 85 47 5 71 4",
"output": "92\n11 12 15 24 37 39 44 46 "
},
{
"input": "70 5\n76 16 20 60 5 96 32 50 35 9 79 42 38 35 72 45 98 33 55 0 86 92 49 87 22 79 35 27 69 35 89 29 31 43 88 1 48 95 3 92 82 97 53 80 79 0 78 58 37 38 45 9 5 38 53 49 71 7 91 3 75 17 76 44 77 31 78 91 59 91",
"output": "92\n6 17 38 40 42 "
},
{
"input": "12 3\n18 64 98 27 36 27 65 43 39 41 69 47",
"output": "65\n3 7 11 "
},
{
"input": "15 13\n6 78 78 78 78 20 78 78 8 3 78 18 32 56 78",
"output": "8\n2 3 4 5 6 7 8 9 11 12 13 14 15 "
},
{
"input": "17 4\n75 52 24 74 70 24 24 53 24 48 24 0 67 47 24 24 6",
"output": "67\n1 4 5 13 "
},
{
"input": "14 2\n31 18 78 90 96 2 90 27 86 9 94 98 94 34",
"output": "96\n5 12 "
},
{
"input": "100 56\n56 64 54 22 46 0 51 27 8 10 5 26 68 37 51 53 4 64 82 23 38 89 97 20 23 31 7 95 55 27 33 23 95 6 64 69 27 54 36 4 96 61 68 26 46 10 61 53 32 19 28 62 7 32 86 84 12 88 92 51 53 23 80 7 36 46 48 29 12 98 72 99 16 0 94 22 83 23 12 37 29 13 93 16 53 21 8 37 67 33 33 67 35 72 3 97 46 30 9 57",
"output": "33\n1 2 3 5 7 13 14 15 16 18 19 21 22 23 28 29 33 35 36 38 39 41 42 43 45 47 48 52 55 56 58 59 60 61 63 65 66 67 70 71 72 75 77 80 83 85 88 89 90 91 92 93 94 96 97 100 "
},
{
"input": "90 41\n43 24 4 69 54 87 33 34 9 77 87 66 66 0 71 43 42 10 78 48 26 40 8 61 80 38 76 63 7 47 99 69 77 43 29 74 86 93 39 28 99 98 11 27 43 58 50 61 1 79 45 17 23 13 10 98 41 28 19 98 87 51 26 28 88 60 42 25 19 3 29 18 0 56 84 27 43 92 93 97 25 90 13 90 75 52 99 6 66 87",
"output": "52\n4 5 6 10 11 12 13 15 19 24 25 27 28 31 32 33 36 37 38 41 42 46 48 50 56 60 61 65 66 74 75 78 79 80 82 84 85 86 87 89 90 "
},
{
"input": "100 71\n29 56 85 57 40 89 93 81 92 38 81 41 18 9 89 21 81 6 95 94 38 11 90 38 6 81 61 43 81 12 36 35 33 10 81 49 59 37 81 61 95 34 43 20 94 88 57 81 42 81 50 24 85 81 1 90 33 8 59 87 17 52 91 54 81 98 28 11 24 51 95 31 98 29 5 81 91 52 41 81 7 9 81 81 13 81 3 81 10 0 37 47 62 50 81 81 81 94 93 38",
"output": "35\n2 3 4 5 6 7 8 9 10 11 12 15 17 19 20 21 23 24 26 27 28 29 31 32 35 36 37 38 39 40 41 43 45 46 47 48 49 50 51 53 54 56 59 60 62 63 64 65 66 70 71 73 76 77 78 79 80 83 84 86 88 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "100 55\n72 70 77 90 86 96 60 60 60 60 87 62 60 87 0 60 82 60 86 74 60 60 60 60 60 60 78 60 60 60 96 60 60 0 60 60 89 99 60 60 60 60 60 60 89 60 88 84 60 93 0 60 60 60 75 60 67 64 65 60 65 60 72 60 76 4 60 60 60 63 96 62 78 71 63 81 89 98 60 60 69 60 61 60 60 60 85 71 82 79 67 60 60 60 79 96 2 60 60 60",
"output": "60\n1 2 3 4 5 6 11 12 14 17 19 20 27 31 37 38 45 47 48 50 55 57 58 59 61 63 65 70 71 72 73 74 75 76 77 78 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 98 99 100 "
},
{
"input": "100 27\n25 87 25 25 77 78 25 73 91 25 25 70 84 25 61 75 82 25 25 25 25 65 25 25 82 63 93 25 93 75 25 25 25 89 98 25 25 72 70 25 72 25 25 25 70 25 25 98 90 25 25 25 25 25 91 25 78 71 63 69 25 25 25 63 25 25 75 94 25 25 25 25 25 97 25 78 66 87 25 89 25 25 73 85 25 91 72 25 25 80 25 70 25 96 25 25 25 25 25 25",
"output": "75\n2 5 6 9 13 16 17 25 27 29 30 34 35 48 49 55 57 67 68 74 76 78 80 84 86 90 94 "
},
{
"input": "100 99\n1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2",
"output": "1\n2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "100 50\n1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2",
"output": "2\n2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100 "
},
{
"input": "100 51\n1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2",
"output": "1\n2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 99 100 "
},
{
"input": "100 75\n1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2",
"output": "1\n2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 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 "
},
{
"input": "100 45\n1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2",
"output": "2\n12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100 "
},
{
"input": "2 2\n2 2",
"output": "2\n1 2 "
},
{
"input": "2 1\n2 1",
"output": "2\n1 "
},
{
"input": "2 1\n1 2",
"output": "2\n2 "
},
{
"input": "3 1\n1 2 0",
"output": "2\n2 "
},
{
"input": "3 2\n0 0 0",
"output": "0\n2 3 "
},
{
"input": "3 3\n0 1 0",
"output": "0\n1 2 3 "
},
{
"input": "3 2\n99 100 99",
"output": "99\n2 3 "
},
{
"input": "1 1\n0",
"output": "0\n1 "
},
{
"input": "4 1\n100 100 0 100",
"output": "100\n4 "
},
{
"input": "4 2\n0 1 0 1",
"output": "1\n2 4 "
},
{
"input": "10 3\n0 1 2 3 0 1 2 3 0 1",
"output": "2\n4 7 8 "
},
{
"input": "10 9\n0 1 0 1 100 100 1 0 1 0",
"output": "0\n2 3 4 5 6 7 8 9 10 "
},
{
"input": "7 7\n1 0 0 0 0 0 0",
"output": "0\n1 2 3 4 5 6 7 "
},
{
"input": "7 6\n0 0 0 0 0 0 1",
"output": "0\n2 3 4 5 6 7 "
},
{
"input": "7 1\n12 33 12 88 10 0 3",
"output": "88\n4 "
},
{
"input": "7 1\n0 1 1 1 1 1 1",
"output": "1\n7 "
},
{
"input": "7 6\n1 1 1 1 1 1 0",
"output": "1\n1 2 3 4 5 6 "
},
{
"input": "7 2\n1 0 1 1 1 1 1",
"output": "1\n6 7 "
},
{
"input": "8 4\n1 1 1 1 0 0 0 0",
"output": "1\n1 2 3 4 "
},
{
"input": "7 5\n1 1 1 2 0 0 0",
"output": "0\n1 2 3 4 7 "
}
] | 1,628,604,588 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 216 | 22,323,200 | n,k=map(int,input().split())
ar=list(map(int,input().split()))
s=sorted(ar)
ammu=[ ]
oh=k
i=0
while(k>0):
ammu.append(ar.index(s[n-i-1])+1)
i+=1
k-=1
print(s[n-oh])
print(*ammu) | Title: Reading
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya is going to the Olympics in the city Ntown by train. The boy wants to read the textbook to prepare for the Olympics. He counted that he needed *k* hours for this. He also found that the light in the train changes every hour. The light is measured on a scale from 0 to 100, where 0 is very dark, and 100 is very light.
Vasya has a train lighting schedule for all *n* hours of the trip — *n* numbers from 0 to 100 each (the light level in the first hour, the second hour and so on). During each of those hours he will either read the whole time, or not read at all. He wants to choose *k* hours to read a book, not necessarily consecutive, so that the minimum level of light among the selected hours were maximum. Vasya is very excited before the upcoming contest, help him choose reading hours.
Input Specification:
The first input line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*k*<=≤<=*n*) — the number of hours on the train and the number of hours to read, correspondingly. The second line contains *n* space-separated integers *a**i* (0<=≤<=*a**i*<=≤<=100), *a**i* is the light level at the *i*-th hour.
Output Specification:
In the first output line print the minimum light level Vasya will read at. In the second line print *k* distinct space-separated integers *b*1,<=*b*2,<=...,<=*b**k*, — the indexes of hours Vasya will read at (1<=≤<=*b**i*<=≤<=*n*). The hours are indexed starting from 1. If there are multiple optimal solutions, print any of them. Print the numbers *b**i* in an arbitrary order.
Demo Input:
['5 3\n20 10 30 40 10\n', '6 5\n90 20 35 40 60 100\n']
Demo Output:
['20\n1 3 4 \n', '35\n1 3 4 5 6 \n']
Note:
In the first sample Vasya should read at the first hour (light 20), third hour (light 30) and at the fourth hour (light 40). The minimum light Vasya will have to read at is 20. | ```python
n,k=map(int,input().split())
ar=list(map(int,input().split()))
s=sorted(ar)
ammu=[ ]
oh=k
i=0
while(k>0):
ammu.append(ar.index(s[n-i-1])+1)
i+=1
k-=1
print(s[n-oh])
print(*ammu)
``` | -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"
}
] | 1,633,339,143 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 62 | 6,758,400 | n, k = map(int,input().split(" "))
if k == 1:
print(n)
xor = 1
while xor <= n:
xor <<= 1
print(xor -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
n, k = map(int,input().split(" "))
if k == 1:
print(n)
xor = 1
while xor <= n:
xor <<= 1
print(xor -1)
``` | 0 | |
26 | A | Almost Prime | PROGRAMMING | 900 | [
"number theory"
] | A. Almost Prime | 2 | 256 | A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive. | Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000). | Output the amount of almost prime numbers between 1 and *n*, inclusive. | [
"10\n",
"21\n"
] | [
"2\n",
"8\n"
] | none | 500 | [
{
"input": "10",
"output": "2"
},
{
"input": "21",
"output": "8"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "8",
"output": "1"
},
{
"input": "19",
"output": "6"
},
{
"input": "40",
"output": "19"
},
{
"input": "77",
"output": "41"
},
{
"input": "222",
"output": "125"
},
{
"input": "987",
"output": "501"
},
{
"input": "1000",
"output": "508"
},
{
"input": "2000",
"output": "958"
},
{
"input": "3000",
"output": "1375"
},
{
"input": "2999",
"output": "1375"
},
{
"input": "2998",
"output": "1375"
},
{
"input": "2997",
"output": "1374"
},
{
"input": "1429",
"output": "706"
},
{
"input": "1673",
"output": "808"
},
{
"input": "1500",
"output": "732"
},
{
"input": "500",
"output": "266"
},
{
"input": "856",
"output": "439"
}
] | 1,680,275,047 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 122 | 0 |
# A number is called almost prime if it has exactly two distinct prime divisors.
def prime_sieve(n: int) -> list[bool]:
is_prime: list[bool] = [True for _ in range(n + 1)]
is_prime[0] = is_prime[1] = False
i = 2
while i <= n:
if is_prime[i]:
j = 2 * i
while j <= n:
is_prime[j] = False
j += i
i += 1
return is_prime
n = int(input())
divisors = [0 for _ in range(n+1)]
isPrime = prime_sieve(n)
for index in range(n):
if isPrime[index]:
i = index * 2
while i < n:
divisors[i] += 1
i += index
count = 1
for divisor in divisors:
if divisor == 2:
count += 1
print(count)
| Title: Almost Prime
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive.
Input Specification:
Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000).
Output Specification:
Output the amount of almost prime numbers between 1 and *n*, inclusive.
Demo Input:
['10\n', '21\n']
Demo Output:
['2\n', '8\n']
Note:
none | ```python
# A number is called almost prime if it has exactly two distinct prime divisors.
def prime_sieve(n: int) -> list[bool]:
is_prime: list[bool] = [True for _ in range(n + 1)]
is_prime[0] = is_prime[1] = False
i = 2
while i <= n:
if is_prime[i]:
j = 2 * i
while j <= n:
is_prime[j] = False
j += i
i += 1
return is_prime
n = int(input())
divisors = [0 for _ in range(n+1)]
isPrime = prime_sieve(n)
for index in range(n):
if isPrime[index]:
i = index * 2
while i < n:
divisors[i] += 1
i += index
count = 1
for divisor in divisors:
if divisor == 2:
count += 1
print(count)
``` | 0 |
318 | A | Even Odds | PROGRAMMING | 900 | [
"math"
] | null | null | Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*. | The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
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 or the %I64d specifier. | Print the number that will stand at the position number *k* after Volodya's manipulations. | [
"10 3\n",
"7 7\n"
] | [
"5",
"6"
] | In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5. | 500 | [
{
"input": "10 3",
"output": "5"
},
{
"input": "7 7",
"output": "6"
},
{
"input": "7 1",
"output": "1"
},
{
"input": "7 2",
"output": "3"
},
{
"input": "8 5",
"output": "2"
},
{
"input": "8 3",
"output": "5"
},
{
"input": "8 4",
"output": "7"
},
{
"input": "1000000000000 500000000001",
"output": "2"
},
{
"input": "999999999997 499999999999",
"output": "999999999997"
},
{
"input": "999999999999 999999999999",
"output": "999999999998"
},
{
"input": "1000000000000 1",
"output": "1"
},
{
"input": "999999999999 1",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000000 1000000000000",
"output": "1000000000000"
},
{
"input": "1000000000000 500000000000",
"output": "999999999999"
},
{
"input": "1000000000000 499999999999",
"output": "999999999997"
},
{
"input": "999999999997 499999999998",
"output": "999999999995"
},
{
"input": "619234238 556154835",
"output": "493075432"
},
{
"input": "38151981 36650624",
"output": "35149266"
},
{
"input": "680402465 442571217",
"output": "204739968"
},
{
"input": "109135284 9408714",
"output": "18817427"
},
{
"input": "603701841 56038951",
"output": "112077901"
},
{
"input": "356764822 321510177",
"output": "286255532"
},
{
"input": "284911189 142190783",
"output": "284381565"
},
{
"input": "91028405 61435545",
"output": "31842684"
}
] | 1,697,220,002 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 1,000 | 224,051,200 | n,k = map(int,input().split())
ls = []
even=0
odd=0
if n%2==0:
even = n//2
odd = n//2
else:
even = (n-1)//2
odd = (n+1)//2
odd_num=1
even_num=2
count_o=0
count_e = 0
while(count_o < odd):
ls.append(odd_num)
odd_num = odd_num+2
count_o = count_o + 1
while(count_e < even):
ls.append(even_num)
even_num = even_num +2
count_e = count_e+1
print(ls[k-1])
| Title: Even Odds
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*.
Input Specification:
The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
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 or the %I64d specifier.
Output Specification:
Print the number that will stand at the position number *k* after Volodya's manipulations.
Demo Input:
['10 3\n', '7 7\n']
Demo Output:
['5', '6']
Note:
In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5. | ```python
n,k = map(int,input().split())
ls = []
even=0
odd=0
if n%2==0:
even = n//2
odd = n//2
else:
even = (n-1)//2
odd = (n+1)//2
odd_num=1
even_num=2
count_o=0
count_e = 0
while(count_o < odd):
ls.append(odd_num)
odd_num = odd_num+2
count_o = count_o + 1
while(count_e < even):
ls.append(even_num)
even_num = even_num +2
count_e = count_e+1
print(ls[k-1])
``` | 0 | |
137 | B | Permutation | PROGRAMMING | 1,000 | [
"greedy"
] | null | null | "Hey, it's homework time" — thought Polycarpus and of course he started with his favourite subject, IT. Polycarpus managed to solve all tasks but for the last one in 20 minutes. However, as he failed to solve the last task after some considerable time, the boy asked you to help him.
The sequence of *n* integers is called a permutation if it contains all integers from 1 to *n* exactly once.
You are given an arbitrary sequence *a*1,<=*a*2,<=...,<=*a**n* containing *n* integers. Each integer is not less than 1 and not greater than 5000. Determine what minimum number of elements Polycarpus needs to change to get a permutation (he should not delete or add numbers). In a single change he can modify any single sequence element (i. e. replace it with another integer). | The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=5000) which represents how many numbers are in the sequence. The second line contains a sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=5000,<=1<=≤<=*i*<=≤<=*n*). | Print the only number — the minimum number of changes needed to get the permutation. | [
"3\n3 1 2\n",
"2\n2 2\n",
"5\n5 3 3 3 1\n"
] | [
"0\n",
"1\n",
"2\n"
] | The first sample contains the permutation, which is why no replacements are required.
In the second sample it is enough to replace the first element with the number 1 and that will make the sequence the needed permutation.
In the third sample we can replace the second element with number 4 and the fourth element with number 2. | 1,000 | [
{
"input": "3\n3 1 2",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "5\n5 3 3 3 1",
"output": "2"
},
{
"input": "5\n6 6 6 6 6",
"output": "5"
},
{
"input": "10\n1 1 2 2 8 8 7 7 9 9",
"output": "5"
},
{
"input": "8\n9 8 7 6 5 4 3 2",
"output": "1"
},
{
"input": "15\n1 2 3 4 5 5 4 3 2 1 1 2 3 4 5",
"output": "10"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n5000",
"output": "1"
},
{
"input": "4\n5000 5000 5000 5000",
"output": "4"
},
{
"input": "5\n3366 3461 4 5 4370",
"output": "3"
},
{
"input": "10\n8 2 10 3 4 6 1 7 9 5",
"output": "0"
},
{
"input": "10\n551 3192 3213 2846 3068 1224 3447 1 10 9",
"output": "7"
},
{
"input": "15\n4 1459 12 4281 3241 2748 10 3590 14 845 3518 1721 2 2880 1974",
"output": "10"
},
{
"input": "15\n15 1 8 2 13 11 12 7 3 14 6 10 9 4 5",
"output": "0"
},
{
"input": "15\n2436 2354 4259 1210 2037 2665 700 3578 2880 973 1317 1024 24 3621 4142",
"output": "15"
},
{
"input": "30\n28 1 3449 9 3242 4735 26 3472 15 21 2698 7 4073 3190 10 3 29 1301 4526 22 345 3876 19 12 4562 2535 2 630 18 27",
"output": "14"
},
{
"input": "100\n50 39 95 30 66 78 2169 4326 81 31 74 34 80 40 19 48 97 63 82 6 88 16 21 57 92 77 10 1213 17 93 32 91 38 4375 29 75 44 22 4 45 14 2395 3254 59 3379 2 85 96 8 83 27 94 1512 2960 100 9 73 79 7 25 55 69 90 99 51 87 98 62 18 35 43 4376 4668 28 72 56 4070 61 65 36 54 4106 11 24 15 86 70 71 4087 23 13 76 20 4694 26 4962 4726 37 14 64",
"output": "18"
},
{
"input": "100\n340 14 3275 2283 2673 1107 817 2243 1226 32 2382 3638 4652 418 68 4962 387 764 4647 159 1846 225 2760 4904 3150 403 3 2439 91 4428 92 4705 75 348 1566 1465 69 6 49 4 62 4643 564 1090 3447 1871 2255 139 24 99 2669 969 86 61 4550 158 4537 3993 1589 872 2907 1888 401 80 1825 1483 63 1 2264 4068 4113 2548 41 885 4806 36 67 167 4447 34 1248 2593 82 202 81 1783 1284 4973 16 43 95 7 865 2091 3008 1793 20 947 4912 3604",
"output": "70"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "2\n5000 5000",
"output": "2"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
},
{
"input": "2\n3 4",
"output": "2"
},
{
"input": "10\n1 2 3 4 5 6 7 1000 10 10",
"output": "2"
}
] | 1,599,826,180 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 154 | 0 | n=int(input())
s=input().split
t=len(set(s))
print(n-t)
| Title: Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Hey, it's homework time" — thought Polycarpus and of course he started with his favourite subject, IT. Polycarpus managed to solve all tasks but for the last one in 20 minutes. However, as he failed to solve the last task after some considerable time, the boy asked you to help him.
The sequence of *n* integers is called a permutation if it contains all integers from 1 to *n* exactly once.
You are given an arbitrary sequence *a*1,<=*a*2,<=...,<=*a**n* containing *n* integers. Each integer is not less than 1 and not greater than 5000. Determine what minimum number of elements Polycarpus needs to change to get a permutation (he should not delete or add numbers). In a single change he can modify any single sequence element (i. e. replace it with another integer).
Input Specification:
The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=5000) which represents how many numbers are in the sequence. The second line contains a sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=5000,<=1<=≤<=*i*<=≤<=*n*).
Output Specification:
Print the only number — the minimum number of changes needed to get the permutation.
Demo Input:
['3\n3 1 2\n', '2\n2 2\n', '5\n5 3 3 3 1\n']
Demo Output:
['0\n', '1\n', '2\n']
Note:
The first sample contains the permutation, which is why no replacements are required.
In the second sample it is enough to replace the first element with the number 1 and that will make the sequence the needed permutation.
In the third sample we can replace the second element with number 4 and the fourth element with number 2. | ```python
n=int(input())
s=input().split
t=len(set(s))
print(n-t)
``` | -1 | |
9 | A | Die Roll | PROGRAMMING | 800 | [
"math",
"probabilities"
] | A. Die Roll | 1 | 64 | Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. | The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. | Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». | [
"4 2\n"
] | [
"1/2\n"
] | Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | 0 | [
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,676,342,274 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 60 | 0 | n,m=map(int,input().split())
print("1/2") | Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | ```python
n,m=map(int,input().split())
print("1/2")
``` | 0 |
979 | A | Pizza, Pizza, Pizza!!! | PROGRAMMING | 1,000 | [
"math"
] | null | null | Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem? | A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces. | A single integer — the number of straight cuts Shiro needs. | [
"3\n",
"4\n"
] | [
"2",
"5"
] | To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts. | 500 | [
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "5"
},
{
"input": "10",
"output": "11"
},
{
"input": "10000000000",
"output": "10000000001"
},
{
"input": "1234567891",
"output": "617283946"
},
{
"input": "7509213957",
"output": "3754606979"
},
{
"input": "99999999999999999",
"output": "50000000000000000"
},
{
"input": "21",
"output": "11"
},
{
"input": "712394453192",
"output": "712394453193"
},
{
"input": "172212168",
"output": "172212169"
},
{
"input": "822981260158260519",
"output": "411490630079130260"
},
{
"input": "28316250877914571",
"output": "14158125438957286"
},
{
"input": "779547116602436424",
"output": "779547116602436425"
},
{
"input": "578223540024979436",
"output": "578223540024979437"
},
{
"input": "335408917861648766",
"output": "335408917861648767"
},
{
"input": "74859962623690078",
"output": "74859962623690079"
},
{
"input": "252509054433933439",
"output": "126254527216966720"
},
{
"input": "760713016476190622",
"output": "760713016476190623"
},
{
"input": "919845426262703496",
"output": "919845426262703497"
},
{
"input": "585335723211047194",
"output": "585335723211047195"
},
{
"input": "522842184971407769",
"output": "261421092485703885"
},
{
"input": "148049062628894320",
"output": "148049062628894321"
},
{
"input": "84324828731963974",
"output": "84324828731963975"
},
{
"input": "354979173822804781",
"output": "177489586911402391"
},
{
"input": "1312150450968413",
"output": "656075225484207"
},
{
"input": "269587449430302150",
"output": "269587449430302151"
},
{
"input": "645762258982631926",
"output": "645762258982631927"
},
{
"input": "615812229161735895",
"output": "307906114580867948"
},
{
"input": "0",
"output": "0"
},
{
"input": "349993004923078531",
"output": "174996502461539266"
},
{
"input": "891351282707723851",
"output": "445675641353861926"
},
{
"input": "563324731189330734",
"output": "563324731189330735"
},
{
"input": "520974001910286909",
"output": "260487000955143455"
},
{
"input": "666729339802329204",
"output": "666729339802329205"
},
{
"input": "856674611404539671",
"output": "428337305702269836"
},
{
"input": "791809296303238499",
"output": "395904648151619250"
},
{
"input": "711066337317063338",
"output": "711066337317063339"
},
{
"input": "931356503492686566",
"output": "931356503492686567"
},
{
"input": "234122432773361866",
"output": "234122432773361867"
},
{
"input": "1000000000000000000",
"output": "1000000000000000001"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "3"
},
{
"input": "7",
"output": "4"
},
{
"input": "63",
"output": "32"
},
{
"input": "24",
"output": "25"
},
{
"input": "8",
"output": "9"
},
{
"input": "15",
"output": "8"
}
] | 1,527,584,974 | 2,147,483,647 | Python 3 | OK | TESTS | 47 | 93 | 0 | n=int(input())
if((n+1)%2==0):
print((n+1)//2)
elif(n==0):
print(0)
else:
print(n+1)
| Title: Pizza, Pizza, Pizza!!!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem?
Input Specification:
A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces.
Output Specification:
A single integer — the number of straight cuts Shiro needs.
Demo Input:
['3\n', '4\n']
Demo Output:
['2', '5']
Note:
To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts. | ```python
n=int(input())
if((n+1)%2==0):
print((n+1)//2)
elif(n==0):
print(0)
else:
print(n+1)
``` | 3 | |
538 | B | Quasi Binary | PROGRAMMING | 1,400 | [
"constructive algorithms",
"dp",
"greedy",
"implementation"
] | null | null | A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106). | In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them. | [
"9\n",
"32\n"
] | [
"9\n1 1 1 1 1 1 1 1 1 \n",
"3\n10 11 11 \n"
] | none | 1,000 | [
{
"input": "9",
"output": "9\n1 1 1 1 1 1 1 1 1 "
},
{
"input": "32",
"output": "3\n10 11 11 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "415",
"output": "5\n1 101 101 101 111 "
},
{
"input": "10011",
"output": "1\n10011 "
},
{
"input": "10201",
"output": "2\n100 10101 "
},
{
"input": "314159",
"output": "9\n1 1 1 1 11 1011 101011 101011 111111 "
},
{
"input": "999999",
"output": "9\n111111 111111 111111 111111 111111 111111 111111 111111 111111 "
},
{
"input": "2",
"output": "2\n1 1 "
},
{
"input": "10",
"output": "1\n10 "
},
{
"input": "21",
"output": "2\n10 11 "
},
{
"input": "98",
"output": "9\n10 11 11 11 11 11 11 11 11 "
},
{
"input": "102030",
"output": "3\n10 1010 101010 "
},
{
"input": "909090",
"output": "9\n101010 101010 101010 101010 101010 101010 101010 101010 101010 "
},
{
"input": "909823",
"output": "9\n101000 101100 101100 101100 101100 101100 101101 101111 101111 "
},
{
"input": "1000000",
"output": "1\n1000000 "
},
{
"input": "111111",
"output": "1\n111111 "
},
{
"input": "123456",
"output": "6\n1 11 111 1111 11111 111111 "
},
{
"input": "987654",
"output": "9\n100000 110000 111000 111100 111110 111111 111111 111111 111111 "
},
{
"input": "908172",
"output": "9\n100000 101000 101010 101010 101010 101010 101010 101011 101111 "
},
{
"input": "8",
"output": "8\n1 1 1 1 1 1 1 1 "
},
{
"input": "100009",
"output": "9\n1 1 1 1 1 1 1 1 100001 "
},
{
"input": "900000",
"output": "9\n100000 100000 100000 100000 100000 100000 100000 100000 100000 "
},
{
"input": "1435",
"output": "5\n1 101 111 111 1111 "
},
{
"input": "1453",
"output": "5\n10 110 111 111 1111 "
}
] | 1,580,626,208 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 124 | 0 | def conv(ele):
return int(''.join(ele))
n = input()
int_n = int(n)
ans = max([int(i) for i in n])
seq = []
for i in n:
seq.append(['1' for i in range(int(i))] + ['0' for i in range(ans-int(i))])
ans_seq = [[i[j] for i in seq] for j in range(ans)]
ans_seq = list(map(conv, ans_seq))
print(*ans_seq)
| Title: Quasi Binary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106).
Output Specification:
In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Demo Input:
['9\n', '32\n']
Demo Output:
['9\n1 1 1 1 1 1 1 1 1 \n', '3\n10 11 11 \n']
Note:
none | ```python
def conv(ele):
return int(''.join(ele))
n = input()
int_n = int(n)
ans = max([int(i) for i in n])
seq = []
for i in n:
seq.append(['1' for i in range(int(i))] + ['0' for i in range(ans-int(i))])
ans_seq = [[i[j] for i in seq] for j in range(ans)]
ans_seq = list(map(conv, ans_seq))
print(*ans_seq)
``` | 0 | |
811 | C | Vladik and Memorable Trip | PROGRAMMING | 1,900 | [
"dp",
"implementation"
] | null | null | Vladik often travels by trains. He remembered some of his trips especially well and I would like to tell you about one of these trips:
Vladik is at initial train station, and now *n* people (including Vladik) want to get on the train. They are already lined up in some order, and for each of them the city code *a**i* is known (the code of the city in which they are going to).
Train chief selects some number of disjoint segments of the original sequence of people (covering entire sequence by segments is not necessary). People who are in the same segment will be in the same train carriage. The segments are selected in such way that if at least one person travels to the city *x*, then all people who are going to city *x* should be in the same railway carriage. This means that they can’t belong to different segments. Note, that all people who travel to the city *x*, either go to it and in the same railway carriage, or do not go anywhere at all.
Comfort of a train trip with people on segment from position *l* to position *r* is equal to XOR of all distinct codes of cities for people on the segment from position *l* to position *r*. XOR operation also known as exclusive OR.
Total comfort of a train trip is equal to sum of comfort for each segment.
Help Vladik to know maximal possible total comfort. | First line contains single integer *n* (1<=≤<=*n*<=≤<=5000) — number of people.
Second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=5000), where *a**i* denotes code of the city to which *i*-th person is going. | The output should contain a single integer — maximal possible total comfort. | [
"6\n4 4 2 5 2 3\n",
"9\n5 1 3 1 5 2 4 2 5\n"
] | [
"14\n",
"9\n"
] | In the first test case best partition into segments is: [4, 4] [2, 5, 2] [3], answer is calculated as follows: 4 + (2 *xor* 5) + 3 = 4 + 7 + 3 = 14
In the second test case best partition into segments is: 5 1 [3] 1 5 [2, 4, 2] 5, answer calculated as follows: 3 + (2 *xor* 4) = 3 + 6 = 9. | 1,500 | [
{
"input": "6\n4 4 2 5 2 3",
"output": "14"
},
{
"input": "9\n5 1 3 1 5 2 4 2 5",
"output": "9"
},
{
"input": "5\n1558 4081 3591 1700 3232",
"output": "14162"
},
{
"input": "10\n3838 1368 4825 2068 4755 2048 1342 4909 2837 4854",
"output": "32844"
},
{
"input": "10\n4764 4867 2346 1449 1063 2002 2577 2089 1566 614",
"output": "23337"
},
{
"input": "10\n689 3996 3974 4778 1740 3481 2916 2744 294 1376",
"output": "25988"
},
{
"input": "100\n1628 4511 4814 3756 4625 1254 906 1033 2420 2622 2640 3225 3570 2925 465 2093 4614 2856 4004 4254 2292 2026 415 2777 905 4452 4737 529 4571 3221 2064 2495 420 1291 493 4073 3207 1217 3463 3047 3627 1783 1723 3586 800 2403 4378 4373 535 64 4014 346 2597 2502 3667 2904 3153 1061 3104 1847 4741 315 1212 501 4504 3947 842 2388 2868 3430 1018 560 2840 4477 2903 2810 3600 4352 1106 1102 4747 433 629 2043 1669 2695 436 403 650 530 1318 1348 4677 3245 2426 1056 702 203 1132 4471",
"output": "238706"
},
{
"input": "100\n2554 1060 1441 4663 301 3629 1245 3214 4623 4909 4283 1596 959 687 2981 1105 122 3820 3205 488 3755 2998 3243 3621 2707 3771 1302 2611 4545 2737 762 173 2513 2204 2433 4483 3095 2620 3265 4215 3085 947 425 144 659 1660 3295 2315 2281 2617 1887 2931 3494 2762 559 3690 3590 3826 3438 2203 101 1316 3688 3532 819 1069 2573 3127 3894 169 547 1305 2085 4753 4292 2116 1623 960 4809 3694 1047 501 1193 4987 1179 1470 647 113 4223 2154 3222 246 3321 1276 2340 1561 4477 665 2256 626",
"output": "233722"
},
{
"input": "100\n931 4584 2116 3004 3813 62 2819 2998 2080 4906 3198 2443 2952 3793 1958 3864 3985 3169 3134 4011 4525 995 4163 308 4362 1148 4906 3092 1647 244 1370 1424 2753 84 2997 1197 2606 425 3501 2606 683 4747 3884 4787 2166 3017 3080 4303 3352 1667 2636 3994 757 2388 870 1788 988 1303 0 1230 1455 4213 2113 2908 871 1997 3878 4604 1575 3385 236 847 2524 3937 1803 2678 4619 1125 3108 1456 3017 1532 3845 3293 2355 2230 4282 2586 2892 4506 3132 4570 1872 2339 2166 3467 3080 2693 1925 2308",
"output": "227685"
},
{
"input": "100\n5 1085 489 2096 1610 108 4005 3869 1826 4145 2450 2546 2719 1030 4443 4222 1 2205 2407 4303 4588 1549 1965 4465 2560 2459 1814 1641 148 728 3566 271 2186 696 1952 4262 2088 4023 4594 1437 4700 2531 1707 1702 1413 4391 4162 3309 1606 4116 1287 1410 3336 2128 3978 1002 552 64 1192 4980 4569 3212 1163 2457 3661 2296 2147 391 550 2540 707 101 4805 2608 4785 4898 1595 1043 4406 3865 1716 4044 1756 4456 1319 4350 4965 2876 4320 4409 3177 671 2596 4308 2253 2962 830 4179 800 1782",
"output": "251690"
},
{
"input": "100\n702 1907 2292 1953 2421 1300 2092 1904 3691 1861 4472 1379 1811 2583 529 3977 4735 997 856 4545 2354 2581 1692 2563 4104 763 1645 4080 3967 3705 4261 448 4854 1903 4449 2768 4214 4815 185 3404 3538 199 4548 4608 46 4673 4406 3379 3790 3567 1139 1236 2755 2242 3723 2118 2716 4824 2770 595 274 840 261 1576 3188 2720 637 4071 2737 2585 4964 4184 120 1622 884 1555 4681 4269 2404 3511 4972 3840 66 4100 1528 1340 1119 2641 1183 3908 1363 28 401 4319 3408 2077 3454 1689 8 3946",
"output": "254107"
},
{
"input": "100\n4 3 5 5 2 0 4 0 1 5 1 2 5 5 2 0 2 3 0 0 0 5 4 4 3 0 5 5 4 0 4 4 1 2 0 4 3 5 4 3 5 1 1 0 0 4 2 0 5 0 1 5 3 3 4 5 1 2 2 5 0 3 3 1 2 0 1 3 0 4 5 4 4 1 5 3 0 2 3 4 1 5 5 0 5 0 0 3 2 1 4 3 4 1 4 5 3 0 5 3",
"output": "1"
},
{
"input": "100\n0 0 0 0 0 1 1 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 0 1 1 1 1 1 1 1 0 1 1 1 0 0 0 0 0 0 1 1 0 0 0 1 0 1 0 0 1 0 1 1 0 0 1 0 0 0 1 1 1 1 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 0",
"output": "1"
},
{
"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\n5 1 12 15 10 0 5 7 12 13 3 11 13 10 0 5 3 1 3 13 1 11 2 6 9 15 8 3 13 3 0 4 11 10 12 10 9 3 13 15 10 11 7 10 1 15 0 7 7 8 12 2 5 2 4 11 7 1 16 14 10 6 14 2 4 15 10 8 6 10 2 7 5 15 9 8 15 6 7 1 5 7 1 15 9 11 2 0 8 12 8 9 4 7 11 2 5 13 12 8",
"output": "16"
},
{
"input": "100\n8 16 16 2 5 7 9 12 14 15 5 11 0 5 9 12 15 13 4 15 10 11 13 2 2 15 15 16 10 7 4 14 9 5 4 10 4 16 2 6 11 0 3 14 12 14 9 5 0 8 11 15 2 14 2 0 3 5 4 4 8 15 14 6 14 5 0 14 12 15 0 15 15 14 2 14 13 7 11 7 2 4 13 11 8 16 9 1 10 13 8 2 7 12 1 14 16 11 15 7",
"output": "16"
},
{
"input": "100\n4 9 4 13 18 17 13 10 28 11 29 32 5 23 14 32 20 17 25 0 18 30 10 17 27 2 13 8 1 20 8 13 6 5 16 1 27 27 24 16 2 18 24 1 0 23 10 21 7 3 21 21 18 27 31 28 10 17 26 27 3 0 6 0 30 9 3 0 3 30 8 3 23 21 18 27 10 16 30 4 1 9 3 8 2 5 20 23 16 22 9 7 11 9 12 30 17 27 14 17",
"output": "145"
},
{
"input": "100\n6 25 23 14 19 5 26 28 5 14 24 2 19 32 4 12 32 12 9 29 23 10 25 31 29 10 3 30 29 13 32 27 13 19 2 24 30 8 11 5 25 32 13 9 28 28 27 1 8 24 15 11 8 6 30 16 29 13 6 11 3 0 8 2 6 9 29 26 11 30 7 21 16 31 23 3 29 18 26 9 26 15 0 31 19 0 0 21 24 15 0 5 19 21 18 32 32 29 5 32",
"output": "51"
},
{
"input": "100\n11 4 31 11 59 23 62 21 49 40 21 1 56 51 22 53 37 28 43 27 15 39 39 33 3 28 60 52 58 21 16 11 10 61 26 59 23 51 26 32 40 21 43 56 55 0 44 48 16 7 26 37 61 19 44 15 63 11 58 62 48 14 38 3 27 50 47 6 46 23 50 16 64 19 45 18 15 30 20 45 50 61 50 57 38 60 61 46 42 39 22 52 7 36 57 23 33 46 29 6",
"output": "598"
},
{
"input": "100\n60 30 6 15 23 15 25 34 55 53 27 23 51 4 47 61 57 62 44 22 18 42 33 29 50 37 62 28 16 4 52 37 33 58 39 36 17 21 59 59 28 26 35 15 37 13 35 29 29 8 56 26 23 18 10 1 3 61 30 11 50 42 48 11 17 47 26 10 46 49 9 29 4 28 40 12 62 33 8 13 26 52 40 30 34 40 40 27 55 42 15 53 53 5 12 47 21 9 23 25",
"output": "656"
},
{
"input": "100\n10 19 72 36 30 38 116 112 65 122 74 62 104 82 64 52 119 109 2 86 114 105 56 12 3 52 35 48 99 68 98 18 68 117 7 76 112 2 57 39 43 2 93 45 1 128 112 90 21 91 61 6 4 53 83 72 120 72 82 111 108 48 12 83 70 78 116 33 22 102 59 31 72 111 33 6 19 91 30 108 110 22 10 93 55 92 20 20 98 10 119 58 17 60 33 4 29 110 127 100",
"output": "2946"
},
{
"input": "100\n83 54 28 107 75 48 55 68 7 33 31 124 22 54 24 83 8 3 10 58 39 106 50 110 17 91 119 87 126 29 40 4 50 44 78 49 41 79 82 6 34 61 80 19 113 67 104 50 15 60 65 97 118 7 48 64 81 5 23 105 64 122 95 25 97 124 97 33 61 20 89 77 24 9 20 84 30 69 12 3 50 122 75 106 41 19 126 112 10 91 42 11 66 20 74 16 120 70 52 43",
"output": "3126"
},
{
"input": "100\n915 7 282 162 24 550 851 240 39 302 538 76 131 150 104 848 507 842 32 453 998 990 1002 225 887 1005 259 199 873 87 258 318 837 511 663 1008 861 516 445 426 335 743 672 345 320 461 650 649 612 9 1017 113 169 722 643 253 562 661 879 522 524 878 600 894 312 1005 283 911 322 509 836 261 424 976 68 606 661 331 830 177 279 772 573 1017 157 250 42 478 582 23 847 119 359 198 839 761 54 1003 270 900",
"output": "45323"
},
{
"input": "100\n139 827 953 669 78 369 980 770 945 509 878 791 550 555 324 682 858 771 525 673 751 746 848 534 573 613 930 135 390 958 60 614 728 444 1018 463 445 662 632 907 536 865 465 974 137 973 386 843 326 314 555 910 258 429 560 559 274 307 409 751 527 724 485 276 18 45 1014 13 321 693 910 397 664 513 110 915 622 76 433 84 704 975 653 716 292 614 218 50 482 620 410 557 862 388 348 1022 663 580 987 149",
"output": "50598"
},
{
"input": "100\n2015 1414 748 1709 110 1094 441 1934 273 1796 451 902 610 914 1613 255 1838 963 1301 1999 393 948 161 510 485 1544 1742 19 12 1036 2007 1394 1898 532 1403 1390 2004 1016 45 675 1264 1696 1511 1523 1335 1997 688 1778 1939 521 222 92 1014 155 135 30 543 1449 229 976 382 654 1827 1158 570 64 1353 1672 295 1573 23 1368 728 597 1263 213 991 1673 1360 183 1256 1539 459 1480 374 1779 1541 858 1470 653 979 342 381 179 388 247 655 198 1762 1249",
"output": "96427"
},
{
"input": "100\n1928 445 1218 1164 1501 1284 973 1503 1132 1999 2046 1259 1604 1279 1044 684 89 733 1431 1133 1141 1954 181 76 997 187 1088 1265 1721 2039 1724 1986 308 402 1777 751 97 484 880 14 936 876 1226 1105 110 1587 588 363 169 296 1087 1490 1640 1378 433 1684 293 153 492 2040 1229 1754 950 1573 771 1052 366 382 88 186 1340 1212 1195 2005 36 2001 248 72 1309 1371 1381 653 1972 1503 571 1490 278 1590 288 183 949 361 1162 639 2003 1271 254 796 987 159",
"output": "93111"
},
{
"input": "100\n3108 2117 3974 3127 3122 796 1234 1269 1723 3313 3522 869 3046 557 334 3085 557 2528 1028 169 2203 595 388 2435 408 2712 2363 2088 2064 1185 3076 2073 2717 492 775 3351 3538 3050 85 3495 2335 1124 2891 3108 284 1123 500 502 808 3352 3988 1318 222 3452 3896 1024 2789 2480 1958 2976 1358 1225 3007 1817 1672 3667 1511 1147 2803 2632 3439 3066 3864 1942 2526 3574 1179 3375 406 782 3866 3157 3396 245 2401 2378 1258 684 2400 2809 3375 1225 1345 3630 2760 2546 1761 3138 2539 1616",
"output": "194223"
},
{
"input": "100\n1599 2642 1471 2093 3813 329 2165 254 3322 629 3286 2332 279 3756 1167 2607 2499 2411 2626 4040 2406 3468 1617 118 2083 2789 1571 333 1815 2600 2579 572 3193 249 1880 2226 1722 1771 3475 4038 951 2942 1135 3348 2785 1947 1937 108 3861 307 3052 2060 50 837 1107 2383 2633 2280 1122 1726 2800 522 714 2322 661 554 2444 3534 1440 2229 718 3311 1834 462 2348 3444 692 17 2866 347 2655 58 483 2298 1074 2163 3007 1858 2435 998 1506 707 1287 3821 2486 1496 3819 3529 1310 3926",
"output": "194571"
}
] | 1,497,628,352 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 93 | 0 | n = int(input())
a = list(map(int,input().split()))
dl = dict()
dr = dict()
for i in range(n):
if a[i] not in dl:
dl[a[i]] = i
dr[a[i]] = i
def XOR(l):
a = list(set(l))
k = len(a)
x = a[0]
for i in range(1,k):
x ^= a[i]
return x
from collections import defaultdict
DP = defaultdict(int)
for i in range(n):
if dr[a[i]] > i:
DP[i] = DP[i-1]
else:
j = dl[a[i]]
DP[i] = DP[j-1] + XOR(a[j:i+1])
print(DP[n-1])
| Title: Vladik and Memorable Trip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vladik often travels by trains. He remembered some of his trips especially well and I would like to tell you about one of these trips:
Vladik is at initial train station, and now *n* people (including Vladik) want to get on the train. They are already lined up in some order, and for each of them the city code *a**i* is known (the code of the city in which they are going to).
Train chief selects some number of disjoint segments of the original sequence of people (covering entire sequence by segments is not necessary). People who are in the same segment will be in the same train carriage. The segments are selected in such way that if at least one person travels to the city *x*, then all people who are going to city *x* should be in the same railway carriage. This means that they can’t belong to different segments. Note, that all people who travel to the city *x*, either go to it and in the same railway carriage, or do not go anywhere at all.
Comfort of a train trip with people on segment from position *l* to position *r* is equal to XOR of all distinct codes of cities for people on the segment from position *l* to position *r*. XOR operation also known as exclusive OR.
Total comfort of a train trip is equal to sum of comfort for each segment.
Help Vladik to know maximal possible total comfort.
Input Specification:
First line contains single integer *n* (1<=≤<=*n*<=≤<=5000) — number of people.
Second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=5000), where *a**i* denotes code of the city to which *i*-th person is going.
Output Specification:
The output should contain a single integer — maximal possible total comfort.
Demo Input:
['6\n4 4 2 5 2 3\n', '9\n5 1 3 1 5 2 4 2 5\n']
Demo Output:
['14\n', '9\n']
Note:
In the first test case best partition into segments is: [4, 4] [2, 5, 2] [3], answer is calculated as follows: 4 + (2 *xor* 5) + 3 = 4 + 7 + 3 = 14
In the second test case best partition into segments is: 5 1 [3] 1 5 [2, 4, 2] 5, answer calculated as follows: 3 + (2 *xor* 4) = 3 + 6 = 9. | ```python
n = int(input())
a = list(map(int,input().split()))
dl = dict()
dr = dict()
for i in range(n):
if a[i] not in dl:
dl[a[i]] = i
dr[a[i]] = i
def XOR(l):
a = list(set(l))
k = len(a)
x = a[0]
for i in range(1,k):
x ^= a[i]
return x
from collections import defaultdict
DP = defaultdict(int)
for i in range(n):
if dr[a[i]] > i:
DP[i] = DP[i-1]
else:
j = dl[a[i]]
DP[i] = DP[j-1] + XOR(a[j:i+1])
print(DP[n-1])
``` | 0 | |
320 | A | Magic Numbers | PROGRAMMING | 900 | [
"brute force",
"greedy"
] | null | null | A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a magic number or not. | The first line of input contains an integer *n*, (1<=≤<=*n*<=≤<=109). This number doesn't contain leading zeros. | Print "YES" if *n* is a magic number or print "NO" if it's not. | [
"114114\n",
"1111\n",
"441231\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "114114",
"output": "YES"
},
{
"input": "1111",
"output": "YES"
},
{
"input": "441231",
"output": "NO"
},
{
"input": "1",
"output": "YES"
},
{
"input": "14",
"output": "YES"
},
{
"input": "114",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "414",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "144144144",
"output": "YES"
},
{
"input": "1444",
"output": "NO"
},
{
"input": "11",
"output": "YES"
},
{
"input": "141414141",
"output": "YES"
},
{
"input": "11110111",
"output": "NO"
},
{
"input": "114114144",
"output": "YES"
},
{
"input": "444",
"output": "NO"
},
{
"input": "9999",
"output": "NO"
},
{
"input": "111444",
"output": "NO"
},
{
"input": "11114",
"output": "YES"
},
{
"input": "41111",
"output": "NO"
},
{
"input": "114414441",
"output": "NO"
},
{
"input": "144414441",
"output": "NO"
},
{
"input": "144244144",
"output": "NO"
},
{
"input": "111111111",
"output": "YES"
},
{
"input": "144444444",
"output": "NO"
},
{
"input": "444444444",
"output": "NO"
},
{
"input": "141441441",
"output": "YES"
},
{
"input": "441",
"output": "NO"
},
{
"input": "15",
"output": "NO"
},
{
"input": "14444",
"output": "NO"
},
{
"input": "11444",
"output": "NO"
},
{
"input": "144",
"output": "YES"
},
{
"input": "1414414",
"output": "YES"
},
{
"input": "141444",
"output": "NO"
},
{
"input": "14144",
"output": "YES"
},
{
"input": "4",
"output": "NO"
},
{
"input": "1144",
"output": "YES"
},
{
"input": "141111444",
"output": "NO"
},
{
"input": "14414414",
"output": "YES"
},
{
"input": "141414144",
"output": "YES"
},
{
"input": "1414",
"output": "YES"
},
{
"input": "1441",
"output": "YES"
},
{
"input": "12",
"output": "NO"
},
{
"input": "144144141",
"output": "YES"
},
{
"input": "144144",
"output": "YES"
},
{
"input": "14414411",
"output": "YES"
},
{
"input": "14414",
"output": "YES"
}
] | 1,627,759,368 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 2,000 | 6,963,200 | s = input()
while len(s) > 3:
current = s[:3]
if current.find("144") != -1:
s = s.replace("144", "")
elif current.find("14") != -1:
s = s.replace("14", "",1)
elif current.find("1") != -1:
s = s.replace("1", "", 1)
s = s.replace("144", "")
s = s.replace("14", "")
s = s.replace("1", "")
if s == "":
print("YES")
else:
print("NO")
| Title: Magic Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a magic number or not.
Input Specification:
The first line of input contains an integer *n*, (1<=≤<=*n*<=≤<=109). This number doesn't contain leading zeros.
Output Specification:
Print "YES" if *n* is a magic number or print "NO" if it's not.
Demo Input:
['114114\n', '1111\n', '441231\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none | ```python
s = input()
while len(s) > 3:
current = s[:3]
if current.find("144") != -1:
s = s.replace("144", "")
elif current.find("14") != -1:
s = s.replace("14", "",1)
elif current.find("1") != -1:
s = s.replace("1", "", 1)
s = s.replace("144", "")
s = s.replace("14", "")
s = s.replace("1", "")
if s == "":
print("YES")
else:
print("NO")
``` | 0 | |
450 | B | Jzzhu and Sequences | PROGRAMMING | 1,300 | [
"implementation",
"math"
] | null | null | Jzzhu has invented a kind of sequences, they meet the following property:
You are given *x* and *y*, please calculate *f**n* modulo 1000000007 (109<=+<=7). | The first line contains two integers *x* and *y* (|*x*|,<=|*y*|<=≤<=109). The second line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109). | Output a single integer representing *f**n* modulo 1000000007 (109<=+<=7). | [
"2 3\n3\n",
"0 -1\n2\n"
] | [
"1\n",
"1000000006\n"
] | In the first sample, *f*<sub class="lower-index">2</sub> = *f*<sub class="lower-index">1</sub> + *f*<sub class="lower-index">3</sub>, 3 = 2 + *f*<sub class="lower-index">3</sub>, *f*<sub class="lower-index">3</sub> = 1.
In the second sample, *f*<sub class="lower-index">2</sub> = - 1; - 1 modulo (10<sup class="upper-index">9</sup> + 7) equals (10<sup class="upper-index">9</sup> + 6). | 1,000 | [
{
"input": "2 3\n3",
"output": "1"
},
{
"input": "0 -1\n2",
"output": "1000000006"
},
{
"input": "-9 -11\n12345",
"output": "1000000005"
},
{
"input": "0 0\n1000000000",
"output": "0"
},
{
"input": "-1000000000 1000000000\n2000000000",
"output": "1000000000"
},
{
"input": "-12345678 12345678\n1912345678",
"output": "12345678"
},
{
"input": "728374857 678374857\n1928374839",
"output": "950000007"
},
{
"input": "278374837 992837483\n1000000000",
"output": "721625170"
},
{
"input": "-693849384 502938493\n982838498",
"output": "502938493"
},
{
"input": "-783928374 983738273\n992837483",
"output": "16261734"
},
{
"input": "-872837483 -682738473\n999999999",
"output": "190099010"
},
{
"input": "-892837483 -998273847\n999283948",
"output": "892837483"
},
{
"input": "-283938494 738473848\n1999999999",
"output": "716061513"
},
{
"input": "-278374857 819283838\n1",
"output": "721625150"
},
{
"input": "-1000000000 123456789\n1",
"output": "7"
},
{
"input": "-529529529 -524524524\n2",
"output": "475475483"
},
{
"input": "1 2\n2000000000",
"output": "2"
},
{
"input": "-1 -2\n2000000000",
"output": "1000000005"
},
{
"input": "1 2\n1999999999",
"output": "1"
},
{
"input": "1 2\n1999999998",
"output": "1000000006"
},
{
"input": "1 2\n1999999997",
"output": "1000000005"
},
{
"input": "1 2\n1999999996",
"output": "1000000006"
},
{
"input": "69975122 366233206\n1189460676",
"output": "703741923"
},
{
"input": "812229413 904420051\n806905621",
"output": "812229413"
},
{
"input": "872099024 962697902\n1505821695",
"output": "90598878"
},
{
"input": "887387283 909670917\n754835014",
"output": "112612724"
},
{
"input": "37759824 131342932\n854621399",
"output": "868657075"
},
{
"input": "-246822123 800496170\n626323615",
"output": "753177884"
},
{
"input": "-861439463 974126967\n349411083",
"output": "835566423"
},
{
"input": "-69811049 258093841\n1412447",
"output": "741906166"
},
{
"input": "844509330 -887335829\n123329059",
"output": "844509330"
},
{
"input": "83712471 -876177148\n1213284777",
"output": "40110388"
},
{
"input": "598730524 -718984219\n1282749880",
"output": "401269483"
},
{
"input": "-474244697 -745885656\n1517883612",
"output": "271640959"
},
{
"input": "-502583588 -894906953\n1154189557",
"output": "497416419"
},
{
"input": "-636523651 -873305815\n154879215",
"output": "763217843"
},
{
"input": "721765550 594845720\n78862386",
"output": "126919830"
},
{
"input": "364141461 158854993\n1337196589",
"output": "364141461"
},
{
"input": "878985260 677031952\n394707801",
"output": "798046699"
},
{
"input": "439527072 -24854079\n1129147002",
"output": "464381151"
},
{
"input": "840435009 -612103127\n565968986",
"output": "387896880"
},
{
"input": "875035447 -826471373\n561914518",
"output": "124964560"
},
{
"input": "-342526698 305357084\n70776744",
"output": "352116225"
},
{
"input": "-903244186 899202229\n1527859274",
"output": "899202229"
},
{
"input": "-839482546 815166320\n1127472130",
"output": "839482546"
},
{
"input": "-976992569 -958313041\n1686580818",
"output": "981320479"
},
{
"input": "-497338894 -51069176\n737081851",
"output": "502661113"
},
{
"input": "-697962643 -143148799\n1287886520",
"output": "856851208"
},
{
"input": "-982572938 -482658433\n1259858332",
"output": "982572938"
},
{
"input": "123123 78817\n2000000000",
"output": "78817"
},
{
"input": "1000000000 -1000000000\n3",
"output": "14"
},
{
"input": "-1000000000 1000000000\n6",
"output": "14"
},
{
"input": "2 3\n6",
"output": "1000000006"
},
{
"input": "0 -1\n6",
"output": "1"
},
{
"input": "500000000 -1000000000\n600000003",
"output": "500000014"
},
{
"input": "-1000000000 1000000000\n3",
"output": "999999993"
},
{
"input": "1 3\n6",
"output": "1000000005"
},
{
"input": "1 2\n12",
"output": "1000000006"
},
{
"input": "7 -1000000000\n3",
"output": "0"
},
{
"input": "-999999997 999999997\n6",
"output": "20"
},
{
"input": "3 4\n6",
"output": "1000000006"
},
{
"input": "-1 2\n6",
"output": "1000000004"
},
{
"input": "2 3\n12",
"output": "1000000006"
},
{
"input": "4 18\n6",
"output": "999999993"
},
{
"input": "1 2\n6",
"output": "1000000006"
},
{
"input": "1000000000 -1000000000\n6",
"output": "999999993"
},
{
"input": "999999999 -999999999\n3",
"output": "16"
},
{
"input": "-1 0\n1",
"output": "1000000006"
},
{
"input": "1000000000 -1000000000\n9",
"output": "14"
},
{
"input": "999999999 -1000000000\n12",
"output": "999999992"
},
{
"input": "1000000000 -7\n3",
"output": "0"
},
{
"input": "-5 5\n6",
"output": "999999997"
},
{
"input": "5 9\n6",
"output": "1000000003"
},
{
"input": "-15 -10\n1",
"output": "999999992"
}
] | 1,662,431,896 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 19 | 62 | 0 | x, y = map(int, input().split())
n = int(input())
n = n % 6
m = 10**9 + 7
if n == 1:
print(x % m)
elif n == 2:
print(y % m)
elif n == 3:
print((y-x) % m)
elif n == 4:
print((-x) % m)
elif n == 5:
print((-y) % m)
else:
print((-x - y) % m) | Title: Jzzhu and Sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jzzhu has invented a kind of sequences, they meet the following property:
You are given *x* and *y*, please calculate *f**n* modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains two integers *x* and *y* (|*x*|,<=|*y*|<=≤<=109). The second line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
Output Specification:
Output a single integer representing *f**n* modulo 1000000007 (109<=+<=7).
Demo Input:
['2 3\n3\n', '0 -1\n2\n']
Demo Output:
['1\n', '1000000006\n']
Note:
In the first sample, *f*<sub class="lower-index">2</sub> = *f*<sub class="lower-index">1</sub> + *f*<sub class="lower-index">3</sub>, 3 = 2 + *f*<sub class="lower-index">3</sub>, *f*<sub class="lower-index">3</sub> = 1.
In the second sample, *f*<sub class="lower-index">2</sub> = - 1; - 1 modulo (10<sup class="upper-index">9</sup> + 7) equals (10<sup class="upper-index">9</sup> + 6). | ```python
x, y = map(int, input().split())
n = int(input())
n = n % 6
m = 10**9 + 7
if n == 1:
print(x % m)
elif n == 2:
print(y % m)
elif n == 3:
print((y-x) % m)
elif n == 4:
print((-x) % m)
elif n == 5:
print((-y) % m)
else:
print((-x - y) % m)
``` | 0 | |
628 | B | New Skateboard | PROGRAMMING | 1,300 | [
"dp"
] | null | null | Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4.
You are given a string *s* consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero.
A substring of a string is a nonempty sequence of consecutive characters.
For example if string *s* is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. | The only line contains string *s* (1<=≤<=|*s*|<=≤<=3·105). The string *s* contains only digits from 0 to 9. | Print integer *a* — the number of substrings of the string *s* that are divisible by 4.
Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type. | [
"124\n",
"04\n",
"5810438174\n"
] | [
"4\n",
"3\n",
"9\n"
] | none | 0 | [
{
"input": "124",
"output": "4"
},
{
"input": "04",
"output": "3"
},
{
"input": "5810438174",
"output": "9"
},
{
"input": "1",
"output": "0"
},
{
"input": "039",
"output": "1"
},
{
"input": "97247",
"output": "6"
},
{
"input": "5810438174",
"output": "9"
},
{
"input": "12883340691714056185860211260984431382156326935244",
"output": "424"
},
{
"input": "2144315253572020279108092911160072328496568665545836825277616363478721946398140227406814602154768031",
"output": "1528"
},
{
"input": "80124649014054971081213608137817466046254652492627741860478258558206397113198232823859870363821007188476405951611069347299689170240023979048198711745011542774268179055311013054073075176122755643483380248999657649211459997766221072399103579977409770898200358240970169892326442892826731631357561876251276209119521202062222947560634301788787748428236988789594458520867663257476744168528121470923031438015546006185059454402637036376247785881323277542968298682307854655591317046086531554595892680980142608",
"output": "30826"
},
{
"input": "123456",
"output": "7"
},
{
"input": "4",
"output": "1"
},
{
"input": "123",
"output": "1"
}
] | 1,587,990,919 | 2,147,483,647 | Python 3 | OK | TESTS | 22 | 436 | 1,024,000 | s=input()
count=0
for i in range(len(s)):
if(int(s[i])%4==0):
count=count+1
for i in range(len(s)-1):
num=s[i]+s[i+1]
if(int(num)%4==0):
count=count+i+1
print(count)
| Title: New Skateboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4.
You are given a string *s* consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero.
A substring of a string is a nonempty sequence of consecutive characters.
For example if string *s* is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input Specification:
The only line contains string *s* (1<=≤<=|*s*|<=≤<=3·105). The string *s* contains only digits from 0 to 9.
Output Specification:
Print integer *a* — the number of substrings of the string *s* that are divisible by 4.
Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Demo Input:
['124\n', '04\n', '5810438174\n']
Demo Output:
['4\n', '3\n', '9\n']
Note:
none | ```python
s=input()
count=0
for i in range(len(s)):
if(int(s[i])%4==0):
count=count+1
for i in range(len(s)-1):
num=s[i]+s[i+1]
if(int(num)%4==0):
count=count+i+1
print(count)
``` | 3 | |
616 | B | Dinner with Emma | PROGRAMMING | 1,000 | [
"games",
"greedy"
] | null | null | Jack decides to invite Emma out for a dinner. Jack is a modest student, he doesn't want to go to an expensive restaurant. Emma is a girl with high taste, she prefers elite places.
Munhattan consists of *n* streets and *m* avenues. There is exactly one restaurant on the intersection of each street and avenue. The streets are numbered with integers from 1 to *n* and the avenues are numbered with integers from 1 to *m*. The cost of dinner in the restaurant at the intersection of the *i*-th street and the *j*-th avenue is *c**ij*.
Jack and Emma decide to choose the restaurant in the following way. Firstly Emma chooses the street to dinner and then Jack chooses the avenue. Emma and Jack makes their choice optimally: Emma wants to maximize the cost of the dinner, Jack wants to minimize it. Emma takes into account that Jack wants to minimize the cost of the dinner. Find the cost of the dinner for the couple in love. | The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of streets and avenues in Munhattan.
Each of the next *n* lines contains *m* integers *c**ij* (1<=≤<=*c**ij*<=≤<=109) — the cost of the dinner in the restaurant on the intersection of the *i*-th street and the *j*-th avenue. | Print the only integer *a* — the cost of the dinner for Jack and Emma. | [
"3 4\n4 1 3 5\n2 2 2 2\n5 4 5 1\n",
"3 3\n1 2 3\n2 3 1\n3 1 2\n"
] | [
"2\n",
"1\n"
] | In the first example if Emma chooses the first or the third streets Jack can choose an avenue with the cost of the dinner 1. So she chooses the second street and Jack chooses any avenue. The cost of the dinner is 2.
In the second example regardless of Emma's choice Jack can choose a restaurant with the cost of the dinner 1. | 0 | [
{
"input": "3 4\n4 1 3 5\n2 2 2 2\n5 4 5 1",
"output": "2"
},
{
"input": "3 3\n1 2 3\n2 3 1\n3 1 2",
"output": "1"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 10\n74 35 82 39 1 84 29 41 70 12",
"output": "1"
},
{
"input": "10 1\n44\n23\n65\n17\n48\n29\n49\n88\n91\n85",
"output": "91"
},
{
"input": "10 10\n256 72 455 45 912 506 235 68 951 92\n246 305 45 212 788 621 449 876 459 899\n732 107 230 357 370 610 997 669 61 192\n131 93 481 527 983 920 825 540 435 54\n777 682 984 20 337 480 264 137 249 502\n51 467 479 228 923 752 714 436 199 973\n3 91 612 571 631 212 751 84 886 948\n252 130 583 23 194 985 234 978 709 16\n636 991 203 469 719 540 184 902 503 652\n826 680 150 284 37 987 360 183 447 51",
"output": "184"
},
{
"input": "1 1\n1000000000",
"output": "1000000000"
},
{
"input": "2 1\n999999999\n1000000000",
"output": "1000000000"
}
] | 1,589,624,551 | 1,051 | Python 3 | OK | TESTS | 16 | 109 | 6,963,200 | n,m=map(int,input().split())
b=[]
for i in range(n):
a=list(map(int,input().split()))
b.append(min(a))
print(max(b)) | Title: Dinner with Emma
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jack decides to invite Emma out for a dinner. Jack is a modest student, he doesn't want to go to an expensive restaurant. Emma is a girl with high taste, she prefers elite places.
Munhattan consists of *n* streets and *m* avenues. There is exactly one restaurant on the intersection of each street and avenue. The streets are numbered with integers from 1 to *n* and the avenues are numbered with integers from 1 to *m*. The cost of dinner in the restaurant at the intersection of the *i*-th street and the *j*-th avenue is *c**ij*.
Jack and Emma decide to choose the restaurant in the following way. Firstly Emma chooses the street to dinner and then Jack chooses the avenue. Emma and Jack makes their choice optimally: Emma wants to maximize the cost of the dinner, Jack wants to minimize it. Emma takes into account that Jack wants to minimize the cost of the dinner. Find the cost of the dinner for the couple in love.
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of streets and avenues in Munhattan.
Each of the next *n* lines contains *m* integers *c**ij* (1<=≤<=*c**ij*<=≤<=109) — the cost of the dinner in the restaurant on the intersection of the *i*-th street and the *j*-th avenue.
Output Specification:
Print the only integer *a* — the cost of the dinner for Jack and Emma.
Demo Input:
['3 4\n4 1 3 5\n2 2 2 2\n5 4 5 1\n', '3 3\n1 2 3\n2 3 1\n3 1 2\n']
Demo Output:
['2\n', '1\n']
Note:
In the first example if Emma chooses the first or the third streets Jack can choose an avenue with the cost of the dinner 1. So she chooses the second street and Jack chooses any avenue. The cost of the dinner is 2.
In the second example regardless of Emma's choice Jack can choose a restaurant with the cost of the dinner 1. | ```python
n,m=map(int,input().split())
b=[]
for i in range(n):
a=list(map(int,input().split()))
b.append(min(a))
print(max(b))
``` | 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"
}
] | 1,612,257,297 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | x=input()
for i in range(len(x)):
if x[i]=='a' or x[i]=='A' or x[i]=='s' or x[i]=='S':
print(x.lower())
break
c=0
a=len(x)/2
for i in x:
if i.isupper()==True:
c+=1
if c==a:
print(x.lower())
break
else:
print(x.upper()) | 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
x=input()
for i in range(len(x)):
if x[i]=='a' or x[i]=='A' or x[i]=='s' or x[i]=='S':
print(x.lower())
break
c=0
a=len(x)/2
for i in x:
if i.isupper()==True:
c+=1
if c==a:
print(x.lower())
break
else:
print(x.upper())
``` | 0 |
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"
}
] | 1,699,546,379 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 92 | 0 | colors = list(map(int, input().split()))
colors2 = [] #without repetitions
counter = 0
for i in range(4):
if colors[i] in colors2:
counter += 1
colors2.append(colors[i])
print(f'{counter}') | 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
colors = list(map(int, input().split()))
colors2 = [] #without repetitions
counter = 0
for i in range(4):
if colors[i] in colors2:
counter += 1
colors2.append(colors[i])
print(f'{counter}')
``` | 3 | |
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"
}
] | 1,615,637,291 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 92 | 0 | a,b=map(int,input().split())
s=str(input())
c=0
for i in range(len(s)-1):
if s[i]=="A" and s[i+1]=="B":
c+=1
elif s[i]=="B" and s[i+1]=="A":
c+=1
else:
c+=0
if b==c:
print("NO")
else:
print("YES")
| 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
a,b=map(int,input().split())
s=str(input())
c=0
for i in range(len(s)-1):
if s[i]=="A" and s[i+1]=="B":
c+=1
elif s[i]=="B" and s[i+1]=="A":
c+=1
else:
c+=0
if b==c:
print("NO")
else:
print("YES")
``` | 0 | |
438 | D | The Child and Sequence | PROGRAMMING | 2,300 | [
"data structures",
"math"
] | null | null | At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.
Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array *a*[1],<=*a*[2],<=...,<=*a*[*n*]. Then he should perform a sequence of *m* operations. An operation can be one of the following:
1. Print operation *l*,<=*r*. Picks should write down the value of . 1. Modulo operation *l*,<=*r*,<=*x*. Picks should perform assignment *a*[*i*]<==<=*a*[*i*] *mod* *x* for each *i* (*l*<=≤<=*i*<=≤<=*r*). 1. Set operation *k*,<=*x*. Picks should set the value of *a*[*k*] to *x* (in other words perform an assignment *a*[*k*]<==<=*x*).
Can you help Picks to perform the whole sequence of operations? | The first line of input contains two integer: *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers, separated by space: *a*[1],<=*a*[2],<=...,<=*a*[*n*] (1<=≤<=*a*[*i*]<=≤<=109) — initial value of array elements.
Each of the next *m* lines begins with a number *type* .
- If *type*<==<=1, there will be two integers more in the line: *l*,<=*r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), which correspond the operation 1. - If *type*<==<=2, there will be three integers more in the line: *l*,<=*r*,<=*x* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*x*<=≤<=109), which correspond the operation 2. - If *type*<==<=3, there will be two integers more in the line: *k*,<=*x* (1<=≤<=*k*<=≤<=*n*; 1<=≤<=*x*<=≤<=109), which correspond the operation 3. | For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer. | [
"5 5\n1 2 3 4 5\n2 3 5 4\n3 3 5\n1 2 5\n2 1 3 3\n1 1 3\n",
"10 10\n6 9 6 7 6 1 10 10 9 5\n1 3 9\n2 7 10 9\n2 5 10 8\n1 4 7\n3 3 7\n2 7 9 9\n1 2 4\n1 6 6\n1 5 9\n3 1 10\n"
] | [
"8\n5\n",
"49\n15\n23\n1\n9\n"
] | Consider the first testcase:
- At first, *a* = {1, 2, 3, 4, 5}. - After operation 1, *a* = {1, 2, 3, 0, 1}. - After operation 2, *a* = {1, 2, 5, 0, 1}. - At operation 3, 2 + 5 + 0 + 1 = 8. - After operation 4, *a* = {1, 2, 2, 0, 1}. <li> At operation 5, 1 + 2 + 2 = 5. <ul> | 2,000 | [
{
"input": "5 5\n1 2 3 4 5\n2 3 5 4\n3 3 5\n1 2 5\n2 1 3 3\n1 1 3",
"output": "8\n5"
},
{
"input": "10 10\n6 9 6 7 6 1 10 10 9 5\n1 3 9\n2 7 10 9\n2 5 10 8\n1 4 7\n3 3 7\n2 7 9 9\n1 2 4\n1 6 6\n1 5 9\n3 1 10",
"output": "49\n15\n23\n1\n9"
},
{
"input": "1 1\n1\n1 1 1",
"output": "1"
},
{
"input": "1 1\n1000000000\n1 1 1",
"output": "1000000000"
}
] | 1,690,707,656 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | def create_tree(nums, l, r):
if l == r:
return nums[l]
mid = (l + r) // 2
return create_tree(nums, l, mid) + create_tree(nums, mid + 1, r)
def update_value(tree, lazy, root, i, val):
n = len(tree) // 2
i += n
tree[i] = val
while i > 1:
i //= 2
tree[i] = tree[2 * i] + tree[2 * i + 1] + lazy[i] * (root[i + 1] - root[i])
def mod_value(tree, lazy, root, i, j, mod):
n = len(tree) // 2
i += n
j += n
while i <= j:
if i % 2 == 1:
lazy[i] = (lazy[i] + mod) % mod
tree[i] = (tree[i] + mod * (root[i + 1] - root[i])) % mod
i += 1
if j % 2 == 0:
lazy[j] = (lazy[j] + mod) % mod
tree[j] = (tree[j] + mod * (root[j + 1] - root[j])) % mod
j -= 1
i //= 2
j //= 2
def range_sum(tree, lazy, root, i, j):
n = len(tree) // 2
i += n
j += n
total = 0
while i <= j:
if i % 2 == 1:
total += tree[i] + lazy[i] * (root[i + 1] - root[i])
i += 1
if j % 2 == 0:
total += tree[j] + lazy[j] * (root[j + 1] - root[j])
j -= 1
i //= 2
j //= 2
return total
def perform_operations(n, m, arr, operations):
unique_vals = list(set(arr))
unique_vals.sort()
val_to_index = {val: i for i, val in enumerate(unique_vals)}
root = [unique_vals[i] for i in range(len(unique_vals))] # Store the unique values as the root of the segment tree
tree = [0] * (2 * len(unique_vals))
lazy = [0] * (2 * len(unique_vals))
for i in range(n):
arr[i] = val_to_index[arr[i]]
for i in range(len(root) - 1, 0, -1):
root.insert(i, (root[i] + root[i - 1]) // 2)
for op in operations:
operation_type = op[0]
if operation_type == 1:
l, r = op[1], op[2]
ans = range_sum(tree, lazy, root, l - 1, r - 1)
print(ans)
elif operation_type == 2:
l, r, x = op[1], op[2], op[3]
mod_value(tree, lazy, root, l - 1, r - 1, val_to_index[x])
elif operation_type == 3:
k, x = op[1], op[2]
update_value(tree, lazy, root, k - 1, val_to_index[x])
if __name__ == "__main__":
n, m = map(int, input().split())
arr = list(map(int, input().split()))
operations = []
for _ in range(m):
operation = list(map(int, input().split()))
operations.append(operation)
perform_operations(n, m, arr, operations) | Title: The Child and Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.
Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array *a*[1],<=*a*[2],<=...,<=*a*[*n*]. Then he should perform a sequence of *m* operations. An operation can be one of the following:
1. Print operation *l*,<=*r*. Picks should write down the value of . 1. Modulo operation *l*,<=*r*,<=*x*. Picks should perform assignment *a*[*i*]<==<=*a*[*i*] *mod* *x* for each *i* (*l*<=≤<=*i*<=≤<=*r*). 1. Set operation *k*,<=*x*. Picks should set the value of *a*[*k*] to *x* (in other words perform an assignment *a*[*k*]<==<=*x*).
Can you help Picks to perform the whole sequence of operations?
Input Specification:
The first line of input contains two integer: *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers, separated by space: *a*[1],<=*a*[2],<=...,<=*a*[*n*] (1<=≤<=*a*[*i*]<=≤<=109) — initial value of array elements.
Each of the next *m* lines begins with a number *type* .
- If *type*<==<=1, there will be two integers more in the line: *l*,<=*r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), which correspond the operation 1. - If *type*<==<=2, there will be three integers more in the line: *l*,<=*r*,<=*x* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*x*<=≤<=109), which correspond the operation 2. - If *type*<==<=3, there will be two integers more in the line: *k*,<=*x* (1<=≤<=*k*<=≤<=*n*; 1<=≤<=*x*<=≤<=109), which correspond the operation 3.
Output Specification:
For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.
Demo Input:
['5 5\n1 2 3 4 5\n2 3 5 4\n3 3 5\n1 2 5\n2 1 3 3\n1 1 3\n', '10 10\n6 9 6 7 6 1 10 10 9 5\n1 3 9\n2 7 10 9\n2 5 10 8\n1 4 7\n3 3 7\n2 7 9 9\n1 2 4\n1 6 6\n1 5 9\n3 1 10\n']
Demo Output:
['8\n5\n', '49\n15\n23\n1\n9\n']
Note:
Consider the first testcase:
- At first, *a* = {1, 2, 3, 4, 5}. - After operation 1, *a* = {1, 2, 3, 0, 1}. - After operation 2, *a* = {1, 2, 5, 0, 1}. - At operation 3, 2 + 5 + 0 + 1 = 8. - After operation 4, *a* = {1, 2, 2, 0, 1}. <li> At operation 5, 1 + 2 + 2 = 5. <ul> | ```python
def create_tree(nums, l, r):
if l == r:
return nums[l]
mid = (l + r) // 2
return create_tree(nums, l, mid) + create_tree(nums, mid + 1, r)
def update_value(tree, lazy, root, i, val):
n = len(tree) // 2
i += n
tree[i] = val
while i > 1:
i //= 2
tree[i] = tree[2 * i] + tree[2 * i + 1] + lazy[i] * (root[i + 1] - root[i])
def mod_value(tree, lazy, root, i, j, mod):
n = len(tree) // 2
i += n
j += n
while i <= j:
if i % 2 == 1:
lazy[i] = (lazy[i] + mod) % mod
tree[i] = (tree[i] + mod * (root[i + 1] - root[i])) % mod
i += 1
if j % 2 == 0:
lazy[j] = (lazy[j] + mod) % mod
tree[j] = (tree[j] + mod * (root[j + 1] - root[j])) % mod
j -= 1
i //= 2
j //= 2
def range_sum(tree, lazy, root, i, j):
n = len(tree) // 2
i += n
j += n
total = 0
while i <= j:
if i % 2 == 1:
total += tree[i] + lazy[i] * (root[i + 1] - root[i])
i += 1
if j % 2 == 0:
total += tree[j] + lazy[j] * (root[j + 1] - root[j])
j -= 1
i //= 2
j //= 2
return total
def perform_operations(n, m, arr, operations):
unique_vals = list(set(arr))
unique_vals.sort()
val_to_index = {val: i for i, val in enumerate(unique_vals)}
root = [unique_vals[i] for i in range(len(unique_vals))] # Store the unique values as the root of the segment tree
tree = [0] * (2 * len(unique_vals))
lazy = [0] * (2 * len(unique_vals))
for i in range(n):
arr[i] = val_to_index[arr[i]]
for i in range(len(root) - 1, 0, -1):
root.insert(i, (root[i] + root[i - 1]) // 2)
for op in operations:
operation_type = op[0]
if operation_type == 1:
l, r = op[1], op[2]
ans = range_sum(tree, lazy, root, l - 1, r - 1)
print(ans)
elif operation_type == 2:
l, r, x = op[1], op[2], op[3]
mod_value(tree, lazy, root, l - 1, r - 1, val_to_index[x])
elif operation_type == 3:
k, x = op[1], op[2]
update_value(tree, lazy, root, k - 1, val_to_index[x])
if __name__ == "__main__":
n, m = map(int, input().split())
arr = list(map(int, input().split()))
operations = []
for _ in range(m):
operation = list(map(int, input().split()))
operations.append(operation)
perform_operations(n, m, arr, operations)
``` | 0 | |
461 | A | Appleman and Toastman | PROGRAMMING | 1,200 | [
"greedy",
"sortings"
] | null | null | Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman. | Print a single integer — the largest possible score. | [
"3\n3 1 5\n",
"1\n10\n"
] | [
"26\n",
"10\n"
] | Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | 500 | [
{
"input": "3\n3 1 5",
"output": "26"
},
{
"input": "1\n10",
"output": "10"
},
{
"input": "10\n8 10 2 5 6 2 4 7 2 1",
"output": "376"
},
{
"input": "10\n171308 397870 724672 431255 228496 892002 542924 718337 888642 161821",
"output": "40204082"
},
{
"input": "10\n1 2 2 2 4 5 6 7 8 10",
"output": "376"
},
{
"input": "10\n161821 171308 228496 397870 431255 542924 718337 724672 888642 892002",
"output": "40204082"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "1\n1000000",
"output": "1000000"
},
{
"input": "10\n10 8 7 6 5 4 2 2 2 1",
"output": "376"
},
{
"input": "10\n892002 888642 724672 718337 542924 431255 397870 228496 171308 161821",
"output": "40204082"
},
{
"input": "10\n5 2 6 10 10 10 10 2 2 5",
"output": "485"
},
{
"input": "10\n431255 724672 228496 397870 397870 397870 397870 724672 888642 431255",
"output": "36742665"
},
{
"input": "10\n2 2 2 5 5 6 10 10 10 10",
"output": "485"
},
{
"input": "10\n228496 397870 397870 397870 397870 431255 431255 724672 724672 888642",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 6 5 5 2 2 2",
"output": "485"
},
{
"input": "10\n888642 724672 724672 431255 431255 397870 397870 397870 397870 228496",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 10 10 10 10 10 10",
"output": "640"
},
{
"input": "10\n1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000",
"output": "64000000"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "2\n1 2",
"output": "6"
},
{
"input": "2\n2 3",
"output": "10"
},
{
"input": "2\n1 1",
"output": "4"
}
] | 1,634,859,922 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 4,505,600 | x=input()
n=input().split()
n.sort(reverse=True)
su=0
while len(n)>1:
for i in n :
su+=int(i)
su+=int(n[-1])
n.pop()
su+=int(n[0])
print(su)
| Title: Appleman and Toastman
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman.
Output Specification:
Print a single integer — the largest possible score.
Demo Input:
['3\n3 1 5\n', '1\n10\n']
Demo Output:
['26\n', '10\n']
Note:
Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | ```python
x=input()
n=input().split()
n.sort(reverse=True)
su=0
while len(n)>1:
for i in n :
su+=int(i)
su+=int(n[-1])
n.pop()
su+=int(n[0])
print(su)
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn! | The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length. | Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise. | [
"aaba\nabaa\n",
"aabb\nabab\n"
] | [
"YES\n",
"NO\n"
] | In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | 0 | [
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "ab\nbb",
"output": "NO"
},
{
"input": "zzaa\naazz",
"output": "YES"
},
{
"input": "azza\nzaaz",
"output": "YES"
},
{
"input": "abc\nabc",
"output": "YES"
},
{
"input": "abc\nacb",
"output": "NO"
},
{
"input": "azzz\nzzaz",
"output": "YES"
},
{
"input": "abcd\ndcab",
"output": "YES"
},
{
"input": "abcd\ncdab",
"output": "YES"
},
{
"input": "abcd\ndcba",
"output": "YES"
},
{
"input": "abcd\nacbd",
"output": "NO"
},
{
"input": "oloaxgddgujq\noloaxgujqddg",
"output": "YES"
},
{
"input": "uwzwdxfmosmqatyv\ndxfmzwwusomqvyta",
"output": "YES"
},
{
"input": "hagnzomowtledfdotnll\nledfdotnllomowthagnz",
"output": "YES"
},
{
"input": "snyaydaeobufdg\nsnyaydaeobufdg",
"output": "YES"
},
{
"input": "baaaaa\nabaaaa",
"output": "NO"
},
{
"input": "hhiisug\nmzdjwju",
"output": "NO"
},
{
"input": "bbbabbabaaab\naaaabbabbbbb",
"output": "NO"
},
{
"input": "bbaaab\naababb",
"output": "NO"
},
{
"input": "aabbaaaa\naaaaabab",
"output": "NO"
},
{
"input": "aab\naba",
"output": "NO"
},
{
"input": "abcddd\nbacddd",
"output": "NO"
},
{
"input": "qgiufelsfhanx\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "aabaababaaba\naababaaababa",
"output": "NO"
},
{
"input": "nocdqzdriyyil\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "zdmctxl\nkojqhgw",
"output": "NO"
},
{
"input": "yhwepqwyhwepqwyhwepqweahnqtueahnqtueahnqtuyhwepqwyhwepqwyhwepqwyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtueahnqtueahnqtueahnqtueahnqtu\neahnqtueahnqtueahnqtuyhwepqweahnqtuyhwepqwyhwepqweahnqtuyhwepqweahnqtuyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtuyhwepqwyhwepqwyhwepqw",
"output": "NO"
},
{
"input": "abc\nbac",
"output": "NO"
},
{
"input": "ottceez\npcstdvz",
"output": "NO"
}
] | 1,619,032,783 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 61 | 0 | def isEquiv(s1, s2):
if s1 == s2 or s1[::-1] == s2:
return True
if len(s1)%2 != 0 or len(s2)%2 != 0:
return False
else:
mid = len(s1)//2
a1 = s1[:mid]
a2 = s1[mid:]
b1 = s2[mid:]
b2 = s2[:mid]
return isEquiv(a1, b1) and isEquiv(a2, b2) or isEquiv(a1, b2) and isEquiv(a2, b1)
s1 = input()
s2 = input()
print("YES" if isEquil(s1, s2) else "NO") | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | ```python
def isEquiv(s1, s2):
if s1 == s2 or s1[::-1] == s2:
return True
if len(s1)%2 != 0 or len(s2)%2 != 0:
return False
else:
mid = len(s1)//2
a1 = s1[:mid]
a2 = s1[mid:]
b1 = s2[mid:]
b2 = s2[:mid]
return isEquiv(a1, b1) and isEquiv(a2, b2) or isEquiv(a1, b2) and isEquiv(a2, b1)
s1 = input()
s2 = input()
print("YES" if isEquil(s1, s2) else "NO")
``` | -1 | |
841 | A | Generous Kefa | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all. | The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends.
Next line contains string *s* — colors of baloons. | Answer to the task — «YES» or «NO» in a single line.
You can choose the case (lower or upper) for each letter arbitrary. | [
"4 2\naabb\n",
"6 3\naacaab\n"
] | [
"YES\n",
"NO\n"
] | In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.
In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO». | 500 | [
{
"input": "4 2\naabb",
"output": "YES"
},
{
"input": "6 3\naacaab",
"output": "NO"
},
{
"input": "2 2\nlu",
"output": "YES"
},
{
"input": "5 3\novvoo",
"output": "YES"
},
{
"input": "36 13\nbzbzcffczzcbcbzzfzbbfzfzzbfbbcbfccbf",
"output": "YES"
},
{
"input": "81 3\nooycgmvvrophvcvpoupepqllqttwcocuilvyxbyumdmmfapvpnxhjhxfuagpnntonibicaqjvwfhwxhbv",
"output": "NO"
},
{
"input": "100 100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "YES"
},
{
"input": "100 1\nnubcvvjvbjgnjsdkajimdcxvewbcytvfkihunycdrlconddlwgzjasjlsrttlrzsumzpyumpveglfqzmaofbshbojmwuwoxxvrod",
"output": "NO"
},
{
"input": "100 13\nvyldolgryldqrvoldvzvrdrgorlorszddtgqvrlisxxrxdxlqtvtgsrqlzixoyrozxzogqxlsgzdddzqrgitxxritoolzolgrtvl",
"output": "YES"
},
{
"input": "18 6\njzwtnkvmscqhmdlsxy",
"output": "YES"
},
{
"input": "21 2\nfscegcqgzesefghhwcexs",
"output": "NO"
},
{
"input": "32 22\ncduamsptaklqtxlyoutlzepxgyfkvngc",
"output": "YES"
},
{
"input": "49 27\noxyorfnkzwsfllnyvdhdanppuzrnbxehugvmlkgeymqjlmfxd",
"output": "YES"
},
{
"input": "50 24\nxxutzjwbggcwvxztttkmzovtmuwttzcbwoztttohzzxghuuthv",
"output": "YES"
},
{
"input": "57 35\nglxshztrqqfyxthqamagvtmrdparhelnzrqvcwqxjytkbuitovkdxueul",
"output": "YES"
},
{
"input": "75 23\nittttiiuitutuiiuuututiuttiuiuutuuuiuiuuuuttuuttuutuiiuiuiiuiitttuututuiuuii",
"output": "NO"
},
{
"input": "81 66\nfeqevfqfebhvubhuuvfuqheuqhbeeuebehuvhffvbqvqvfbqqvvhevqffbqqhvvqhfeehuhqeqhueuqqq",
"output": "YES"
},
{
"input": "93 42\npqeiafraiavfcteumflpcbpozcomlvpovlzdbldvoopnhdoeqaopzthiuzbzmeieiatthdeqovaqfipqlddllmfcrrnhb",
"output": "YES"
},
{
"input": "100 53\nizszyqyndzwzyzgsdagdwdazadiawizinagqqgczaqqnawgijziziawzszdjdcqjdjqiwgadydcnqisaayjiqqsscwwzjzaycwwc",
"output": "YES"
},
{
"input": "100 14\nvkrdcqbvkwuckpmnbydmczdxoagdsgtqxvhaxntdcxhjcrjyvukhugoglbmyoaqexgtcfdgemmizoniwtmisqqwcwfusmygollab",
"output": "YES"
},
{
"input": "100 42\naaaaaiiiiaiiiaaiaiiaaiiiiiaaaaaiaiiiaiiiiaiiiaaaaaiiiaaaiiaaiiiaiiiaiaaaiaiiiiaaiiiaiiaiaiiaiiiaaaia",
"output": "NO"
},
{
"input": "100 89\ntjbkmydejporbqhcbztkcumxjjgsrvxpuulbhzeeckkbchpbxwhedrlhjsabcexcohgdzouvsgphjdthpuqrlkgzxvqbuhqxdsmf",
"output": "YES"
},
{
"input": "100 100\njhpyiuuzizhubhhpxbbhpyxzhbpjphzppuhiahihiappbhuypyauhizpbibzixjbzxzpbphuiaypyujappuxiyuyaajaxjupbahb",
"output": "YES"
},
{
"input": "100 3\nsszoovvzysavsvzsozzvoozvysozsaszayaszasaysszzzysosyayyvzozovavzoyavsooaoyvoozvvozsaosvayyovazzszzssa",
"output": "NO"
},
{
"input": "100 44\ndluthkxwnorabqsukgnxnvhmsmzilyulpursnxkdsavgemiuizbyzebhyjejgqrvuckhaqtuvdmpziesmpmewpvozdanjyvwcdgo",
"output": "YES"
},
{
"input": "100 90\ntljonbnwnqounictqqctgonktiqoqlocgoblngijqokuquoolciqwnctgoggcbojtwjlculoikbggquqncittwnjbkgkgubnioib",
"output": "YES"
},
{
"input": "100 79\nykxptzgvbqxlregvkvucewtydvnhqhuggdsyqlvcfiuaiddnrrnstityyehiamrggftsqyduwxpuldztyzgmfkehprrneyvtknmf",
"output": "YES"
},
{
"input": "100 79\naagwekyovbviiqeuakbqbqifwavkfkutoriovgfmittulhwojaptacekdirgqoovlleeoqkkdukpadygfwavppohgdrmymmulgci",
"output": "YES"
},
{
"input": "100 93\nearrehrehenaddhdnrdddhdahnadndheeennrearrhraharddreaeraddhehhhrdnredanndneheddrraaneerreedhnadnerhdn",
"output": "YES"
},
{
"input": "100 48\nbmmaebaebmmmbbmxvmammbvvebvaemvbbaxvbvmaxvvmveaxmbbxaaemxmxvxxxvxbmmxaaaevvaxmvamvvmaxaxavexbmmbmmev",
"output": "YES"
},
{
"input": "100 55\nhsavbkehaaesffaeeffakhkhfehbbvbeasahbbbvkesbfvkefeesesevbsvfkbffakvshsbkahfkfakebsvafkbvsskfhfvaasss",
"output": "YES"
},
{
"input": "100 2\ncscffcffsccffsfsfffccssfsscfsfsssffcffsscfccssfffcfscfsscsccccfsssffffcfcfsfffcsfsccffscffcfccccfffs",
"output": "NO"
},
{
"input": "100 3\nzrgznxgdpgfoiifrrrsjfuhvtqxjlgochhyemismjnanfvvpzzvsgajcbsulxyeoepjfwvhkqogiiwqxjkrpsyaqdlwffoockxnc",
"output": "NO"
},
{
"input": "100 5\njbltyyfjakrjeodqepxpkjideulofbhqzxjwlarufwzwsoxhaexpydpqjvhybmvjvntuvhvflokhshpicbnfgsqsmrkrfzcrswwi",
"output": "NO"
},
{
"input": "100 1\nfnslnqktlbmxqpvcvnemxcutebdwepoxikifkzaaixzzydffpdxodmsxjribmxuqhueifdlwzytxkklwhljswqvlejedyrgguvah",
"output": "NO"
},
{
"input": "100 21\nddjenetwgwmdtjbpzssyoqrtirvoygkjlqhhdcjgeurqpunxpupwaepcqkbjjfhnvgpyqnozhhrmhfwararmlcvpgtnopvjqsrka",
"output": "YES"
},
{
"input": "100 100\nnjrhiauqlgkkpkuvciwzivjbbplipvhslqgdkfnmqrxuxnycmpheenmnrglotzuyxycosfediqcuadklsnzjqzfxnbjwvfljnlvq",
"output": "YES"
},
{
"input": "100 100\nbbbbbbbtbbttbtbbbttbttbtbbttttbbbtbttbbbtbttbtbbttttbbbbbtbbttbtbbtbttbbbtbtbtbtbtbtbbbttbbtbtbtbbtb",
"output": "YES"
},
{
"input": "14 5\nfssmmsfffmfmmm",
"output": "NO"
},
{
"input": "2 1\nff",
"output": "NO"
},
{
"input": "2 1\nhw",
"output": "YES"
},
{
"input": "2 2\nss",
"output": "YES"
},
{
"input": "1 1\nl",
"output": "YES"
},
{
"input": "100 50\nfffffttttttjjjuuuvvvvvdddxxxxwwwwgggbsssncccczzyyyyyhhhhhkrreeeeeeaaaaaiiillllllllooooqqqqqqmmpppppp",
"output": "YES"
},
{
"input": "100 50\nbbbbbbbbgggggggggggaaaaaaaahhhhhhhhhhpppppppppsssssssrrrrrrrrllzzzzzzzeeeeeeekkkkkkkwwwwwwwwjjjjjjjj",
"output": "YES"
},
{
"input": "100 50\nwwwwwwwwwwwwwwxxxxxxxxxxxxxxxxxxxxxxxxzzzzzzzzzzzzzzzzzzbbbbbbbbbbbbbbbbbbbbjjjjjjjjjjjjjjjjjjjjjjjj",
"output": "YES"
},
{
"input": "100 80\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm",
"output": "YES"
},
{
"input": "100 10\nbbttthhhhiiiiiiijjjjjvvvvpppssssseeeeeeewwwwgggkkkkkkkkmmmddddduuuzzzzllllnnnnnxxyyyffffccraaaaooooq",
"output": "YES"
},
{
"input": "100 20\nssssssssssbbbbbbbhhhhhhhyyyyyyyzzzzzzzzzzzzcccccxxxxxxxxxxddddmmmmmmmeeeeeeejjjjjjjjjwwwwwwwtttttttt",
"output": "YES"
},
{
"input": "1 2\na",
"output": "YES"
},
{
"input": "3 1\nabb",
"output": "NO"
},
{
"input": "2 1\naa",
"output": "NO"
},
{
"input": "2 1\nab",
"output": "YES"
},
{
"input": "6 2\naaaaaa",
"output": "NO"
},
{
"input": "8 4\naaaaaaaa",
"output": "NO"
},
{
"input": "4 2\naaaa",
"output": "NO"
},
{
"input": "4 3\naaaa",
"output": "NO"
},
{
"input": "1 3\na",
"output": "YES"
},
{
"input": "4 3\nzzzz",
"output": "NO"
},
{
"input": "4 1\naaaa",
"output": "NO"
},
{
"input": "3 4\nabc",
"output": "YES"
},
{
"input": "2 5\nab",
"output": "YES"
},
{
"input": "2 4\nab",
"output": "YES"
},
{
"input": "1 10\na",
"output": "YES"
},
{
"input": "5 2\nzzzzz",
"output": "NO"
},
{
"input": "53 26\naaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "NO"
},
{
"input": "4 1\nabab",
"output": "NO"
},
{
"input": "4 1\nabcb",
"output": "NO"
},
{
"input": "4 2\nabbb",
"output": "NO"
},
{
"input": "5 2\nabccc",
"output": "NO"
},
{
"input": "2 3\nab",
"output": "YES"
},
{
"input": "4 3\nbbbs",
"output": "YES"
},
{
"input": "10 2\nazzzzzzzzz",
"output": "NO"
},
{
"input": "1 2\nb",
"output": "YES"
},
{
"input": "1 3\nb",
"output": "YES"
},
{
"input": "4 5\nabcd",
"output": "YES"
},
{
"input": "4 6\naabb",
"output": "YES"
},
{
"input": "5 2\naaaab",
"output": "NO"
},
{
"input": "3 5\naaa",
"output": "YES"
},
{
"input": "5 3\nazzzz",
"output": "NO"
},
{
"input": "4 100\naabb",
"output": "YES"
},
{
"input": "3 10\naaa",
"output": "YES"
},
{
"input": "3 4\naaa",
"output": "YES"
},
{
"input": "12 5\naaaaabbbbbbb",
"output": "NO"
},
{
"input": "5 2\naabbb",
"output": "NO"
},
{
"input": "10 5\nzzzzzzzzzz",
"output": "NO"
},
{
"input": "2 4\naa",
"output": "YES"
},
{
"input": "1 5\na",
"output": "YES"
},
{
"input": "10 5\naaaaaaaaaa",
"output": "NO"
},
{
"input": "6 3\naaaaaa",
"output": "NO"
},
{
"input": "7 1\nabcdeee",
"output": "NO"
},
{
"input": "18 3\naaaaaabbbbbbcccccc",
"output": "NO"
},
{
"input": "8 2\naabbccdd",
"output": "YES"
},
{
"input": "4 2\nzzzz",
"output": "NO"
},
{
"input": "4 2\nabaa",
"output": "NO"
},
{
"input": "3 2\naaa",
"output": "NO"
},
{
"input": "3 1\nzzz",
"output": "NO"
},
{
"input": "5 4\nzzzzz",
"output": "NO"
},
{
"input": "6 2\naabbbc",
"output": "NO"
},
{
"input": "3 6\naaa",
"output": "YES"
},
{
"input": "2 1\nzz",
"output": "NO"
},
{
"input": "10 3\naaaeeeeeee",
"output": "NO"
},
{
"input": "4 5\naabb",
"output": "YES"
},
{
"input": "3 1\naaa",
"output": "NO"
},
{
"input": "5 2\naazzz",
"output": "NO"
},
{
"input": "6 2\nabbbbc",
"output": "NO"
},
{
"input": "4 2\nxxxx",
"output": "NO"
},
{
"input": "6 3\nzzzzzz",
"output": "NO"
},
{
"input": "3 2\nabb",
"output": "YES"
},
{
"input": "3 2\nzzz",
"output": "NO"
},
{
"input": "6 5\nzzzzzz",
"output": "NO"
},
{
"input": "6 3\nbcaaaa",
"output": "NO"
},
{
"input": "100 100\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "YES"
},
{
"input": "3 6\nabc",
"output": "YES"
}
] | 1,623,732,021 | 2,147,483,647 | Python 3 | OK | TESTS | 114 | 77 | 102,400 | from collections import Counter
n,k=map(int,input().split())
s=input()
l=list(Counter(s).values())
if max(l)<=k:
print("YES")
else:
print("NO") | Title: Generous Kefa
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends.
Next line contains string *s* — colors of baloons.
Output Specification:
Answer to the task — «YES» or «NO» in a single line.
You can choose the case (lower or upper) for each letter arbitrary.
Demo Input:
['4 2\naabb\n', '6 3\naacaab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.
In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO». | ```python
from collections import Counter
n,k=map(int,input().split())
s=input()
l=list(Counter(s).values())
if max(l)<=k:
print("YES")
else:
print("NO")
``` | 3 | |
630 | A | Again Twenty Five! | PROGRAMMING | 800 | [
"number theory"
] | null | null | The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City? | The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5. | Output the last two digits of 5*n* without spaces between them. | [
"2\n"
] | [
"25"
] | none | 0 | [
{
"input": "2",
"output": "25"
},
{
"input": "7",
"output": "25"
},
{
"input": "1000000000000000000",
"output": "25"
},
{
"input": "2000000000000000000",
"output": "25"
},
{
"input": "987654321012345678",
"output": "25"
}
] | 1,691,766,508 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 5 | 61 | 0 | '''
#list(map(int,input().split()))'''
# before you win, you have to fail
input();print(25)
| Title: Again Twenty Five!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.
Output Specification:
Output the last two digits of 5*n* without spaces between them.
Demo Input:
['2\n']
Demo Output:
['25']
Note:
none | ```python
'''
#list(map(int,input().split()))'''
# before you win, you have to fail
input();print(25)
``` | 3 | |
471 | D | MUH and Cube Walls | PROGRAMMING | 1,800 | [
"string suffix structures",
"strings"
] | null | null | Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got hold of lots of wooden cubes somewhere. They started making cube towers by placing the cubes one on top of the other. They defined multiple towers standing in a line as a wall. A wall can consist of towers of different heights.
Horace was the first to finish making his wall. He called his wall an elephant. The wall consists of *w* towers. The bears also finished making their wall but they didn't give it a name. Their wall consists of *n* towers. Horace looked at the bears' tower and wondered: in how many parts of the wall can he "see an elephant"? He can "see an elephant" on a segment of *w* contiguous towers if the heights of the towers on the segment match as a sequence the heights of the towers in Horace's wall. In order to see as many elephants as possible, Horace can raise and lower his wall. He even can lower the wall below the ground level (see the pictures to the samples for clarification).
Your task is to count the number of segments where Horace can "see an elephant". | The first line contains two integers *n* and *w* (1<=≤<=*n*,<=*w*<=≤<=2·105) — the number of towers in the bears' and the elephant's walls correspondingly. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the heights of the towers in the bears' wall. The third line contains *w* integers *b**i* (1<=≤<=*b**i*<=≤<=109) — the heights of the towers in the elephant's wall. | Print the number of segments in the bears' wall where Horace can "see an elephant". | [
"13 5\n2 4 5 5 4 3 2 2 2 3 3 2 1\n3 4 4 3 2\n"
] | [
"2"
] | The picture to the left shows Horace's wall from the sample, the picture to the right shows the bears' wall. The segments where Horace can "see an elephant" are in gray. | 2,000 | [
{
"input": "13 5\n2 4 5 5 4 3 2 2 2 3 3 2 1\n3 4 4 3 2",
"output": "2"
},
{
"input": "5 1\n8 71 1 24 2\n31",
"output": "5"
},
{
"input": "6 3\n2 2 2 2 2 2\n5 5 5",
"output": "4"
},
{
"input": "1 1\n576560149\n691846236",
"output": "1"
},
{
"input": "10 5\n5 10 8 10 11 9 11 12 10 15\n4 2 4 5 3",
"output": "2"
},
{
"input": "10 10\n6 8 1 2 5 1 4 24 2 4\n6 8 1 2 5 1 4 24 2 4",
"output": "1"
},
{
"input": "10 10\n6 8 1 2 5 1 14 24 12 4\n7 9 2 3 6 2 15 25 13 5",
"output": "1"
},
{
"input": "8 4\n1 2 3 4 5 6 7 8\n10 11 12 13",
"output": "5"
},
{
"input": "10 5\n172960147 951061917 502625539 319177159 720665763 402410416 880790711 734191412 452846733 449904402\n640219326 792464591 173792179 691347674 125427306",
"output": "0"
},
{
"input": "10 3\n2 3 3 2 1 1 3 1 3 1\n2 1 2",
"output": "0"
},
{
"input": "10 5\n260725416 260725506 260725422 260725512 260725428 260725518 260725434 260725524 260725440 260725530\n925033135 925033225 925033141 925033231 925033147",
"output": "3"
},
{
"input": "2 2\n1000000000 10\n1 20",
"output": "0"
},
{
"input": "7 3\n1 1 1 1 1 1 1\n1000 1256 1512",
"output": "0"
},
{
"input": "3 3\n1 132 3\n2 1 3",
"output": "0"
},
{
"input": "53 3\n1 3 4 4 5 7 10 14 19 25 32 40 49 59 70 82 95 109 124 140 157 175 194 214 235 257 280 304 329 355 382 410 439 469 500 532 565 599 634 670 707 745 784 824 865 907 950 994 1039 1085 1132 1180 1229\n1 2 40",
"output": "0"
}
] | 1,621,413,097 | 2,147,483,647 | PyPy 3 | OK | TESTS | 30 | 405 | 21,811,200 | from sys import stdin
# input=stdin.buffer.readline
input=lambda : stdin.readline().strip()
lin=lambda :list(map(int,input().split()))
iin=lambda :int(input())
main=lambda :map(int,input().split())
from math import ceil,sqrt,factorial,log
from collections import deque
from bisect import bisect_left
mod=998244353
mod=1000000007
def kmp_pi(n,w,l,h):
h=h+[10**9]+l
pi=[0 for i in range(len(h))]
ans=0
for i in range(1,len(h)):
j=pi[i-1]
while(j>0 and h[i]!=h[j]):
# print(j)
j=pi[j-1]
if(h[i]==h[j]):
j+=1
pi[i]=j
if j==w:
ans+=1
# # print(i)
# print(pi)
return ans
def dif(l):
x=[]
for i in range(1,len(l)):
x.append(l[i-1]-l[i])
return x
def solve(we):
n,w=main()
l=dif(lin())
h=dif(lin())
if w==1:
print(n)
else:
x=kmp_pi(n-1,w-1,l,h)
print(x)
qwe=1
# qwe=iin()
for _ in range(qwe):
solve(_+1) | Title: MUH and Cube Walls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got hold of lots of wooden cubes somewhere. They started making cube towers by placing the cubes one on top of the other. They defined multiple towers standing in a line as a wall. A wall can consist of towers of different heights.
Horace was the first to finish making his wall. He called his wall an elephant. The wall consists of *w* towers. The bears also finished making their wall but they didn't give it a name. Their wall consists of *n* towers. Horace looked at the bears' tower and wondered: in how many parts of the wall can he "see an elephant"? He can "see an elephant" on a segment of *w* contiguous towers if the heights of the towers on the segment match as a sequence the heights of the towers in Horace's wall. In order to see as many elephants as possible, Horace can raise and lower his wall. He even can lower the wall below the ground level (see the pictures to the samples for clarification).
Your task is to count the number of segments where Horace can "see an elephant".
Input Specification:
The first line contains two integers *n* and *w* (1<=≤<=*n*,<=*w*<=≤<=2·105) — the number of towers in the bears' and the elephant's walls correspondingly. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the heights of the towers in the bears' wall. The third line contains *w* integers *b**i* (1<=≤<=*b**i*<=≤<=109) — the heights of the towers in the elephant's wall.
Output Specification:
Print the number of segments in the bears' wall where Horace can "see an elephant".
Demo Input:
['13 5\n2 4 5 5 4 3 2 2 2 3 3 2 1\n3 4 4 3 2\n']
Demo Output:
['2']
Note:
The picture to the left shows Horace's wall from the sample, the picture to the right shows the bears' wall. The segments where Horace can "see an elephant" are in gray. | ```python
from sys import stdin
# input=stdin.buffer.readline
input=lambda : stdin.readline().strip()
lin=lambda :list(map(int,input().split()))
iin=lambda :int(input())
main=lambda :map(int,input().split())
from math import ceil,sqrt,factorial,log
from collections import deque
from bisect import bisect_left
mod=998244353
mod=1000000007
def kmp_pi(n,w,l,h):
h=h+[10**9]+l
pi=[0 for i in range(len(h))]
ans=0
for i in range(1,len(h)):
j=pi[i-1]
while(j>0 and h[i]!=h[j]):
# print(j)
j=pi[j-1]
if(h[i]==h[j]):
j+=1
pi[i]=j
if j==w:
ans+=1
# # print(i)
# print(pi)
return ans
def dif(l):
x=[]
for i in range(1,len(l)):
x.append(l[i-1]-l[i])
return x
def solve(we):
n,w=main()
l=dif(lin())
h=dif(lin())
if w==1:
print(n)
else:
x=kmp_pi(n-1,w-1,l,h)
print(x)
qwe=1
# qwe=iin()
for _ in range(qwe):
solve(_+1)
``` | 3 | |
439 | B | Devu, the Dumb Guy | PROGRAMMING | 1,200 | [
"implementation",
"sortings"
] | null | null | Devu is a dumb guy, his learning curve is very slow. You are supposed to teach him *n* subjects, the *i**th* subject has *c**i* chapters. When you teach him, you are supposed to teach all the chapters of a subject continuously.
Let us say that his initial per chapter learning power of a subject is *x* hours. In other words he can learn a chapter of a particular subject in *x* hours.
Well Devu is not complete dumb, there is a good thing about him too. If you teach him a subject, then time required to teach any chapter of the next subject will require exactly 1 hour less than previously required (see the examples to understand it more clearly). Note that his per chapter learning power can not be less than 1 hour.
You can teach him the *n* subjects in any possible order. Find out minimum amount of time (in hours) Devu will take to understand all the subjects and you will be free to do some enjoying task rather than teaching a dumb guy.
Please be careful that answer might not fit in 32 bit data type. | The first line will contain two space separated integers *n*, *x* (1<=≤<=*n*,<=*x*<=≤<=105). The next line will contain *n* space separated integers: *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105). | Output a single integer representing the answer to the problem. | [
"2 3\n4 1\n",
"4 2\n5 1 2 1\n",
"3 3\n1 1 1\n"
] | [
"11\n",
"10\n",
"6\n"
] | Look at the first example. Consider the order of subjects: 1, 2. When you teach Devu the first subject, it will take him 3 hours per chapter, so it will take 12 hours to teach first subject. After teaching first subject, his per chapter learning time will be 2 hours. Now teaching him second subject will take 2 × 1 = 2 hours. Hence you will need to spend 12 + 2 = 14 hours.
Consider the order of subjects: 2, 1. When you teach Devu the second subject, then it will take him 3 hours per chapter, so it will take 3 × 1 = 3 hours to teach the second subject. After teaching the second subject, his per chapter learning time will be 2 hours. Now teaching him the first subject will take 2 × 4 = 8 hours. Hence you will need to spend 11 hours.
So overall, minimum of both the cases is 11 hours.
Look at the third example. The order in this example doesn't matter. When you teach Devu the first subject, it will take him 3 hours per chapter. When you teach Devu the second subject, it will take him 2 hours per chapter. When you teach Devu the third subject, it will take him 1 hours per chapter. In total it takes 6 hours. | 1,000 | [
{
"input": "2 3\n4 1",
"output": "11"
},
{
"input": "4 2\n5 1 2 1",
"output": "10"
},
{
"input": "3 3\n1 1 1",
"output": "6"
},
{
"input": "20 4\n1 1 3 5 5 1 3 4 2 5 2 4 3 1 3 3 3 3 4 3",
"output": "65"
},
{
"input": "20 10\n6 6 1 2 6 4 5 3 6 5 4 5 6 5 4 6 6 2 3 3",
"output": "196"
},
{
"input": "1 1\n9273",
"output": "9273"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 2\n1",
"output": "2"
},
{
"input": "1 2\n2",
"output": "4"
},
{
"input": "2 1\n1 2",
"output": "3"
}
] | 1,538,157,816 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 217 | 7,372,800 | n, x = map(int, input().split())
l = list(map(int, input().split()))
# Sap xep lai phan tu
l.sort()
# Thoi gian ngan nhat
s = 0
for i in l:
if x <= 1:
s += i
else:
s += i*x
x -= 1
print(s) | Title: Devu, the Dumb Guy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu is a dumb guy, his learning curve is very slow. You are supposed to teach him *n* subjects, the *i**th* subject has *c**i* chapters. When you teach him, you are supposed to teach all the chapters of a subject continuously.
Let us say that his initial per chapter learning power of a subject is *x* hours. In other words he can learn a chapter of a particular subject in *x* hours.
Well Devu is not complete dumb, there is a good thing about him too. If you teach him a subject, then time required to teach any chapter of the next subject will require exactly 1 hour less than previously required (see the examples to understand it more clearly). Note that his per chapter learning power can not be less than 1 hour.
You can teach him the *n* subjects in any possible order. Find out minimum amount of time (in hours) Devu will take to understand all the subjects and you will be free to do some enjoying task rather than teaching a dumb guy.
Please be careful that answer might not fit in 32 bit data type.
Input Specification:
The first line will contain two space separated integers *n*, *x* (1<=≤<=*n*,<=*x*<=≤<=105). The next line will contain *n* space separated integers: *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105).
Output Specification:
Output a single integer representing the answer to the problem.
Demo Input:
['2 3\n4 1\n', '4 2\n5 1 2 1\n', '3 3\n1 1 1\n']
Demo Output:
['11\n', '10\n', '6\n']
Note:
Look at the first example. Consider the order of subjects: 1, 2. When you teach Devu the first subject, it will take him 3 hours per chapter, so it will take 12 hours to teach first subject. After teaching first subject, his per chapter learning time will be 2 hours. Now teaching him second subject will take 2 × 1 = 2 hours. Hence you will need to spend 12 + 2 = 14 hours.
Consider the order of subjects: 2, 1. When you teach Devu the second subject, then it will take him 3 hours per chapter, so it will take 3 × 1 = 3 hours to teach the second subject. After teaching the second subject, his per chapter learning time will be 2 hours. Now teaching him the first subject will take 2 × 4 = 8 hours. Hence you will need to spend 11 hours.
So overall, minimum of both the cases is 11 hours.
Look at the third example. The order in this example doesn't matter. When you teach Devu the first subject, it will take him 3 hours per chapter. When you teach Devu the second subject, it will take him 2 hours per chapter. When you teach Devu the third subject, it will take him 1 hours per chapter. In total it takes 6 hours. | ```python
n, x = map(int, input().split())
l = list(map(int, input().split()))
# Sap xep lai phan tu
l.sort()
# Thoi gian ngan nhat
s = 0
for i in l:
if x <= 1:
s += i
else:
s += i*x
x -= 1
print(s)
``` | 3 | |
460 | A | Vasya and Socks | PROGRAMMING | 900 | [
"brute force",
"implementation",
"math"
] | null | null | Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks? | The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space. | Print a single integer — the answer to the problem. | [
"2 2\n",
"9 3\n"
] | [
"3\n",
"13\n"
] | In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day. | 500 | [
{
"input": "2 2",
"output": "3"
},
{
"input": "9 3",
"output": "13"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "1 99",
"output": "1"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "10 2",
"output": "19"
},
{
"input": "10 9",
"output": "11"
},
{
"input": "100 100",
"output": "101"
},
{
"input": "2 27",
"output": "2"
},
{
"input": "99 100",
"output": "99"
},
{
"input": "99 2",
"output": "197"
},
{
"input": "100 3",
"output": "149"
},
{
"input": "98 3",
"output": "146"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "100 2",
"output": "199"
},
{
"input": "62 4",
"output": "82"
},
{
"input": "99 10",
"output": "109"
},
{
"input": "100 5",
"output": "124"
},
{
"input": "80 80",
"output": "81"
},
{
"input": "95 16",
"output": "101"
},
{
"input": "75 16",
"output": "79"
},
{
"input": "99 74",
"output": "100"
},
{
"input": "20 21",
"output": "20"
},
{
"input": "52 96",
"output": "52"
},
{
"input": "24 5",
"output": "29"
}
] | 1,639,673,446 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 46 | 0 | n, m = [int(x) for x in input().split()]
x = n + n // m
if x - n < m:
print(x)
elif x - n == m:
print(x+1)
elif x - n > m:
x = x + (x - n) // m
print(x+1)
| Title: Vasya and Socks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
Input Specification:
The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['2 2\n', '9 3\n']
Demo Output:
['3\n', '13\n']
Note:
In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day. | ```python
n, m = [int(x) for x in input().split()]
x = n + n // m
if x - n < m:
print(x)
elif x - n == m:
print(x+1)
elif x - n > m:
x = x + (x - n) // m
print(x+1)
``` | 0 | |
495 | A | Digital Counter | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Malek lives in an apartment block with 100 floors numbered from 0 to 99. The apartment has an elevator with a digital counter showing the floor that the elevator is currently on. The elevator shows each digit of a number with 7 light sticks by turning them on or off. The picture below shows how the elevator shows each digit.
One day when Malek wanted to go from floor 88 to floor 0 using the elevator he noticed that the counter shows number 89 instead of 88. Then when the elevator started moving the number on the counter changed to 87. After a little thinking Malek came to the conclusion that there is only one explanation for this: One of the sticks of the counter was broken. Later that day Malek was thinking about the broken stick and suddenly he came up with the following problem.
Suppose the digital counter is showing number *n*. Malek calls an integer *x* (0<=≤<=*x*<=≤<=99) good if it's possible that the digital counter was supposed to show *x* but because of some(possibly none) broken sticks it's showing *n* instead. Malek wants to know number of good integers for a specific *n*. So you must write a program that calculates this number. Please note that the counter always shows two digits. | The only line of input contains exactly two digits representing number *n* (0<=≤<=*n*<=≤<=99). Note that *n* may have a leading zero. | In the only line of the output print the number of good integers. | [
"89\n",
"00\n",
"73\n"
] | [
"2\n",
"4\n",
"15\n"
] | In the first sample the counter may be supposed to show 88 or 89.
In the second sample the good integers are 00, 08, 80 and 88.
In the third sample the good integers are 03, 08, 09, 33, 38, 39, 73, 78, 79, 83, 88, 89, 93, 98, 99. | 500 | [
{
"input": "89",
"output": "2"
},
{
"input": "00",
"output": "4"
},
{
"input": "73",
"output": "15"
},
{
"input": "08",
"output": "2"
},
{
"input": "26",
"output": "4"
},
{
"input": "49",
"output": "6"
},
{
"input": "88",
"output": "1"
},
{
"input": "04",
"output": "6"
},
{
"input": "60",
"output": "4"
},
{
"input": "11",
"output": "49"
},
{
"input": "22",
"output": "4"
},
{
"input": "33",
"output": "9"
},
{
"input": "44",
"output": "9"
},
{
"input": "55",
"output": "16"
},
{
"input": "66",
"output": "4"
},
{
"input": "77",
"output": "25"
},
{
"input": "88",
"output": "1"
},
{
"input": "99",
"output": "4"
},
{
"input": "80",
"output": "2"
},
{
"input": "78",
"output": "5"
},
{
"input": "67",
"output": "10"
},
{
"input": "89",
"output": "2"
},
{
"input": "46",
"output": "6"
},
{
"input": "90",
"output": "4"
},
{
"input": "92",
"output": "4"
},
{
"input": "35",
"output": "12"
},
{
"input": "05",
"output": "8"
},
{
"input": "57",
"output": "20"
},
{
"input": "20",
"output": "4"
}
] | 1,427,207,138 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 13 | 62 | 0 | # coding: utf-8
li = [2, 7, 2, 3, 3, 3, 2, 5, 1, 2]
s = input()
print(li[int(s[0])]*li[int(s[1])])
| Title: Digital Counter
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Malek lives in an apartment block with 100 floors numbered from 0 to 99. The apartment has an elevator with a digital counter showing the floor that the elevator is currently on. The elevator shows each digit of a number with 7 light sticks by turning them on or off. The picture below shows how the elevator shows each digit.
One day when Malek wanted to go from floor 88 to floor 0 using the elevator he noticed that the counter shows number 89 instead of 88. Then when the elevator started moving the number on the counter changed to 87. After a little thinking Malek came to the conclusion that there is only one explanation for this: One of the sticks of the counter was broken. Later that day Malek was thinking about the broken stick and suddenly he came up with the following problem.
Suppose the digital counter is showing number *n*. Malek calls an integer *x* (0<=≤<=*x*<=≤<=99) good if it's possible that the digital counter was supposed to show *x* but because of some(possibly none) broken sticks it's showing *n* instead. Malek wants to know number of good integers for a specific *n*. So you must write a program that calculates this number. Please note that the counter always shows two digits.
Input Specification:
The only line of input contains exactly two digits representing number *n* (0<=≤<=*n*<=≤<=99). Note that *n* may have a leading zero.
Output Specification:
In the only line of the output print the number of good integers.
Demo Input:
['89\n', '00\n', '73\n']
Demo Output:
['2\n', '4\n', '15\n']
Note:
In the first sample the counter may be supposed to show 88 or 89.
In the second sample the good integers are 00, 08, 80 and 88.
In the third sample the good integers are 03, 08, 09, 33, 38, 39, 73, 78, 79, 83, 88, 89, 93, 98, 99. | ```python
# coding: utf-8
li = [2, 7, 2, 3, 3, 3, 2, 5, 1, 2]
s = input()
print(li[int(s[0])]*li[int(s[1])])
``` | 0 | |
227 | B | Effective Approach | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. | Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | [
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] | [
"1 2\n",
"2 1\n",
"6 6\n"
] | In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | 1,000 | [
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4",
"output": "9 2"
},
{
"input": "10\n5 2 10 8 3 1 9 7 6 4\n9\n2 5 9 2 3 2 5 9 8",
"output": "31 68"
},
{
"input": "9\n3 8 4 7 1 2 5 6 9\n3\n2 7 1",
"output": "15 15"
},
{
"input": "9\n5 3 8 4 2 6 1 7 9\n4\n6 1 9 2",
"output": "27 13"
},
{
"input": "4\n1 3 2 4\n4\n3 1 2 3",
"output": "8 12"
},
{
"input": "3\n1 2 3\n8\n3 2 1 1 2 3 1 2",
"output": "15 17"
}
] | 1,670,949,601 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 34 | 372 | 28,876,800 | n = int(input())
sozluk = {}
sayac = 0
sayi = 1
for i in input().split():
sozluk[i] = sayi
sayi += 1
m = int(input())
sorgu = list(map(int, input().split()))
for t in range(len(sorgu)):
sayac += sozluk[str(sorgu[t])]
print(sayac, (n+1)*m-sayac)
| Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | ```python
n = int(input())
sozluk = {}
sayac = 0
sayi = 1
for i in input().split():
sozluk[i] = sayi
sayi += 1
m = int(input())
sorgu = list(map(int, input().split()))
for t in range(len(sorgu)):
sayac += sozluk[str(sorgu[t])]
print(sayac, (n+1)*m-sayac)
``` | 3 | |
807 | A | Is it rated? | PROGRAMMING | 900 | [
"implementation",
"sortings"
] | null | null | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings. | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | [
"6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n",
"4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n",
"5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n"
] | [
"rated\n",
"unrated\n",
"maybe\n"
] | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not. | 500 | [
{
"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884",
"output": "rated"
},
{
"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400",
"output": "unrated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699",
"output": "maybe"
},
{
"input": "2\n1 1\n1 1",
"output": "maybe"
},
{
"input": "2\n4126 4126\n4126 4126",
"output": "maybe"
},
{
"input": "10\n446 446\n1331 1331\n3594 3594\n1346 1902\n91 91\n3590 3590\n2437 2437\n4007 3871\n2797 699\n1423 1423",
"output": "rated"
},
{
"input": "10\n4078 4078\n2876 2876\n1061 1061\n3721 3721\n143 143\n2992 2992\n3279 3279\n3389 3389\n1702 1702\n1110 1110",
"output": "unrated"
},
{
"input": "10\n4078 4078\n3721 3721\n3389 3389\n3279 3279\n2992 2992\n2876 2876\n1702 1702\n1110 1110\n1061 1061\n143 143",
"output": "maybe"
},
{
"input": "2\n3936 3936\n2967 2967",
"output": "maybe"
},
{
"input": "2\n1 1\n2 2",
"output": "unrated"
},
{
"input": "2\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n2 1\n1 2",
"output": "rated"
},
{
"input": "2\n2967 2967\n3936 3936",
"output": "unrated"
},
{
"input": "3\n1200 1200\n1200 1200\n1300 1300",
"output": "unrated"
},
{
"input": "3\n3 3\n2 2\n1 1",
"output": "maybe"
},
{
"input": "3\n1 1\n1 1\n2 2",
"output": "unrated"
},
{
"input": "2\n3 2\n3 2",
"output": "rated"
},
{
"input": "3\n5 5\n4 4\n3 4",
"output": "rated"
},
{
"input": "3\n200 200\n200 200\n300 300",
"output": "unrated"
},
{
"input": "3\n1 1\n2 2\n3 3",
"output": "unrated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2245 2245\n1699 1699",
"output": "maybe"
},
{
"input": "2\n10 10\n8 8",
"output": "maybe"
},
{
"input": "3\n1500 1500\n1500 1500\n1600 1600",
"output": "unrated"
},
{
"input": "3\n1500 1500\n1500 1500\n1700 1700",
"output": "unrated"
},
{
"input": "4\n100 100\n100 100\n70 70\n80 80",
"output": "unrated"
},
{
"input": "2\n1 2\n2 1",
"output": "rated"
},
{
"input": "3\n5 5\n4 3\n3 3",
"output": "rated"
},
{
"input": "3\n1600 1650\n1500 1550\n1400 1450",
"output": "rated"
},
{
"input": "4\n2000 2000\n1500 1500\n1500 1500\n1700 1700",
"output": "unrated"
},
{
"input": "4\n1500 1500\n1400 1400\n1400 1400\n1700 1700",
"output": "unrated"
},
{
"input": "2\n1600 1600\n1400 1400",
"output": "maybe"
},
{
"input": "2\n3 1\n9 8",
"output": "rated"
},
{
"input": "2\n2 1\n1 1",
"output": "rated"
},
{
"input": "4\n4123 4123\n4123 4123\n2670 2670\n3670 3670",
"output": "unrated"
},
{
"input": "2\n2 2\n3 3",
"output": "unrated"
},
{
"input": "2\n10 11\n5 4",
"output": "rated"
},
{
"input": "2\n15 14\n13 12",
"output": "rated"
},
{
"input": "2\n2 1\n2 2",
"output": "rated"
},
{
"input": "3\n2670 2670\n3670 3670\n4106 4106",
"output": "unrated"
},
{
"input": "3\n4 5\n3 3\n2 2",
"output": "rated"
},
{
"input": "2\n10 9\n10 10",
"output": "rated"
},
{
"input": "3\n1011 1011\n1011 999\n2200 2100",
"output": "rated"
},
{
"input": "2\n3 3\n5 5",
"output": "unrated"
},
{
"input": "2\n1500 1500\n3000 2000",
"output": "rated"
},
{
"input": "2\n5 6\n5 5",
"output": "rated"
},
{
"input": "3\n2000 2000\n1500 1501\n500 500",
"output": "rated"
},
{
"input": "2\n2 3\n2 2",
"output": "rated"
},
{
"input": "2\n3 3\n2 2",
"output": "maybe"
},
{
"input": "2\n1 2\n1 1",
"output": "rated"
},
{
"input": "4\n3123 3123\n2777 2777\n2246 2246\n1699 1699",
"output": "maybe"
},
{
"input": "2\n15 14\n14 13",
"output": "rated"
},
{
"input": "4\n3000 3000\n2900 2900\n3000 3000\n2900 2900",
"output": "unrated"
},
{
"input": "6\n30 3060\n24 2194\n26 2903\n24 2624\n37 2991\n24 2884",
"output": "rated"
},
{
"input": "2\n100 99\n100 100",
"output": "rated"
},
{
"input": "4\n2 2\n1 1\n1 1\n2 2",
"output": "unrated"
},
{
"input": "3\n100 101\n100 100\n100 100",
"output": "rated"
},
{
"input": "4\n1000 1001\n900 900\n950 950\n890 890",
"output": "rated"
},
{
"input": "2\n2 3\n1 1",
"output": "rated"
},
{
"input": "2\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n3 2\n2 2",
"output": "rated"
},
{
"input": "2\n3 2\n3 3",
"output": "rated"
},
{
"input": "2\n1 1\n2 2",
"output": "unrated"
},
{
"input": "3\n3 2\n3 3\n3 3",
"output": "rated"
},
{
"input": "4\n1500 1501\n1300 1300\n1200 1200\n1400 1400",
"output": "rated"
},
{
"input": "3\n1000 1000\n500 500\n400 300",
"output": "rated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n3000 3000",
"output": "unrated"
},
{
"input": "2\n1 1\n2 3",
"output": "rated"
},
{
"input": "2\n6 2\n6 2",
"output": "rated"
},
{
"input": "5\n3123 3123\n1699 1699\n2777 2777\n2246 2246\n2246 2246",
"output": "unrated"
},
{
"input": "2\n1500 1500\n1600 1600",
"output": "unrated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2241 2241\n1699 1699",
"output": "maybe"
},
{
"input": "2\n20 30\n10 5",
"output": "rated"
},
{
"input": "3\n1 1\n2 2\n1 1",
"output": "unrated"
},
{
"input": "2\n1 2\n3 3",
"output": "rated"
},
{
"input": "5\n5 5\n4 4\n3 3\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n2 2\n2 1",
"output": "rated"
},
{
"input": "2\n100 100\n90 89",
"output": "rated"
},
{
"input": "2\n1000 900\n2000 2000",
"output": "rated"
},
{
"input": "2\n50 10\n10 50",
"output": "rated"
},
{
"input": "2\n200 200\n100 100",
"output": "maybe"
},
{
"input": "3\n2 2\n2 2\n3 3",
"output": "unrated"
},
{
"input": "3\n1000 1000\n300 300\n100 100",
"output": "maybe"
},
{
"input": "4\n2 2\n2 2\n3 3\n4 4",
"output": "unrated"
},
{
"input": "2\n5 3\n6 3",
"output": "rated"
},
{
"input": "2\n1200 1100\n1200 1000",
"output": "rated"
},
{
"input": "2\n5 5\n4 4",
"output": "maybe"
},
{
"input": "2\n5 5\n3 3",
"output": "maybe"
},
{
"input": "5\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n1100 1100",
"output": "unrated"
},
{
"input": "5\n10 10\n9 9\n8 8\n7 7\n6 6",
"output": "maybe"
},
{
"input": "3\n1000 1000\n300 300\n10 10",
"output": "maybe"
},
{
"input": "5\n6 6\n5 5\n4 4\n3 3\n2 2",
"output": "maybe"
},
{
"input": "2\n3 3\n1 1",
"output": "maybe"
},
{
"input": "4\n2 2\n2 2\n2 2\n3 3",
"output": "unrated"
},
{
"input": "2\n1000 1000\n700 700",
"output": "maybe"
},
{
"input": "2\n4 3\n5 3",
"output": "rated"
},
{
"input": "2\n1000 1000\n1100 1100",
"output": "unrated"
},
{
"input": "4\n5 5\n4 4\n3 3\n2 2",
"output": "maybe"
},
{
"input": "3\n1 1\n2 3\n2 2",
"output": "rated"
},
{
"input": "2\n1 2\n1 3",
"output": "rated"
},
{
"input": "2\n3 3\n1 2",
"output": "rated"
},
{
"input": "4\n1501 1500\n1300 1300\n1200 1200\n1400 1400",
"output": "rated"
},
{
"input": "5\n1 1\n2 2\n3 3\n4 4\n5 5",
"output": "unrated"
},
{
"input": "2\n10 10\n1 2",
"output": "rated"
},
{
"input": "6\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n1900 1900",
"output": "unrated"
},
{
"input": "6\n3123 3123\n2777 2777\n3000 3000\n2246 2246\n2246 2246\n1699 1699",
"output": "unrated"
},
{
"input": "2\n100 100\n110 110",
"output": "unrated"
},
{
"input": "3\n3 3\n3 3\n4 4",
"output": "unrated"
},
{
"input": "3\n3 3\n3 2\n4 4",
"output": "rated"
},
{
"input": "3\n5 2\n4 4\n3 3",
"output": "rated"
},
{
"input": "4\n4 4\n3 3\n2 2\n1 1",
"output": "maybe"
},
{
"input": "2\n1 1\n3 2",
"output": "rated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n2699 2699",
"output": "unrated"
},
{
"input": "3\n3 3\n3 3\n3 4",
"output": "rated"
},
{
"input": "3\n1 2\n2 2\n3 3",
"output": "rated"
},
{
"input": "3\n1 2\n1 2\n1 2",
"output": "rated"
},
{
"input": "2\n2 1\n2 1",
"output": "rated"
},
{
"input": "2\n1 2\n3 4",
"output": "rated"
},
{
"input": "2\n3 2\n2 3",
"output": "rated"
},
{
"input": "3\n1500 1500\n1600 1600\n1600 1600",
"output": "unrated"
},
{
"input": "3\n1 1\n3 3\n4 4",
"output": "unrated"
},
{
"input": "3\n1 1\n2 2\n2 2",
"output": "unrated"
},
{
"input": "2\n10 12\n8 8",
"output": "rated"
},
{
"input": "5\n1200 1200\n1500 1500\n1500 1500\n1500 1500\n1500 1500",
"output": "unrated"
},
{
"input": "2\n1 2\n2 2",
"output": "rated"
},
{
"input": "3\n1500 1400\n1200 1200\n1100 1100",
"output": "rated"
},
{
"input": "2\n10 12\n10 10",
"output": "rated"
},
{
"input": "3\n1500 1500\n1400 1400\n1300 1300",
"output": "maybe"
},
{
"input": "3\n3 3\n4 4\n5 5",
"output": "unrated"
},
{
"input": "3\n2 6\n3 5\n4 4",
"output": "rated"
},
{
"input": "2\n5 6\n4 6",
"output": "rated"
},
{
"input": "4\n10 10\n10 10\n7 7\n8 8",
"output": "unrated"
},
{
"input": "2\n4 4\n3 3",
"output": "maybe"
}
] | 1,565,360,851 | 2,147,483,647 | Python 3 | OK | TESTS | 150 | 124 | 307,200 | n = int(input())
rate = list(list(map(int, input().split())) for _ in range(n))
if any(rate[i][0] != rate[i][1] for i in range(n)):
print('rated')
elif rate == sorted(rate, reverse=True):
print('maybe')
else:
print('unrated')
| Title: Is it rated?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output Specification:
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Demo Input:
['6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n', '4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n', '5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n']
Demo Output:
['rated\n', 'unrated\n', 'maybe\n']
Note:
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not. | ```python
n = int(input())
rate = list(list(map(int, input().split())) for _ in range(n))
if any(rate[i][0] != rate[i][1] for i in range(n)):
print('rated')
elif rate == sorted(rate, reverse=True):
print('maybe')
else:
print('unrated')
``` | 3 | |
374 | B | Inna and Nine | PROGRAMMING | 1,500 | [
"combinatorics",
"greedy"
] | null | null | Inna loves digit 9 very much. That's why she asked Dima to write a small number consisting of nines. But Dima must have misunderstood her and he wrote a very large number *a*, consisting of digits from 1 to 9.
Inna wants to slightly alter the number Dima wrote so that in the end the number contained as many digits nine as possible. In one move, Inna can choose two adjacent digits in a number which sum equals 9 and replace them by a single digit 9.
For instance, Inna can alter number 14545181 like this: 14545181<=→<=1945181<=→<=194519<=→<=19919. Also, she can use this method to transform number 14545181 into number 19991. Inna will not transform it into 149591 as she can get numbers 19919 and 19991 which contain more digits nine.
Dima is a programmer so he wants to find out how many distinct numbers containing as many digits nine as possible Inna can get from the written number. Help him with this challenging task. | The first line of the input contains integer *a* (1<=≤<=*a*<=≤<=10100000). Number *a* doesn't have any zeroes. | In a single line print a single integer — the answer to the problem. It is guaranteed that the answer to the problem doesn't exceed 263<=-<=1.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | [
"369727\n",
"123456789987654321\n",
"1\n"
] | [
"2\n",
"1\n",
"1\n"
] | Notes to the samples
In the first sample Inna can get the following numbers: 369727 → 99727 → 9997, 369727 → 99727 → 9979.
In the second sample, Inna can act like this: 123456789987654321 → 12396789987654321 → 1239678998769321. | 1,000 | [
{
"input": "369727",
"output": "2"
},
{
"input": "123456789987654321",
"output": "1"
},
{
"input": "1",
"output": "1"
},
{
"input": "3636363636363454545454543636363636454545452727272727218181818181999111777",
"output": "1512"
},
{
"input": "1188",
"output": "1"
},
{
"input": "121212912121291299129191219",
"output": "1"
},
{
"input": "181818918181891918918181918189181818181891818191818191819189",
"output": "54"
},
{
"input": "12191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219121912191219",
"output": "1"
},
{
"input": "444444444444445444444444454444444444444444444444445544444444444444444444444444444444444444444554444444444444444444444444444444444444445454444444444444444444444444444444444444454444444444444444444444444444444444445444444444444444444444444444444444444444444445444444444444444444444444444444444456666666666666666666663555555555555555888888888882333333312567312389542179415242164512341234213443123412341293412341234123412746129342154796124123459123491238471234213451692341278451234125934",
"output": "96"
},
{
"input": "123456789",
"output": "1"
},
{
"input": "3639272918194549",
"output": "16"
},
{
"input": "1121314151617181921222324252627282931323334353637383941424344454647484951525354556575859616263646566768697172737475767787981828384858687888991929394959696979899",
"output": "256"
},
{
"input": "14545181",
"output": "2"
},
{
"input": "272727272",
"output": "5"
},
{
"input": "1212121217272727121",
"output": "4"
},
{
"input": "1212172727",
"output": "3"
},
{
"input": "181817272727",
"output": "12"
}
] | 1,387,450,628 | 2,147,483,647 | Python 3 | OK | TESTS | 24 | 187 | 102,400 | #!/usr/bin/python3
def readln(): return list(map(int, input().split()))
ans = 1
l = 1
p = -1
a = input()
for i in range(len(a)):
c = int(a[i])
if c + p == 9:
l += 1
if c + p != 9 or i == len(a) - 1:
if l % 2:
ans *= (l + 1) // 2
l = 1
p = c
print(ans)
| Title: Inna and Nine
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Inna loves digit 9 very much. That's why she asked Dima to write a small number consisting of nines. But Dima must have misunderstood her and he wrote a very large number *a*, consisting of digits from 1 to 9.
Inna wants to slightly alter the number Dima wrote so that in the end the number contained as many digits nine as possible. In one move, Inna can choose two adjacent digits in a number which sum equals 9 and replace them by a single digit 9.
For instance, Inna can alter number 14545181 like this: 14545181<=→<=1945181<=→<=194519<=→<=19919. Also, she can use this method to transform number 14545181 into number 19991. Inna will not transform it into 149591 as she can get numbers 19919 and 19991 which contain more digits nine.
Dima is a programmer so he wants to find out how many distinct numbers containing as many digits nine as possible Inna can get from the written number. Help him with this challenging task.
Input Specification:
The first line of the input contains integer *a* (1<=≤<=*a*<=≤<=10100000). Number *a* doesn't have any zeroes.
Output Specification:
In a single line print a single integer — the answer to the problem. It is guaranteed that the answer to the problem doesn't exceed 263<=-<=1.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['369727\n', '123456789987654321\n', '1\n']
Demo Output:
['2\n', '1\n', '1\n']
Note:
Notes to the samples
In the first sample Inna can get the following numbers: 369727 → 99727 → 9997, 369727 → 99727 → 9979.
In the second sample, Inna can act like this: 123456789987654321 → 12396789987654321 → 1239678998769321. | ```python
#!/usr/bin/python3
def readln(): return list(map(int, input().split()))
ans = 1
l = 1
p = -1
a = input()
for i in range(len(a)):
c = int(a[i])
if c + p == 9:
l += 1
if c + p != 9 or i == len(a) - 1:
if l % 2:
ans *= (l + 1) // 2
l = 1
p = c
print(ans)
``` | 3 | |
791 | A | Bear and Big Brother | PROGRAMMING | 800 | [
"implementation"
] | null | null | Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob? | The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively. | Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob. | [
"4 7\n",
"4 9\n",
"1 1\n"
] | [
"2\n",
"3\n",
"1\n"
] | In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then. | 500 | [
{
"input": "4 7",
"output": "2"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "1 10",
"output": "6"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "4"
},
{
"input": "1 6",
"output": "5"
},
{
"input": "1 7",
"output": "5"
},
{
"input": "1 8",
"output": "6"
},
{
"input": "1 9",
"output": "6"
},
{
"input": "1 10",
"output": "6"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "2 4",
"output": "2"
},
{
"input": "2 5",
"output": "3"
},
{
"input": "2 6",
"output": "3"
},
{
"input": "2 7",
"output": "4"
},
{
"input": "2 8",
"output": "4"
},
{
"input": "2 9",
"output": "4"
},
{
"input": "2 10",
"output": "4"
},
{
"input": "3 3",
"output": "1"
},
{
"input": "3 4",
"output": "1"
},
{
"input": "3 5",
"output": "2"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "3 7",
"output": "3"
},
{
"input": "3 8",
"output": "3"
},
{
"input": "3 9",
"output": "3"
},
{
"input": "3 10",
"output": "3"
},
{
"input": "4 4",
"output": "1"
},
{
"input": "4 5",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "4 7",
"output": "2"
},
{
"input": "4 8",
"output": "2"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "4 10",
"output": "3"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "5 6",
"output": "1"
},
{
"input": "5 7",
"output": "1"
},
{
"input": "5 8",
"output": "2"
},
{
"input": "5 9",
"output": "2"
},
{
"input": "5 10",
"output": "2"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "6 7",
"output": "1"
},
{
"input": "6 8",
"output": "1"
},
{
"input": "6 9",
"output": "2"
},
{
"input": "6 10",
"output": "2"
},
{
"input": "7 7",
"output": "1"
},
{
"input": "7 8",
"output": "1"
},
{
"input": "7 9",
"output": "1"
},
{
"input": "7 10",
"output": "1"
},
{
"input": "8 8",
"output": "1"
},
{
"input": "8 9",
"output": "1"
},
{
"input": "8 10",
"output": "1"
},
{
"input": "9 9",
"output": "1"
},
{
"input": "9 10",
"output": "1"
},
{
"input": "10 10",
"output": "1"
},
{
"input": "10 10",
"output": "1"
},
{
"input": "1 2",
"output": "2"
}
] | 1,699,965,386 | 2,147,483,647 | Python 3 | OK | TESTS | 62 | 46 | 0 | a,b = map(int, input().split())
year = 0
tempa = a
tempb = b
while True:
tempa *= 3
tempb *= 2
year += 1
if tempa > tempb:
break
print(year) | Title: Bear and Big Brother
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob?
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively.
Output Specification:
Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.
Demo Input:
['4 7\n', '4 9\n', '1 1\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then. | ```python
a,b = map(int, input().split())
year = 0
tempa = a
tempb = b
while True:
tempa *= 3
tempb *= 2
year += 1
if tempa > tempb:
break
print(year)
``` | 3 | |
500 | A | New Year Transportation | PROGRAMMING | 1,000 | [
"dfs and similar",
"graphs",
"implementation"
] | null | null | New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system. | The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World. | If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO". | [
"8 4\n1 2 1 2 1 2 1\n",
"8 5\n1 2 1 2 1 1 1\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | 500 | [
{
"input": "8 4\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "8 5\n1 2 1 2 1 1 1",
"output": "NO"
},
{
"input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1",
"output": "YES"
},
{
"input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 22 8 8 9 1 7 5 12 5 6 13 6 11 2 6 3 1 5 1 1 2 2 1",
"output": "YES"
},
{
"input": "120 104\n41 15 95 85 34 11 25 42 65 39 77 80 74 17 66 73 21 14 36 63 63 79 45 24 65 7 63 80 51 21 2 19 78 28 71 2 15 23 17 68 62 18 54 39 43 70 3 46 34 23 41 65 32 10 13 18 10 3 16 48 54 18 57 28 3 24 44 50 15 2 20 22 45 44 3 29 2 27 11 2 12 25 25 31 1 2 32 4 11 30 13 16 26 21 1 13 21 8 15 5 18 13 5 15 3 8 13 6 5 1 9 7 1 2 4 1 1 2 1",
"output": "NO"
},
{
"input": "10 3\n8 3 5 4 2 3 2 2 1",
"output": "NO"
},
{
"input": "10 9\n8 3 5 4 2 3 2 2 1",
"output": "YES"
},
{
"input": "3 2\n1 1",
"output": "YES"
},
{
"input": "3 2\n2 1",
"output": "NO"
},
{
"input": "4 2\n2 1 1",
"output": "NO"
},
{
"input": "4 4\n2 2 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "3 3\n1 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 1 1",
"output": "YES"
},
{
"input": "3 3\n2 1",
"output": "YES"
},
{
"input": "4 4\n1 1 1",
"output": "YES"
},
{
"input": "8 8\n1 1 1 1 1 1 1",
"output": "YES"
},
{
"input": "5 5\n1 1 1 1",
"output": "YES"
}
] | 1,696,577,643 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 46 | 3,584,000 | c,d = map(int,input().split())
t = list(map(int,input().split()))
curr = 1
i = 0
while curr < d and i < len(t):
curr += t[i]
i = curr-1
if curr == d:
print('YES')
else:
print("NO") | Title: New Year Transportation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output Specification:
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
Demo Input:
['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | ```python
c,d = map(int,input().split())
t = list(map(int,input().split()))
curr = 1
i = 0
while curr < d and i < len(t):
curr += t[i]
i = curr-1
if curr == d:
print('YES')
else:
print("NO")
``` | 3 | |
785 | A | Anton and Polyhedrons | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number! | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron. | Output one number — the total number of faces in all the polyhedrons in Anton's collection. | [
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] | [
"42\n",
"28\n"
] | In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | 500 | [
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosahedron\nTetrahedron\nOctahedron\nDodecahedron\nIcosahedron\nOctahedron\nIcosahedron\nTetrahedron\nDodecahedron\nTetrahedron\nOctahedron\nCube\nCube\nDodecahedron\nTetrahedron",
"output": "256"
},
{
"input": "1\nTetrahedron",
"output": "4"
},
{
"input": "1\nCube",
"output": "6"
},
{
"input": "1\nOctahedron",
"output": "8"
},
{
"input": "1\nDodecahedron",
"output": "12"
},
{
"input": "1\nIcosahedron",
"output": "20"
},
{
"input": "28\nOctahedron\nDodecahedron\nOctahedron\nOctahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nDodecahedron\nDodecahedron\nCube\nDodecahedron\nCube\nTetrahedron\nCube\nCube\nTetrahedron\nDodecahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nIcosahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron",
"output": "340"
}
] | 1,693,133,169 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 217 | 0 | n = int(input())
faces = {"Tetrahedron":4 , "Cube": 6, "Octahedron":8 , "Dodecahedron":12, "Icosahedron":20}
total = 0
while n:
s = input()
total += faces[s]
n -= 1
print(total) | Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | ```python
n = int(input())
faces = {"Tetrahedron":4 , "Cube": 6, "Octahedron":8 , "Dodecahedron":12, "Icosahedron":20}
total = 0
while n:
s = input()
total += faces[s]
n -= 1
print(total)
``` | 3 | |
292 | D | Connected Components | PROGRAMMING | 1,900 | [
"data structures",
"dfs and similar",
"dp",
"dsu"
] | null | null | We already know of the large corporation where Polycarpus works as a system administrator. The computer network there consists of *n* computers and *m* cables that connect some pairs of computers. In other words, the computer network can be represented as some non-directed graph with *n* nodes and *m* edges. Let's index the computers with integers from 1 to *n*, let's index the cables with integers from 1 to *m*.
Polycarpus was given an important task — check the reliability of his company's network. For that Polycarpus decided to carry out a series of *k* experiments on the computer network, where the *i*-th experiment goes as follows:
1. Temporarily disconnect the cables with indexes from *l**i* to *r**i*, inclusive (the other cables remain connected). 1. Count the number of connected components in the graph that is defining the computer network at that moment. 1. Re-connect the disconnected cables with indexes from *l**i* to *r**i* (that is, restore the initial network).
Help Polycarpus carry out all experiments and for each print the number of connected components in the graph that defines the computer network through the given experiment. Isolated vertex should be counted as single component. | The first line contains two space-separated integers *n*, *m* (2<=≤<=*n*<=≤<=500; 1<=≤<=*m*<=≤<=104) — the number of computers and the number of cables, correspondingly.
The following *m* lines contain the cables' description. The *i*-th line contains space-separated pair of integers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*; *x**i*<=≠<=*y**i*) — the numbers of the computers that are connected by the *i*-th cable. Note that a pair of computers can be connected by multiple cables.
The next line contains integer *k* (1<=≤<=*k*<=≤<=2·104) — the number of experiments. Next *k* lines contain the experiments' descriptions. The *i*-th line contains space-separated integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*m*) — the numbers of the cables that Polycarpus disconnects during the *i*-th experiment. | Print *k* numbers, the *i*-th number represents the number of connected components of the graph that defines the computer network during the *i*-th experiment. | [
"6 5\n1 2\n5 4\n2 3\n3 1\n3 6\n6\n1 3\n2 5\n1 5\n5 5\n2 4\n3 3\n"
] | [
"4\n5\n6\n3\n4\n2\n"
] | none | 2,000 | [
{
"input": "6 5\n1 2\n5 4\n2 3\n3 1\n3 6\n6\n1 3\n2 5\n1 5\n5 5\n2 4\n3 3",
"output": "4\n5\n6\n3\n4\n2"
},
{
"input": "2 1\n2 1\n2\n1 1\n1 1",
"output": "2\n2"
},
{
"input": "3 2\n3 2\n3 1\n4\n1 1\n1 2\n2 2\n2 2",
"output": "2\n3\n2\n2"
},
{
"input": "3 3\n2 3\n3 1\n2 1\n5\n2 3\n3 3\n2 2\n2 2\n2 2",
"output": "2\n1\n1\n1\n1"
},
{
"input": "4 5\n1 4\n2 1\n4 3\n2 1\n3 4\n5\n4 5\n2 4\n4 4\n1 3\n4 4",
"output": "1\n2\n1\n2\n1"
},
{
"input": "5 4\n3 2\n5 2\n5 3\n2 3\n8\n4 4\n1 1\n3 4\n1 1\n3 3\n3 4\n3 4\n4 4",
"output": "3\n3\n3\n3\n3\n3\n3\n3"
},
{
"input": "8 10\n8 6\n8 7\n8 3\n3 7\n4 8\n1 6\n5 1\n8 7\n6 8\n1 6\n13\n1 10\n2 6\n3 3\n5 5\n2 2\n1 3\n10 10\n7 7\n2 4\n3 6\n2 7\n9 9\n3 6",
"output": "8\n4\n2\n3\n2\n2\n2\n3\n3\n4\n5\n2\n4"
},
{
"input": "10 10\n7 5\n5 9\n10 9\n8 7\n5 10\n4 2\n8 2\n9 1\n2 8\n10 7\n10\n10 10\n7 9\n2 6\n1 5\n4 7\n9 9\n7 7\n2 6\n6 9\n10 10",
"output": "3\n5\n6\n6\n5\n3\n3\n6\n6\n3"
},
{
"input": "7 14\n7 1\n1 5\n6 4\n7 6\n2 4\n2 4\n7 2\n3 1\n7 6\n6 7\n5 3\n5 4\n1 3\n6 2\n40\n2 3\n14 14\n13 14\n13 13\n7 9\n1 13\n12 14\n14 14\n12 12\n6 10\n6 14\n8 8\n14 14\n9 10\n8 9\n8 11\n9 9\n2 3\n1 11\n13 14\n4 11\n2 9\n1 10\n6 11\n3 3\n4 12\n5 11\n8 8\n7 14\n13 13\n14 14\n14 14\n8 12\n14 14\n8 8\n7 7\n2 11\n10 12\n4 5\n9 10",
"output": "1\n1\n1\n1\n1\n6\n1\n1\n1\n1\n2\n1\n1\n1\n1\n1\n1\n1\n4\n1\n1\n1\n3\n1\n1\n2\n1\n1\n2\n1\n1\n1\n1\n1\n1\n1\n3\n1\n1\n1"
}
] | 1,686,834,073 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 11 | 2,000 | 11,264,000 | import sys
input = sys.stdin.readline
from copy import deepcopy
class UnionFind:
def __init__(self, n: int) -> None:
self.fa = list(range(n))
self.rank = [0] * n
self.set_count = n
def find(self, x: int) -> int:
cur = x
while x != self.fa[x]:
x = self.fa[x]
# while fa[cur] != x:
# tmp = fa[cur]
# fa[cur] = x
# cur = tmp
while cur != x:
self.fa[cur], cur = x, self.fa[cur]
return x
def union(self, x: int, y: int):
x, y = self.find(x), self.find(y)
if x == y: return
# if x < y: x, y = y, x
if self.rank[x] < self.rank[y]:
x, y = y, x
elif self.rank[x] == self.rank[y]:
self.rank[x] += 1
# self.rank[x] += self.rank[y]
self.fa[y] = x
self.set_count -= 1
n, m = list(map(int, input().split()))
edges = []
for _ in range(m):
u, v = list(map(int, input().split()))
edges.append((u - 1, v - 1))
dsu = UnionFind(n)
pre = []
for i in range(m):
pre.append(deepcopy(dsu))
x, y = edges[i]
dsu.union(x, y)
dsu = UnionFind(n)
suf = []
for i in range(m - 1, -1, -1):
suf.append(dsu.fa[:])
x, y = edges[i]
dsu.union(x, y)
# print(pre,suf)
q = int(input())
for _ in range(q):
l, r = list(map(int, input().split()))
# 合并两个并查集
s = suf[m - r]
dsu = deepcopy(pre[l - 1])
for i in range(n):
dsu.union(i, s[i])
print(dsu.set_count) | Title: Connected Components
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We already know of the large corporation where Polycarpus works as a system administrator. The computer network there consists of *n* computers and *m* cables that connect some pairs of computers. In other words, the computer network can be represented as some non-directed graph with *n* nodes and *m* edges. Let's index the computers with integers from 1 to *n*, let's index the cables with integers from 1 to *m*.
Polycarpus was given an important task — check the reliability of his company's network. For that Polycarpus decided to carry out a series of *k* experiments on the computer network, where the *i*-th experiment goes as follows:
1. Temporarily disconnect the cables with indexes from *l**i* to *r**i*, inclusive (the other cables remain connected). 1. Count the number of connected components in the graph that is defining the computer network at that moment. 1. Re-connect the disconnected cables with indexes from *l**i* to *r**i* (that is, restore the initial network).
Help Polycarpus carry out all experiments and for each print the number of connected components in the graph that defines the computer network through the given experiment. Isolated vertex should be counted as single component.
Input Specification:
The first line contains two space-separated integers *n*, *m* (2<=≤<=*n*<=≤<=500; 1<=≤<=*m*<=≤<=104) — the number of computers and the number of cables, correspondingly.
The following *m* lines contain the cables' description. The *i*-th line contains space-separated pair of integers *x**i*, *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*; *x**i*<=≠<=*y**i*) — the numbers of the computers that are connected by the *i*-th cable. Note that a pair of computers can be connected by multiple cables.
The next line contains integer *k* (1<=≤<=*k*<=≤<=2·104) — the number of experiments. Next *k* lines contain the experiments' descriptions. The *i*-th line contains space-separated integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*m*) — the numbers of the cables that Polycarpus disconnects during the *i*-th experiment.
Output Specification:
Print *k* numbers, the *i*-th number represents the number of connected components of the graph that defines the computer network during the *i*-th experiment.
Demo Input:
['6 5\n1 2\n5 4\n2 3\n3 1\n3 6\n6\n1 3\n2 5\n1 5\n5 5\n2 4\n3 3\n']
Demo Output:
['4\n5\n6\n3\n4\n2\n']
Note:
none | ```python
import sys
input = sys.stdin.readline
from copy import deepcopy
class UnionFind:
def __init__(self, n: int) -> None:
self.fa = list(range(n))
self.rank = [0] * n
self.set_count = n
def find(self, x: int) -> int:
cur = x
while x != self.fa[x]:
x = self.fa[x]
# while fa[cur] != x:
# tmp = fa[cur]
# fa[cur] = x
# cur = tmp
while cur != x:
self.fa[cur], cur = x, self.fa[cur]
return x
def union(self, x: int, y: int):
x, y = self.find(x), self.find(y)
if x == y: return
# if x < y: x, y = y, x
if self.rank[x] < self.rank[y]:
x, y = y, x
elif self.rank[x] == self.rank[y]:
self.rank[x] += 1
# self.rank[x] += self.rank[y]
self.fa[y] = x
self.set_count -= 1
n, m = list(map(int, input().split()))
edges = []
for _ in range(m):
u, v = list(map(int, input().split()))
edges.append((u - 1, v - 1))
dsu = UnionFind(n)
pre = []
for i in range(m):
pre.append(deepcopy(dsu))
x, y = edges[i]
dsu.union(x, y)
dsu = UnionFind(n)
suf = []
for i in range(m - 1, -1, -1):
suf.append(dsu.fa[:])
x, y = edges[i]
dsu.union(x, y)
# print(pre,suf)
q = int(input())
for _ in range(q):
l, r = list(map(int, input().split()))
# 合并两个并查集
s = suf[m - r]
dsu = deepcopy(pre[l - 1])
for i in range(n):
dsu.union(i, s[i])
print(dsu.set_count)
``` | 0 | |
44 | E | Anfisa the Monkey | PROGRAMMING | 1,400 | [
"dp"
] | E. Anfisa the Monkey | 2 | 256 | Anfisa the monkey learns to type. She is yet unfamiliar with the "space" key and can only type in lower-case Latin letters. Having typed for a fairly long line, Anfisa understood that it would be great to divide what she has written into *k* lines not shorter than *a* and not longer than *b*, for the text to resemble human speech more. Help Anfisa. | The first line contains three integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=200, 1<=≤<=*a*<=≤<=*b*<=≤<=200). The second line contains a sequence of lowercase Latin letters — the text typed by Anfisa. It is guaranteed that the given line is not empty and its length does not exceed 200 symbols. | Print *k* lines, each of which contains no less than *a* and no more than *b* symbols — Anfisa's text divided into lines. It is not allowed to perform any changes in the text, such as: deleting or adding symbols, changing their order, etc. If the solution is not unique, print any of them. If there is no solution, print "No solution" (without quotes). | [
"3 2 5\nabrakadabra\n",
"4 1 2\nabrakadabra\n"
] | [
"ab\nrakad\nabra\n",
"No solution\n"
] | none | 0 | [
{
"input": "3 2 5\nabrakadabra",
"output": "abra\nkada\nbra"
},
{
"input": "4 1 2\nabrakadabra",
"output": "No solution"
},
{
"input": "3 1 2\nvgnfpo",
"output": "vg\nnf\npo"
},
{
"input": "5 3 4\nvrrdnhazvexzjfv",
"output": "vrr\ndnh\nazv\nexz\njfv"
},
{
"input": "10 12 15\nctxgddcfdtllmpuxsjkubuqpldznulsilueakbwwlzgeyudyrjachmitfdcgyzszoejphrubpxzpdtgexaqpxgnoxwfjoikljudnoucirussumyhetfwgaoxfbugfiyjmp",
"output": "ctxgddcfdtllm\npuxsjkubuqpld\nznulsilueakbw\nwlzgeyudyrjac\nhmitfdcgyzszo\nejphrubpxzpdt\ngexaqpxgnoxwf\njoikljudnouci\nrussumyhetfwg\naoxfbugfiyjmp"
},
{
"input": "10 20 30\nbvdqvlxiyogiyimdlwdyxsummjgqxaxsucfeuegleetybsylpnepkqzbutibtlgqrbjbwqnvkysxftmsjqkczoploxoqfuwyrufzwwsxpcqfuckjainpphpbvvtllgkljnnoibsvwnxvaksxjrffakpoxwkhjjjemqatbfkmmlmjhhroetlqvfaumctbicqkuxaabpsh",
"output": "bvdqvlxiyogiyimdlwdy\nxsummjgqxaxsucfeuegl\neetybsylpnepkqzbutib\ntlgqrbjbwqnvkysxftms\njqkczoploxoqfuwyrufz\nwwsxpcqfuckjainpphpb\nvvtllgkljnnoibsvwnxv\naksxjrffakpoxwkhjjje\nmqatbfkmmlmjhhroetlq\nvfaumctbicqkuxaabpsh"
},
{
"input": "10 1 200\nolahgjusovchbowjxtwzvjakrktyjqcgkqmcxknjchzxcvbnkbakwnxdouebomyhjsrfsicmzsgdweabbuipbzrhuqfpynybaohzquqbbsqpoaskccszzsmnfleevtasmjuwqgcqtvysohvyutqipnvuhjumwwyytkeuebbncxsnpavwdkoxyycqrhcidf",
"output": "olahgjusovchbowjxtw\nzvjakrktyjqcgkqmcxk\nnjchzxcvbnkbakwnxdo\nuebomyhjsrfsicmzsgd\nweabbuipbzrhuqfpyny\nbaohzquqbbsqpoaskcc\nszzsmnfleevtasmjuwq\ngcqtvysohvyutqipnvu\nhjumwwyytkeuebbncxs\nnpavwdkoxyycqrhcidf"
},
{
"input": "30 3 6\nebdgacrmhfldirwrcfadurngearrfyjiqkmfqmgzpnzcpprkjyeuuppzvmibzzwyouhxclcgqtjhjmucypqnhdaqke",
"output": "ebd\ngac\nrmh\nfld\nirw\nrcf\nadu\nrng\near\nrfy\njiq\nkmf\nqmg\nzpn\nzcp\nprk\njye\nuup\npzv\nmib\nzzw\nyou\nhxc\nlcg\nqtj\nhjm\nucy\npqn\nhda\nqke"
},
{
"input": "200 1 200\nlycjpjrpkgxrkfvutlcwglghxadttpihmlpphwfttegfpimjxintjdxgqfhzrmxfcfojnxruhyfynlzgpxjeobjyxarsfxaqeogxfzvdlwsimupkwujudtfenryulzvsiazneyibqtweeuxpzrbumqqswjasliyjnnzfzuvthhzcsgfljikkajqkpjftztrzpjneaxqg",
"output": "l\ny\nc\nj\np\nj\nr\np\nk\ng\nx\nr\nk\nf\nv\nu\nt\nl\nc\nw\ng\nl\ng\nh\nx\na\nd\nt\nt\np\ni\nh\nm\nl\np\np\nh\nw\nf\nt\nt\ne\ng\nf\np\ni\nm\nj\nx\ni\nn\nt\nj\nd\nx\ng\nq\nf\nh\nz\nr\nm\nx\nf\nc\nf\no\nj\nn\nx\nr\nu\nh\ny\nf\ny\nn\nl\nz\ng\np\nx\nj\ne\no\nb\nj\ny\nx\na\nr\ns\nf\nx\na\nq\ne\no\ng\nx\nf\nz\nv\nd\nl\nw\ns\ni\nm\nu\np\nk\nw\nu\nj\nu\nd\nt\nf\ne\nn\nr\ny\nu\nl\nz\nv\ns\ni\na\nz\nn\ne\ny\ni\nb\nq\nt\nw\ne\ne\nu\nx\np\nz\nr\nb\nu\nm\nq\nq\ns\nw\nj\na\ns\nl\ni\ny\nj\nn\nn\nz\nf\nz\nu\nv\nt\nh\nh\nz..."
},
{
"input": "15 3 4\naronayjutjdhjcelgexgalnyiruevjelvcvzaihgbwwrc",
"output": "aro\nnay\njut\njdh\njce\nlge\nxga\nlny\niru\nevj\nelv\ncvz\naih\ngbw\nwrc"
},
{
"input": "7 3 4\nweoghhroclwslkfcsszplh",
"output": "weog\nhhr\nocl\nwsl\nkfc\nssz\nplh"
},
{
"input": "12 2 5\nozgscnrddhejkhllokmafxcuorxryhvqnkikauclhfbddfoxl",
"output": "ozgsc\nnrdd\nhejk\nhllo\nkmaf\nxcuo\nrxry\nhvqn\nkika\nuclh\nfbdd\nfoxl"
},
{
"input": "3 1 2\nfpos",
"output": "fp\no\ns"
},
{
"input": "5 3 4\nvrrdnhazvexzjfvs",
"output": "vrrd\nnha\nzve\nxzj\nfvs"
},
{
"input": "10 12 15\nllmpuxsjkubuqpldznulsilueakbwwlzgeyudyrjachmitfdcgyzszoejphrubpxzpdtgexaqpxgnoxwfjoikljudnoucirussumyhetfwgaoxfbugfiyjmpm",
"output": "llmpuxsjkubuq\npldznulsilue\nakbwwlzgeyud\nyrjachmitfdc\ngyzszoejphru\nbpxzpdtgexaq\npxgnoxwfjoik\nljudnoucirus\nsumyhetfwgao\nxfbugfiyjmpm"
},
{
"input": "10 20 30\nvdqvlxiyogiyimdlwdyxsummjgqxaxsucfeuegleetybsylpnepkqzbutibtlgqrbjbwqnvkysxftmsjqkczoploxoqfuwyrufzwwsxpcqfuckjainpphpbvvtllgkljnnoibsvwnxvaksxjrffakpoxwkhjjjemqatbfkmmlmjhhroetlqvfaumctbicqkuxaabpshu",
"output": "vdqvlxiyogiyimdlwdyx\nsummjgqxaxsucfeuegle\netybsylpnepkqzbutibt\nlgqrbjbwqnvkysxftmsj\nqkczoploxoqfuwyrufzw\nwsxpcqfuckjainpphpbv\nvtllgkljnnoibsvwnxva\nksxjrffakpoxwkhjjjem\nqatbfkmmlmjhhroetlqv\nfaumctbicqkuxaabpshu"
},
{
"input": "10 1 200\nolahgjusovchbowjxtwzvjakrktyjqcgkqmcxknjchzxcvbnkbakwnxdouebomyhjsrfsicmzsgdweabbuipbzrhuqfpynybaohzquqbbsqpoaskccszzsmnfleevtasmjuwqgcqtvysohvyutqipnvuhjumwwyytkeuebbncxsnpavwdkoxyycqrhcidfd",
"output": "olahgjusovchbowjxtwz\nvjakrktyjqcgkqmcxkn\njchzxcvbnkbakwnxdou\nebomyhjsrfsicmzsgdw\neabbuipbzrhuqfpynyb\naohzquqbbsqpoaskccs\nzzsmnfleevtasmjuwqg\ncqtvysohvyutqipnvuh\njumwwyytkeuebbncxsn\npavwdkoxyycqrhcidfd"
},
{
"input": "30 3 6\nhstvoyuksbbsbgatemzmvbhbjdmnzpluefgzlcqgfsmkdydadsonaryzskleebdgacrmhfldirwrcfadurngearrfyjiqkmfqmgzpnzcpprkjyeuuppzvmibzzwyouhxclcgqtjhjmucypqnhdaqkea",
"output": "hstvoy\nuksbb\nsbgat\nemzmv\nbhbjd\nmnzpl\nuefgz\nlcqgf\nsmkdy\ndadso\nnaryz\nsklee\nbdgac\nrmhfl\ndirwr\ncfadu\nrngea\nrrfyj\niqkmf\nqmgzp\nnzcpp\nrkjye\nuuppz\nvmibz\nzwyou\nhxclc\ngqtjh\njmucy\npqnhd\naqkea"
},
{
"input": "200 1 200\nycjpjrpkgxrkfvutlcwglghxadttpihmlpphwfttegfpimjxintjdxgqfhzrmxfcfojnxruhyfynlzgpxjeobjyxarsfxaqeogxfzvdlwsimupkwujudtfenryulzvsiazneyibqtweeuxpzrbumqqswjasliyjnnzfzuvthhzcsgfljikkajqkpjftztrzpjneaxqgn",
"output": "y\nc\nj\np\nj\nr\np\nk\ng\nx\nr\nk\nf\nv\nu\nt\nl\nc\nw\ng\nl\ng\nh\nx\na\nd\nt\nt\np\ni\nh\nm\nl\np\np\nh\nw\nf\nt\nt\ne\ng\nf\np\ni\nm\nj\nx\ni\nn\nt\nj\nd\nx\ng\nq\nf\nh\nz\nr\nm\nx\nf\nc\nf\no\nj\nn\nx\nr\nu\nh\ny\nf\ny\nn\nl\nz\ng\np\nx\nj\ne\no\nb\nj\ny\nx\na\nr\ns\nf\nx\na\nq\ne\no\ng\nx\nf\nz\nv\nd\nl\nw\ns\ni\nm\nu\np\nk\nw\nu\nj\nu\nd\nt\nf\ne\nn\nr\ny\nu\nl\nz\nv\ns\ni\na\nz\nn\ne\ny\ni\nb\nq\nt\nw\ne\ne\nu\nx\np\nz\nr\nb\nu\nm\nq\nq\ns\nw\nj\na\ns\nl\ni\ny\nj\nn\nn\nz\nf\nz\nu\nv\nt\nh\nh\nz\nc..."
},
{
"input": "15 3 4\naronayjutjdhjcelgexgalnyiruevjelvcvzaihgbwwrcq",
"output": "aron\nayj\nutj\ndhj\ncel\ngex\ngal\nnyi\nrue\nvje\nlvc\nvza\nihg\nbww\nrcq"
},
{
"input": "200 1 10\njtlykeyfekfrzbpzrhvrxagzywzlsktyzoriwiyatoetikfnhyhlrhuogyhjrxdmlqvpfsmqiqkivtodligzerymdtnqahuprhbfefbjwuavmpkurtfzmwediq",
"output": "No solution"
},
{
"input": "15 2 3\ndplkzxpsxodehcj",
"output": "No solution"
},
{
"input": "100 100 200\nximcxraplfjygtrpxrgjhqagrojixizlogaqfvwvqjaiqvcimelxtmtcsqluvcrdzhihgmwhywfgxmzmikdqdytfrlpzqmvhaexrtflwacsuxhkuzbukgvbdcmwpcvxwznupsmmryxwexlevjlonpipuxjgagxtcgqjdczrnmktgcaagmiumnbcxuafmysisahaqnngc",
"output": "No solution"
},
{
"input": "7 2 3\nggzkinj",
"output": "No solution"
},
{
"input": "17 2 4\npgyujupquzenuldnt",
"output": "No solution"
},
{
"input": "100 1 1\nratfdjnvjmaqgcttjtenixeocyxrtuwhpmejhpxjcqhzjsujqolgcccmvnpoomkrforsdtvhgrcpakibozhgqotcrctzozhggrufk",
"output": "No solution"
},
{
"input": "50 2 3\nizlszyucwjarrrgxzbfzyoxapozmunxuygfjynslcjnxitimjjklucjowtkccbnfsuwtyroxirhxzosbyhvnrroaxryhcvvcjvwfcpvnpdaqwzaiuzycyrtvkgkjfbdqnzrmritaonptpvncdifushrquywzykybhjdplbmsrgibpknxkxkqqywmkeljpxrrmufpkubv",
"output": "No solution"
},
{
"input": "15 2 5\nkddainaviqrjsesrhhdnbuisennbgcxseeyxqtmautpoobtpfigcpgagcixmyzsntmgzwmiczsfp",
"output": "No solution"
},
{
"input": "3 1 50\nhcdonseimahtfmtejvxebwctfkjsrcqjrunpcofrapijvwmmbbbrohkskjomeknlwkdxscybxkintcaynwyjfaghwcofpsbwruzqqqkhyndbxbdpgqokjqitznnnrfuaciriqmyuvktpdxewkrycjefkmjwglhoggpgvztvqndbhiajryxqlrqdb",
"output": "No solution"
},
{
"input": "5 1 30\nxmuatgstrlkerxzezenrauupxiskpfugncncatcgtffhuwzojuapgrevnwzfkpyzbzljbzwvfoeuqhinyravsfqrjmgidjoszvkkhxrdstmydvbertvzltpipmcuakzqflldztzdjqlicvadgpicqio",
"output": "No solution"
},
{
"input": "5 2 3\nabacababb",
"output": "No solution"
},
{
"input": "5 6 6\nabacabadabacabaabacabadabacab",
"output": "No solution"
}
] | 1,595,570,687 | 2,147,483,647 | PyPy 3 | OK | TESTS | 32 | 310 | 20,172,800 | from sys import stdin
def solution():
input = stdin.readline
k, a, b = map(int, input().split())
s = input().rstrip()
n = len(s)
minn = k * a
maxx = k * b
if minn <= n <= maxx:
diff1 = n - minn
diff2 = b - a
try:
l1 = diff1 // diff2
except:
l1 = 0
m = n - minn - (diff2 * l1)
j = 0
for i in range(k):
if l1:
print(s[j:j + b])
l1 -= 1
j += b
elif m:
print(s[j:j + a + m])
j = j + a + m
m = 0
else:
print(s[j:j + a])
j += a
else:
print("No solution")
if __name__ == "__main__":
solution()
| Title: Anfisa the Monkey
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Anfisa the monkey learns to type. She is yet unfamiliar with the "space" key and can only type in lower-case Latin letters. Having typed for a fairly long line, Anfisa understood that it would be great to divide what she has written into *k* lines not shorter than *a* and not longer than *b*, for the text to resemble human speech more. Help Anfisa.
Input Specification:
The first line contains three integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=200, 1<=≤<=*a*<=≤<=*b*<=≤<=200). The second line contains a sequence of lowercase Latin letters — the text typed by Anfisa. It is guaranteed that the given line is not empty and its length does not exceed 200 symbols.
Output Specification:
Print *k* lines, each of which contains no less than *a* and no more than *b* symbols — Anfisa's text divided into lines. It is not allowed to perform any changes in the text, such as: deleting or adding symbols, changing their order, etc. If the solution is not unique, print any of them. If there is no solution, print "No solution" (without quotes).
Demo Input:
['3 2 5\nabrakadabra\n', '4 1 2\nabrakadabra\n']
Demo Output:
['ab\nrakad\nabra\n', 'No solution\n']
Note:
none | ```python
from sys import stdin
def solution():
input = stdin.readline
k, a, b = map(int, input().split())
s = input().rstrip()
n = len(s)
minn = k * a
maxx = k * b
if minn <= n <= maxx:
diff1 = n - minn
diff2 = b - a
try:
l1 = diff1 // diff2
except:
l1 = 0
m = n - minn - (diff2 * l1)
j = 0
for i in range(k):
if l1:
print(s[j:j + b])
l1 -= 1
j += b
elif m:
print(s[j:j + a + m])
j = j + a + m
m = 0
else:
print(s[j:j + a])
j += a
else:
print("No solution")
if __name__ == "__main__":
solution()
``` | 3.884925 |
137 | B | Permutation | PROGRAMMING | 1,000 | [
"greedy"
] | null | null | "Hey, it's homework time" — thought Polycarpus and of course he started with his favourite subject, IT. Polycarpus managed to solve all tasks but for the last one in 20 minutes. However, as he failed to solve the last task after some considerable time, the boy asked you to help him.
The sequence of *n* integers is called a permutation if it contains all integers from 1 to *n* exactly once.
You are given an arbitrary sequence *a*1,<=*a*2,<=...,<=*a**n* containing *n* integers. Each integer is not less than 1 and not greater than 5000. Determine what minimum number of elements Polycarpus needs to change to get a permutation (he should not delete or add numbers). In a single change he can modify any single sequence element (i. e. replace it with another integer). | The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=5000) which represents how many numbers are in the sequence. The second line contains a sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=5000,<=1<=≤<=*i*<=≤<=*n*). | Print the only number — the minimum number of changes needed to get the permutation. | [
"3\n3 1 2\n",
"2\n2 2\n",
"5\n5 3 3 3 1\n"
] | [
"0\n",
"1\n",
"2\n"
] | The first sample contains the permutation, which is why no replacements are required.
In the second sample it is enough to replace the first element with the number 1 and that will make the sequence the needed permutation.
In the third sample we can replace the second element with number 4 and the fourth element with number 2. | 1,000 | [
{
"input": "3\n3 1 2",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "5\n5 3 3 3 1",
"output": "2"
},
{
"input": "5\n6 6 6 6 6",
"output": "5"
},
{
"input": "10\n1 1 2 2 8 8 7 7 9 9",
"output": "5"
},
{
"input": "8\n9 8 7 6 5 4 3 2",
"output": "1"
},
{
"input": "15\n1 2 3 4 5 5 4 3 2 1 1 2 3 4 5",
"output": "10"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n5000",
"output": "1"
},
{
"input": "4\n5000 5000 5000 5000",
"output": "4"
},
{
"input": "5\n3366 3461 4 5 4370",
"output": "3"
},
{
"input": "10\n8 2 10 3 4 6 1 7 9 5",
"output": "0"
},
{
"input": "10\n551 3192 3213 2846 3068 1224 3447 1 10 9",
"output": "7"
},
{
"input": "15\n4 1459 12 4281 3241 2748 10 3590 14 845 3518 1721 2 2880 1974",
"output": "10"
},
{
"input": "15\n15 1 8 2 13 11 12 7 3 14 6 10 9 4 5",
"output": "0"
},
{
"input": "15\n2436 2354 4259 1210 2037 2665 700 3578 2880 973 1317 1024 24 3621 4142",
"output": "15"
},
{
"input": "30\n28 1 3449 9 3242 4735 26 3472 15 21 2698 7 4073 3190 10 3 29 1301 4526 22 345 3876 19 12 4562 2535 2 630 18 27",
"output": "14"
},
{
"input": "100\n50 39 95 30 66 78 2169 4326 81 31 74 34 80 40 19 48 97 63 82 6 88 16 21 57 92 77 10 1213 17 93 32 91 38 4375 29 75 44 22 4 45 14 2395 3254 59 3379 2 85 96 8 83 27 94 1512 2960 100 9 73 79 7 25 55 69 90 99 51 87 98 62 18 35 43 4376 4668 28 72 56 4070 61 65 36 54 4106 11 24 15 86 70 71 4087 23 13 76 20 4694 26 4962 4726 37 14 64",
"output": "18"
},
{
"input": "100\n340 14 3275 2283 2673 1107 817 2243 1226 32 2382 3638 4652 418 68 4962 387 764 4647 159 1846 225 2760 4904 3150 403 3 2439 91 4428 92 4705 75 348 1566 1465 69 6 49 4 62 4643 564 1090 3447 1871 2255 139 24 99 2669 969 86 61 4550 158 4537 3993 1589 872 2907 1888 401 80 1825 1483 63 1 2264 4068 4113 2548 41 885 4806 36 67 167 4447 34 1248 2593 82 202 81 1783 1284 4973 16 43 95 7 865 2091 3008 1793 20 947 4912 3604",
"output": "70"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "2\n5000 5000",
"output": "2"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
},
{
"input": "2\n3 4",
"output": "2"
},
{
"input": "10\n1 2 3 4 5 6 7 1000 10 10",
"output": "2"
}
] | 1,591,998,358 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 18 | 2,000 | 1,945,600 |
from collections import Counter
def main():
n = int(input())
arr = list(map(int,input().split()))
c = 0
for i in range(1,n+1):
if arr.count(i) >= 1:
c += 1
print(n - c)
main() | Title: Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Hey, it's homework time" — thought Polycarpus and of course he started with his favourite subject, IT. Polycarpus managed to solve all tasks but for the last one in 20 minutes. However, as he failed to solve the last task after some considerable time, the boy asked you to help him.
The sequence of *n* integers is called a permutation if it contains all integers from 1 to *n* exactly once.
You are given an arbitrary sequence *a*1,<=*a*2,<=...,<=*a**n* containing *n* integers. Each integer is not less than 1 and not greater than 5000. Determine what minimum number of elements Polycarpus needs to change to get a permutation (he should not delete or add numbers). In a single change he can modify any single sequence element (i. e. replace it with another integer).
Input Specification:
The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=5000) which represents how many numbers are in the sequence. The second line contains a sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=5000,<=1<=≤<=*i*<=≤<=*n*).
Output Specification:
Print the only number — the minimum number of changes needed to get the permutation.
Demo Input:
['3\n3 1 2\n', '2\n2 2\n', '5\n5 3 3 3 1\n']
Demo Output:
['0\n', '1\n', '2\n']
Note:
The first sample contains the permutation, which is why no replacements are required.
In the second sample it is enough to replace the first element with the number 1 and that will make the sequence the needed permutation.
In the third sample we can replace the second element with number 4 and the fourth element with number 2. | ```python
from collections import Counter
def main():
n = int(input())
arr = list(map(int,input().split()))
c = 0
for i in range(1,n+1):
if arr.count(i) >= 1:
c += 1
print(n - c)
main()
``` | 0 | |
710 | B | Optimal Point on a Line | PROGRAMMING | 1,400 | [
"brute force",
"sortings"
] | null | null | You are given *n* points on a line with their coordinates *x**i*. Find the point *x* so the sum of distances to the given points is minimal. | The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of points on the line.
The second line contains *n* integers *x**i* (<=-<=109<=≤<=*x**i*<=≤<=109) — the coordinates of the given *n* points. | Print the only integer *x* — the position of the optimal point on the line. If there are several optimal points print the position of the leftmost one. It is guaranteed that the answer is always the integer. | [
"4\n1 2 3 4\n"
] | [
"2\n"
] | none | 0 | [
{
"input": "4\n1 2 3 4",
"output": "2"
},
{
"input": "5\n-1 -10 2 6 7",
"output": "2"
},
{
"input": "10\n-68 10 87 22 30 89 82 -97 -52 25",
"output": "22"
},
{
"input": "100\n457 827 807 17 871 935 907 -415 536 170 551 -988 865 758 -457 -892 -875 -488 684 19 0 555 -807 -624 -239 826 318 811 20 -732 -91 460 551 -610 555 -493 -154 442 -141 946 -913 -104 704 -380 699 32 106 -455 -518 214 -464 -861 243 -798 -472 559 529 -844 -32 871 -459 236 387 626 -318 -580 -611 -842 790 486 64 951 81 78 -693 403 -731 309 678 696 891 846 -106 918 212 -44 994 606 -829 -454 243 -477 -402 -818 -819 -310 -837 -209 736 424",
"output": "64"
},
{
"input": "2\n-1 0",
"output": "-1"
},
{
"input": "48\n-777 -767 -764 -713 -688 -682 -606 -586 -585 -483 -465 -440 -433 -397 -390 -377 -299 -252 -159 -147 -96 -29 -15 15 52 109 124 129 142 218 231 314 320 339 442 496 505 548 575 576 594 624 694 827 891 979 981 981",
"output": "15"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 1 1 1 1 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "4\n-1 -1 0 1",
"output": "-1"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 1000000000",
"output": "0"
},
{
"input": "2\n1 -1",
"output": "-1"
},
{
"input": "2\n100 50",
"output": "50"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "1\n10",
"output": "10"
},
{
"input": "3\n606194955 -856471310 117647402",
"output": "117647402"
},
{
"input": "2\n615002717 -843553590",
"output": "-843553590"
},
{
"input": "2\n-1 2",
"output": "-1"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "5\n-638512131 348325781 -550537933 -618161835 -567935532",
"output": "-567935532"
},
{
"input": "1\n120",
"output": "120"
},
{
"input": "2\n-1000000000 1000000000",
"output": "-1000000000"
},
{
"input": "1\n618309368",
"output": "618309368"
}
] | 1,597,502,192 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 124 | 0 | n=int(input())
a=[int(i) for i in input().split()]
a=sorted(a)
b=a[0]
c=a[len(a)-1]
d=(c-b)
print(a[0]+(d//2)) | Title: Optimal Point on a Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* points on a line with their coordinates *x**i*. Find the point *x* so the sum of distances to the given points is minimal.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of points on the line.
The second line contains *n* integers *x**i* (<=-<=109<=≤<=*x**i*<=≤<=109) — the coordinates of the given *n* points.
Output Specification:
Print the only integer *x* — the position of the optimal point on the line. If there are several optimal points print the position of the leftmost one. It is guaranteed that the answer is always the integer.
Demo Input:
['4\n1 2 3 4\n']
Demo Output:
['2\n']
Note:
none | ```python
n=int(input())
a=[int(i) for i in input().split()]
a=sorted(a)
b=a[0]
c=a[len(a)-1]
d=(c-b)
print(a[0]+(d//2))
``` | 0 | |
397 | A | On Segment's Own Points | PROGRAMMING | 0 | [
"implementation"
] | null | null | Our old friend Alexey has finally entered the University of City N — the Berland capital. Alexey expected his father to get him a place to live in but his father said it was high time for Alexey to practice some financial independence. So, Alexey is living in a dorm.
The dorm has exactly one straight dryer — a 100 centimeter long rope to hang clothes on. The dryer has got a coordinate system installed: the leftmost end of the dryer has coordinate 0, and the opposite end has coordinate 100. Overall, the university has *n* students. Dean's office allows *i*-th student to use the segment (*l**i*,<=*r**i*) of the dryer. However, the dean's office actions are contradictory and now one part of the dryer can belong to multiple students!
Alexey don't like when someone touch his clothes. That's why he want make it impossible to someone clothes touch his ones. So Alexey wonders: what is the total length of the parts of the dryer that he may use in a such way that clothes of the others (*n*<=-<=1) students aren't drying there. Help him! Note that Alexey, as the most respected student, has number 1. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100). The (*i*<=+<=1)-th line contains integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=100) — the endpoints of the corresponding segment for the *i*-th student. | On a single line print a single number *k*, equal to the sum of lengths of the parts of the dryer which are inside Alexey's segment and are outside all other segments. | [
"3\n0 5\n2 8\n1 6\n",
"3\n0 10\n1 5\n7 15\n"
] | [
"1\n",
"3\n"
] | Note that it's not important are clothes drying on the touching segments (e.g. (0, 1) and (1, 2)) considered to be touching or not because you need to find the length of segments.
In the first test sample Alexey may use the only segment (0, 1). In such case his clothes will not touch clothes on the segments (1, 6) and (2, 8). The length of segment (0, 1) is 1.
In the second test sample Alexey may dry his clothes on segments (0, 1) and (5, 7). Overall length of these segments is 3. | 500 | [
{
"input": "3\n0 5\n2 8\n1 6",
"output": "1"
},
{
"input": "3\n0 10\n1 5\n7 15",
"output": "3"
},
{
"input": "1\n0 100",
"output": "100"
},
{
"input": "2\n1 9\n1 9",
"output": "0"
},
{
"input": "2\n1 9\n5 10",
"output": "4"
},
{
"input": "2\n1 9\n3 5",
"output": "6"
},
{
"input": "2\n3 5\n1 9",
"output": "0"
},
{
"input": "10\n43 80\n39 75\n26 71\n4 17\n11 57\n31 42\n1 62\n9 19\n27 76\n34 53",
"output": "4"
},
{
"input": "50\n33 35\n98 99\n1 2\n4 6\n17 18\n63 66\n29 30\n35 37\n44 45\n73 75\n4 5\n39 40\n92 93\n96 97\n23 27\n49 50\n2 3\n60 61\n43 44\n69 70\n7 8\n45 46\n21 22\n85 86\n48 49\n41 43\n70 71\n10 11\n27 28\n71 72\n6 7\n15 16\n46 47\n89 91\n54 55\n19 21\n86 87\n37 38\n77 82\n84 85\n54 55\n93 94\n45 46\n37 38\n75 76\n22 23\n50 52\n38 39\n1 2\n66 67",
"output": "2"
},
{
"input": "2\n1 5\n7 9",
"output": "4"
},
{
"input": "2\n1 5\n3 5",
"output": "2"
},
{
"input": "2\n1 5\n1 2",
"output": "3"
},
{
"input": "5\n5 10\n5 10\n5 10\n5 10\n5 10",
"output": "0"
},
{
"input": "6\n1 99\n33 94\n68 69\n3 35\n93 94\n5 98",
"output": "3"
},
{
"input": "11\n2 98\n63 97\n4 33\n12 34\n34 65\n23 31\n43 54\n82 99\n15 84\n23 52\n4 50",
"output": "2"
},
{
"input": "10\n95 96\n19 20\n72 73\n1 2\n25 26\n48 49\n90 91\n22 23\n16 17\n16 17",
"output": "1"
},
{
"input": "11\n1 100\n63 97\n4 33\n12 34\n34 65\n23 31\n43 54\n82 99\n15 84\n23 52\n4 50",
"output": "4"
},
{
"input": "21\n0 100\n81 90\n11 68\n18 23\n75 78\n45 86\n37 58\n15 21\n40 98\n53 100\n10 70\n14 75\n1 92\n23 81\n13 66\n93 100\n6 34\n22 87\n27 84\n15 63\n54 91",
"output": "1"
},
{
"input": "10\n60 66\n5 14\n1 3\n55 56\n70 87\n34 35\n16 21\n23 24\n30 31\n25 27",
"output": "6"
},
{
"input": "40\n29 31\n22 23\n59 60\n70 71\n42 43\n13 15\n11 12\n64 65\n1 2\n62 63\n54 56\n8 9\n2 3\n53 54\n27 28\n48 49\n72 73\n17 18\n46 47\n18 19\n43 44\n39 40\n83 84\n63 64\n52 53\n33 34\n3 4\n24 25\n74 75\n0 1\n61 62\n68 69\n80 81\n5 6\n36 37\n81 82\n50 51\n66 67\n69 70\n20 21",
"output": "2"
},
{
"input": "15\n22 31\n0 4\n31 40\n77 80\n81 83\n11 13\n59 61\n53 59\n51 53\n87 88\n14 22\n43 45\n8 10\n45 47\n68 71",
"output": "9"
},
{
"input": "31\n0 100\n2 97\n8 94\n9 94\n14 94\n15 93\n15 90\n17 88\n19 88\n19 87\n20 86\n25 86\n30 85\n32 85\n35 82\n35 81\n36 80\n37 78\n38 74\n38 74\n39 71\n40 69\n40 68\n41 65\n43 62\n44 62\n45 61\n45 59\n46 57\n49 54\n50 52",
"output": "5"
},
{
"input": "21\n0 97\n46 59\n64 95\n3 16\n86 95\n55 71\n51 77\n26 28\n47 88\n30 40\n26 34\n2 12\n9 10\n4 19\n35 36\n41 92\n1 16\n41 78\n56 81\n23 35\n40 68",
"output": "7"
},
{
"input": "27\n0 97\n7 9\n6 9\n12 33\n12 26\n15 27\n10 46\n33 50\n31 47\n15 38\n12 44\n21 35\n24 37\n51 52\n65 67\n58 63\n53 60\n63 68\n57 63\n60 68\n55 58\n74 80\n70 75\n89 90\n81 85\n93 99\n93 98",
"output": "19"
},
{
"input": "20\n23 24\n22 23\n84 86\n6 10\n40 45\n11 13\n24 27\n81 82\n53 58\n87 90\n14 15\n49 50\n70 75\n75 78\n98 100\n66 68\n18 21\n1 2\n92 93\n34 37",
"output": "1"
},
{
"input": "11\n2 100\n34 65\n4 50\n63 97\n82 99\n43 54\n23 52\n4 33\n15 84\n23 31\n12 34",
"output": "3"
},
{
"input": "60\n73 75\n6 7\n69 70\n15 16\n54 55\n66 67\n7 8\n39 40\n38 39\n37 38\n1 2\n46 47\n7 8\n21 22\n23 27\n15 16\n45 46\n37 38\n60 61\n4 6\n63 66\n10 11\n33 35\n43 44\n2 3\n4 6\n10 11\n93 94\n45 46\n7 8\n44 45\n41 43\n35 37\n17 18\n48 49\n89 91\n27 28\n46 47\n71 72\n1 2\n75 76\n49 50\n84 85\n17 18\n98 99\n54 55\n46 47\n19 21\n77 82\n29 30\n4 5\n70 71\n85 86\n96 97\n86 87\n92 93\n22 23\n50 52\n44 45\n63 66",
"output": "2"
},
{
"input": "40\n47 48\n42 44\n92 94\n15 17\n20 22\n11 13\n37 39\n6 8\n39 40\n35 37\n21 22\n41 42\n77 78\n76 78\n69 71\n17 19\n18 19\n17 18\n84 85\n9 10\n11 12\n51 52\n99 100\n7 8\n97 99\n22 23\n60 62\n7 8\n67 69\n20 22\n13 14\n89 91\n15 17\n12 13\n56 57\n37 39\n29 30\n24 26\n37 38\n25 27",
"output": "1"
},
{
"input": "10\n28 36\n18 26\n28 35\n95 100\n68 72\n41 42\n76 84\n99 100\n6 8\n58 60",
"output": "1"
},
{
"input": "20\n69 72\n88 92\n77 80\n64 69\n66 67\n79 81\n91 96\n78 83\n81 86\n11 12\n48 53\n22 23\n81 84\n89 92\n56 60\n1 4\n1 5\n60 62\n20 23\n63 66",
"output": "3"
},
{
"input": "71\n1 99\n11 69\n86 92\n7 49\n31 70\n42 53\n48 81\n86 96\n36 91\n19 38\n39 91\n41 64\n87 93\n83 97\n40 41\n3 32\n15 18\n58 65\n22 32\n1 71\n58 86\n64 77\n15 69\n4 34\n42 89\n9 66\n15 18\n58 65\n59 96\n39 89\n19 38\n6 63\n26 73\n29 47\n55 88\n5 78\n41 74\n48 81\n20 71\n59 96\n42 49\n4 69\n41 74\n87 93\n0 65\n2 34\n15 18\n10 56\n55 88\n33 56\n42 89\n86 92\n42 81\n65 82\n5 78\n13 52\n32 85\n7 65\n59 96\n4 65\n46 69\n10 56\n42 89\n4 69\n0 65\n32 35\n5 78\n32 75\n42 53\n55 59\n64 77",
"output": "2"
},
{
"input": "1\n1 2",
"output": "1"
}
] | 1,563,285,363 | 2,147,483,647 | PyPy 3 | OK | TESTS | 37 | 139 | 1,228,800 | n = int(input())
h = [1] * 100
a, b = map(int, input().split())
for i in range(n - 1):
x, y = map(int, input().split())
for j in range(x, y):
h[j] = 0
print(sum(h[a : b])) | Title: On Segment's Own Points
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our old friend Alexey has finally entered the University of City N — the Berland capital. Alexey expected his father to get him a place to live in but his father said it was high time for Alexey to practice some financial independence. So, Alexey is living in a dorm.
The dorm has exactly one straight dryer — a 100 centimeter long rope to hang clothes on. The dryer has got a coordinate system installed: the leftmost end of the dryer has coordinate 0, and the opposite end has coordinate 100. Overall, the university has *n* students. Dean's office allows *i*-th student to use the segment (*l**i*,<=*r**i*) of the dryer. However, the dean's office actions are contradictory and now one part of the dryer can belong to multiple students!
Alexey don't like when someone touch his clothes. That's why he want make it impossible to someone clothes touch his ones. So Alexey wonders: what is the total length of the parts of the dryer that he may use in a such way that clothes of the others (*n*<=-<=1) students aren't drying there. Help him! Note that Alexey, as the most respected student, has number 1.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100). The (*i*<=+<=1)-th line contains integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=100) — the endpoints of the corresponding segment for the *i*-th student.
Output Specification:
On a single line print a single number *k*, equal to the sum of lengths of the parts of the dryer which are inside Alexey's segment and are outside all other segments.
Demo Input:
['3\n0 5\n2 8\n1 6\n', '3\n0 10\n1 5\n7 15\n']
Demo Output:
['1\n', '3\n']
Note:
Note that it's not important are clothes drying on the touching segments (e.g. (0, 1) and (1, 2)) considered to be touching or not because you need to find the length of segments.
In the first test sample Alexey may use the only segment (0, 1). In such case his clothes will not touch clothes on the segments (1, 6) and (2, 8). The length of segment (0, 1) is 1.
In the second test sample Alexey may dry his clothes on segments (0, 1) and (5, 7). Overall length of these segments is 3. | ```python
n = int(input())
h = [1] * 100
a, b = map(int, input().split())
for i in range(n - 1):
x, y = map(int, input().split())
for j in range(x, y):
h[j] = 0
print(sum(h[a : b]))
``` | 3 | |
876 | B | Divisiblity of Differences | PROGRAMMING | 1,300 | [
"implementation",
"math",
"number theory"
] | null | null | You are given a multiset of *n* integers. You should select exactly *k* of them in a such way that the difference between any two of them is divisible by *m*, or tell that it is impossible.
Numbers can be repeated in the original multiset and in the multiset of selected numbers, but number of occurrences of any number in multiset of selected numbers should not exceed the number of its occurrences in the original multiset. | First line contains three integers *n*, *k* and *m* (2<=≤<=*k*<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — number of integers in the multiset, number of integers you should select and the required divisor of any pair of selected integers.
Second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — the numbers in the multiset. | If it is not possible to select *k* numbers in the desired way, output «No» (without the quotes).
Otherwise, in the first line of output print «Yes» (without the quotes). In the second line print *k* integers *b*1,<=*b*2,<=...,<=*b**k* — the selected numbers. If there are multiple possible solutions, print any of them. | [
"3 2 3\n1 8 4\n",
"3 3 3\n1 8 4\n",
"4 3 5\n2 7 7 7\n"
] | [
"Yes\n1 4 ",
"No",
"Yes\n2 7 7 "
] | none | 1,000 | [
{
"input": "3 2 3\n1 8 4",
"output": "Yes\n1 4 "
},
{
"input": "3 3 3\n1 8 4",
"output": "No"
},
{
"input": "4 3 5\n2 7 7 7",
"output": "Yes\n2 7 7 "
},
{
"input": "9 9 5\n389149775 833127990 969340400 364457730 48649145 316121525 640054660 924273385 973207825",
"output": "Yes\n389149775 833127990 969340400 364457730 48649145 316121525 640054660 924273385 973207825 "
},
{
"input": "15 8 10\n216175135 15241965 611723934 987180005 151601897 403701727 533996295 207637446 875331635 46172555 604086315 350146655 401084142 156540458 982110455",
"output": "Yes\n216175135 15241965 987180005 533996295 875331635 46172555 604086315 350146655 "
},
{
"input": "2 2 100000\n0 1",
"output": "No"
},
{
"input": "101 25 64\n451 230 14 53 7 520 709 102 678 358 166 870 807 230 230 279 166 230 765 176 742 358 924 976 647 806 870 473 976 994 750 146 802 224 503 801 105 614 882 203 390 338 29 587 214 213 405 806 102 102 621 358 521 742 678 205 309 871 796 326 162 693 268 486 68 627 304 829 806 623 748 934 714 672 712 614 587 589 846 260 593 85 839 257 711 395 336 358 472 133 324 527 599 5 845 920 989 494 358 70 882",
"output": "Yes\n230 102 678 358 166 870 230 230 166 230 742 358 806 870 614 806 102 102 358 742 678 486 806 934 614 "
},
{
"input": "108 29 72\n738 619 711 235 288 288 679 36 785 233 706 71 216 144 216 781 338 583 495 648 144 432 72 720 541 288 158 328 154 202 10 533 635 176 707 216 314 397 440 142 326 458 568 701 745 144 61 634 520 720 744 144 409 127 526 476 101 469 72 432 738 432 235 641 695 276 144 144 231 555 630 9 109 319 437 288 288 317 453 432 601 0 449 576 743 352 333 504 504 369 228 288 381 142 500 72 297 359 230 773 216 576 144 244 437 772 483 51",
"output": "Yes\n288 288 216 144 216 648 144 432 72 720 288 216 144 720 144 72 432 432 144 144 288 288 432 0 576 504 504 288 72 "
},
{
"input": "8 2 6\n750462183 165947982 770714338 368445737 363145692 966611485 376672869 678687947",
"output": "Yes\n165947982 363145692 "
},
{
"input": "12 2 1\n512497388 499105388 575265677 864726520 678272195 667107176 809432109 439696443 770034376 873126825 690514828 541499950",
"output": "Yes\n512497388 499105388 "
},
{
"input": "9 3 1\n506004039 471451660 614118177 518013571 43210072 454727076 285905913 543002174 298515615",
"output": "Yes\n506004039 471451660 614118177 "
},
{
"input": "8 4 6\n344417267 377591123 938158786 682031413 804153975 89006697 275945670 735510539",
"output": "No"
},
{
"input": "8 8 1\n314088413 315795280 271532387 241073087 961218399 884234132 419866508 286799253",
"output": "Yes\n314088413 315795280 271532387 241073087 961218399 884234132 419866508 286799253 "
},
{
"input": "7 7 1\n0 0 0 0 0 0 0",
"output": "Yes\n0 0 0 0 0 0 0 "
},
{
"input": "11 4 3\n0 1 0 1 1 0 0 0 0 0 0",
"output": "Yes\n0 0 0 0 "
},
{
"input": "13 4 4\n1 1 0 3 2 4 1 0 3 4 2 4 3",
"output": "Yes\n0 4 0 4 "
},
{
"input": "5 5 1\n6 4 6 0 4",
"output": "Yes\n6 4 6 0 4 "
},
{
"input": "3 2 3\n1 2 3",
"output": "No"
},
{
"input": "6 3 4\n5 9 10 6 7 8",
"output": "No"
},
{
"input": "4 4 3\n4 7 2 5",
"output": "No"
},
{
"input": "4 2 6\n5 5 5 5",
"output": "Yes\n5 5 "
},
{
"input": "4 3 3\n5 1 4 7",
"output": "Yes\n1 4 7 "
},
{
"input": "4 3 10\n0 1 11 11",
"output": "Yes\n1 11 11 "
},
{
"input": "5 2 4\n1 2 2 2 2",
"output": "Yes\n2 2 "
},
{
"input": "5 5 9\n8 17 26 35 44",
"output": "Yes\n8 17 26 35 44 "
},
{
"input": "3 2 2\n1 2 4",
"output": "Yes\n2 4 "
},
{
"input": "5 4 2\n1 3 6 10 12",
"output": "No"
},
{
"input": "3 2 4\n1 2 3",
"output": "No"
},
{
"input": "8 3 5\n1 2 3 4 1 2 3 4",
"output": "No"
},
{
"input": "5 2 9\n8 8 8 8 8",
"output": "Yes\n8 8 "
},
{
"input": "3 3 3\n1 3 6",
"output": "No"
},
{
"input": "5 2 10\n4 5 6 19 29",
"output": "Yes\n19 29 "
},
{
"input": "5 3 3\n3 6 4 7 10",
"output": "Yes\n4 7 10 "
}
] | 1,508,154,484 | 2,584 | Python 3 | WRONG_ANSWER | PRETESTS | 0 | 46 | 0 | n, k, m=map(int, input().split())
arr=list(map(int, input().split()))
mod=[]
for i in arr:
mod.append(i%m)
r=0
for i in set(mod):
if mod.count(i)>=k:
print("Yes")
r=mod.index(i)
break
else:
print("No")
quit()
i=0
while k:
if mod[i]==mod[r]:
print(arr[i],end=" ")
i+=1
k-=1 | Title: Divisiblity of Differences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a multiset of *n* integers. You should select exactly *k* of them in a such way that the difference between any two of them is divisible by *m*, or tell that it is impossible.
Numbers can be repeated in the original multiset and in the multiset of selected numbers, but number of occurrences of any number in multiset of selected numbers should not exceed the number of its occurrences in the original multiset.
Input Specification:
First line contains three integers *n*, *k* and *m* (2<=≤<=*k*<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — number of integers in the multiset, number of integers you should select and the required divisor of any pair of selected integers.
Second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — the numbers in the multiset.
Output Specification:
If it is not possible to select *k* numbers in the desired way, output «No» (without the quotes).
Otherwise, in the first line of output print «Yes» (without the quotes). In the second line print *k* integers *b*1,<=*b*2,<=...,<=*b**k* — the selected numbers. If there are multiple possible solutions, print any of them.
Demo Input:
['3 2 3\n1 8 4\n', '3 3 3\n1 8 4\n', '4 3 5\n2 7 7 7\n']
Demo Output:
['Yes\n1 4 ', 'No', 'Yes\n2 7 7 ']
Note:
none | ```python
n, k, m=map(int, input().split())
arr=list(map(int, input().split()))
mod=[]
for i in arr:
mod.append(i%m)
r=0
for i in set(mod):
if mod.count(i)>=k:
print("Yes")
r=mod.index(i)
break
else:
print("No")
quit()
i=0
while k:
if mod[i]==mod[r]:
print(arr[i],end=" ")
i+=1
k-=1
``` | 0 | |
227 | B | Effective Approach | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. | Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | [
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] | [
"1 2\n",
"2 1\n",
"6 6\n"
] | In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | 1,000 | [
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4",
"output": "9 2"
},
{
"input": "10\n5 2 10 8 3 1 9 7 6 4\n9\n2 5 9 2 3 2 5 9 8",
"output": "31 68"
},
{
"input": "9\n3 8 4 7 1 2 5 6 9\n3\n2 7 1",
"output": "15 15"
},
{
"input": "9\n5 3 8 4 2 6 1 7 9\n4\n6 1 9 2",
"output": "27 13"
},
{
"input": "4\n1 3 2 4\n4\n3 1 2 3",
"output": "8 12"
},
{
"input": "3\n1 2 3\n8\n3 2 1 1 2 3 1 2",
"output": "15 17"
}
] | 1,625,690,993 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 2,000 | 7,372,800 | def main():
i = input()
arr = list(map(int, input().split()))
r = input()
req = list(map(int, input().split()))
print(V(arr, req), P(arr, req))
def V(arr, n):
z = 0
for i in n:
z += arr.index(i) + 1
return z
def P(arr, n):
arr = list(reversed(arr))
z = 0
for i in n:
z += arr.index(i) + 1
return z
if __name__ == '__main__':
main()
| Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | ```python
def main():
i = input()
arr = list(map(int, input().split()))
r = input()
req = list(map(int, input().split()))
print(V(arr, req), P(arr, req))
def V(arr, n):
z = 0
for i in n:
z += arr.index(i) + 1
return z
def P(arr, n):
arr = list(reversed(arr))
z = 0
for i in n:
z += arr.index(i) + 1
return z
if __name__ == '__main__':
main()
``` | 0 | |
734 | A | Anton and Danik | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Anton likes to play chess, and so does his friend Danik.
Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played.
The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game. | If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes). | [
"6\nADAAAA\n",
"7\nDDDAADA\n",
"6\nDADADA\n"
] | [
"Anton\n",
"Danik\n",
"Friendship\n"
] | In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship". | 500 | [
{
"input": "6\nADAAAA",
"output": "Anton"
},
{
"input": "7\nDDDAADA",
"output": "Danik"
},
{
"input": "6\nDADADA",
"output": "Friendship"
},
{
"input": "10\nDDDDADDADD",
"output": "Danik"
},
{
"input": "40\nAAAAAAAAADDAAAAAAAAAAADADDAAAAAAAAAAADAA",
"output": "Anton"
},
{
"input": "200\nDDDDDDDADDDDDDAADADAADAAADAADADAAADDDADDDDDDADDDAADDDAADADDDDDADDDAAAADAAADDDDDAAADAADDDAAAADDADADDDAADDAADAAADAADAAAADDAADDADAAAADADDDAAAAAADDAADAADAADADDDAAADAAAADADDADAAAAAADADADDDADDDAADDADDDAAAAD",
"output": "Friendship"
},
{
"input": "1\nA",
"output": "Anton"
},
{
"input": "1\nD",
"output": "Danik"
},
{
"input": "2\nDA",
"output": "Friendship"
},
{
"input": "4\nDADA",
"output": "Friendship"
},
{
"input": "4\nDAAD",
"output": "Friendship"
},
{
"input": "3\nADD",
"output": "Danik"
},
{
"input": "3\nDAD",
"output": "Danik"
},
{
"input": "2\nDA",
"output": "Friendship"
},
{
"input": "379\nAADAAAAAADDAAAAAADAADADADDAAAAADADDAADAAAADDDADAAAAAAADAADAAAAAAADAAAAAAAAADAAAAAAADAAAAAAAAAAADDDADAAAAAAAADAADADAAAADAAAAAAAAAAAAAAAAADAAAADDDAADAAAAAAADAAADAAADAADDDADDAAADAAAAAADDDADDDAAADAAAADAAAAAAAAADAAADAAAAAAAAADAAAAAAAAAAAAAAAAAADADAAAAAAAAAAADAAAAADAAAADAAAAAAAAAAAAADADAADAAAAAAAADAADAAAAAAAADAAAAAAAADDDAAAAAADAAADAAAAAADAADAAAAAADAAAADADAADAAAAAADAAAADAADDAADAADAAA",
"output": "Anton"
}
] | 1,693,639,396 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | a=int(input())
b=int(input())
ac=b.count('A')
bc=b.count('D')
if ac>bc:
print("Anton")
if ac<bc:
print("Danik")
if ac==bc:
print("Frienship") | Title: Anton and Danik
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton likes to play chess, and so does his friend Danik.
Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played.
The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game.
Output Specification:
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Demo Input:
['6\nADAAAA\n', '7\nDDDAADA\n', '6\nDADADA\n']
Demo Output:
['Anton\n', 'Danik\n', 'Friendship\n']
Note:
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship". | ```python
a=int(input())
b=int(input())
ac=b.count('A')
bc=b.count('D')
if ac>bc:
print("Anton")
if ac<bc:
print("Danik")
if ac==bc:
print("Frienship")
``` | -1 | |
219 | A | k-String | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string. | The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*. | Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes). | [
"2\naazz\n",
"3\nabcabcabz\n"
] | [
"azaz\n",
"-1\n"
] | none | 500 | [
{
"input": "2\naazz",
"output": "azaz"
},
{
"input": "3\nabcabcabz",
"output": "-1"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "2\nabba",
"output": "abab"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "7\nabacaba",
"output": "-1"
},
{
"input": "5\naaaaa",
"output": "aaaaa"
},
{
"input": "3\naabaaaaabb",
"output": "-1"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "2\nbabac",
"output": "-1"
},
{
"input": "3\nbbbccc",
"output": "bcbcbc"
},
{
"input": "2\naa",
"output": "aa"
},
{
"input": "250\ncececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece",
"output": "cececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece"
},
{
"input": "15\nabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaa",
"output": "aaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbc"
},
{
"input": "1\naaa",
"output": "aaa"
},
{
"input": "1\naabaab",
"output": "aaaabb"
},
{
"input": "2\naabbbbccccccdddddddd",
"output": "abbcccddddabbcccdddd"
},
{
"input": "3\naaaaaaaaacccdddddd",
"output": "aaacddaaacddaaacdd"
},
{
"input": "3\naaaaaaaaacccbbbbbb",
"output": "aaabbcaaabbcaaabbc"
},
{
"input": "2\naazzzz",
"output": "azzazz"
},
{
"input": "2\naaaa",
"output": "aaaa"
},
{
"input": "2\naaaazz",
"output": "aazaaz"
},
{
"input": "2\naaaabb",
"output": "aabaab"
},
{
"input": "2\naabbbb",
"output": "abbabb"
},
{
"input": "2\naaaazzzz",
"output": "aazzaazz"
},
{
"input": "2\naaazz",
"output": "-1"
},
{
"input": "2\nbbbbaa",
"output": "abbabb"
},
{
"input": "2\naaaaaabbbb",
"output": "aaabbaaabb"
},
{
"input": "2\naaaaaazz",
"output": "aaazaaaz"
},
{
"input": "2\naaaabbbb",
"output": "aabbaabb"
},
{
"input": "2\naabaab",
"output": "aabaab"
},
{
"input": "1\naaaaa",
"output": "aaaaa"
},
{
"input": "2\nbbaaaa",
"output": "aabaab"
},
{
"input": "2\nacaccc",
"output": "accacc"
}
] | 1,622,537,848 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 186 | 0 | def string_to_list(s):
ls = []
for ele in s:
ls.append(ele)
return ls
def freq(a, list1):
count = 0
for ele in list1:
if a is ele:
count = count + 1
return count
n = int(input())
s = input()
ls = string_to_list(s)
unique = []
no_of_uniques = int(len(ls)/n)
for ele in ls:
if ele not in unique:
x = freq(ele, ls)/no_of_uniques
print(x)
if x - x//1 != 0 :
print(-1)
exit()
for ele in ls:
if ele not in unique:
x = [ele] * int(freq(ele, ls)/no_of_uniques)
unique = unique + x
final = unique * no_of_uniques
#print(final)
string = ""
for ele in final:
string = string + str(ele)
print(string)
#4
#abcffffbaffaccb
| Title: k-String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string.
Input Specification:
The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*.
Output Specification:
Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes).
Demo Input:
['2\naazz\n', '3\nabcabcabz\n']
Demo Output:
['azaz\n', '-1\n']
Note:
none | ```python
def string_to_list(s):
ls = []
for ele in s:
ls.append(ele)
return ls
def freq(a, list1):
count = 0
for ele in list1:
if a is ele:
count = count + 1
return count
n = int(input())
s = input()
ls = string_to_list(s)
unique = []
no_of_uniques = int(len(ls)/n)
for ele in ls:
if ele not in unique:
x = freq(ele, ls)/no_of_uniques
print(x)
if x - x//1 != 0 :
print(-1)
exit()
for ele in ls:
if ele not in unique:
x = [ele] * int(freq(ele, ls)/no_of_uniques)
unique = unique + x
final = unique * no_of_uniques
#print(final)
string = ""
for ele in final:
string = string + str(ele)
print(string)
#4
#abcffffbaffaccb
``` | 0 | |
621 | A | Wet Shark and Odd and Even | PROGRAMMING | 900 | [
"implementation"
] | null | null | Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0. | The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive. | Print the maximum possible even sum that can be obtained if we use some of the given integers. | [
"3\n1 2 3\n",
"5\n999999999 999999999 999999999 999999999 999999999\n"
] | [
"6",
"3999999996"
] | In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999. | 500 | [
{
"input": "3\n1 2 3",
"output": "6"
},
{
"input": "5\n999999999 999999999 999999999 999999999 999999999",
"output": "3999999996"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "15\n39 52 88 78 46 95 84 98 55 3 68 42 6 18 98",
"output": "870"
},
{
"input": "15\n59 96 34 48 8 72 67 90 15 85 7 90 97 47 25",
"output": "840"
},
{
"input": "15\n87 37 91 29 58 45 51 74 70 71 47 38 91 89 44",
"output": "922"
},
{
"input": "15\n11 81 49 7 11 14 30 67 29 50 90 81 77 18 59",
"output": "674"
},
{
"input": "15\n39 21 95 89 73 90 9 55 85 32 30 21 68 59 82",
"output": "848"
},
{
"input": "15\n59 70 48 54 26 67 84 39 40 18 77 69 70 88 93",
"output": "902"
},
{
"input": "15\n87 22 98 32 88 36 72 31 100 97 17 16 60 22 20",
"output": "798"
},
{
"input": "15\n15 63 51 13 37 9 43 19 55 79 57 60 50 59 31",
"output": "632"
},
{
"input": "1\n4",
"output": "4"
},
{
"input": "2\n1 4",
"output": "4"
},
{
"input": "3\n1 2 4",
"output": "6"
},
{
"input": "2\n9 3",
"output": "12"
},
{
"input": "2\n1000000000 1001",
"output": "1000000000"
},
{
"input": "3\n1 8 4",
"output": "12"
},
{
"input": "3\n7 4 4",
"output": "8"
},
{
"input": "5\n2 3 4 5 3",
"output": "14"
},
{
"input": "2\n4 5",
"output": "4"
},
{
"input": "3\n2 4 5",
"output": "6"
},
{
"input": "3\n2 2 3",
"output": "4"
},
{
"input": "2\n2 3",
"output": "2"
},
{
"input": "4\n2 3 7 7",
"output": "16"
},
{
"input": "2\n999999999 2",
"output": "2"
},
{
"input": "2\n2 5",
"output": "2"
},
{
"input": "3\n5 3 1",
"output": "8"
},
{
"input": "4\n3 2 5 7",
"output": "14"
}
] | 1,664,989,462 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | l = [1, 2, 3]
print(l)
print(*l)
for i in range (len(l)):
print(l[i] , end= "*")
| Title: Wet Shark and Odd and Even
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.
Input Specification:
The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
Output Specification:
Print the maximum possible even sum that can be obtained if we use some of the given integers.
Demo Input:
['3\n1 2 3\n', '5\n999999999 999999999 999999999 999999999 999999999\n']
Demo Output:
['6', '3999999996']
Note:
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999. | ```python
l = [1, 2, 3]
print(l)
print(*l)
for i in range (len(l)):
print(l[i] , end= "*")
``` | 0 | |
34 | B | Sale | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | B. Sale | 2 | 256 | Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets. | Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets. | [
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] | [
"8\n",
"7\n"
] | none | 1,000 | [
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "6"
},
{
"input": "5 1\n998 997 985 937 998",
"output": "0"
},
{
"input": "2 2\n-742 -187",
"output": "929"
},
{
"input": "3 3\n522 597 384",
"output": "0"
},
{
"input": "4 2\n-215 -620 192 647",
"output": "835"
},
{
"input": "10 6\n557 605 685 231 910 633 130 838 -564 -85",
"output": "649"
},
{
"input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59",
"output": "10"
},
{
"input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000",
"output": "0"
},
{
"input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886",
"output": "19441"
},
{
"input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999",
"output": "0"
},
{
"input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993",
"output": "16984"
},
{
"input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981",
"output": "21743"
},
{
"input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654",
"output": "81283"
},
{
"input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782",
"output": "0"
},
{
"input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823",
"output": "0"
},
{
"input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992",
"output": "75068"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 1\n555",
"output": "0"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n-24",
"output": "24"
}
] | 1,610,268,940 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 122 | 0 | n,m = map(int,input().split())
a = sorted([x for x in map(int,input().split()) if x < 0])
print(-sum(a[:m])) | Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none | ```python
n,m = map(int,input().split())
a = sorted([x for x in map(int,input().split()) if x < 0])
print(-sum(a[:m]))
``` | 3.9695 |
620 | E | New Year Tree | PROGRAMMING | 2,100 | [
"bitmasks",
"data structures",
"trees"
] | null | null | The New Year holidays are over, but Resha doesn't want to throw away the New Year tree. He invited his best friends Kerim and Gural to help him to redecorate the New Year tree.
The New Year tree is an undirected tree with *n* vertices and root in the vertex 1.
You should process the queries of the two types:
1. Change the colours of all vertices in the subtree of the vertex *v* to the colour *c*. 1. Find the number of different colours in the subtree of the vertex *v*. | The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=4·105) — the number of vertices in the tree and the number of the queries.
The second line contains *n* integers *c**i* (1<=≤<=*c**i*<=≤<=60) — the colour of the *i*-th vertex.
Each of the next *n*<=-<=1 lines contains two integers *x**j*,<=*y**j* (1<=≤<=*x**j*,<=*y**j*<=≤<=*n*) — the vertices of the *j*-th edge. It is guaranteed that you are given correct undirected tree.
The last *m* lines contains the description of the queries. Each description starts with the integer *t**k* (1<=≤<=*t**k*<=≤<=2) — the type of the *k*-th query. For the queries of the first type then follows two integers *v**k*,<=*c**k* (1<=≤<=*v**k*<=≤<=*n*,<=1<=≤<=*c**k*<=≤<=60) — the number of the vertex whose subtree will be recoloured with the colour *c**k*. For the queries of the second type then follows integer *v**k* (1<=≤<=*v**k*<=≤<=*n*) — the number of the vertex for which subtree you should find the number of different colours. | For each query of the second type print the integer *a* — the number of different colours in the subtree of the vertex given in the query.
Each of the numbers should be printed on a separate line in order of query appearing in the input. | [
"7 10\n1 1 1 1 1 1 1\n1 2\n1 3\n1 4\n3 5\n3 6\n3 7\n1 3 2\n2 1\n1 4 3\n2 1\n1 2 5\n2 1\n1 6 4\n2 1\n2 2\n2 3\n",
"23 30\n1 2 2 6 5 3 2 1 1 1 2 4 5 3 4 4 3 3 3 3 3 4 6\n1 2\n1 3\n1 4\n2 5\n2 6\n3 7\n3 8\n4 9\n4 10\n4 11\n6 12\n6 13\n7 14\n7 15\n7 16\n8 17\n8 18\n10 19\n10 20\n10 21\n11 22\n11 23\n2 1\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 4\n1 12 1\n1 13 1\n1 14 1\n1 15 1\n1 16 1\n1 17 1\n1 18 1\n1 19 1\n1 20 1\n1 21 1\n1 22 1\n1 23 1\n2 1\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 4\n"
] | [
"2\n3\n4\n5\n1\n2\n",
"6\n1\n3\n3\n2\n1\n2\n3\n5\n5\n1\n2\n2\n1\n1\n1\n2\n3\n"
] | none | 0 | [
{
"input": "7 10\n1 1 1 1 1 1 1\n1 2\n1 3\n1 4\n3 5\n3 6\n3 7\n1 3 2\n2 1\n1 4 3\n2 1\n1 2 5\n2 1\n1 6 4\n2 1\n2 2\n2 3",
"output": "2\n3\n4\n5\n1\n2"
},
{
"input": "23 30\n1 2 2 6 5 3 2 1 1 1 2 4 5 3 4 4 3 3 3 3 3 4 6\n1 2\n1 3\n1 4\n2 5\n2 6\n3 7\n3 8\n4 9\n4 10\n4 11\n6 12\n6 13\n7 14\n7 15\n7 16\n8 17\n8 18\n10 19\n10 20\n10 21\n11 22\n11 23\n2 1\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 4\n1 12 1\n1 13 1\n1 14 1\n1 15 1\n1 16 1\n1 17 1\n1 18 1\n1 19 1\n1 20 1\n1 21 1\n1 22 1\n1 23 1\n2 1\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 4",
"output": "6\n1\n3\n3\n2\n1\n2\n3\n5\n5\n1\n2\n2\n1\n1\n1\n2\n3"
},
{
"input": "1 1\n14\n2 1",
"output": "1"
},
{
"input": "1 1\n36\n2 1",
"output": "1"
},
{
"input": "1 1\n3\n2 1",
"output": "1"
},
{
"input": "1 1\n43\n2 1",
"output": "1"
},
{
"input": "1 1\n41\n2 1",
"output": "1"
},
{
"input": "10 10\n59 59 59 59 59 59 59 59 59 59\n6 8\n6 10\n2 6\n2 5\n7 2\n10 1\n4 2\n7 3\n9 1\n1 8 59\n2 8\n1 3 59\n1 4 59\n1 8 59\n1 2 59\n1 5 59\n1 10 59\n2 2\n2 5",
"output": "1\n1\n1"
},
{
"input": "10 10\n8 8 14 32 14 8 32 8 14 32\n4 5\n4 1\n4 8\n4 9\n7 4\n2 5\n3 5\n4 6\n10 4\n2 2\n1 9 8\n1 1 40\n1 7 32\n1 4 8\n2 8\n1 1 8\n2 2\n2 8\n2 4",
"output": "1\n1\n1\n1\n1"
},
{
"input": "10 10\n39 50 50 7 39 7 46 7 39 7\n10 7\n7 3\n3 5\n3 4\n6 4\n1 4\n1 8\n8 2\n2 9\n2 8\n1 6 50\n2 4\n2 6\n1 7 39\n1 3 39\n2 9\n1 1 15\n2 7\n1 10 7",
"output": "3\n4\n1\n1\n1"
},
{
"input": "10 10\n23 25 23 42 23 53 49 40 28 44\n1 7\n1 2\n2 4\n4 10\n8 10\n6 8\n3 8\n5 3\n9 5\n2 10\n1 6 52\n1 8 43\n2 3\n1 4 39\n1 8 44\n1 9 39\n2 1\n2 4\n1 6 36",
"output": "5\n1\n5\n2"
},
{
"input": "10 10\n16 25 25 27 39 29 29 58 50 30\n8 2\n2 10\n4 2\n2 1\n6 2\n2 3\n9 2\n5 2\n2 7\n2 4\n1 3 31\n2 5\n1 7 27\n1 4 56\n1 4 52\n1 5 25\n1 6 32\n1 6 22\n1 7 42",
"output": "1\n1"
},
{
"input": "10 10\n60 46 56 7 4 27 43 28 4 9\n1 5\n5 8\n10 8\n10 6\n7 6\n2 10\n4 2\n9 4\n9 3\n2 3\n1 9 57\n2 2\n1 6 50\n1 5 34\n1 8 45\n1 9 39\n2 2\n1 10 1\n2 4",
"output": "1\n3\n2\n1"
},
{
"input": "10 10\n15 39 52 24 36 30 46 21 40 24\n5 9\n5 3\n5 10\n1 3\n9 4\n9 8\n9 7\n7 2\n3 6\n1 4 47\n1 7 25\n1 10 42\n2 10\n1 2 18\n1 1 60\n1 7 56\n2 7\n2 9\n2 10",
"output": "1\n1\n2\n1"
},
{
"input": "10 10\n39 28 21 20 11 11 40 30 42 14\n7 1\n10 1\n6 1\n1 9\n5 1\n8 1\n1 3\n1 4\n2 10\n1 7 55\n2 3\n1 8 18\n1 10 48\n2 7\n1 6 26\n2 2\n1 1 4\n2 9\n1 5 31",
"output": "1\n1\n1\n1"
}
] | 1,585,396,738 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 42 | 3,000 | 36,352,000 | n,m = [int(x) for x in input().split()]
c = [int(x) for x in input().split()]
class Node:
parent = None
colors = None
children = None
def __init__(self, i, c):
self.x = i
self.colors = set((c,))
self.color = c
self.children = set()
def connect(self, j):
self.children.add(nodes[j])
def __repr__(self):
return f'Node({self.x}), {self.colors}'
nodes = [Node(i,c[i]) for i in range(n)]
for _ in range(n-1):
i,j = [int(x)-1 for x in input().split()]
nodes[i].connect(j)
nodes[j].connect(i)
dfs = [nodes[0]]
while dfs:
parent = dfs.pop()
for c in list(parent.children):
c.children.remove(nodes[parent.x])
c.parent = nodes[parent.x]
dfs.append(c)
def set_colors(node):
for c in node.children:
set_colors(c)
node.colors.update(c.colors)
set_colors(nodes[0])
visited = set()
while dfs:
node = dfs.pop()
if node in visited:
node.colors = node.c
# bfs = [nodes[0]]
# while bfs:
# node = bfs.pop(0)
# print(f'{node}: {node.children}')
# bfs.extend(node.children)
def ncolors(i):
return len(nodes[i].colors)
def paint(i,c):
dfs = [nodes[i]]
# print(f'painsting {i} with {c}')
while dfs:
node = dfs.pop()
node.colors = set((c,))
node.color = c
dfs.extend(node.children)
node = nodes[i]
while node.parent is not None:
node = node.parent
node.colors = set((node.color,))
for c in node.children:
node.colors.update(c.colors)
# bfs = [nodes[0]]
# while bfs:
# node = bfs.pop(0)
# print(f'{node}: {node.children}')
# bfs.extend(node.children)
for _ in range(m):
command = [int(x) for x in input().split()]
if len(command) == 3:
paint(command[1]-1, command[2])
else:
print(ncolors(command[1]-1))
| Title: New Year Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The New Year holidays are over, but Resha doesn't want to throw away the New Year tree. He invited his best friends Kerim and Gural to help him to redecorate the New Year tree.
The New Year tree is an undirected tree with *n* vertices and root in the vertex 1.
You should process the queries of the two types:
1. Change the colours of all vertices in the subtree of the vertex *v* to the colour *c*. 1. Find the number of different colours in the subtree of the vertex *v*.
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=4·105) — the number of vertices in the tree and the number of the queries.
The second line contains *n* integers *c**i* (1<=≤<=*c**i*<=≤<=60) — the colour of the *i*-th vertex.
Each of the next *n*<=-<=1 lines contains two integers *x**j*,<=*y**j* (1<=≤<=*x**j*,<=*y**j*<=≤<=*n*) — the vertices of the *j*-th edge. It is guaranteed that you are given correct undirected tree.
The last *m* lines contains the description of the queries. Each description starts with the integer *t**k* (1<=≤<=*t**k*<=≤<=2) — the type of the *k*-th query. For the queries of the first type then follows two integers *v**k*,<=*c**k* (1<=≤<=*v**k*<=≤<=*n*,<=1<=≤<=*c**k*<=≤<=60) — the number of the vertex whose subtree will be recoloured with the colour *c**k*. For the queries of the second type then follows integer *v**k* (1<=≤<=*v**k*<=≤<=*n*) — the number of the vertex for which subtree you should find the number of different colours.
Output Specification:
For each query of the second type print the integer *a* — the number of different colours in the subtree of the vertex given in the query.
Each of the numbers should be printed on a separate line in order of query appearing in the input.
Demo Input:
['7 10\n1 1 1 1 1 1 1\n1 2\n1 3\n1 4\n3 5\n3 6\n3 7\n1 3 2\n2 1\n1 4 3\n2 1\n1 2 5\n2 1\n1 6 4\n2 1\n2 2\n2 3\n', '23 30\n1 2 2 6 5 3 2 1 1 1 2 4 5 3 4 4 3 3 3 3 3 4 6\n1 2\n1 3\n1 4\n2 5\n2 6\n3 7\n3 8\n4 9\n4 10\n4 11\n6 12\n6 13\n7 14\n7 15\n7 16\n8 17\n8 18\n10 19\n10 20\n10 21\n11 22\n11 23\n2 1\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 4\n1 12 1\n1 13 1\n1 14 1\n1 15 1\n1 16 1\n1 17 1\n1 18 1\n1 19 1\n1 20 1\n1 21 1\n1 22 1\n1 23 1\n2 1\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 4\n']
Demo Output:
['2\n3\n4\n5\n1\n2\n', '6\n1\n3\n3\n2\n1\n2\n3\n5\n5\n1\n2\n2\n1\n1\n1\n2\n3\n']
Note:
none | ```python
n,m = [int(x) for x in input().split()]
c = [int(x) for x in input().split()]
class Node:
parent = None
colors = None
children = None
def __init__(self, i, c):
self.x = i
self.colors = set((c,))
self.color = c
self.children = set()
def connect(self, j):
self.children.add(nodes[j])
def __repr__(self):
return f'Node({self.x}), {self.colors}'
nodes = [Node(i,c[i]) for i in range(n)]
for _ in range(n-1):
i,j = [int(x)-1 for x in input().split()]
nodes[i].connect(j)
nodes[j].connect(i)
dfs = [nodes[0]]
while dfs:
parent = dfs.pop()
for c in list(parent.children):
c.children.remove(nodes[parent.x])
c.parent = nodes[parent.x]
dfs.append(c)
def set_colors(node):
for c in node.children:
set_colors(c)
node.colors.update(c.colors)
set_colors(nodes[0])
visited = set()
while dfs:
node = dfs.pop()
if node in visited:
node.colors = node.c
# bfs = [nodes[0]]
# while bfs:
# node = bfs.pop(0)
# print(f'{node}: {node.children}')
# bfs.extend(node.children)
def ncolors(i):
return len(nodes[i].colors)
def paint(i,c):
dfs = [nodes[i]]
# print(f'painsting {i} with {c}')
while dfs:
node = dfs.pop()
node.colors = set((c,))
node.color = c
dfs.extend(node.children)
node = nodes[i]
while node.parent is not None:
node = node.parent
node.colors = set((node.color,))
for c in node.children:
node.colors.update(c.colors)
# bfs = [nodes[0]]
# while bfs:
# node = bfs.pop(0)
# print(f'{node}: {node.children}')
# bfs.extend(node.children)
for _ in range(m):
command = [int(x) for x in input().split()]
if len(command) == 3:
paint(command[1]-1, command[2])
else:
print(ncolors(command[1]-1))
``` | 0 | |
784 | E | Twisted Circuit | PROGRAMMING | 2,000 | [
"*special",
"brute force",
"implementation"
] | null | null | The input consists of four lines, each line containing a single digit 0 or 1. | Output a single digit, 0 or 1. | [
"0\n1\n1\n0\n"
] | [
"0\n"
] | none | 0 | [
{
"input": "0\n1\n1\n0",
"output": "0"
},
{
"input": "0\n0\n0\n0",
"output": "0"
},
{
"input": "1\n0\n0\n0",
"output": "1"
},
{
"input": "0\n1\n0\n0",
"output": "0"
},
{
"input": "1\n1\n0\n0",
"output": "1"
},
{
"input": "0\n0\n1\n0",
"output": "0"
},
{
"input": "1\n0\n1\n0",
"output": "0"
},
{
"input": "1\n1\n1\n0",
"output": "1"
},
{
"input": "0\n0\n0\n1",
"output": "1"
},
{
"input": "1\n0\n0\n1",
"output": "1"
},
{
"input": "0\n1\n0\n1",
"output": "0"
},
{
"input": "1\n1\n0\n1",
"output": "0"
},
{
"input": "0\n0\n1\n1",
"output": "1"
},
{
"input": "1\n0\n1\n1",
"output": "1"
},
{
"input": "0\n1\n1\n1",
"output": "0"
},
{
"input": "1\n1\n1\n1",
"output": "1"
}
] | 1,522,544,015 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 9 | 77 | 7,065,600 | a = int(input())
b = int(input())
c = int(input())
d = int(input())
x = a or b
y = c ^ d
z = b and c
t = d or a
one = x and y
two = z ^ t
last = one ^ two
print(last) | Title: Twisted Circuit
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Input Specification:
The input consists of four lines, each line containing a single digit 0 or 1.
Output Specification:
Output a single digit, 0 or 1.
Demo Input:
['0\n1\n1\n0\n']
Demo Output:
['0\n']
Note:
none | ```python
a = int(input())
b = int(input())
c = int(input())
d = int(input())
x = a or b
y = c ^ d
z = b and c
t = d or a
one = x and y
two = z ^ t
last = one ^ two
print(last)
``` | 0 | ||
459 | B | Pashmak and Flowers | PROGRAMMING | 1,300 | [
"combinatorics",
"implementation",
"sortings"
] | null | null | Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way. | The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109). | The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively. | [
"2\n1 2\n",
"3\n1 4 5\n",
"5\n3 1 2 3 1\n"
] | [
"1 1",
"4 1",
"2 4"
] | In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers. | 500 | [
{
"input": "2\n1 2",
"output": "1 1"
},
{
"input": "3\n1 4 5",
"output": "4 1"
},
{
"input": "5\n3 1 2 3 1",
"output": "2 4"
},
{
"input": "2\n1 1",
"output": "0 1"
},
{
"input": "3\n1 1 1",
"output": "0 3"
},
{
"input": "4\n1 1 1 1",
"output": "0 6"
},
{
"input": "5\n1 1 1 1 1",
"output": "0 10"
},
{
"input": "5\n2 2 2 2 2",
"output": "0 10"
},
{
"input": "10\n2 2 2 2 2 2 2 2 2 2",
"output": "0 45"
},
{
"input": "3\n2 2 2",
"output": "0 3"
},
{
"input": "3\n3 3 3",
"output": "0 3"
},
{
"input": "2\n10000000 100000000",
"output": "90000000 1"
},
{
"input": "5\n5 5 5 5 5",
"output": "0 10"
},
{
"input": "5\n3 3 3 3 3",
"output": "0 10"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "0 15"
},
{
"input": "2\n5 6",
"output": "1 1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "0 45"
},
{
"input": "10\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "0 45"
},
{
"input": "4\n4 4 4 4",
"output": "0 6"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "0 21"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "0 55"
},
{
"input": "8\n8 8 8 8 8 8 8 8",
"output": "0 28"
},
{
"input": "2\n3 2",
"output": "1 1"
}
] | 1,691,528,832 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 93 | 14,131,200 | n = int(input())
b = [int(e) for e in input().split(' ')]
bs = {}
for e in b:
if e in bs:
bs[e]+=1
else:
bs[e]=1
bsmin = min(bs.keys())
bsmax = max(bs.keys())
print(bsmax-bsmin, bs[bsmin]*bs[bsmax], sep=" ")
| Title: Pashmak and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109).
Output Specification:
The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.
Demo Input:
['2\n1 2\n', '3\n1 4 5\n', '5\n3 1 2 3 1\n']
Demo Output:
['1 1', '4 1', '2 4']
Note:
In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers. | ```python
n = int(input())
b = [int(e) for e in input().split(' ')]
bs = {}
for e in b:
if e in bs:
bs[e]+=1
else:
bs[e]=1
bsmin = min(bs.keys())
bsmax = max(bs.keys())
print(bsmax-bsmin, bs[bsmin]*bs[bsmax], sep=" ")
``` | 0 | |
160 | D | Edges in MST | PROGRAMMING | 2,300 | [
"dfs and similar",
"dsu",
"graphs",
"sortings"
] | null | null | You are given a connected weighted undirected graph without any loops and multiple edges.
Let us remind you that a graph's spanning tree is defined as an acyclic connected subgraph of the given graph that includes all of the graph's vertexes. The weight of a tree is defined as the sum of weights of the edges that the given tree contains. The minimum spanning tree (MST) of a graph is defined as the graph's spanning tree having the minimum possible weight. For any connected graph obviously exists the minimum spanning tree, but in the general case, a graph's minimum spanning tree is not unique.
Your task is to determine the following for each edge of the given graph: whether it is either included in any MST, or included at least in one MST, or not included in any MST. | The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105, ) — the number of the graph's vertexes and edges, correspondingly. Then follow *m* lines, each of them contains three integers — the description of the graph's edges as "*a**i* *b**i* *w**i*" (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106,<=*a**i*<=≠<=*b**i*), where *a**i* and *b**i* are the numbers of vertexes connected by the *i*-th edge, *w**i* is the edge's weight. It is guaranteed that the graph is connected and doesn't contain loops or multiple edges. | Print *m* lines — the answers for all edges. If the *i*-th edge is included in any MST, print "any"; if the *i*-th edge is included at least in one MST, print "at least one"; if the *i*-th edge isn't included in any MST, print "none". Print the answers for the edges in the order in which the edges are specified in the input. | [
"4 5\n1 2 101\n1 3 100\n2 3 2\n2 4 2\n3 4 1\n",
"3 3\n1 2 1\n2 3 1\n1 3 2\n",
"3 3\n1 2 1\n2 3 1\n1 3 1\n"
] | [
"none\nany\nat least one\nat least one\nany\n",
"any\nany\nnone\n",
"at least one\nat least one\nat least one\n"
] | In the second sample the MST is unique for the given graph: it contains two first edges.
In the third sample any two edges form the MST for the given graph. That means that each edge is included at least in one MST. | 2,000 | [] | 1,696,917,959 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 102,400 | class UnionFind:
def __init__(self, n):
self.parent = [i for i in range(n + 1)]
self.rank = [0] * (n + 1)
def find(self, x):
if self.parent[x] != x:
self.parent[x] = self.find(self.parent[x])
return self.parent[x]
def union(self, x, y):
root_x = self.find(x)
root_y = self.find(y)
if root_x != root_y:
if self.rank[root_x] < self.rank[root_y]:
root_x, root_y = root_y, root_x
self.parent[root_y] = root_x
if self.rank[root_x] == self.rank[root_y]:
self.rank[root_x] += 1
def kruskal(n, edges, excluded_edge):
edges.sort(key=lambda x: x[2])
uf = UnionFind(n)
total_weight = 0
included_edges = set()
for edge in edges:
u, v, w = edge
if (u, v, w) != excluded_edge and uf.find(u) != uf.find(v):
uf.union(u, v)
total_weight += w
included_edges.add((u, v, w))
return total_weight, included_edges
def main():
n, m = map(int, input().split())
edges = [tuple(map(int, input().split())) for _ in range(m)]
for i in range(m):
excluded_edge = edges[i]
weight, included_edges = kruskal(n, edges, excluded_edge)
if len(included_edges) == n - 1:
print("any")
elif weight == 0:
print("none")
else:
print("at least one")
if __name__ == "__main__":
main()
| Title: Edges in MST
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a connected weighted undirected graph without any loops and multiple edges.
Let us remind you that a graph's spanning tree is defined as an acyclic connected subgraph of the given graph that includes all of the graph's vertexes. The weight of a tree is defined as the sum of weights of the edges that the given tree contains. The minimum spanning tree (MST) of a graph is defined as the graph's spanning tree having the minimum possible weight. For any connected graph obviously exists the minimum spanning tree, but in the general case, a graph's minimum spanning tree is not unique.
Your task is to determine the following for each edge of the given graph: whether it is either included in any MST, or included at least in one MST, or not included in any MST.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105, ) — the number of the graph's vertexes and edges, correspondingly. Then follow *m* lines, each of them contains three integers — the description of the graph's edges as "*a**i* *b**i* *w**i*" (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106,<=*a**i*<=≠<=*b**i*), where *a**i* and *b**i* are the numbers of vertexes connected by the *i*-th edge, *w**i* is the edge's weight. It is guaranteed that the graph is connected and doesn't contain loops or multiple edges.
Output Specification:
Print *m* lines — the answers for all edges. If the *i*-th edge is included in any MST, print "any"; if the *i*-th edge is included at least in one MST, print "at least one"; if the *i*-th edge isn't included in any MST, print "none". Print the answers for the edges in the order in which the edges are specified in the input.
Demo Input:
['4 5\n1 2 101\n1 3 100\n2 3 2\n2 4 2\n3 4 1\n', '3 3\n1 2 1\n2 3 1\n1 3 2\n', '3 3\n1 2 1\n2 3 1\n1 3 1\n']
Demo Output:
['none\nany\nat least one\nat least one\nany\n', 'any\nany\nnone\n', 'at least one\nat least one\nat least one\n']
Note:
In the second sample the MST is unique for the given graph: it contains two first edges.
In the third sample any two edges form the MST for the given graph. That means that each edge is included at least in one MST. | ```python
class UnionFind:
def __init__(self, n):
self.parent = [i for i in range(n + 1)]
self.rank = [0] * (n + 1)
def find(self, x):
if self.parent[x] != x:
self.parent[x] = self.find(self.parent[x])
return self.parent[x]
def union(self, x, y):
root_x = self.find(x)
root_y = self.find(y)
if root_x != root_y:
if self.rank[root_x] < self.rank[root_y]:
root_x, root_y = root_y, root_x
self.parent[root_y] = root_x
if self.rank[root_x] == self.rank[root_y]:
self.rank[root_x] += 1
def kruskal(n, edges, excluded_edge):
edges.sort(key=lambda x: x[2])
uf = UnionFind(n)
total_weight = 0
included_edges = set()
for edge in edges:
u, v, w = edge
if (u, v, w) != excluded_edge and uf.find(u) != uf.find(v):
uf.union(u, v)
total_weight += w
included_edges.add((u, v, w))
return total_weight, included_edges
def main():
n, m = map(int, input().split())
edges = [tuple(map(int, input().split())) for _ in range(m)]
for i in range(m):
excluded_edge = edges[i]
weight, included_edges = kruskal(n, edges, excluded_edge)
if len(included_edges) == n - 1:
print("any")
elif weight == 0:
print("none")
else:
print("at least one")
if __name__ == "__main__":
main()
``` | 0 | |
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"
}
] | 1,693,502,391 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 280 | 1,740,800 | n=int(input())
a=[]
for i in range(n):
a.append(int(input()))
ctr=1
for i in range(n-1):
if a[i]!=a[i+1]:
ctr+=1
print(ctr)
| 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())
a=[]
for i in range(n):
a.append(int(input()))
ctr=1
for i in range(n-1):
if a[i]!=a[i+1]:
ctr+=1
print(ctr)
``` | 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"
}
] | 1,678,712,214 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 62 | 0 | s = input().strip()
a = "hello"
i = 0
for c in s:
if c == a[i]:
i += 1
if i == len(a):
print("YES")
break
else:
print("NO") | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
s = input().strip()
a = "hello"
i = 0
for c in s:
if c == a[i]:
i += 1
if i == len(a):
print("YES")
break
else:
print("NO")
``` | 3.969 |
719 | B | Anatoly and Cockroaches | PROGRAMMING | 1,400 | [
"greedy"
] | null | null | Anatoly lives in the university dorm as many other students do. As you know, cockroaches are also living there together with students. Cockroaches might be of two colors: black and red. There are *n* cockroaches living in Anatoly's room.
Anatoly just made all his cockroaches to form a single line. As he is a perfectionist, he would like the colors of cockroaches in the line to alternate. He has a can of black paint and a can of red paint. In one turn he can either swap any two cockroaches, or take any single cockroach and change it's color.
Help Anatoly find out the minimum number of turns he needs to make the colors of cockroaches in the line alternate. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cockroaches.
The second line contains a string of length *n*, consisting of characters 'b' and 'r' that denote black cockroach and red cockroach respectively. | Print one integer — the minimum number of moves Anatoly has to perform in order to make the colors of cockroaches in the line to alternate. | [
"5\nrbbrr\n",
"5\nbbbbb\n",
"3\nrbr\n"
] | [
"1\n",
"2\n",
"0\n"
] | In the first sample, Anatoly has to swap third and fourth cockroaches. He needs 1 turn to do this.
In the second sample, the optimum answer is to paint the second and the fourth cockroaches red. This requires 2 turns.
In the third sample, the colors of cockroaches in the line are alternating already, thus the answer is 0. | 1,000 | [
{
"input": "5\nrbbrr",
"output": "1"
},
{
"input": "5\nbbbbb",
"output": "2"
},
{
"input": "3\nrbr",
"output": "0"
},
{
"input": "13\nrbbbrbrrbrrbb",
"output": "3"
},
{
"input": "18\nrrrrrrrrrrrrrrrrrb",
"output": "8"
},
{
"input": "100\nbrbbbrrrbbrbrbbrbbrbbbbrbbrrbbbrrbbbbrbrbbbbbbbbbbbbbbbbrrrrbbbbrrrbbbbbbbrbrrbrbbbbrrrbbbbrbbrbbbrb",
"output": "34"
},
{
"input": "166\nrbbbbbbbbbbbbrbrrbbrbbbrbbbbbbbbbbrbbbbbbrbbbrbbbbbrbbbbbbbrbbbbbbbrbbrbbbbbbbbrbbbbbbbbbbbbbbrrbbbrbbbbbbbbbbbbbbrbrbbbbbbbbbbbrbbbbbbbbbbbbbbrbbbbbbbbbbbbbbbbbbbbbb",
"output": "70"
},
{
"input": "1\nr",
"output": "0"
},
{
"input": "1\nb",
"output": "0"
},
{
"input": "2\nrb",
"output": "0"
},
{
"input": "2\nbr",
"output": "0"
},
{
"input": "2\nrr",
"output": "1"
},
{
"input": "2\nbb",
"output": "1"
},
{
"input": "8\nrbbrbrbr",
"output": "1"
},
{
"input": "7\nrrbrbrb",
"output": "1"
}
] | 1,475,672,789 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 0 | a = int(input())
mas = input()
count1 = 0
count2 = 0
count3 = 0
count4 = 0
for i in range(0, (a-1), 2): #rbrb
if (mas[i] == 'b'):
count1 += 1
for i in range(0, (a-1), 2):
if (mas[i] == 'r'):
count2 += 1
for i in range(1, (a-1), 2): #brbr
if (mas[i] == 'r'):
count3 += 1
for i in range(1, (a-1), 2):
if (mas[i] == 'b'):
count4 += 1
# print(count1)
# print(count2)
# print(count3)
# print(count4)
print(min(max(count3,count1),max(count4,count2))) | Title: Anatoly and Cockroaches
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anatoly lives in the university dorm as many other students do. As you know, cockroaches are also living there together with students. Cockroaches might be of two colors: black and red. There are *n* cockroaches living in Anatoly's room.
Anatoly just made all his cockroaches to form a single line. As he is a perfectionist, he would like the colors of cockroaches in the line to alternate. He has a can of black paint and a can of red paint. In one turn he can either swap any two cockroaches, or take any single cockroach and change it's color.
Help Anatoly find out the minimum number of turns he needs to make the colors of cockroaches in the line alternate.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cockroaches.
The second line contains a string of length *n*, consisting of characters 'b' and 'r' that denote black cockroach and red cockroach respectively.
Output Specification:
Print one integer — the minimum number of moves Anatoly has to perform in order to make the colors of cockroaches in the line to alternate.
Demo Input:
['5\nrbbrr\n', '5\nbbbbb\n', '3\nrbr\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first sample, Anatoly has to swap third and fourth cockroaches. He needs 1 turn to do this.
In the second sample, the optimum answer is to paint the second and the fourth cockroaches red. This requires 2 turns.
In the third sample, the colors of cockroaches in the line are alternating already, thus the answer is 0. | ```python
a = int(input())
mas = input()
count1 = 0
count2 = 0
count3 = 0
count4 = 0
for i in range(0, (a-1), 2): #rbrb
if (mas[i] == 'b'):
count1 += 1
for i in range(0, (a-1), 2):
if (mas[i] == 'r'):
count2 += 1
for i in range(1, (a-1), 2): #brbr
if (mas[i] == 'r'):
count3 += 1
for i in range(1, (a-1), 2):
if (mas[i] == 'b'):
count4 += 1
# print(count1)
# print(count2)
# print(count3)
# print(count4)
print(min(max(count3,count1),max(count4,count2)))
``` | 0 | |
441 | A | Valera and Antique Items | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Valera is a collector. Once he wanted to expand his collection with exactly one antique item.
Valera knows *n* sellers of antiques, the *i*-th of them auctioned *k**i* items. Currently the auction price of the *j*-th object of the *i*-th seller is *s**ij*. Valera gets on well with each of the *n* sellers. He is perfectly sure that if he outbids the current price of one of the items in the auction (in other words, offers the seller the money that is strictly greater than the current price of the item at the auction), the seller of the object will immediately sign a contract with him.
Unfortunately, Valera has only *v* units of money. Help him to determine which of the *n* sellers he can make a deal with. | The first line contains two space-separated integers *n*,<=*v* (1<=≤<=*n*<=≤<=50; 104<=≤<=*v*<=≤<=106) — the number of sellers and the units of money the Valera has.
Then *n* lines follow. The *i*-th line first contains integer *k**i* (1<=≤<=*k**i*<=≤<=50) the number of items of the *i*-th seller. Then go *k**i* space-separated integers *s**i*1,<=*s**i*2,<=...,<=*s**ik**i* (104<=≤<=*s**ij*<=≤<=106) — the current prices of the items of the *i*-th seller. | In the first line, print integer *p* — the number of sellers with who Valera can make a deal.
In the second line print *p* space-separated integers *q*1,<=*q*2,<=...,<=*q**p* (1<=≤<=*q**i*<=≤<=*n*) — the numbers of the sellers with who Valera can make a deal. Print the numbers of the sellers in the increasing order. | [
"3 50000\n1 40000\n2 20000 60000\n3 10000 70000 190000\n",
"3 50000\n1 50000\n3 100000 120000 110000\n3 120000 110000 120000\n"
] | [
"3\n1 2 3\n",
"0\n\n"
] | In the first sample Valera can bargain with each of the sellers. He can outbid the following items: a 40000 item from the first seller, a 20000 item from the second seller, and a 10000 item from the third seller.
In the second sample Valera can not make a deal with any of the sellers, as the prices of all items in the auction too big for him. | 500 | [
{
"input": "3 50000\n1 40000\n2 20000 60000\n3 10000 70000 190000",
"output": "3\n1 2 3"
},
{
"input": "3 50000\n1 50000\n3 100000 120000 110000\n3 120000 110000 120000",
"output": "0"
},
{
"input": "2 100001\n1 895737\n1 541571",
"output": "0"
},
{
"input": "1 1000000\n1 1000000",
"output": "0"
},
{
"input": "1 1000000\n1 561774",
"output": "1\n1"
},
{
"input": "3 1000000\n5 1000000 568832 1000000 1000000 1000000\n13 1000000 1000000 1000000 596527 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000\n20 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000",
"output": "2\n1 2"
},
{
"input": "3 999999\n7 1000000 1000000 1000000 999999 1000000 999999 1000000\n6 999999 1000000 999999 1000000 999999 999999\n7 999999 1000000 1000000 999999 1000000 1000000 1000000",
"output": "0"
},
{
"input": "3 999999\n22 1000000 1000000 999999 999999 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 999999 1000000 1000000 999999 1000000 1000000 1000000 352800 999999 1000000\n14 999999 999999 999999 999999 999999 1000000 999999 999999 999999 999999 702638 999999 1000000 999999\n5 999999 1000000 1000000 999999 363236",
"output": "3\n1 2 3"
},
{
"input": "1 50001\n1 50000",
"output": "1\n1"
}
] | 1,573,129,593 | 2,147,483,647 | PyPy 3 | OK | TESTS | 26 | 171 | 1,638,400 | n, v = map(int, input().split())
ans = list()
for j in range(n):
l = list(map(int, input().split()))
for i in range(1, l[0]+1):
if l[i] < v:
ans.append(j+1)
break
print(len(ans))
print(*ans) | Title: Valera and Antique Items
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a collector. Once he wanted to expand his collection with exactly one antique item.
Valera knows *n* sellers of antiques, the *i*-th of them auctioned *k**i* items. Currently the auction price of the *j*-th object of the *i*-th seller is *s**ij*. Valera gets on well with each of the *n* sellers. He is perfectly sure that if he outbids the current price of one of the items in the auction (in other words, offers the seller the money that is strictly greater than the current price of the item at the auction), the seller of the object will immediately sign a contract with him.
Unfortunately, Valera has only *v* units of money. Help him to determine which of the *n* sellers he can make a deal with.
Input Specification:
The first line contains two space-separated integers *n*,<=*v* (1<=≤<=*n*<=≤<=50; 104<=≤<=*v*<=≤<=106) — the number of sellers and the units of money the Valera has.
Then *n* lines follow. The *i*-th line first contains integer *k**i* (1<=≤<=*k**i*<=≤<=50) the number of items of the *i*-th seller. Then go *k**i* space-separated integers *s**i*1,<=*s**i*2,<=...,<=*s**ik**i* (104<=≤<=*s**ij*<=≤<=106) — the current prices of the items of the *i*-th seller.
Output Specification:
In the first line, print integer *p* — the number of sellers with who Valera can make a deal.
In the second line print *p* space-separated integers *q*1,<=*q*2,<=...,<=*q**p* (1<=≤<=*q**i*<=≤<=*n*) — the numbers of the sellers with who Valera can make a deal. Print the numbers of the sellers in the increasing order.
Demo Input:
['3 50000\n1 40000\n2 20000 60000\n3 10000 70000 190000\n', '3 50000\n1 50000\n3 100000 120000 110000\n3 120000 110000 120000\n']
Demo Output:
['3\n1 2 3\n', '0\n\n']
Note:
In the first sample Valera can bargain with each of the sellers. He can outbid the following items: a 40000 item from the first seller, a 20000 item from the second seller, and a 10000 item from the third seller.
In the second sample Valera can not make a deal with any of the sellers, as the prices of all items in the auction too big for him. | ```python
n, v = map(int, input().split())
ans = list()
for j in range(n):
l = list(map(int, input().split()))
for i in range(1, l[0]+1):
if l[i] < v:
ans.append(j+1)
break
print(len(ans))
print(*ans)
``` | 3 | |
918 | A | Eleven | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters.
Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where
- *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2).
As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name. | The first and only line of input contains an integer *n* (1<=≤<=*n*<=≤<=1000). | Print Eleven's new name on the first and only line of output. | [
"8\n",
"15\n"
] | [
"OOOoOooO\n",
"OOOoOooOooooOoo\n"
] | none | 500 | [
{
"input": "8",
"output": "OOOoOooO"
},
{
"input": "15",
"output": "OOOoOooOooooOoo"
},
{
"input": "85",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooo"
},
{
"input": "381",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooo"
},
{
"input": "805",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1000",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1",
"output": "O"
},
{
"input": "2",
"output": "OO"
},
{
"input": "3",
"output": "OOO"
},
{
"input": "5",
"output": "OOOoO"
},
{
"input": "17",
"output": "OOOoOooOooooOoooo"
},
{
"input": "49",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooo"
},
{
"input": "256",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooo"
},
{
"input": "512",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "933",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "61",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooo"
},
{
"input": "781",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "999",
"output": "OOOoOooOooooOoooooooOooooooooooooOooooooooooooooooooooOoooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooOooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooOoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
}
] | 1,637,909,269 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | n = int(input())
a, b = 1, 1
S = {1, 1}
for i in range(1, 1001):
a, b = b, a + b
S.add(b)
result = 'OO'
for i in range(2, n):
if i+1 in S:
print(i)
result = result + 'O'
else :
result = result + 'o'
print(result) | Title: Eleven
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly *n* characters.
Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the *i*-th letter of her name should be 'O' (uppercase) if *i* is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to *n*. Fibonacci sequence is the sequence *f* where
- *f*1<==<=1, - *f*2<==<=1, - *f**n*<==<=*f**n*<=-<=2<=+<=*f**n*<=-<=1 (*n*<=><=2).
As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.
Input Specification:
The first and only line of input contains an integer *n* (1<=≤<=*n*<=≤<=1000).
Output Specification:
Print Eleven's new name on the first and only line of output.
Demo Input:
['8\n', '15\n']
Demo Output:
['OOOoOooO\n', 'OOOoOooOooooOoo\n']
Note:
none | ```python
n = int(input())
a, b = 1, 1
S = {1, 1}
for i in range(1, 1001):
a, b = b, a + b
S.add(b)
result = 'OO'
for i in range(2, n):
if i+1 in S:
print(i)
result = result + 'O'
else :
result = result + 'o'
print(result)
``` | 0 | |
961 | B | Lecture Sleep | PROGRAMMING | 1,200 | [
"data structures",
"dp",
"implementation",
"two pointers"
] | null | null | Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. | The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture. | Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. | [
"6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n"
] | [
"16\n"
] | In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16. | 0 | [
{
"input": "6 3\n1 3 5 2 5 4\n1 1 0 1 0 0",
"output": "16"
},
{
"input": "5 3\n1 9999 10000 10000 10000\n0 0 0 0 0",
"output": "30000"
},
{
"input": "3 3\n10 10 10\n1 1 0",
"output": "30"
},
{
"input": "1 1\n423\n0",
"output": "423"
},
{
"input": "6 6\n1 3 5 2 5 4\n1 1 0 1 0 0",
"output": "20"
},
{
"input": "5 2\n1 2 3 4 20\n0 0 0 1 0",
"output": "24"
},
{
"input": "3 1\n1 2 3\n0 0 1",
"output": "5"
},
{
"input": "4 2\n4 5 6 8\n1 0 1 0",
"output": "18"
},
{
"input": "6 3\n1 3 5 2 1 15\n1 1 0 1 0 0",
"output": "22"
},
{
"input": "5 5\n1 2 3 4 5\n1 1 1 0 1",
"output": "15"
},
{
"input": "3 3\n3 3 3\n1 0 1",
"output": "9"
},
{
"input": "5 5\n500 44 3 4 50\n1 0 0 0 0",
"output": "601"
},
{
"input": "2 2\n3 2\n1 0",
"output": "5"
},
{
"input": "7 6\n4 9 1 7 1 8 4\n0 0 0 1 0 1 0",
"output": "30"
},
{
"input": "4 3\n6 5 9 6\n1 1 0 1",
"output": "26"
},
{
"input": "2 1\n3 2\n0 0",
"output": "3"
},
{
"input": "1 1\n10\n0",
"output": "10"
},
{
"input": "2 1\n3 2\n1 0",
"output": "5"
},
{
"input": "4 2\n3 6 7 2\n0 0 1 1",
"output": "18"
},
{
"input": "10 5\n3 5 9 2 5 9 3 8 8 1\n0 1 1 1 0 1 0 0 0 0",
"output": "49"
},
{
"input": "10 4\n9 5 6 4 3 9 5 1 10 7\n0 0 0 0 0 0 1 0 0 1",
"output": "36"
},
{
"input": "9 8\n3 3 7 7 1 9 10 7 1\n1 1 1 1 1 1 1 1 1",
"output": "48"
},
{
"input": "2 1\n3 4\n0 0",
"output": "4"
},
{
"input": "2 1\n3 2\n0 1",
"output": "5"
},
{
"input": "10 1\n6 6 8 7 6 6 3 2 5 6\n0 0 1 0 0 1 0 1 1 1",
"output": "34"
},
{
"input": "3 2\n10 10 6\n0 0 0",
"output": "20"
},
{
"input": "6 3\n1 3 5 2 5 4\n1 1 1 1 1 1",
"output": "20"
},
{
"input": "10 5\n1 1 1 1 1 1 1 1 10000 1\n1 1 1 1 1 1 1 1 0 1",
"output": "10009"
}
] | 1,612,807,243 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 186 | 7,680,000 | n,k=map(int,input().split(" "))
v=list(map(int,input().split(" ")))
condition =list(map(int,input().split(" ")))
sum=0
for i in range(n):
if condition[i]==1:
sum+=v[i]
v[i]=0
sum2=0
for i in range(k):
if condition[i]==0:
sum2+=v[i]
temp=sum2
for i in range(0,n-k):
temp=temp-v[i]+v[i+k]
if temp>sum2:
sum2=temp
print(sum2+sum)
| Title: Lecture Sleep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input Specification:
The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture.
Output Specification:
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Demo Input:
['6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n']
Demo Output:
['16\n']
Note:
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16. | ```python
n,k=map(int,input().split(" "))
v=list(map(int,input().split(" ")))
condition =list(map(int,input().split(" ")))
sum=0
for i in range(n):
if condition[i]==1:
sum+=v[i]
v[i]=0
sum2=0
for i in range(k):
if condition[i]==0:
sum2+=v[i]
temp=sum2
for i in range(0,n-k):
temp=temp-v[i]+v[i+k]
if temp>sum2:
sum2=temp
print(sum2+sum)
``` | 3 | |
222 | A | Shooshuns and Sequence | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same. | The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found. | Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1. | [
"3 2\n3 1 1\n",
"3 1\n3 1 1\n"
] | [
"1\n",
"-1\n"
] | In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1. | 500 | [
{
"input": "3 2\n3 1 1",
"output": "1"
},
{
"input": "3 1\n3 1 1",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "0"
},
{
"input": "2 1\n2 1",
"output": "-1"
},
{
"input": "4 4\n1 2 3 4",
"output": "3"
},
{
"input": "4 3\n1 2 3 4",
"output": "-1"
},
{
"input": "5 3\n2 1 1 1 1",
"output": "1"
},
{
"input": "5 3\n1 1 1 1 1",
"output": "0"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "0"
},
{
"input": "10 1\n1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "10 1\n1 2 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 1\n2 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 2\n2 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "10 2\n1 2 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 3\n3 2 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10 1\n1 2 3 4 5 6 7 8 9 10",
"output": "-1"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
},
{
"input": "20 10\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 11\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "10"
},
{
"input": "20 9\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 10\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "20 20\n20 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 11 10",
"output": "19"
}
] | 1,586,126,252 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 153,804,800 | n,k = map(int,input().split(' '))
a = list(input().split(' '))
k -= 1
seen = dict()
t = 0
end = 1
while True:
end = 1
for x in a:
if x != a[0]:
end = 0
break
if end:
break
if ''.join(a) in seen:
break
seen[''.join(a)] = 1
a = a[1:] + [a[k]]
t += 1
if end:
print(t)
else:
print(-1)
| Title: Shooshuns and Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found.
Output Specification:
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Demo Input:
['3 2\n3 1 1\n', '3 1\n3 1 1\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1. | ```python
n,k = map(int,input().split(' '))
a = list(input().split(' '))
k -= 1
seen = dict()
t = 0
end = 1
while True:
end = 1
for x in a:
if x != a[0]:
end = 0
break
if end:
break
if ''.join(a) in seen:
break
seen[''.join(a)] = 1
a = a[1:] + [a[k]]
t += 1
if end:
print(t)
else:
print(-1)
``` | 0 | |
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"
}
] | 1,586,321,792 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 248 | 0 | #http://codeforces.com/problemset/problem/69/A
if __name__ == '__main__':
n = int(input())
x, y, z = 0, 0, 0
for i in range(n):
ins = list(map(int, input().split()))
x += ins[0]
y += ins[1]
z += ins[2]
if x != 0 or y !=0 or z != 0:
print('NO')
else:
print('YES')
| 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
#http://codeforces.com/problemset/problem/69/A
if __name__ == '__main__':
n = int(input())
x, y, z = 0, 0, 0
for i in range(n):
ins = list(map(int, input().split()))
x += ins[0]
y += ins[1]
z += ins[2]
if x != 0 or y !=0 or z != 0:
print('NO')
else:
print('YES')
``` | 3.938 |
118 | A | String Task | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task. | The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive. | Print the resulting string. It is guaranteed that this string is not empty. | [
"tour\n",
"Codeforces\n",
"aBAcAba\n"
] | [
".t.r\n",
".c.d.f.r.c.s\n",
".b.c.b\n"
] | none | 500 | [
{
"input": "tour",
"output": ".t.r"
},
{
"input": "Codeforces",
"output": ".c.d.f.r.c.s"
},
{
"input": "aBAcAba",
"output": ".b.c.b"
},
{
"input": "obn",
"output": ".b.n"
},
{
"input": "wpwl",
"output": ".w.p.w.l"
},
{
"input": "ggdvq",
"output": ".g.g.d.v.q"
},
{
"input": "pumesz",
"output": ".p.m.s.z"
},
{
"input": "g",
"output": ".g"
},
{
"input": "zjuotps",
"output": ".z.j.t.p.s"
},
{
"input": "jzbwuehe",
"output": ".j.z.b.w.h"
},
{
"input": "tnkgwuugu",
"output": ".t.n.k.g.w.g"
},
{
"input": "kincenvizh",
"output": ".k.n.c.n.v.z.h"
},
{
"input": "xattxjenual",
"output": ".x.t.t.x.j.n.l"
},
{
"input": "ktajqhpqsvhw",
"output": ".k.t.j.q.h.p.q.s.v.h.w"
},
{
"input": "xnhcigytnqcmy",
"output": ".x.n.h.c.g.t.n.q.c.m"
},
{
"input": "jfmtbejyilxcec",
"output": ".j.f.m.t.b.j.l.x.c.c"
},
{
"input": "D",
"output": ".d"
},
{
"input": "ab",
"output": ".b"
},
{
"input": "Ab",
"output": ".b"
},
{
"input": "aB",
"output": ".b"
},
{
"input": "AB",
"output": ".b"
},
{
"input": "ba",
"output": ".b"
},
{
"input": "bA",
"output": ".b"
},
{
"input": "Ba",
"output": ".b"
},
{
"input": "BA",
"output": ".b"
},
{
"input": "aab",
"output": ".b"
},
{
"input": "baa",
"output": ".b"
},
{
"input": "femOZeCArKCpUiHYnbBPTIOFmsHmcpObtPYcLCdjFrUMIyqYzAokKUiiKZRouZiNMoiOuGVoQzaaCAOkquRjmmKKElLNqCnhGdQM",
"output": ".f.m.z.c.r.k.c.p.h.n.b.b.p.t.f.m.s.h.m.c.p.b.t.p.c.l.c.d.j.f.r.m.q.z.k.k.k.z.r.z.n.m.g.v.q.z.c.k.q.r.j.m.m.k.k.l.l.n.q.c.n.h.g.d.q.m"
},
{
"input": "VMBPMCmMDCLFELLIISUJDWQRXYRDGKMXJXJHXVZADRZWVWJRKFRRNSAWKKDPZZLFLNSGUNIVJFBEQsMDHSBJVDTOCSCgZWWKvZZN",
"output": ".v.m.b.p.m.c.m.m.d.c.l.f.l.l.s.j.d.w.q.r.x.r.d.g.k.m.x.j.x.j.h.x.v.z.d.r.z.w.v.w.j.r.k.f.r.r.n.s.w.k.k.d.p.z.z.l.f.l.n.s.g.n.v.j.f.b.q.s.m.d.h.s.b.j.v.d.t.c.s.c.g.z.w.w.k.v.z.z.n"
},
{
"input": "MCGFQQJNUKuAEXrLXibVjClSHjSxmlkQGTKZrRaDNDomIPOmtSgjJAjNVIVLeUGUAOHNkCBwNObVCHOWvNkLFQQbFnugYVMkJruJ",
"output": ".m.c.g.f.q.q.j.n.k.x.r.l.x.b.v.j.c.l.s.h.j.s.x.m.l.k.q.g.t.k.z.r.r.d.n.d.m.p.m.t.s.g.j.j.j.n.v.v.l.g.h.n.k.c.b.w.n.b.v.c.h.w.v.n.k.l.f.q.q.b.f.n.g.v.m.k.j.r.j"
},
{
"input": "iyaiuiwioOyzUaOtAeuEYcevvUyveuyioeeueoeiaoeiavizeeoeyYYaaAOuouueaUioueauayoiuuyiuovyOyiyoyioaoyuoyea",
"output": ".w.z.t.c.v.v.v.v.z.v"
},
{
"input": "yjnckpfyLtzwjsgpcrgCfpljnjwqzgVcufnOvhxplvflxJzqxnhrwgfJmPzifgubvspffmqrwbzivatlmdiBaddiaktdsfPwsevl",
"output": ".j.n.c.k.p.f.l.t.z.w.j.s.g.p.c.r.g.c.f.p.l.j.n.j.w.q.z.g.v.c.f.n.v.h.x.p.l.v.f.l.x.j.z.q.x.n.h.r.w.g.f.j.m.p.z.f.g.b.v.s.p.f.f.m.q.r.w.b.z.v.t.l.m.d.b.d.d.k.t.d.s.f.p.w.s.v.l"
},
{
"input": "RIIIUaAIYJOiuYIUWFPOOAIuaUEZeIooyUEUEAoIyIHYOEAlVAAIiLUAUAeiUIEiUMuuOiAgEUOIAoOUYYEYFEoOIIVeOOAOIIEg",
"output": ".r.j.w.f.p.z.h.l.v.l.m.g.f.v.g"
},
{
"input": "VBKQCFBMQHDMGNSGBQVJTGQCNHHRJMNKGKDPPSQRRVQTZNKBZGSXBPBRXPMVFTXCHZMSJVBRNFNTHBHGJLMDZJSVPZZBCCZNVLMQ",
"output": ".v.b.k.q.c.f.b.m.q.h.d.m.g.n.s.g.b.q.v.j.t.g.q.c.n.h.h.r.j.m.n.k.g.k.d.p.p.s.q.r.r.v.q.t.z.n.k.b.z.g.s.x.b.p.b.r.x.p.m.v.f.t.x.c.h.z.m.s.j.v.b.r.n.f.n.t.h.b.h.g.j.l.m.d.z.j.s.v.p.z.z.b.c.c.z.n.v.l.m.q"
},
{
"input": "iioyoaayeuyoolyiyoeuouiayiiuyTueyiaoiueyioiouyuauouayyiaeoeiiigmioiououeieeeyuyyaYyioiiooaiuouyoeoeg",
"output": ".l.t.g.m.g"
},
{
"input": "ueyiuiauuyyeueykeioouiiauzoyoeyeuyiaoaiiaaoaueyaeydaoauexuueafouiyioueeaaeyoeuaueiyiuiaeeayaioeouiuy",
"output": ".k.z.d.x.f"
},
{
"input": "FSNRBXLFQHZXGVMKLQDVHWLDSLKGKFMDRQWMWSSKPKKQBNDZRSCBLRSKCKKFFKRDMZFZGCNSMXNPMZVDLKXGNXGZQCLRTTDXLMXQ",
"output": ".f.s.n.r.b.x.l.f.q.h.z.x.g.v.m.k.l.q.d.v.h.w.l.d.s.l.k.g.k.f.m.d.r.q.w.m.w.s.s.k.p.k.k.q.b.n.d.z.r.s.c.b.l.r.s.k.c.k.k.f.f.k.r.d.m.z.f.z.g.c.n.s.m.x.n.p.m.z.v.d.l.k.x.g.n.x.g.z.q.c.l.r.t.t.d.x.l.m.x.q"
},
{
"input": "EYAYAYIOIOYOOAUOEUEUOUUYIYUUMOEOIIIAOIUOAAOIYOIOEUIERCEYYAOIOIGYUIAOYUEOEUAEAYPOYEYUUAUOAOEIYIEYUEEY",
"output": ".m.r.c.g.p"
},
{
"input": "jvvzcdcxjstbbksmqjsngxkgtttdxrljjxtwptgwwqzpvqchvgrkqlzxmptzblxhhsmrkmzzmgdfskhtmmnqzzflpmqdctvrfgtx",
"output": ".j.v.v.z.c.d.c.x.j.s.t.b.b.k.s.m.q.j.s.n.g.x.k.g.t.t.t.d.x.r.l.j.j.x.t.w.p.t.g.w.w.q.z.p.v.q.c.h.v.g.r.k.q.l.z.x.m.p.t.z.b.l.x.h.h.s.m.r.k.m.z.z.m.g.d.f.s.k.h.t.m.m.n.q.z.z.f.l.p.m.q.d.c.t.v.r.f.g.t.x"
},
{
"input": "YB",
"output": ".b"
},
{
"input": "fly",
"output": ".f.l"
},
{
"input": "YyyYYYyyYxdwdawdDAWDdaddYYYY",
"output": ".x.d.w.d.w.d.d.w.d.d.d.d"
}
] | 1,697,376,764 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 42 | 124 | 0 | word = input()
vowels = ['a', 'e','i', 'o', 'u', 'y', 'A', 'E', 'I', 'O','U','Y']
new_word = ""
for i in range(len(word)):
if word[i] not in vowels:
new_word += "."
new_word += word[i]
print(new_word.lower())
| Title: String Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
Input Specification:
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Output Specification:
Print the resulting string. It is guaranteed that this string is not empty.
Demo Input:
['tour\n', 'Codeforces\n', 'aBAcAba\n']
Demo Output:
['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n']
Note:
none | ```python
word = input()
vowels = ['a', 'e','i', 'o', 'u', 'y', 'A', 'E', 'I', 'O','U','Y']
new_word = ""
for i in range(len(word)):
if word[i] not in vowels:
new_word += "."
new_word += word[i]
print(new_word.lower())
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question. | The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants | Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES"). | [
"1 3 2 1 2 1\n",
"1 1 1 1 1 99\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater. | 0 | [
{
"input": "1 3 2 1 2 1",
"output": "YES"
},
{
"input": "1 1 1 1 1 99",
"output": "NO"
},
{
"input": "1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "0 0 0 0 0 0",
"output": "YES"
},
{
"input": "633 609 369 704 573 416",
"output": "NO"
},
{
"input": "353 313 327 470 597 31",
"output": "NO"
},
{
"input": "835 638 673 624 232 266",
"output": "NO"
},
{
"input": "936 342 19 398 247 874",
"output": "NO"
},
{
"input": "417 666 978 553 271 488",
"output": "NO"
},
{
"input": "71 66 124 199 67 147",
"output": "YES"
},
{
"input": "54 26 0 171 239 12",
"output": "YES"
},
{
"input": "72 8 186 92 267 69",
"output": "YES"
},
{
"input": "180 179 188 50 75 214",
"output": "YES"
},
{
"input": "16 169 110 136 404 277",
"output": "YES"
},
{
"input": "101 400 9 200 300 10",
"output": "YES"
},
{
"input": "101 400 200 9 300 10",
"output": "YES"
},
{
"input": "101 200 400 9 300 10",
"output": "YES"
},
{
"input": "101 400 200 300 9 10",
"output": "YES"
},
{
"input": "101 200 400 300 9 10",
"output": "YES"
},
{
"input": "4 4 4 4 5 4",
"output": "NO"
},
{
"input": "2 2 2 2 2 1",
"output": "NO"
},
{
"input": "1000 1000 999 1000 1000 1000",
"output": "NO"
},
{
"input": "129 1 10 29 8 111",
"output": "NO"
},
{
"input": "1000 1000 1000 999 999 1000",
"output": "YES"
},
{
"input": "101 200 300 400 9 10",
"output": "YES"
},
{
"input": "101 400 200 300 10 9",
"output": "YES"
},
{
"input": "101 200 400 300 10 9",
"output": "YES"
},
{
"input": "101 200 300 400 10 9",
"output": "YES"
},
{
"input": "101 200 300 10 400 9",
"output": "YES"
},
{
"input": "1 1 1 1 1 5",
"output": "NO"
},
{
"input": "8 1 1 3 3 0",
"output": "NO"
},
{
"input": "1 1 2 2 3 3",
"output": "YES"
},
{
"input": "1 2 2 5 2 5",
"output": "NO"
},
{
"input": "1 2 3 6 6 6",
"output": "NO"
},
{
"input": "36 91 7 86 51 89",
"output": "NO"
},
{
"input": "10 1 1 1 23 24",
"output": "NO"
},
{
"input": "1 1 1 10 23 24",
"output": "NO"
},
{
"input": "20 10 1 2 3 44",
"output": "NO"
},
{
"input": "7 0 14 11 8 6",
"output": "NO"
},
{
"input": "100 496 1 1 1 1",
"output": "NO"
},
{
"input": "5 4 2 5 11 3",
"output": "NO"
},
{
"input": "1 3 7 8 8 9",
"output": "YES"
},
{
"input": "1 3 4 5 18 19",
"output": "YES"
},
{
"input": "5 5 1 2 2 15",
"output": "NO"
},
{
"input": "2 1 0 0 0 5",
"output": "NO"
},
{
"input": "1 6 6 1 20 2",
"output": "NO"
},
{
"input": "2 10 0 0 0 0",
"output": "NO"
},
{
"input": "1 1 3 1 1 11",
"output": "NO"
},
{
"input": "10 10 1 1 1 37",
"output": "NO"
},
{
"input": "1 1 0 0 0 4",
"output": "NO"
},
{
"input": "1 1 10 1 1 28",
"output": "NO"
},
{
"input": "1 5 5 5 6 8",
"output": "YES"
},
{
"input": "0 2 3 4 4 5",
"output": "YES"
}
] | 1,510,503,009 | 309 | Python 3 | WRONG_ANSWER | TESTS | 30 | 77 | 0 | a = list( map( int, input().split() ) )
n = len(a)
tot = sum( a )
for i in range( n ):
for j in range( n ):
for k in range( n ):
if i != j and j != k:
tmp = a[i] + a[j] + a[k]
if tmp+tmp == tot:
print( "YES" )
exit(0)
print( "NO" ) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
Input Specification:
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
Output Specification:
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['1 3 2 1 2 1\n', '1 1 1 1 1 99\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater. | ```python
a = list( map( int, input().split() ) )
n = len(a)
tot = sum( a )
for i in range( n ):
for j in range( n ):
for k in range( n ):
if i != j and j != k:
tmp = a[i] + a[j] + a[k]
if tmp+tmp == tot:
print( "YES" )
exit(0)
print( "NO" )
``` | 0 | |
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"
}
] | 1,678,117,291 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 25 | 155 | 8,396,800 | a,b,n=map(int,input().split())
a=str(a)
flag=True
for i in range(10):
if int(a+str(i))%b==0:
a=int(a+str(i))
flag=False
break
if flag:
print(-1)
else:
a=int (str(a)+"0"*(n-1))
print(a) | 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())
a=str(a)
flag=True
for i in range(10):
if int(a+str(i))%b==0:
a=int(a+str(i))
flag=False
break
if flag:
print(-1)
else:
a=int (str(a)+"0"*(n-1))
print(a)
``` | 3 | |
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"
}
] | 1,679,461,301 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 124 | 0 | m,n=map(int,input().split())
a=m*n
b=a//2
print(b)
| 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
m,n=map(int,input().split())
a=m*n
b=a//2
print(b)
``` | 3.969 |
4 | A | Watermelon | PROGRAMMING | 800 | [
"brute force",
"math"
] | A. Watermelon | 1 | 64 | One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. | The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. | Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. | [
"8\n"
] | [
"YES\n"
] | For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | 0 | [
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "10",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "53",
"output": "NO"
},
{
"input": "77",
"output": "NO"
},
{
"input": "32",
"output": "YES"
},
{
"input": "44",
"output": "YES"
},
{
"input": "98",
"output": "YES"
},
{
"input": "99",
"output": "NO"
},
{
"input": "90",
"output": "YES"
},
{
"input": "67",
"output": "NO"
},
{
"input": "100",
"output": "YES"
},
{
"input": "88",
"output": "YES"
}
] | 1,693,307,125 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 92 | 0 | we = input()
wr=int(we)
if wr < 1:
quit()
if wr ==2 :
print("NO")
quit()
if wr % 2 == 0:
print("YES")
else :
print("NO") | Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | ```python
we = input()
wr=int(we)
if wr < 1:
quit()
if wr ==2 :
print("NO")
quit()
if wr % 2 == 0:
print("YES")
else :
print("NO")
``` | 3.954 |
81 | A | Plug-in | PROGRAMMING | 1,400 | [
"implementation"
] | A. Plug-in | 1 | 256 | Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result. | The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters. | Print the given string after it is processed. It is guaranteed that the result will contain at least one character. | [
"hhoowaaaareyyoouu\n",
"reallazy\n",
"abacabaabacabaa\n"
] | [
"wre",
"rezy",
"a"
] | none | 500 | [
{
"input": "hhoowaaaareyyoouu",
"output": "wre"
},
{
"input": "reallazy",
"output": "rezy"
},
{
"input": "abacabaabacabaa",
"output": "a"
},
{
"input": "xraccabccbry",
"output": "xy"
},
{
"input": "a",
"output": "a"
},
{
"input": "b",
"output": "b"
},
{
"input": "il",
"output": "il"
},
{
"input": "gfj",
"output": "gfj"
},
{
"input": "babbbbbababa",
"output": "babababa"
},
{
"input": "babbbbabbabbbababbabbbbbbabaabaababaaabbbbbabbbbaaaaabbaaabbaabaabbbbabbbababbabaaabbababaaababbbaaa",
"output": "babababababababababababa"
},
{
"input": "aab",
"output": "b"
},
{
"input": "abb",
"output": "a"
},
{
"input": "aba",
"output": "aba"
},
{
"input": "ab",
"output": "ab"
}
] | 1,572,796,810 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 186 | 0 | input = list(input())
output = []
for x in input:
if not input:
output.append(x)
else:
if x == output[-1]:
output.pop()
else:
output.append(x)
print(''.join(output))
| Title: Plug-in
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input Specification:
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output Specification:
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Demo Input:
['hhoowaaaareyyoouu\n', 'reallazy\n', 'abacabaabacabaa\n']
Demo Output:
['wre', 'rezy', 'a']
Note:
none | ```python
input = list(input())
output = []
for x in input:
if not input:
output.append(x)
else:
if x == output[-1]:
output.pop()
else:
output.append(x)
print(''.join(output))
``` | -1 |
461 | A | Appleman and Toastman | PROGRAMMING | 1,200 | [
"greedy",
"sortings"
] | null | null | Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman. | Print a single integer — the largest possible score. | [
"3\n3 1 5\n",
"1\n10\n"
] | [
"26\n",
"10\n"
] | Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | 500 | [
{
"input": "3\n3 1 5",
"output": "26"
},
{
"input": "1\n10",
"output": "10"
},
{
"input": "10\n8 10 2 5 6 2 4 7 2 1",
"output": "376"
},
{
"input": "10\n171308 397870 724672 431255 228496 892002 542924 718337 888642 161821",
"output": "40204082"
},
{
"input": "10\n1 2 2 2 4 5 6 7 8 10",
"output": "376"
},
{
"input": "10\n161821 171308 228496 397870 431255 542924 718337 724672 888642 892002",
"output": "40204082"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "1\n1000000",
"output": "1000000"
},
{
"input": "10\n10 8 7 6 5 4 2 2 2 1",
"output": "376"
},
{
"input": "10\n892002 888642 724672 718337 542924 431255 397870 228496 171308 161821",
"output": "40204082"
},
{
"input": "10\n5 2 6 10 10 10 10 2 2 5",
"output": "485"
},
{
"input": "10\n431255 724672 228496 397870 397870 397870 397870 724672 888642 431255",
"output": "36742665"
},
{
"input": "10\n2 2 2 5 5 6 10 10 10 10",
"output": "485"
},
{
"input": "10\n228496 397870 397870 397870 397870 431255 431255 724672 724672 888642",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 6 5 5 2 2 2",
"output": "485"
},
{
"input": "10\n888642 724672 724672 431255 431255 397870 397870 397870 397870 228496",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 10 10 10 10 10 10",
"output": "640"
},
{
"input": "10\n1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000",
"output": "64000000"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "2\n1 2",
"output": "6"
},
{
"input": "2\n2 3",
"output": "10"
},
{
"input": "2\n1 1",
"output": "4"
}
] | 1,512,980,602 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 10,854,400 | n=int(input())
Li=input().split()
score=0
for a in range(n):
Li[a]=int(Li[a])
score=score+Li[a]
if n>1:
i=1
S=n*score
while i<n-1:
S=S-(n-1-i)*min(Li)
Li.remove(min(Li))
i+=1
else:
S=score
# if n>1:
# k=n
# S=score+score
# while k>2:
# MIN_=min(Li)
# score=score-MIN_
# S=S+score
# Li.remove(min(Li))
# k-=1
# else:
# S=score
# print(S)
print(S)
| Title: Appleman and Toastman
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman.
Output Specification:
Print a single integer — the largest possible score.
Demo Input:
['3\n3 1 5\n', '1\n10\n']
Demo Output:
['26\n', '10\n']
Note:
Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | ```python
n=int(input())
Li=input().split()
score=0
for a in range(n):
Li[a]=int(Li[a])
score=score+Li[a]
if n>1:
i=1
S=n*score
while i<n-1:
S=S-(n-1-i)*min(Li)
Li.remove(min(Li))
i+=1
else:
S=score
# if n>1:
# k=n
# S=score+score
# while k>2:
# MIN_=min(Li)
# score=score-MIN_
# S=S+score
# Li.remove(min(Li))
# k-=1
# else:
# S=score
# print(S)
print(S)
``` | 0 | |
849 | A | Odds and Ends | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Where do odds begin, and where do they end? Where does hope emerge, and will they ever break?
Given an integer sequence *a*1,<=*a*2,<=...,<=*a**n* of length *n*. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers.
A subsegment is a contiguous slice of the whole sequence. For example, {3,<=4,<=5} and {1} are subsegments of sequence {1,<=2,<=3,<=4,<=5,<=6}, while {1,<=2,<=4} and {7} are not. | The first line of input contains a non-negative integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence.
The second line contains *n* space-separated non-negative integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — the elements of the sequence. | Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise.
You can output each letter in any case (upper or lower). | [
"3\n1 3 5\n",
"5\n1 0 1 5 1\n",
"3\n4 3 1\n",
"4\n3 9 9 3\n"
] | [
"Yes\n",
"Yes\n",
"No\n",
"No\n"
] | In the first example, divide the sequence into 1 subsegment: {1, 3, 5} and the requirements will be met.
In the second example, divide the sequence into 3 subsegments: {1, 0, 1}, {5}, {1}.
In the third example, one of the subsegments must start with 4 which is an even number, thus the requirements cannot be met.
In the fourth example, the sequence can be divided into 2 subsegments: {3, 9, 9}, {3}, but this is not a valid solution because 2 is an even number. | 500 | [
{
"input": "3\n1 3 5",
"output": "Yes"
},
{
"input": "5\n1 0 1 5 1",
"output": "Yes"
},
{
"input": "3\n4 3 1",
"output": "No"
},
{
"input": "4\n3 9 9 3",
"output": "No"
},
{
"input": "1\n1",
"output": "Yes"
},
{
"input": "5\n100 99 100 99 99",
"output": "No"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "No"
},
{
"input": "1\n0",
"output": "No"
},
{
"input": "2\n1 1",
"output": "No"
},
{
"input": "2\n10 10",
"output": "No"
},
{
"input": "2\n54 21",
"output": "No"
},
{
"input": "5\n0 0 0 0 0",
"output": "No"
},
{
"input": "5\n67 92 0 26 43",
"output": "Yes"
},
{
"input": "15\n45 52 35 80 68 80 93 57 47 32 69 23 63 90 43",
"output": "Yes"
},
{
"input": "15\n81 28 0 82 71 64 63 89 87 92 38 30 76 72 36",
"output": "No"
},
{
"input": "50\n49 32 17 59 77 98 65 50 85 10 40 84 65 34 52 25 1 31 61 45 48 24 41 14 76 12 33 76 44 86 53 33 92 58 63 93 50 24 31 79 67 50 72 93 2 38 32 14 87 99",
"output": "No"
},
{
"input": "55\n65 69 53 66 11 100 68 44 43 17 6 66 24 2 6 6 61 72 91 53 93 61 52 96 56 42 6 8 79 49 76 36 83 58 8 43 2 90 71 49 80 21 75 13 76 54 95 61 58 82 40 33 73 61 46",
"output": "No"
},
{
"input": "99\n73 89 51 85 42 67 22 80 75 3 90 0 52 100 90 48 7 15 41 1 54 2 23 62 86 68 2 87 57 12 45 34 68 54 36 49 27 46 22 70 95 90 57 91 90 79 48 89 67 92 28 27 25 37 73 66 13 89 7 99 62 53 48 24 73 82 62 88 26 39 21 86 50 95 26 27 60 6 56 14 27 90 55 80 97 18 37 36 70 2 28 53 36 77 39 79 82 42 69",
"output": "Yes"
},
{
"input": "99\n99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99",
"output": "Yes"
},
{
"input": "100\n61 63 34 45 20 91 31 28 40 27 94 1 73 5 69 10 56 94 80 23 79 99 59 58 13 56 91 59 77 78 88 72 80 72 70 71 63 60 41 41 41 27 83 10 43 14 35 48 0 78 69 29 63 33 42 67 1 74 51 46 79 41 37 61 16 29 82 28 22 14 64 49 86 92 82 55 54 24 75 58 95 31 3 34 26 23 78 91 49 6 30 57 27 69 29 54 42 0 61 83",
"output": "No"
},
{
"input": "6\n1 2 2 2 2 1",
"output": "No"
},
{
"input": "3\n1 2 1",
"output": "Yes"
},
{
"input": "4\n1 3 2 3",
"output": "No"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "No"
},
{
"input": "6\n1 1 0 0 1 1",
"output": "No"
},
{
"input": "4\n1 4 9 3",
"output": "No"
},
{
"input": "4\n1 0 1 1",
"output": "No"
},
{
"input": "10\n1 0 0 1 1 1 1 1 1 1",
"output": "No"
},
{
"input": "10\n9 2 5 7 8 3 1 9 4 9",
"output": "No"
},
{
"input": "99\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 2",
"output": "No"
},
{
"input": "6\n1 2 1 2 2 1",
"output": "No"
},
{
"input": "6\n1 0 1 0 0 1",
"output": "No"
},
{
"input": "4\n1 3 4 7",
"output": "No"
},
{
"input": "8\n1 1 1 2 1 1 1 1",
"output": "No"
},
{
"input": "3\n1 1 2",
"output": "No"
},
{
"input": "5\n1 2 1 2 1",
"output": "Yes"
},
{
"input": "5\n5 4 4 2 1",
"output": "Yes"
},
{
"input": "6\n1 3 3 3 3 1",
"output": "No"
},
{
"input": "7\n1 2 1 2 2 2 1",
"output": "Yes"
},
{
"input": "4\n1 2 2 1",
"output": "No"
},
{
"input": "6\n1 2 3 4 6 5",
"output": "No"
},
{
"input": "5\n1 1 2 2 2",
"output": "No"
},
{
"input": "5\n1 0 0 1 1",
"output": "Yes"
},
{
"input": "3\n1 2 4",
"output": "No"
},
{
"input": "3\n1 0 2",
"output": "No"
},
{
"input": "5\n1 1 1 0 1",
"output": "Yes"
},
{
"input": "4\n3 9 2 3",
"output": "No"
},
{
"input": "6\n1 1 1 4 4 1",
"output": "No"
},
{
"input": "6\n1 2 3 5 6 7",
"output": "No"
},
{
"input": "6\n1 1 1 2 2 1",
"output": "No"
},
{
"input": "6\n1 1 1 0 0 1",
"output": "No"
},
{
"input": "5\n1 2 2 5 5",
"output": "Yes"
},
{
"input": "5\n1 3 2 4 5",
"output": "Yes"
},
{
"input": "8\n1 2 3 5 7 8 8 5",
"output": "No"
},
{
"input": "10\n1 1 1 2 1 1 1 1 1 1",
"output": "No"
},
{
"input": "4\n1 0 0 1",
"output": "No"
},
{
"input": "7\n1 0 1 1 0 0 1",
"output": "Yes"
},
{
"input": "7\n1 4 5 7 6 6 3",
"output": "Yes"
},
{
"input": "4\n2 2 2 2",
"output": "No"
},
{
"input": "5\n2 3 4 5 6",
"output": "No"
},
{
"input": "4\n1 1 2 1",
"output": "No"
},
{
"input": "3\n1 2 3",
"output": "Yes"
},
{
"input": "6\n1 3 3 2 2 3",
"output": "No"
},
{
"input": "4\n1 1 2 3",
"output": "No"
},
{
"input": "4\n1 2 3 5",
"output": "No"
},
{
"input": "5\n3 4 4 3 3",
"output": "Yes"
},
{
"input": "4\n3 2 2 3",
"output": "No"
},
{
"input": "6\n1 1 1 1 2 1",
"output": "No"
},
{
"input": "6\n1 1 2 2 1 1",
"output": "No"
},
{
"input": "10\n3 4 2 4 3 2 2 4 4 3",
"output": "No"
},
{
"input": "7\n1 2 4 3 2 4 5",
"output": "Yes"
},
{
"input": "28\n75 51 25 52 13 7 34 29 5 59 68 56 13 2 9 37 59 83 18 32 36 30 20 43 92 76 78 67",
"output": "No"
},
{
"input": "79\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 18",
"output": "No"
},
{
"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": "No"
}
] | 1,504,369,663 | 2,147,483,647 | Python 3 | OK | TESTS | 74 | 62 | 0 | n = int(input())
p = list(map(int, input().split()))
if n % 2 != 0 and p[0] % 2 != 0 and p[-1] % 2 != 0:
print("Yes")
else:
print("No") | Title: Odds and Ends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Where do odds begin, and where do they end? Where does hope emerge, and will they ever break?
Given an integer sequence *a*1,<=*a*2,<=...,<=*a**n* of length *n*. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers.
A subsegment is a contiguous slice of the whole sequence. For example, {3,<=4,<=5} and {1} are subsegments of sequence {1,<=2,<=3,<=4,<=5,<=6}, while {1,<=2,<=4} and {7} are not.
Input Specification:
The first line of input contains a non-negative integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence.
The second line contains *n* space-separated non-negative integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — the elements of the sequence.
Output Specification:
Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise.
You can output each letter in any case (upper or lower).
Demo Input:
['3\n1 3 5\n', '5\n1 0 1 5 1\n', '3\n4 3 1\n', '4\n3 9 9 3\n']
Demo Output:
['Yes\n', 'Yes\n', 'No\n', 'No\n']
Note:
In the first example, divide the sequence into 1 subsegment: {1, 3, 5} and the requirements will be met.
In the second example, divide the sequence into 3 subsegments: {1, 0, 1}, {5}, {1}.
In the third example, one of the subsegments must start with 4 which is an even number, thus the requirements cannot be met.
In the fourth example, the sequence can be divided into 2 subsegments: {3, 9, 9}, {3}, but this is not a valid solution because 2 is an even number. | ```python
n = int(input())
p = list(map(int, input().split()))
if n % 2 != 0 and p[0] % 2 != 0 and p[-1] % 2 != 0:
print("Yes")
else:
print("No")
``` | 3 | |
9 | A | Die Roll | PROGRAMMING | 800 | [
"math",
"probabilities"
] | A. Die Roll | 1 | 64 | Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. | The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. | Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». | [
"4 2\n"
] | [
"1/2\n"
] | Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | 0 | [
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,679,335,816 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | x,y = map(int, input().split())
ans = str()
if max([x,y]) == 6:
ans = '0/1'
elif x == y:
ans = f'1/{6-x+1}'
else:
ans = f'1/{6-x}'
print(ans) | Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | ```python
x,y = map(int, input().split())
ans = str()
if max([x,y]) == 6:
ans = '0/1'
elif x == y:
ans = f'1/{6-x+1}'
else:
ans = f'1/{6-x}'
print(ans)
``` | 0 |
632 | C | The Smallest String Concatenation | PROGRAMMING | 1,700 | [
"sortings",
"strings"
] | null | null | You're given a list of *n* strings *a*1,<=*a*2,<=...,<=*a**n*. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.
Given the list of strings, output the lexicographically smallest concatenation. | The first line contains integer *n* — the number of strings (1<=≤<=*n*<=≤<=5·104).
Each of the next *n* lines contains one string *a**i* (1<=≤<=|*a**i*|<=≤<=50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5·104. | Print the only string *a* — the lexicographically smallest string concatenation. | [
"4\nabba\nabacaba\nbcd\ner\n",
"5\nx\nxx\nxxa\nxxaa\nxxaaa\n",
"3\nc\ncb\ncba\n"
] | [
"abacabaabbabcder\n",
"xxaaaxxaaxxaxxx\n",
"cbacbc\n"
] | none | 0 | [
{
"input": "4\nabba\nabacaba\nbcd\ner",
"output": "abacabaabbabcder"
},
{
"input": "5\nx\nxx\nxxa\nxxaa\nxxaaa",
"output": "xxaaaxxaaxxaxxx"
},
{
"input": "3\nc\ncb\ncba",
"output": "cbacbc"
},
{
"input": "10\naba\nabaaca\naba\nacaaaabbac\nabaacac\nb\ncabbcccaab\nbaacbb\nbcab\ncc",
"output": "abaabaabaacaabaacacacaaaabbacbaacbbbbcabcabbcccaabcc"
},
{
"input": "13\nclgknjjojyuvdtv\nclgknjjojyuvdtvzxz\nclgknjjojyuvdtvzxzxradqhm\ngvzpnckalbaubfviyhijosiixvxaydxagvymq\nclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymt\nbvwzpuut\nclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymth\nclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymtlkdl\nrbwvaayhhunlvmwexvkmqbquypsnjbdkkeytthekttywypud\nqlvkfkshkxyrfefigtngifxsuhpjgkb\nlhxtkfjrrwcvzuamizotupjqckaq\nenngvhxvpaoiyhctddyscasyqtilvbvgfphayykjeoylcvmjz\nxkazwk",
"output": "bvwzpuutclgknjjojyuvdtvclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymtclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymthclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymtlkdlclgknjjojyuvdtvzxzclgknjjojyuvdtvzxzxradqhmenngvhxvpaoiyhctddyscasyqtilvbvgfphayykjeoylcvmjzgvzpnckalbaubfviyhijosiixvxaydxagvymqlhxtkfjrrwcvzuamizotupjqckaqqlvkfkshkxyrfefigtngifxsuhpjgkbrbwvaayhhunlvmwexvkmqbquypsnjbdkkeytthekttywypudxkazwk"
}
] | 1,679,283,609 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | # LUOGU_RID: 105290817
// Problem: CF632C The Smallest String Concatenation
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/CF632C
// Memory Limit: 250 MB
// Time Limit: 3000 ms
// Date:2023-03-20 11:38:47
// By:SmallBlack
//
// Powered by CP Editor (https://cpeditor.org)
#include<bits/stdc++.h>
using namespace std;
inline long long read()
{
long long s=0,k=1;
char c=getchar();
while(!isdigit(c))
{
k=(c=='-')?-1:1;
c=getchar();
}
while(isdigit(c))
{
s=s*10+c-'0';
c=getchar();
}
return s*k;
}
#define d read()
#define ll long long
#define Maxn 10010
#define Size 50010
#define mp make_pair
#define pb push_back
#define fi first
#define se second
string a[Size];
bool cmp(string i,string j)
{return i+j<j+i;}
int main()
{
ll n=d;
for(int i=1;i<=n;i++) cin>>a[i];
sort(a+1,a+n+1,cmp);
for(int i=1;i<=n;i++) cout<<a[i];
} | Title: The Smallest String Concatenation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You're given a list of *n* strings *a*1,<=*a*2,<=...,<=*a**n*. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.
Given the list of strings, output the lexicographically smallest concatenation.
Input Specification:
The first line contains integer *n* — the number of strings (1<=≤<=*n*<=≤<=5·104).
Each of the next *n* lines contains one string *a**i* (1<=≤<=|*a**i*|<=≤<=50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5·104.
Output Specification:
Print the only string *a* — the lexicographically smallest string concatenation.
Demo Input:
['4\nabba\nabacaba\nbcd\ner\n', '5\nx\nxx\nxxa\nxxaa\nxxaaa\n', '3\nc\ncb\ncba\n']
Demo Output:
['abacabaabbabcder\n', 'xxaaaxxaaxxaxxx\n', 'cbacbc\n']
Note:
none | ```python
# LUOGU_RID: 105290817
// Problem: CF632C The Smallest String Concatenation
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/CF632C
// Memory Limit: 250 MB
// Time Limit: 3000 ms
// Date:2023-03-20 11:38:47
// By:SmallBlack
//
// Powered by CP Editor (https://cpeditor.org)
#include<bits/stdc++.h>
using namespace std;
inline long long read()
{
long long s=0,k=1;
char c=getchar();
while(!isdigit(c))
{
k=(c=='-')?-1:1;
c=getchar();
}
while(isdigit(c))
{
s=s*10+c-'0';
c=getchar();
}
return s*k;
}
#define d read()
#define ll long long
#define Maxn 10010
#define Size 50010
#define mp make_pair
#define pb push_back
#define fi first
#define se second
string a[Size];
bool cmp(string i,string j)
{return i+j<j+i;}
int main()
{
ll n=d;
for(int i=1;i<=n;i++) cin>>a[i];
sort(a+1,a+n+1,cmp);
for(int i=1;i<=n;i++) cout<<a[i];
}
``` | -1 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Galya is playing one-dimensional Sea Battle on a 1<=×<=*n* grid. In this game *a* ships are placed on the grid. Each of the ships consists of *b* consecutive cells. No cell can be part of two ships, however, the ships can touch each other.
Galya doesn't know the ships location. She can shoot to some cells and after each shot she is told if that cell was a part of some ship (this case is called "hit") or not (this case is called "miss").
Galya has already made *k* shots, all of them were misses.
Your task is to calculate the minimum number of cells such that if Galya shoot at all of them, she would hit at least one ship.
It is guaranteed that there is at least one valid ships placement. | The first line contains four positive integers *n*, *a*, *b*, *k* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=*n*, 0<=≤<=*k*<=≤<=*n*<=-<=1) — the length of the grid, the number of ships on the grid, the length of each ship and the number of shots Galya has already made.
The second line contains a string of length *n*, consisting of zeros and ones. If the *i*-th character is one, Galya has already made a shot to this cell. Otherwise, she hasn't. It is guaranteed that there are exactly *k* ones in this string. | In the first line print the minimum number of cells such that if Galya shoot at all of them, she would hit at least one ship.
In the second line print the cells Galya should shoot at.
Each cell should be printed exactly once. You can print the cells in arbitrary order. The cells are numbered from 1 to *n*, starting from the left.
If there are multiple answers, you can print any of them. | [
"5 1 2 1\n00100\n",
"13 3 2 3\n1000000010001\n"
] | [
"2\n4 2\n",
"2\n7 11\n"
] | There is one ship in the first sample. It can be either to the left or to the right from the shot Galya has already made (the "1" character). So, it is necessary to make two shots: one at the left part, and one at the right part. | 0 | [
{
"input": "5 1 2 1\n00100",
"output": "2\n2 5 "
},
{
"input": "13 3 2 3\n1000000010001",
"output": "2\n3 5 "
},
{
"input": "1 1 1 0\n0",
"output": "1\n1 "
},
{
"input": "2 2 1 0\n00",
"output": "1\n1 "
},
{
"input": "5 4 1 0\n00000",
"output": "2\n1 2 "
},
{
"input": "10 2 2 0\n0000000000",
"output": "4\n2 4 6 8 "
},
{
"input": "20 1 3 5\n01001010000000010010",
"output": "2\n10 13 "
},
{
"input": "100 17 4 11\n0100000100000000000000001000000000010001100000000000101000000000000000000000001000001000010000000000",
"output": "2\n6 12 "
}
] | 1,479,910,206 | 3,186 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 93 | 0 | n,a,b,k = map(int,input().split())
l = list(map(int,input()))
#print(l)
cnt = 0
shots = []
for i in range(n):
if l[i]==0:
cnt+=1
if cnt>=b:
shots.append(str(i+1))
count = 0
else:
cnt = 0
print(len(shots)-a+1)
if a==1:
print(" ".join(shots))
else:
print(" ".join(shots[0:-(a-1)])) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Galya is playing one-dimensional Sea Battle on a 1<=×<=*n* grid. In this game *a* ships are placed on the grid. Each of the ships consists of *b* consecutive cells. No cell can be part of two ships, however, the ships can touch each other.
Galya doesn't know the ships location. She can shoot to some cells and after each shot she is told if that cell was a part of some ship (this case is called "hit") or not (this case is called "miss").
Galya has already made *k* shots, all of them were misses.
Your task is to calculate the minimum number of cells such that if Galya shoot at all of them, she would hit at least one ship.
It is guaranteed that there is at least one valid ships placement.
Input Specification:
The first line contains four positive integers *n*, *a*, *b*, *k* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=*n*, 0<=≤<=*k*<=≤<=*n*<=-<=1) — the length of the grid, the number of ships on the grid, the length of each ship and the number of shots Galya has already made.
The second line contains a string of length *n*, consisting of zeros and ones. If the *i*-th character is one, Galya has already made a shot to this cell. Otherwise, she hasn't. It is guaranteed that there are exactly *k* ones in this string.
Output Specification:
In the first line print the minimum number of cells such that if Galya shoot at all of them, she would hit at least one ship.
In the second line print the cells Galya should shoot at.
Each cell should be printed exactly once. You can print the cells in arbitrary order. The cells are numbered from 1 to *n*, starting from the left.
If there are multiple answers, you can print any of them.
Demo Input:
['5 1 2 1\n00100\n', '13 3 2 3\n1000000010001\n']
Demo Output:
['2\n4 2\n', '2\n7 11\n']
Note:
There is one ship in the first sample. It can be either to the left or to the right from the shot Galya has already made (the "1" character). So, it is necessary to make two shots: one at the left part, and one at the right part. | ```python
n,a,b,k = map(int,input().split())
l = list(map(int,input()))
#print(l)
cnt = 0
shots = []
for i in range(n):
if l[i]==0:
cnt+=1
if cnt>=b:
shots.append(str(i+1))
count = 0
else:
cnt = 0
print(len(shots)-a+1)
if a==1:
print(" ".join(shots))
else:
print(" ".join(shots[0:-(a-1)]))
``` | 0 | |
766 | B | Mahmoud and a Triangle | PROGRAMMING | 1,000 | [
"constructive algorithms",
"geometry",
"greedy",
"math",
"number theory",
"sortings"
] | null | null | Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area. | The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has. | In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise. | [
"5\n1 5 3 2 4\n",
"3\n4 1 2\n"
] | [
"YES\n",
"NO\n"
] | For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle. | 1,000 | [
{
"input": "5\n1 5 3 2 4",
"output": "YES"
},
{
"input": "3\n4 1 2",
"output": "NO"
},
{
"input": "30\n197 75 517 39724 7906061 1153471 3 15166 168284 3019844 272293 316 16 24548 42 118 5792 5 9373 1866366 4886214 24 2206 712886 104005 1363 836 64273 440585 3576",
"output": "NO"
},
{
"input": "30\n229017064 335281886 247217656 670601882 743442492 615491486 544941439 911270108 474843964 803323771 177115397 62179276 390270885 754889875 881720571 902691435 154083299 328505383 761264351 182674686 94104683 357622370 573909964 320060691 33548810 247029007 812823597 946798893 813659359 710111761",
"output": "YES"
},
{
"input": "40\n740553458 532562042 138583675 75471987 487348843 476240280 972115023 103690894 546736371 915774563 35356828 819948191 138721993 24257926 761587264 767176616 608310208 78275645 386063134 227581756 672567198 177797611 87579917 941781518 274774331 843623616 981221615 630282032 118843963 749160513 354134861 132333165 405839062 522698334 29698277 541005920 856214146 167344951 398332403 68622974",
"output": "YES"
},
{
"input": "40\n155 1470176 7384 765965701 1075 4 561554 6227772 93 16304522 1744 662 3 292572860 19335 908613 42685804 347058 20 132560 3848974 69067081 58 2819 111752888 408 81925 30 11951 4564 251 26381275 473392832 50628 180819969 2378797 10076746 9 214492 31291",
"output": "NO"
},
{
"input": "3\n1 1000000000 1000000000",
"output": "YES"
},
{
"input": "4\n1 1000000000 1000000000 1000000000",
"output": "YES"
},
{
"input": "3\n1 1000000000 1",
"output": "NO"
},
{
"input": "5\n1 2 3 5 2",
"output": "YES"
},
{
"input": "41\n19 161 4090221 118757367 2 45361275 1562319 596751 140871 97 1844 310910829 10708344 6618115 698 1 87059 33 2527892 12703 73396090 17326460 3 368811 20550 813975131 10 53804 28034805 7847 2992 33254 1139 227930 965568 261 4846 503064297 192153458 57 431",
"output": "NO"
},
{
"input": "42\n4317083 530966905 202811311 104 389267 35 1203 18287479 125344279 21690 859122498 65 859122508 56790 1951 148683 457 1 22 2668100 8283 2 77467028 13405 11302280 47877251 328155592 35095 29589769 240574 4 10 1019123 6985189 629846 5118 169 1648973 91891 741 282 3159",
"output": "YES"
},
{
"input": "43\n729551585 11379 5931704 330557 1653 15529406 729551578 278663905 1 729551584 2683 40656510 29802 147 1400284 2 126260 865419 51 17 172223763 86 1 534861 450887671 32 234 25127103 9597697 48226 7034 389 204294 2265706 65783617 4343 3665990 626 78034 106440137 5 18421 1023",
"output": "YES"
},
{
"input": "44\n719528276 2 235 444692918 24781885 169857576 18164 47558 15316043 9465834 64879816 2234575 1631 853530 8 1001 621 719528259 84 6933 31 1 3615623 719528266 40097928 274835337 1381044 11225 2642 5850203 6 527506 18 104977753 76959 29393 49 4283 141 201482 380 1 124523 326015",
"output": "YES"
},
{
"input": "45\n28237 82 62327732 506757 691225170 5 970 4118 264024506 313192 367 14713577 73933 691225154 6660 599 691225145 3473403 51 427200630 1326718 2146678 100848386 1569 27 163176119 193562 10784 45687 819951 38520653 225 119620 1 3 691225169 691225164 17445 23807072 1 9093493 5620082 2542 139 14",
"output": "YES"
},
{
"input": "44\n165580141 21 34 55 1 89 144 17711 2 377 610 987 2584 13 5 4181 6765 10946 1597 8 28657 3 233 75025 121393 196418 317811 9227465 832040 1346269 2178309 3524578 5702887 1 14930352 102334155 24157817 39088169 63245986 701408733 267914296 433494437 514229 46368",
"output": "NO"
},
{
"input": "3\n1 1000000000 999999999",
"output": "NO"
},
{
"input": "5\n1 1 1 1 1",
"output": "YES"
},
{
"input": "10\n1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000",
"output": "NO"
},
{
"input": "5\n2 3 4 10 20",
"output": "YES"
},
{
"input": "6\n18 23 40 80 160 161",
"output": "YES"
},
{
"input": "4\n5 6 7 888",
"output": "YES"
},
{
"input": "9\n1 1 2 2 4 5 10 10 20",
"output": "YES"
},
{
"input": "7\n3 150 900 4 500 1500 5",
"output": "YES"
},
{
"input": "3\n2 2 3",
"output": "YES"
},
{
"input": "7\n1 2 100 200 250 1000000 2000000",
"output": "YES"
},
{
"input": "8\n2 3 5 5 5 6 6 13",
"output": "YES"
},
{
"input": "3\n2 3 4",
"output": "YES"
},
{
"input": "6\n1 1 1 4 5 100",
"output": "YES"
},
{
"input": "13\n1 2 3 5 8 13 22 34 55 89 144 233 377",
"output": "YES"
},
{
"input": "4\n2 3 4 8",
"output": "YES"
},
{
"input": "3\n5 6 7",
"output": "YES"
},
{
"input": "5\n1 4 5 6 1000000",
"output": "YES"
},
{
"input": "4\n5 6 7 20",
"output": "YES"
},
{
"input": "6\n1 1 1 1 1 65",
"output": "YES"
},
{
"input": "4\n3 4 5 100",
"output": "YES"
},
{
"input": "3\n2 4 5",
"output": "YES"
},
{
"input": "7\n1 1 1 1 1 10 1000",
"output": "YES"
},
{
"input": "4\n1 1 2 3",
"output": "NO"
},
{
"input": "11\n1 2 5 6 7 8 9 17 18 19 100",
"output": "YES"
},
{
"input": "4\n5 16 20 200",
"output": "YES"
},
{
"input": "5\n17 6 3 3 1",
"output": "YES"
},
{
"input": "3\n1 1 1",
"output": "YES"
},
{
"input": "6\n1 1 1 2 3 5",
"output": "YES"
},
{
"input": "4\n2 4 6 6",
"output": "YES"
},
{
"input": "9\n1 2 4 4 4 4 7 8 20",
"output": "YES"
},
{
"input": "9\n1 1 2 5 5 5 10 10 20",
"output": "YES"
},
{
"input": "7\n3 150 600 4 1700 6000 5",
"output": "YES"
},
{
"input": "5\n5761 20966 27841 28800 29399",
"output": "YES"
},
{
"input": "9\n1 2 3 6 7 10 11 12 24",
"output": "YES"
},
{
"input": "4\n1 2 1 1",
"output": "YES"
},
{
"input": "5\n1 1 2 3 4",
"output": "YES"
}
] | 1,668,150,387 | 2,147,483,647 | Python 3 | OK | TESTS | 87 | 78 | 7,884,800 | n = int(input())
a = list(map(int, input().split()))
b = False
if n > 45:
print("YES")
else:
a = sorted(a, reverse = True)
for i in range(len(a) - 2):
if (a[i] < a[i + 1] + a[i + 2]):
b = True
break
if b:
print("YES")
else:
print("NO") | Title: Mahmoud and a Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
Input Specification:
The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.
Output Specification:
In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.
Demo Input:
['5\n1 5 3 2 4\n', '3\n4 1 2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle. | ```python
n = int(input())
a = list(map(int, input().split()))
b = False
if n > 45:
print("YES")
else:
a = sorted(a, reverse = True)
for i in range(len(a) - 2):
if (a[i] < a[i + 1] + a[i + 2]):
b = True
break
if b:
print("YES")
else:
print("NO")
``` | 3 | |
722 | E | Research Rover | PROGRAMMING | 2,900 | [
"combinatorics",
"dp"
] | null | null | Unfortunately, the formal description of the task turned out to be too long, so here is the legend.
Research rover finally reached the surface of Mars and is ready to complete its mission. Unfortunately, due to the mistake in the navigation system design, the rover is located in the wrong place.
The rover will operate on the grid consisting of *n* rows and *m* columns. We will define as (*r*,<=*c*) the cell located in the row *r* and column *c*. From each cell the rover is able to move to any cell that share a side with the current one.
The rover is currently located at cell (1,<=1) and has to move to the cell (*n*,<=*m*). It will randomly follow some shortest path between these two cells. Each possible way is chosen equiprobably.
The cargo section of the rover contains the battery required to conduct the research. Initially, the battery charge is equal to *s* units of energy.
Some of the cells contain anomaly. Each time the rover gets to the cell with anomaly, the battery looses half of its charge rounded down. Formally, if the charge was equal to *x* before the rover gets to the cell with anomaly, the charge will change to .
While the rover picks a random shortest path to proceed, compute the expected value of the battery charge after it reaches cell (*n*,<=*m*). If the cells (1,<=1) and (*n*,<=*m*) contain anomaly, they also affect the charge of the battery. | The first line of the input contains four integers *n*, *m*, *k* and *s* (1<=≤<=*n*,<=*m*<=≤<=100<=000, 0<=≤<=*k*<=≤<=2000, 1<=≤<=*s*<=≤<=1<=000<=000) — the number of rows and columns of the field, the number of cells with anomaly and the initial charge of the battery respectively.
The follow *k* lines containing two integers *r**i* and *c**i* (1<=≤<=*r**i*<=≤<=*n*, 1<=≤<=*c**i*<=≤<=*m*) — coordinates of the cells, containing anomaly. It's guaranteed that each cell appears in this list no more than once. | The answer can always be represented as an irreducible fraction . Print the only integer *P*·*Q*<=-<=1 modulo 109<=+<=7. | [
"3 3 2 11\n2 1\n2 3\n",
"4 5 3 17\n1 2\n3 3\n4 1\n",
"1 6 2 15\n1 1\n1 5\n"
] | [
"333333342\n",
"514285727\n",
"4\n"
] | In the first sample, the rover picks one of the following six routes:
1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c9ec158c30775d6289140a3854e05168b09af399.png" style="max-width: 100.0%;max-height: 100.0%;"/>, after passing cell (2, 3) charge is equal to 6. 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/fcc490c05d2acb731046a7c4c861f4c9ebff3633.png" style="max-width: 100.0%;max-height: 100.0%;"/>, after passing cell (2, 3) charge is equal to 6. 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/8d5828aadc35714d7a3453c40de81ad186e87ab3.png" style="max-width: 100.0%;max-height: 100.0%;"/>, charge remains unchanged and equals 11. 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3bc680a61ca3712bbbec0eb682f3af16ab7664a2.png" style="max-width: 100.0%;max-height: 100.0%;"/>, after passing cells (2, 1) and (2, 3) charge equals 6 and then 3. 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9a87005ef2b3eb1efc9e486e608fadf3a5b557fa.png" style="max-width: 100.0%;max-height: 100.0%;"/>, after passing cell (2, 1) charge is equal to 6. 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/342ddbff927874c07e9d307d73383aa50f2117b6.png" style="max-width: 100.0%;max-height: 100.0%;"/>, after passing cell (2, 1) charge is equal to 6.
Expected value of the battery charge is calculated by the following formula:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7ddca3ad80b71df649b7beb01944a4ad8f863265.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
Thus *P* = 19, and *Q* = 3.
3<sup class="upper-index"> - 1</sup> modulo 10<sup class="upper-index">9</sup> + 7 equals 333333336.
19·333333336 = 333333342 (*mod* 10<sup class="upper-index">9</sup> + 7) | 2,000 | [] | 1,689,416,105 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | print("_RANDOM_GUESS_1689416105.399758")# 1689416105.3997796 | Title: Research Rover
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Unfortunately, the formal description of the task turned out to be too long, so here is the legend.
Research rover finally reached the surface of Mars and is ready to complete its mission. Unfortunately, due to the mistake in the navigation system design, the rover is located in the wrong place.
The rover will operate on the grid consisting of *n* rows and *m* columns. We will define as (*r*,<=*c*) the cell located in the row *r* and column *c*. From each cell the rover is able to move to any cell that share a side with the current one.
The rover is currently located at cell (1,<=1) and has to move to the cell (*n*,<=*m*). It will randomly follow some shortest path between these two cells. Each possible way is chosen equiprobably.
The cargo section of the rover contains the battery required to conduct the research. Initially, the battery charge is equal to *s* units of energy.
Some of the cells contain anomaly. Each time the rover gets to the cell with anomaly, the battery looses half of its charge rounded down. Formally, if the charge was equal to *x* before the rover gets to the cell with anomaly, the charge will change to .
While the rover picks a random shortest path to proceed, compute the expected value of the battery charge after it reaches cell (*n*,<=*m*). If the cells (1,<=1) and (*n*,<=*m*) contain anomaly, they also affect the charge of the battery.
Input Specification:
The first line of the input contains four integers *n*, *m*, *k* and *s* (1<=≤<=*n*,<=*m*<=≤<=100<=000, 0<=≤<=*k*<=≤<=2000, 1<=≤<=*s*<=≤<=1<=000<=000) — the number of rows and columns of the field, the number of cells with anomaly and the initial charge of the battery respectively.
The follow *k* lines containing two integers *r**i* and *c**i* (1<=≤<=*r**i*<=≤<=*n*, 1<=≤<=*c**i*<=≤<=*m*) — coordinates of the cells, containing anomaly. It's guaranteed that each cell appears in this list no more than once.
Output Specification:
The answer can always be represented as an irreducible fraction . Print the only integer *P*·*Q*<=-<=1 modulo 109<=+<=7.
Demo Input:
['3 3 2 11\n2 1\n2 3\n', '4 5 3 17\n1 2\n3 3\n4 1\n', '1 6 2 15\n1 1\n1 5\n']
Demo Output:
['333333342\n', '514285727\n', '4\n']
Note:
In the first sample, the rover picks one of the following six routes:
1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c9ec158c30775d6289140a3854e05168b09af399.png" style="max-width: 100.0%;max-height: 100.0%;"/>, after passing cell (2, 3) charge is equal to 6. 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/fcc490c05d2acb731046a7c4c861f4c9ebff3633.png" style="max-width: 100.0%;max-height: 100.0%;"/>, after passing cell (2, 3) charge is equal to 6. 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/8d5828aadc35714d7a3453c40de81ad186e87ab3.png" style="max-width: 100.0%;max-height: 100.0%;"/>, charge remains unchanged and equals 11. 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3bc680a61ca3712bbbec0eb682f3af16ab7664a2.png" style="max-width: 100.0%;max-height: 100.0%;"/>, after passing cells (2, 1) and (2, 3) charge equals 6 and then 3. 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9a87005ef2b3eb1efc9e486e608fadf3a5b557fa.png" style="max-width: 100.0%;max-height: 100.0%;"/>, after passing cell (2, 1) charge is equal to 6. 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/342ddbff927874c07e9d307d73383aa50f2117b6.png" style="max-width: 100.0%;max-height: 100.0%;"/>, after passing cell (2, 1) charge is equal to 6.
Expected value of the battery charge is calculated by the following formula:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7ddca3ad80b71df649b7beb01944a4ad8f863265.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
Thus *P* = 19, and *Q* = 3.
3<sup class="upper-index"> - 1</sup> modulo 10<sup class="upper-index">9</sup> + 7 equals 333333336.
19·333333336 = 333333342 (*mod* 10<sup class="upper-index">9</sup> + 7) | ```python
print("_RANDOM_GUESS_1689416105.399758")# 1689416105.3997796
``` | 0 | |
102 | B | Sum of Digits | PROGRAMMING | 1,000 | [
"implementation"
] | B. Sum of Digits | 2 | 265 | Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit? | The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes. | Print the number of times a number can be replaced by the sum of its digits until it only contains one digit. | [
"0\n",
"10\n",
"991\n"
] | [
"0\n",
"1\n",
"3\n"
] | In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit. | 1,000 | [
{
"input": "0",
"output": "0"
},
{
"input": "10",
"output": "1"
},
{
"input": "991",
"output": "3"
},
{
"input": "99",
"output": "2"
},
{
"input": "100",
"output": "1"
},
{
"input": "123456789",
"output": "2"
},
{
"input": "32",
"output": "1"
},
{
"input": "86",
"output": "2"
},
{
"input": "2",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "34",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "28",
"output": "2"
},
{
"input": "23",
"output": "1"
},
{
"input": "57",
"output": "2"
},
{
"input": "29",
"output": "2"
},
{
"input": "353869285",
"output": "3"
},
{
"input": "549548646",
"output": "2"
},
{
"input": "858893424",
"output": "2"
},
{
"input": "644818852",
"output": "3"
},
{
"input": "360322525",
"output": "3"
},
{
"input": "122937520",
"output": "2"
},
{
"input": "288403032",
"output": "2"
},
{
"input": "677257481",
"output": "3"
},
{
"input": "58059561",
"output": "3"
},
{
"input": "211288847",
"output": "2"
},
{
"input": "339900034079539584",
"output": "2"
},
{
"input": "784084029310269952",
"output": "3"
},
{
"input": "774730557264864000",
"output": "3"
},
{
"input": "478233807148352256",
"output": "3"
},
{
"input": "165646874056812544",
"output": "3"
},
{
"input": "477533739511673792",
"output": "3"
},
{
"input": "660119944377002240",
"output": "3"
},
{
"input": "448375218506078080",
"output": "3"
},
{
"input": "77213334785795616",
"output": "3"
},
{
"input": "165767221702271872",
"output": "3"
},
{
"input": "1",
"output": "0"
},
{
"input": "5",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561",
"output": "3"
},
{
"input": "9",
"output": "0"
}
] | 1,621,346,396 | 2,147,483,647 | Python 3 | OK | TESTS | 51 | 1,870 | 0 | # Problem Statement - https://codeforces.com/problemset/problem/102/B
n = int(input())
c = 0
while n > 9:
n = sum(map(int, str(n)))
c += 1
if n == 0:
print(0)
else:
print(c) | Title: Sum of Digits
Time Limit: 2 seconds
Memory Limit: 265 megabytes
Problem Description:
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
Input Specification:
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Output Specification:
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
Demo Input:
['0\n', '10\n', '991\n']
Demo Output:
['0\n', '1\n', '3\n']
Note:
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit. | ```python
# Problem Statement - https://codeforces.com/problemset/problem/102/B
n = int(input())
c = 0
while n > 9:
n = sum(map(int, str(n)))
c += 1
if n == 0:
print(0)
else:
print(c)
``` | 3.5325 |
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"
}
] | 1,694,068,315 | 2,147,483,647 | Python 3 | OK | TESTS | 52 | 296 | 8,499,200 | n,m,k=map(int,input().split())
matrix=[[0]*(m+2) for _ in range(n+2)]
#method2:创建矩阵
#import numpy as np
#matrix=np.zeros((n,m))
for i in range(k):
row,column=map(int,input().split())
matrix[row][column]+=1
a=matrix[row-1][column-1]!=0 and matrix[row][column-1]!=0 and matrix[row-1][column]!=0
b=matrix[row+1][column-1]!=0 and matrix[row][column-1]!=0 and matrix[row+1][column]!=0
c=matrix[row-1][column+1]!=0 and matrix[row-1][column]!=0 and matrix[row][column+1]!=0
d=matrix[row+1][column+1]!=0 and matrix[row+1][column]!=0 and matrix[row][column+1]!=0
if a or b or c or d:
print(i+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())
matrix=[[0]*(m+2) for _ in range(n+2)]
#method2:创建矩阵
#import numpy as np
#matrix=np.zeros((n,m))
for i in range(k):
row,column=map(int,input().split())
matrix[row][column]+=1
a=matrix[row-1][column-1]!=0 and matrix[row][column-1]!=0 and matrix[row-1][column]!=0
b=matrix[row+1][column-1]!=0 and matrix[row][column-1]!=0 and matrix[row+1][column]!=0
c=matrix[row-1][column+1]!=0 and matrix[row-1][column]!=0 and matrix[row][column+1]!=0
d=matrix[row+1][column+1]!=0 and matrix[row+1][column]!=0 and matrix[row][column+1]!=0
if a or b or c or d:
print(i+1)
break
else:
print('0')
``` | 3 | |
501 | B | Misha and Changing Handles | PROGRAMMING | 1,100 | [
"data structures",
"dsu",
"strings"
] | null | null | Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that. | The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone. | In the first line output the integer *n* — the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description. | [
"5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n"
] | [
"3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n"
] | none | 500 | [
{
"input": "5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov",
"output": "3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123"
},
{
"input": "1\nMisha Vasya",
"output": "1\nMisha Vasya"
},
{
"input": "10\na b\nb c\nc d\nd e\ne f\nf g\ng h\nh i\ni j\nj k",
"output": "1\na k"
},
{
"input": "5\n123abc abc123\nabc123 a1b2c3\na1b2c3 1A2B3C\n1 2\n2 Misha",
"output": "2\n123abc 1A2B3C\n1 Misha"
},
{
"input": "8\nM F\nS D\n1 2\nF G\n2 R\nD Q\nQ W\nW e",
"output": "3\nM G\n1 R\nS e"
},
{
"input": "17\nn5WhQ VCczxtxKwFio5U\nVCczxtxKwFio5U 1WMVGA17cd1LRcp4r\n1WMVGA17cd1LRcp4r SJl\nSJl D8bPUoIft5v1\nNAvvUgunbPZNCL9ZY2 jnLkarKYsotz\nD8bPUoIft5v1 DnDkHi7\njnLkarKYsotz GfjX109HSQ81gFEBJc\nGfjX109HSQ81gFEBJc kBJ0zrH78mveJ\nkBJ0zrH78mveJ 9DrAypYW\nDnDkHi7 3Wkho2PglMDaFQw\n3Wkho2PglMDaFQw pOqW\n9DrAypYW G3y0cXXGsWAh\npOqW yr1Ec\nG3y0cXXGsWAh HrmWWg5u4Hsy\nyr1Ec GkFeivXjQ01\nGkFeivXjQ01 mSsWgbCCZcotV4goiA\nHrmWWg5u4Hsy zkCmEV",
"output": "2\nn5WhQ mSsWgbCCZcotV4goiA\nNAvvUgunbPZNCL9ZY2 zkCmEV"
},
{
"input": "10\nH1nauWCJOImtVqXk gWPMQ9DHv5CtkYp9lwm9\nSEj 2knOMLyzr\n0v69ijnAc S7d7zGTjmlku01Gv\n2knOMLyzr otGmEd\nacwr3TfMV7oCIp RUSVFa9TIWlLsd7SB\nS7d7zGTjmlku01Gv Gd6ZufVmQnBpi\nS1 WOJLpk\nWOJLpk Gu\nRUSVFa9TIWlLsd7SB RFawatGnbVB\notGmEd OTB1zKiOI",
"output": "5\n0v69ijnAc Gd6ZufVmQnBpi\nS1 Gu\nSEj OTB1zKiOI\nacwr3TfMV7oCIp RFawatGnbVB\nH1nauWCJOImtVqXk gWPMQ9DHv5CtkYp9lwm9"
},
{
"input": "14\nTPdoztSZROpjZe z6F8bYFvnER4V5SP0n\n8Aa3PQY3hzHZTPEUz fhrZZPJ3iUS\nm9p888KaZAoQaO KNmdRSAlUVn8zXOM0\nAO s1VGWTCbHzM\ni 4F\nfhrZZPJ3iUS j0OVZQF6MvNcKN9xDZFJ\nDnlkXtaKNlYEI2ApBuwu DMA9i8ScKRxwhe72a3\nj0OVZQF6MvNcKN9xDZFJ DzjmeNqN0H4Teq0Awr\n4F wJcdxt1kwqfDeJ\nqxXlsa5t RHCL1K6aUyns\nr6WYbDaXt hEHw\nJ0Usg DKdKMFJ6tK8XA\nz6F8bYFvnER4V5SP0n 0alJ\nMijh2O6 qic8kXWuR6",
"output": "10\nTPdoztSZROpjZe 0alJ\nJ0Usg DKdKMFJ6tK8XA\nDnlkXtaKNlYEI2ApBuwu DMA9i8ScKRxwhe72a3\n8Aa3PQY3hzHZTPEUz DzjmeNqN0H4Teq0Awr\nm9p888KaZAoQaO KNmdRSAlUVn8zXOM0\nqxXlsa5t RHCL1K6aUyns\nr6WYbDaXt hEHw\nMijh2O6 qic8kXWuR6\nAO s1VGWTCbHzM\ni wJcdxt1kwqfDeJ"
},
{
"input": "14\nHAXRxayyf1Dj1F0mT hjR4A8IQMb0nyBtqG\nWNuMJa5Jg05qkqZOrL noNkWXrSidHGwxgbQ\nmOitVy6W52s0FENMz6 oLUkLNfojssvLvb1t\nhjR4A8IQMb0nyBtqG oA7uBFu4Oo\noA7uBFu4Oo M450\nM450 LXEzO4\noLUkLNfojssvLvb1t YG5\nnoNkWXrSidHGwxgbQ L\nL YBWzu4W\nYBWzu4W ML\nML scVZE9m8JnH\nLXEzO4 Ne0oBPY0Iy\nscVZE9m8JnH GXhznv\nYG5 UY08abilYF1LaXj49hQ",
"output": "3\nWNuMJa5Jg05qkqZOrL GXhznv\nHAXRxayyf1Dj1F0mT Ne0oBPY0Iy\nmOitVy6W52s0FENMz6 UY08abilYF1LaXj49hQ"
}
] | 1,544,014,654 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 124 | 409,600 | import sys
data = sys.stdin.readlines()
archivo = [line.split() for line in data]
cambios = int(archivo[0][0])
dic = {}
contador=0
lista=[]
for i in range(cambios):
dic[archivo[i+1][0]] = archivo[i+1][1]
for key, value in dic.items():
viejo = key
nuevo = value
eliminar = nuevo
while True:
if nuevo in dic.keys():
eliminar = nuevo
nuevo = dic.get(nuevo)
dic[eliminar] = 0
else:
if eliminar != 0 and nuevo != 0:
cambio = viejo + " " + nuevo
lista.append(cambio)
break
print(len(lista))
for i in lista:
print(i)
| Title: Misha and Changing Handles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input Specification:
The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone.
Output Specification:
In the first line output the integer *n* — the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Demo Input:
['5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n']
Demo Output:
['3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n']
Note:
none | ```python
import sys
data = sys.stdin.readlines()
archivo = [line.split() for line in data]
cambios = int(archivo[0][0])
dic = {}
contador=0
lista=[]
for i in range(cambios):
dic[archivo[i+1][0]] = archivo[i+1][1]
for key, value in dic.items():
viejo = key
nuevo = value
eliminar = nuevo
while True:
if nuevo in dic.keys():
eliminar = nuevo
nuevo = dic.get(nuevo)
dic[eliminar] = 0
else:
if eliminar != 0 and nuevo != 0:
cambio = viejo + " " + nuevo
lista.append(cambio)
break
print(len(lista))
for i in lista:
print(i)
``` | 3 | |
570 | B | Simple Game | PROGRAMMING | 1,300 | [
"constructive algorithms",
"games",
"greedy",
"implementation",
"math"
] | null | null | One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to *n*. Let's assume that Misha chose number *m*, and Andrew chose number *a*.
Then, by using a random generator they choose a random integer *c* in the range between 1 and *n* (any integer from 1 to *n* is chosen with the same probability), after which the winner is the player, whose number was closer to *c*. The boys agreed that if *m* and *a* are located on the same distance from *c*, Misha wins.
Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number *n*. You need to determine which value of *a* Andrew must choose, so that the probability of his victory is the highest possible.
More formally, you need to find such integer *a* (1<=≤<=*a*<=≤<=*n*), that the probability that is maximal, where *c* is the equiprobably chosen integer from 1 to *n* (inclusive). | The first line contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the range of numbers in the game, and the number selected by Misha respectively. | Print a single number — such value *a*, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them. | [
"3 1\n",
"4 3\n"
] | [
"2",
"2"
] | In the first sample test: Andrew wins if *c* is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses *a* = 3, the probability of winning will be 1 / 3. If *a* = 1, the probability of winning is 0.
In the second sample test: Andrew wins if *c* is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of *a* the probability of winning is less. | 1,000 | [
{
"input": "3 1",
"output": "2"
},
{
"input": "4 3",
"output": "2"
},
{
"input": "5 5",
"output": "4"
},
{
"input": "10 5",
"output": "6"
},
{
"input": "20 13",
"output": "12"
},
{
"input": "51 1",
"output": "2"
},
{
"input": "100 50",
"output": "51"
},
{
"input": "100 51",
"output": "50"
},
{
"input": "100 49",
"output": "50"
},
{
"input": "1000000000 1000000000",
"output": "999999999"
},
{
"input": "1000000000 1",
"output": "2"
},
{
"input": "1000000000 100000000",
"output": "100000001"
},
{
"input": "1000000000 500000000",
"output": "500000001"
},
{
"input": "1000000000 123124",
"output": "123125"
},
{
"input": "12412523 125123",
"output": "125124"
},
{
"input": "54645723 432423",
"output": "432424"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "262833325 131416663",
"output": "131416662"
},
{
"input": "477667530 238833766",
"output": "238833765"
},
{
"input": "692501734 346250868",
"output": "346250867"
},
{
"input": "907335939 453667970",
"output": "453667969"
},
{
"input": "746085224 373042613",
"output": "373042612"
},
{
"input": "189520699 94760350",
"output": "94760349"
},
{
"input": "404354904 202177453",
"output": "202177452"
},
{
"input": "619189108 309594555",
"output": "309594554"
},
{
"input": "81813292 40906647",
"output": "40906646"
},
{
"input": "296647497 148323750",
"output": "148323749"
},
{
"input": "511481701 255740851",
"output": "255740850"
},
{
"input": "726315905 363157953",
"output": "363157952"
},
{
"input": "496110970 201868357",
"output": "201868358"
},
{
"input": "710945175 173165570",
"output": "173165571"
},
{
"input": "925779379 720443954",
"output": "720443953"
},
{
"input": "140613583 93171580",
"output": "93171579"
},
{
"input": "355447788 85890184",
"output": "85890185"
},
{
"input": "570281992 291648263",
"output": "291648262"
},
{
"input": "541904957 459371829",
"output": "459371828"
},
{
"input": "756739161 125332525",
"output": "125332526"
},
{
"input": "971573366 216791157",
"output": "216791158"
},
{
"input": "186407570 160453970",
"output": "160453969"
},
{
"input": "401241775 170032078",
"output": "170032079"
},
{
"input": "616075979 207073797",
"output": "207073798"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "3 1",
"output": "2"
},
{
"input": "3 2",
"output": "1"
},
{
"input": "3 3",
"output": "2"
},
{
"input": "4 1",
"output": "2"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "4 3",
"output": "2"
},
{
"input": "4 4",
"output": "3"
},
{
"input": "5 1",
"output": "2"
},
{
"input": "5 2",
"output": "3"
},
{
"input": "5 3",
"output": "2"
},
{
"input": "5 4",
"output": "3"
},
{
"input": "5 5",
"output": "4"
},
{
"input": "3 2",
"output": "1"
},
{
"input": "7 4",
"output": "3"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "7 3",
"output": "4"
}
] | 1,563,368,149 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 93 | 0 | n,m=map(int, input().split())
if abs(m-n)>=2:
print(n-1)
else:
print(min(m,n)-1) | Title: Simple Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to *n*. Let's assume that Misha chose number *m*, and Andrew chose number *a*.
Then, by using a random generator they choose a random integer *c* in the range between 1 and *n* (any integer from 1 to *n* is chosen with the same probability), after which the winner is the player, whose number was closer to *c*. The boys agreed that if *m* and *a* are located on the same distance from *c*, Misha wins.
Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number *n*. You need to determine which value of *a* Andrew must choose, so that the probability of his victory is the highest possible.
More formally, you need to find such integer *a* (1<=≤<=*a*<=≤<=*n*), that the probability that is maximal, where *c* is the equiprobably chosen integer from 1 to *n* (inclusive).
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the range of numbers in the game, and the number selected by Misha respectively.
Output Specification:
Print a single number — such value *a*, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them.
Demo Input:
['3 1\n', '4 3\n']
Demo Output:
['2', '2']
Note:
In the first sample test: Andrew wins if *c* is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses *a* = 3, the probability of winning will be 1 / 3. If *a* = 1, the probability of winning is 0.
In the second sample test: Andrew wins if *c* is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of *a* the probability of winning is less. | ```python
n,m=map(int, input().split())
if abs(m-n)>=2:
print(n-1)
else:
print(min(m,n)-1)
``` | 0 | |
242 | B | Big Segment | PROGRAMMING | 1,100 | [
"implementation",
"sortings"
] | null | null | A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment.
It is guaranteed that no two segments coincide. | Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input. | [
"3\n1 1\n2 2\n3 3\n",
"6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n"
] | [
"-1\n",
"3\n"
] | none | 1,000 | [
{
"input": "3\n1 1\n2 2\n3 3",
"output": "-1"
},
{
"input": "6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10",
"output": "3"
},
{
"input": "4\n1 5\n2 2\n2 4\n2 5",
"output": "1"
},
{
"input": "5\n3 3\n1 3\n2 2\n2 3\n1 2",
"output": "2"
},
{
"input": "7\n7 7\n8 8\n3 7\n1 6\n1 7\n4 7\n2 8",
"output": "-1"
},
{
"input": "3\n2 5\n3 4\n2 3",
"output": "1"
},
{
"input": "16\n15 15\n8 12\n6 9\n15 16\n8 14\n3 12\n7 19\n9 13\n5 16\n9 17\n10 15\n9 14\n9 9\n18 19\n5 15\n6 19",
"output": "-1"
},
{
"input": "9\n1 10\n7 8\n6 7\n1 4\n5 9\n2 8\n3 10\n1 1\n2 3",
"output": "1"
},
{
"input": "1\n1 100000",
"output": "1"
},
{
"input": "6\n2 2\n3 3\n3 5\n4 5\n1 1\n1 5",
"output": "6"
},
{
"input": "33\n2 18\n4 14\n2 16\n10 12\n4 6\n9 17\n2 8\n4 12\n8 20\n1 10\n11 14\n11 17\n8 15\n3 16\n3 4\n6 9\n6 19\n4 17\n17 19\n6 16\n3 12\n1 7\n6 20\n8 16\n12 19\n1 3\n12 18\n6 11\n7 20\n16 18\n4 15\n3 15\n15 19",
"output": "-1"
},
{
"input": "34\n3 8\n5 9\n2 9\n1 4\n3 7\n3 3\n8 9\n6 10\n4 7\n6 7\n5 8\n5 10\n1 5\n8 8\n2 5\n3 5\n7 7\n2 8\n4 5\n1 1\n7 9\n5 6\n2 3\n1 2\n2 4\n8 10\n7 8\n1 3\n4 8\n9 10\n1 7\n10 10\n2 2\n1 8",
"output": "-1"
},
{
"input": "55\n3 4\n6 8\n9 10\n3 9\n9 9\n2 5\n4 8\n3 8\n8 10\n1 1\n4 9\n10 10\n6 6\n8 8\n1 8\n5 5\n4 5\n5 9\n2 2\n3 10\n4 6\n3 6\n1 6\n1 7\n6 10\n2 6\n3 7\n2 4\n4 4\n5 10\n1 4\n2 9\n1 3\n7 9\n7 8\n1 9\n1 10\n2 8\n8 9\n6 7\n1 2\n6 9\n7 7\n4 7\n3 3\n2 7\n4 10\n7 10\n2 3\n2 10\n5 7\n3 5\n5 8\n1 5\n5 6",
"output": "37"
},
{
"input": "1\n999999999 1000000000",
"output": "1"
},
{
"input": "3\n1 20\n2 22\n3 18",
"output": "-1"
},
{
"input": "1\n1000000000 1000000000",
"output": "1"
},
{
"input": "2\n100001 100008\n100005 100006",
"output": "1"
},
{
"input": "1\n1000000 10000000",
"output": "1"
},
{
"input": "3\n3 6\n2 4\n1 5",
"output": "-1"
},
{
"input": "2\n3 5\n1 2",
"output": "-1"
}
] | 1,540,580,206 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 154 | 0 | def solve(lines):
#just see if min, max of start, end is together.
minidx = set()
maxidx = set()
mincand = 0
maxcand = 0
for i in range(1, len(lines)):
if lines[i][0] < lines[mincand][0]:
minidx = set()
minidx.add(i)
mincand = i
elif lines[i][0] <= lines[mincand][0]:
minidx.add(i)
if lines[i][1] > lines[maxcand][1]:
maxidx = set()
maxidx.add(i)
maxcand = i
elif lines[i][1] >= lines[maxcand][1]:
maxidx.add(i)
res = list(minidx.intersection(maxidx))
if res:
return res[0] + 1
return -1
n = input()
lines = []
for i in range(n):
nums = map(int, input().split())
lines.append(nums)
print (solve(lines))
| Title: Big Segment
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment.
It is guaranteed that no two segments coincide.
Output Specification:
Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n']
Demo Output:
['-1\n', '3\n']
Note:
none | ```python
def solve(lines):
#just see if min, max of start, end is together.
minidx = set()
maxidx = set()
mincand = 0
maxcand = 0
for i in range(1, len(lines)):
if lines[i][0] < lines[mincand][0]:
minidx = set()
minidx.add(i)
mincand = i
elif lines[i][0] <= lines[mincand][0]:
minidx.add(i)
if lines[i][1] > lines[maxcand][1]:
maxidx = set()
maxidx.add(i)
maxcand = i
elif lines[i][1] >= lines[maxcand][1]:
maxidx.add(i)
res = list(minidx.intersection(maxidx))
if res:
return res[0] + 1
return -1
n = input()
lines = []
for i in range(n):
nums = map(int, input().split())
lines.append(nums)
print (solve(lines))
``` | -1 | |
558 | E | A Simple Task | PROGRAMMING | 2,300 | [
"data structures",
"sortings",
"strings"
] | null | null | This task is very simple. Given a string *S* of length *n* and *q* queries each query is on the format *i* *j* *k* which means sort the substring consisting of the characters from *i* to *j* in non-decreasing order if *k*<==<=1 or in non-increasing order if *k*<==<=0.
Output the final string after applying the queries. | The first line will contain two integers *n*,<=*q* (1<=≤<=*n*<=≤<=105, 0<=≤<=*q*<=≤<=50<=000), the length of the string and the number of queries respectively.
Next line contains a string *S* itself. It contains only lowercase English letters.
Next *q* lines will contain three integers each *i*,<=*j*,<=*k* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*, ). | Output one line, the string *S* after applying the queries. | [
"10 5\nabacdabcda\n7 10 0\n5 8 1\n1 4 0\n3 6 0\n7 10 1\n",
"10 1\nagjucbvdfk\n1 10 1\n"
] | [
"cbcaaaabdd",
"abcdfgjkuv"
] | First sample test explanation:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3ac4e8cc7e335675a4a2b7b4758bfb3865377cea.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a90b5b03cf59288d8861f0142ecbdf6b12f69e5c.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1f482a91a275b6bce07eaed85312eac0cfcc6ccf.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/33b1a4a924f4bd562551ba4e40309f180dbe22e0.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bddc77fd5b02858eb2ff29819cd16a93dbd241e6.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 2,500 | [
{
"input": "10 5\nabacdabcda\n7 10 0\n5 8 1\n1 4 0\n3 6 0\n7 10 1",
"output": "cbcaaaabdd"
},
{
"input": "10 1\nagjucbvdfk\n1 10 1",
"output": "abcdfgjkuv"
},
{
"input": "10 6\nrmaahmdmuo\n1 3 1\n4 6 0\n5 6 1\n7 8 0\n8 10 0\n8 9 1",
"output": "amrmahmoud"
},
{
"input": "10 5\nhbtngdflmj\n1 10 1\n2 9 0\n3 8 1\n4 7 0\n5 6 1",
"output": "bnflhjgmdt"
},
{
"input": "13 0\nokaywearedone",
"output": "okaywearedone"
}
] | 1,558,065,602 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 5,000 | 8,499,200 | d=[int(n) for n in input().split()]
s=[ord(n) for n in input()]
for n in range(d[1]):
z=[int(n) for n in input().split()]
x=s[z[0]-1:z[1]]
y=s[z[1]:]
x.sort()
if z[2]==0:
x=x[::-1]
del s[z[0]-1:]
s+=x
s+=y
x=[]
y=[]
for n in s:
print(chr(n),end='') | Title: A Simple Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This task is very simple. Given a string *S* of length *n* and *q* queries each query is on the format *i* *j* *k* which means sort the substring consisting of the characters from *i* to *j* in non-decreasing order if *k*<==<=1 or in non-increasing order if *k*<==<=0.
Output the final string after applying the queries.
Input Specification:
The first line will contain two integers *n*,<=*q* (1<=≤<=*n*<=≤<=105, 0<=≤<=*q*<=≤<=50<=000), the length of the string and the number of queries respectively.
Next line contains a string *S* itself. It contains only lowercase English letters.
Next *q* lines will contain three integers each *i*,<=*j*,<=*k* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*, ).
Output Specification:
Output one line, the string *S* after applying the queries.
Demo Input:
['10 5\nabacdabcda\n7 10 0\n5 8 1\n1 4 0\n3 6 0\n7 10 1\n', '10 1\nagjucbvdfk\n1 10 1\n']
Demo Output:
['cbcaaaabdd', 'abcdfgjkuv']
Note:
First sample test explanation:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3ac4e8cc7e335675a4a2b7b4758bfb3865377cea.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a90b5b03cf59288d8861f0142ecbdf6b12f69e5c.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1f482a91a275b6bce07eaed85312eac0cfcc6ccf.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/33b1a4a924f4bd562551ba4e40309f180dbe22e0.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bddc77fd5b02858eb2ff29819cd16a93dbd241e6.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
d=[int(n) for n in input().split()]
s=[ord(n) for n in input()]
for n in range(d[1]):
z=[int(n) for n in input().split()]
x=s[z[0]-1:z[1]]
y=s[z[1]:]
x.sort()
if z[2]==0:
x=x[::-1]
del s[z[0]-1:]
s+=x
s+=y
x=[]
y=[]
for n in s:
print(chr(n),end='')
``` | 0 | |
903 | A | Hungry Student Problem | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken.
CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly *x* chunks. Now he wonders whether he can buy exactly this amount of chicken.
Formally, Ivan wants to know if he can choose two non-negative integers *a* and *b* in such a way that *a* small portions and *b* large ones contain exactly *x* chunks.
Help Ivan to answer this question for several values of *x*! | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of testcases.
The *i*-th of the following *n* lines contains one integer *x**i* (1<=≤<=*x**i*<=≤<=100) — the number of chicken chunks Ivan wants to eat. | Print *n* lines, in *i*-th line output YES if Ivan can buy exactly *x**i* chunks. Otherwise, print NO. | [
"2\n6\n5\n"
] | [
"YES\nNO\n"
] | In the first example Ivan can buy two small portions.
In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much. | 0 | [
{
"input": "2\n6\n5",
"output": "YES\nNO"
},
{
"input": "100\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\n19\n20\n21\n22\n23\n24\n25\n26\n27\n28\n29\n30\n31\n32\n33\n34\n35\n36\n37\n38\n39\n40\n41\n42\n43\n44\n45\n46\n47\n48\n49\n50\n51\n52\n53\n54\n55\n56\n57\n58\n59\n60\n61\n62\n63\n64\n65\n66\n67\n68\n69\n70\n71\n72\n73\n74\n75\n76\n77\n78\n79\n80\n81\n82\n83\n84\n85\n86\n87\n88\n89\n90\n91\n92\n93\n94\n95\n96\n97\n98\n99\n100",
"output": "NO\nNO\nYES\nNO\nNO\nYES\nYES\nNO\nYES\nYES\nNO\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\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\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
},
{
"input": "3\n6\n6\n6",
"output": "YES\nYES\nYES"
},
{
"input": "47\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "3\n1\n52\n76",
"output": "NO\nYES\nYES"
},
{
"input": "87\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "YES\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\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\nYES\nYES\nYES\nYES"
},
{
"input": "3\n3\n2\n1",
"output": "YES\nNO\nNO"
},
{
"input": "100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "YES\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\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\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
}
] | 1,580,375,955 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 108 | 0 | for _ in range(int(input())):a=int(input());print("YES" if a%3==0 or a%7==0 or (a-3)%7==0 or (a-7)%3==0 else "NO") | Title: Hungry Student Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken.
CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly *x* chunks. Now he wonders whether he can buy exactly this amount of chicken.
Formally, Ivan wants to know if he can choose two non-negative integers *a* and *b* in such a way that *a* small portions and *b* large ones contain exactly *x* chunks.
Help Ivan to answer this question for several values of *x*!
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of testcases.
The *i*-th of the following *n* lines contains one integer *x**i* (1<=≤<=*x**i*<=≤<=100) — the number of chicken chunks Ivan wants to eat.
Output Specification:
Print *n* lines, in *i*-th line output YES if Ivan can buy exactly *x**i* chunks. Otherwise, print NO.
Demo Input:
['2\n6\n5\n']
Demo Output:
['YES\nNO\n']
Note:
In the first example Ivan can buy two small portions.
In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much. | ```python
for _ in range(int(input())):a=int(input());print("YES" if a%3==0 or a%7==0 or (a-3)%7==0 or (a-7)%3==0 else "NO")
``` | 0 | |
368 | B | Sereja and Suffixes | PROGRAMMING | 1,100 | [
"data structures",
"dp"
] | null | null | Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.?
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements.
Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*). | Print *m* lines — on the *i*-th line print the answer to the number *l**i*. | [
"10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n"
] | [
"6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n"
] | none | 1,000 | [
{
"input": "10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10",
"output": "6\n6\n6\n6\n6\n5\n4\n3\n2\n1"
},
{
"input": "8 3\n8 6 4 3 4 2 4 8\n6\n4\n2",
"output": "3\n4\n5"
},
{
"input": "7 10\n1 3 8 6 2 2 7\n4\n2\n6\n3\n4\n4\n6\n2\n7\n4",
"output": "3\n5\n2\n4\n3\n3\n2\n5\n1\n3"
},
{
"input": "10 2\n2 6 5 7 2 2 3 2 4 8\n1\n2",
"output": "7\n7"
},
{
"input": "7 1\n68346 10956 76708 23018 84063 34833 80407\n1",
"output": "7"
},
{
"input": "2 2\n8 4\n1\n1",
"output": "2\n2"
},
{
"input": "1 5\n5\n1\n1\n1\n1\n1",
"output": "1\n1\n1\n1\n1"
},
{
"input": "4 7\n3 1 4 2\n4\n1\n2\n3\n2\n4\n4",
"output": "1\n4\n3\n2\n3\n1\n1"
},
{
"input": "4 3\n9 1 7 1\n1\n4\n2",
"output": "3\n1\n2"
},
{
"input": "8 3\n9280 6676 2720 6172 8329 10413 3975 1394\n5\n6\n7",
"output": "4\n3\n2"
},
{
"input": "1 1\n1\n1",
"output": "1"
}
] | 1,698,301,972 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 1,000 | 11,980,800 | ip1=input().split()
n,m=int(ip1[0]),int(ip1[1])
nseq=list(map(int,input().split()))
fltr=[1 for i in range(n)]
ans={}
for i in reversed(list(range(n))):
if fltr[i]:
a=nseq[i]
ans[i]=ans.get(i+1,0)+1
for j in reversed(list(range(i))):
if fltr[j]:
if nseq[j]==a:
fltr[j]=0
else:
ans[i]=ans[i+1]
for i in range(m):
print(ans[int(input())-1]) | Title: Sereja and Suffixes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.?
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements.
Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*).
Output Specification:
Print *m* lines — on the *i*-th line print the answer to the number *l**i*.
Demo Input:
['10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n']
Demo Output:
['6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n']
Note:
none | ```python
ip1=input().split()
n,m=int(ip1[0]),int(ip1[1])
nseq=list(map(int,input().split()))
fltr=[1 for i in range(n)]
ans={}
for i in reversed(list(range(n))):
if fltr[i]:
a=nseq[i]
ans[i]=ans.get(i+1,0)+1
for j in reversed(list(range(i))):
if fltr[j]:
if nseq[j]==a:
fltr[j]=0
else:
ans[i]=ans[i+1]
for i in range(m):
print(ans[int(input())-1])
``` | 0 | |
99 | A | Help Far Away Kingdom | PROGRAMMING | 800 | [
"strings"
] | A. Help Far Away Kingdom | 2 | 256 | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Kingdom healers had already had three appointments with the merchants who were asked to sell, say, exactly 0.273549107 beer barrels. To deal with the problem somehow, the King issued an order obliging rounding up all numbers to the closest integer to simplify calculations. Specifically, the order went like this:
- If a number's integer part does not end with digit 9 and its fractional part is strictly less than 0.5, then the rounded up number coincides with the number’s integer part. - If a number's integer part does not end with digit 9 and its fractional part is not less than 0.5, the rounded up number is obtained if we add 1 to the last digit of the number’s integer part.- If the number’s integer part ends with digit 9, to round up the numbers one should go to Vasilisa the Wise. In the whole Kingdom she is the only one who can perform the tricky operation of carrying into the next position.
Merchants found the algorithm very sophisticated and they asked you (the ACMers) to help them. Can you write a program that would perform the rounding according to the King’s order? | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empty set of decimal digits). The number's length does not exceed 1000 characters, including the dot. There are no other characters in the input data. | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | [
"0.0\n",
"1.49\n",
"1.50\n",
"2.71828182845904523536\n",
"3.14159265358979323846\n",
"12345678901234567890.1\n",
"123456789123456789.999\n"
] | [
"0",
"1",
"2",
"3",
"3",
"12345678901234567890",
"GOTO Vasilisa."
] | none | 500 | [
{
"input": "0.0",
"output": "0"
},
{
"input": "1.49",
"output": "1"
},
{
"input": "1.50",
"output": "2"
},
{
"input": "2.71828182845904523536",
"output": "3"
},
{
"input": "3.14159265358979323846",
"output": "3"
},
{
"input": "12345678901234567890.1",
"output": "12345678901234567890"
},
{
"input": "123456789123456789.999",
"output": "GOTO Vasilisa."
},
{
"input": "12345678901234567890.9",
"output": "12345678901234567891"
},
{
"input": "123456789123456788.999",
"output": "123456789123456789"
},
{
"input": "9.000",
"output": "GOTO Vasilisa."
},
{
"input": "0.1",
"output": "0"
},
{
"input": "0.2",
"output": "0"
},
{
"input": "0.3",
"output": "0"
},
{
"input": "0.4",
"output": "0"
},
{
"input": "0.5",
"output": "1"
},
{
"input": "0.6",
"output": "1"
},
{
"input": "0.7",
"output": "1"
},
{
"input": "0.8",
"output": "1"
},
{
"input": "0.9",
"output": "1"
},
{
"input": "1.0",
"output": "1"
},
{
"input": "1.1",
"output": "1"
},
{
"input": "1.2",
"output": "1"
},
{
"input": "1.3",
"output": "1"
},
{
"input": "1.4",
"output": "1"
},
{
"input": "1.5",
"output": "2"
},
{
"input": "1.6",
"output": "2"
},
{
"input": "1.7",
"output": "2"
},
{
"input": "1.8",
"output": "2"
},
{
"input": "1.9",
"output": "2"
},
{
"input": "2.0",
"output": "2"
},
{
"input": "2.1",
"output": "2"
},
{
"input": "2.2",
"output": "2"
},
{
"input": "2.3",
"output": "2"
},
{
"input": "2.4",
"output": "2"
},
{
"input": "2.5",
"output": "3"
},
{
"input": "2.6",
"output": "3"
},
{
"input": "2.7",
"output": "3"
},
{
"input": "2.8",
"output": "3"
},
{
"input": "2.9",
"output": "3"
},
{
"input": "3.0",
"output": "3"
},
{
"input": "3.1",
"output": "3"
},
{
"input": "3.2",
"output": "3"
},
{
"input": "3.3",
"output": "3"
},
{
"input": "3.4",
"output": "3"
},
{
"input": "3.5",
"output": "4"
},
{
"input": "3.6",
"output": "4"
},
{
"input": "3.7",
"output": "4"
},
{
"input": "3.8",
"output": "4"
},
{
"input": "3.9",
"output": "4"
},
{
"input": "4.0",
"output": "4"
},
{
"input": "4.1",
"output": "4"
},
{
"input": "4.2",
"output": "4"
},
{
"input": "4.3",
"output": "4"
},
{
"input": "4.4",
"output": "4"
},
{
"input": "4.5",
"output": "5"
},
{
"input": "4.6",
"output": "5"
},
{
"input": "4.7",
"output": "5"
},
{
"input": "4.8",
"output": "5"
},
{
"input": "4.9",
"output": "5"
},
{
"input": "5.0",
"output": "5"
},
{
"input": "5.1",
"output": "5"
},
{
"input": "5.2",
"output": "5"
},
{
"input": "5.3",
"output": "5"
},
{
"input": "5.4",
"output": "5"
},
{
"input": "5.5",
"output": "6"
},
{
"input": "5.6",
"output": "6"
},
{
"input": "5.7",
"output": "6"
},
{
"input": "5.8",
"output": "6"
},
{
"input": "5.9",
"output": "6"
},
{
"input": "6.0",
"output": "6"
},
{
"input": "6.1",
"output": "6"
},
{
"input": "6.2",
"output": "6"
},
{
"input": "6.3",
"output": "6"
},
{
"input": "6.4",
"output": "6"
},
{
"input": "6.5",
"output": "7"
},
{
"input": "6.6",
"output": "7"
},
{
"input": "6.7",
"output": "7"
},
{
"input": "6.8",
"output": "7"
},
{
"input": "6.9",
"output": "7"
},
{
"input": "7.0",
"output": "7"
},
{
"input": "7.1",
"output": "7"
},
{
"input": "7.2",
"output": "7"
},
{
"input": "7.3",
"output": "7"
},
{
"input": "7.4",
"output": "7"
},
{
"input": "7.5",
"output": "8"
},
{
"input": "7.6",
"output": "8"
},
{
"input": "7.7",
"output": "8"
},
{
"input": "7.8",
"output": "8"
},
{
"input": "7.9",
"output": "8"
},
{
"input": "8.0",
"output": "8"
},
{
"input": "8.1",
"output": "8"
},
{
"input": "8.2",
"output": "8"
},
{
"input": "8.3",
"output": "8"
},
{
"input": "8.4",
"output": "8"
},
{
"input": "8.5",
"output": "9"
},
{
"input": "8.6",
"output": "9"
},
{
"input": "8.7",
"output": "9"
},
{
"input": "8.8",
"output": "9"
},
{
"input": "8.9",
"output": "9"
},
{
"input": "9.0",
"output": "GOTO Vasilisa."
},
{
"input": "9.1",
"output": "GOTO Vasilisa."
},
{
"input": "9.2",
"output": "GOTO Vasilisa."
},
{
"input": "9.3",
"output": "GOTO Vasilisa."
},
{
"input": "9.4",
"output": "GOTO Vasilisa."
},
{
"input": "9.5",
"output": "GOTO Vasilisa."
},
{
"input": "9.6",
"output": "GOTO Vasilisa."
},
{
"input": "9.7",
"output": "GOTO Vasilisa."
},
{
"input": "9.8",
"output": "GOTO Vasilisa."
},
{
"input": "9.9",
"output": "GOTO Vasilisa."
},
{
"input": "609942239104813108618306232517836377583566292129955473517174437591594761209877970062547641606473593416245554763832875919009472288995880898848455284062760160557686724163817329189799336769669146848904803188614226720978399787805489531837751080926098.1664915772983166314490532653577560222779830866949001942720729759794777105570672781798092416748052690224813237139640723361527601154465287615917169132637313918577673651098507390501962",
"output": "609942239104813108618306232517836377583566292129955473517174437591594761209877970062547641606473593416245554763832875919009472288995880898848455284062760160557686724163817329189799336769669146848904803188614226720978399787805489531837751080926098"
},
{
"input": "7002108534951820589946967018226114921984364117669853212254634761258884835434844673935047882480101006606512119541798298905598015607366335061012709906661245805358900665571472645463994925687210711492820804158354236327017974683658305043146543214454877759341394.20211856263503281388748282682120712214711232598021393495443628276945042110862480888110959179019986486690931930108026302665438087068150666835901617457150158918705186964935221768346957536540345814875615118637945520917367155931078965",
"output": "7002108534951820589946967018226114921984364117669853212254634761258884835434844673935047882480101006606512119541798298905598015607366335061012709906661245805358900665571472645463994925687210711492820804158354236327017974683658305043146543214454877759341394"
},
{
"input": "1950583094879039694852660558765931995628486712128191844305265555887022812284005463780616067.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1950583094879039694852660558765931995628486712128191844305265555887022812284005463780616068"
},
{
"input": "718130341896330596635811874410345440628950330.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "718130341896330596635811874410345440628950331"
},
{
"input": "927925904158088313481229162503626281882161630091489367140850985555900173018122871746924067186432044676083646964286435457446768031295712712803570690846298544912543439221596866052681116386179629036945370280722.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "927925904158088313481229162503626281882161630091489367140850985555900173018122871746924067186432044676083646964286435457446768031295712712803570690846298544912543439221596866052681116386179629036945370280723"
},
{
"input": "68289614863244584294178637364598054554769889.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "GOTO Vasilisa."
},
{
"input": "7536521504744364134984603189602839063535643888645969434165019366202558753840519.4999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999",
"output": "GOTO Vasilisa."
},
{
"input": "646188694587964249318078225173.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "646188694587964249318078225174"
},
{
"input": "7017472758334494297677068672222822550374695787171163207025095950296957261530.50",
"output": "7017472758334494297677068672222822550374695787171163207025095950296957261531"
},
{
"input": "259085737066615534998640212505663524594409165063310128108448186246980628179842202905722595400477937071746695941939306735605849342959111887834258250883469840846714848774368.50000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "259085737066615534998640212505663524594409165063310128108448186246980628179842202905722595400477937071746695941939306735605849342959111887834258250883469840846714848774369"
},
{
"input": "682500858233333594535201113441004740771119672961581796618069185960107115823662126812159957094407454522028503739299.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "GOTO Vasilisa."
},
{
"input": "41203422675619090661099806687619.49999999999999999999999999999999999999999999999999999999999999999999999999999999999999999",
"output": "GOTO Vasilisa."
}
] | 1,588,683,613 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 216 | 0 | s = float(input())
if int(s) % 10 == 9:
print('GOTO Vasilisa.')
else:
if s < int(s) + 0.5:
print(int(s))
else:
print(int(s) + 1) | Title: Help Far Away Kingdom
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Kingdom healers had already had three appointments with the merchants who were asked to sell, say, exactly 0.273549107 beer barrels. To deal with the problem somehow, the King issued an order obliging rounding up all numbers to the closest integer to simplify calculations. Specifically, the order went like this:
- If a number's integer part does not end with digit 9 and its fractional part is strictly less than 0.5, then the rounded up number coincides with the number’s integer part. - If a number's integer part does not end with digit 9 and its fractional part is not less than 0.5, the rounded up number is obtained if we add 1 to the last digit of the number’s integer part.- If the number’s integer part ends with digit 9, to round up the numbers one should go to Vasilisa the Wise. In the whole Kingdom she is the only one who can perform the tricky operation of carrying into the next position.
Merchants found the algorithm very sophisticated and they asked you (the ACMers) to help them. Can you write a program that would perform the rounding according to the King’s order?
Input Specification:
The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empty set of decimal digits). The number's length does not exceed 1000 characters, including the dot. There are no other characters in the input data.
Output Specification:
If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes).
Demo Input:
['0.0\n', '1.49\n', '1.50\n', '2.71828182845904523536\n', '3.14159265358979323846\n', '12345678901234567890.1\n', '123456789123456789.999\n']
Demo Output:
['0', '1', '2', '3', '3', '12345678901234567890', 'GOTO Vasilisa.']
Note:
none | ```python
s = float(input())
if int(s) % 10 == 9:
print('GOTO Vasilisa.')
else:
if s < int(s) + 0.5:
print(int(s))
else:
print(int(s) + 1)
``` | 0 |
888 | A | Local Extrema | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | You are given an array *a*. Some element of this array *a**i* is a local minimum iff it is strictly less than both of its neighbours (that is, *a**i*<=<<=*a**i*<=-<=1 and *a**i*<=<<=*a**i*<=+<=1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, *a**i*<=><=*a**i*<=-<=1 and *a**i*<=><=*a**i*<=+<=1). Since *a*1 and *a**n* have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in array *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000) — the elements of array *a*. | Print the number of local extrema in the given array. | [
"3\n1 2 3\n",
"4\n1 5 2 5\n"
] | [
"0\n",
"2\n"
] | none | 0 | [
{
"input": "3\n1 2 3",
"output": "0"
},
{
"input": "4\n1 5 2 5",
"output": "2"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n548",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "3\n3 2 3",
"output": "1"
},
{
"input": "2\n1 1000",
"output": "0"
},
{
"input": "3\n1 2 1",
"output": "1"
}
] | 1,591,726,022 | 2,147,483,647 | PyPy 3 | OK | TESTS | 14 | 140 | 0 | n = int(input())
A = list(map(int, input().split()))
if n <= 2:
print(0)
exit()
ans = 0
for i in range(1, n-1):
if A[i] < A[i-1] and A[i] < A[i+1]:
ans += 1
if A[i] > A[i-1] and A[i] > A[i+1]:
ans += 1
print(ans)
| Title: Local Extrema
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a*. Some element of this array *a**i* is a local minimum iff it is strictly less than both of its neighbours (that is, *a**i*<=<<=*a**i*<=-<=1 and *a**i*<=<<=*a**i*<=+<=1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, *a**i*<=><=*a**i*<=-<=1 and *a**i*<=><=*a**i*<=+<=1). Since *a*1 and *a**n* have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in array *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000) — the elements of array *a*.
Output Specification:
Print the number of local extrema in the given array.
Demo Input:
['3\n1 2 3\n', '4\n1 5 2 5\n']
Demo Output:
['0\n', '2\n']
Note:
none | ```python
n = int(input())
A = list(map(int, input().split()))
if n <= 2:
print(0)
exit()
ans = 0
for i in range(1, n-1):
if A[i] < A[i-1] and A[i] < A[i+1]:
ans += 1
if A[i] > A[i-1] and A[i] > A[i+1]:
ans += 1
print(ans)
``` | 3 | |
935 | A | Fafa and his Company | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly. | The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company. | Print a single integer representing the answer to the problem. | [
"2\n",
"10\n"
] | [
"1\n",
"3\n"
] | In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | 500 | [
{
"input": "2",
"output": "1"
},
{
"input": "10",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "6",
"output": "3"
},
{
"input": "13",
"output": "1"
},
{
"input": "100000",
"output": "35"
},
{
"input": "1024",
"output": "10"
},
{
"input": "99999",
"output": "11"
},
{
"input": "10007",
"output": "1"
},
{
"input": "4096",
"output": "12"
},
{
"input": "65536",
"output": "16"
},
{
"input": "40320",
"output": "95"
},
{
"input": "30030",
"output": "63"
},
{
"input": "161",
"output": "3"
},
{
"input": "1000",
"output": "15"
},
{
"input": "10000",
"output": "24"
},
{
"input": "777",
"output": "7"
},
{
"input": "121",
"output": "2"
},
{
"input": "25",
"output": "2"
},
{
"input": "40000",
"output": "34"
},
{
"input": "99990",
"output": "47"
},
{
"input": "98765",
"output": "3"
},
{
"input": "56789",
"output": "3"
},
{
"input": "13579",
"output": "3"
},
{
"input": "97531",
"output": "3"
},
{
"input": "12345",
"output": "7"
},
{
"input": "54321",
"output": "7"
},
{
"input": "83160",
"output": "127"
},
{
"input": "9",
"output": "2"
},
{
"input": "21",
"output": "3"
},
{
"input": "11",
"output": "1"
},
{
"input": "15",
"output": "3"
}
] | 1,682,697,331 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 62 | 0 | num = int(input())
num_count = 0
for n in range(1, num):
if num % n == 0:
num_count += 1
print(num_count)
| Title: Fafa and his Company
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
Input Specification:
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
Output Specification:
Print a single integer representing the answer to the problem.
Demo Input:
['2\n', '10\n']
Demo Output:
['1\n', '3\n']
Note:
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | ```python
num = int(input())
num_count = 0
for n in range(1, num):
if num % n == 0:
num_count += 1
print(num_count)
``` | 3 | |
63 | A | Sinking Ship | PROGRAMMING | 900 | [
"implementation",
"sortings",
"strings"
] | A. Sinking Ship | 2 | 256 | The ship crashed into a reef and is sinking. Now the entire crew must be evacuated. All *n* crew members have already lined up in a row (for convenience let's label them all from left to right with positive integers from 1 to *n*) and await further instructions. However, one should evacuate the crew properly, in a strict order. Specifically:
The first crew members to leave the ship are rats. Then women and children (both groups have the same priority) leave the ship. After that all men are evacuated from the ship. The captain leaves the sinking ship last.
If we cannot determine exactly who should leave the ship first for any two members of the crew by the rules from the previous paragraph, then the one who stands to the left in the line leaves the ship first (or in other words, the one whose number in the line is less).
For each crew member we know his status as a crew member, and also his name. All crew members have different names. Determine the order in which to evacuate the crew. | The first line contains an integer *n*, which is the number of people in the crew (1<=≤<=*n*<=≤<=100). Then follow *n* lines. The *i*-th of those lines contains two words — the name of the crew member who is *i*-th in line, and his status on the ship. The words are separated by exactly one space. There are no other spaces in the line. The names consist of Latin letters, the first letter is uppercase, the rest are lowercase. The length of any name is from 1 to 10 characters. The status can have the following values: rat for a rat, woman for a woman, child for a child, man for a man, captain for the captain. The crew contains exactly one captain. | Print *n* lines. The *i*-th of them should contain the name of the crew member who must be the *i*-th one to leave the ship. | [
"6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman\n"
] | [
"Teddy\nAlice\nBob\nJulia\nCharlie\nJack\n"
] | none | 500 | [
{
"input": "6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman",
"output": "Teddy\nAlice\nBob\nJulia\nCharlie\nJack"
},
{
"input": "1\nA captain",
"output": "A"
},
{
"input": "1\nAbcdefjhij captain",
"output": "Abcdefjhij"
},
{
"input": "5\nA captain\nB man\nD woman\nC child\nE rat",
"output": "E\nD\nC\nB\nA"
},
{
"input": "10\nCap captain\nD child\nC woman\nA woman\nE child\nMan man\nB child\nF woman\nRat rat\nRatt rat",
"output": "Rat\nRatt\nD\nC\nA\nE\nB\nF\nMan\nCap"
},
{
"input": "5\nJoyxnkypf captain\nDxssgr woman\nKeojmnpd rat\nGdv man\nHnw man",
"output": "Keojmnpd\nDxssgr\nGdv\nHnw\nJoyxnkypf"
},
{
"input": "11\nJue rat\nWyglbyphk rat\nGjlgu child\nGi man\nAttx rat\nTheorpkgx man\nYm rat\nX child\nB captain\nEnualf rat\nKktsgyuyv woman",
"output": "Jue\nWyglbyphk\nAttx\nYm\nEnualf\nGjlgu\nX\nKktsgyuyv\nGi\nTheorpkgx\nB"
},
{
"input": "22\nWswwcvvm woman\nBtmfats rat\nI rat\nOcmtsnwx man\nUrcqv rat\nYghnogt woman\nWtyfc man\nWqle child\nUjfrelpu rat\nDstixj man\nAhksnio woman\nKhkvaap woman\nSjppvwm rat\nEgdmsv rat\nDank rat\nNquicjnw rat\nLh captain\nTdyaqaqln rat\nQtj rat\nTfgwijvq rat\nNbiso child\nNqthvbf woman",
"output": "Btmfats\nI\nUrcqv\nUjfrelpu\nSjppvwm\nEgdmsv\nDank\nNquicjnw\nTdyaqaqln\nQtj\nTfgwijvq\nWswwcvvm\nYghnogt\nWqle\nAhksnio\nKhkvaap\nNbiso\nNqthvbf\nOcmtsnwx\nWtyfc\nDstixj\nLh"
},
{
"input": "36\nKqxmtwmsf child\nIze woman\nDlpr child\nK woman\nF captain\nRjwfeuhba rat\nBbv rat\nS rat\nMnmg woman\nSmzyx woman\nSr man\nQmhroracn rat\nSoqpuqock rat\nPibdq man\nIlrkrptx rat\nZaecfyqka man\nMmersfs child\nVvvocqi man\nHjeqxvq rat\nMpmb woman\nWmgu woman\nCerelmhoxi child\nA man\nDylv man\nXrdgmmtcpq woman\nXj woman\nCeh child\nOfccguo man\nNevr man\nAknkwdkq child\nBmo man\nLqo woman\nVej child\nT child\nRlijc man\nNsdtpdg woman",
"output": "Rjwfeuhba\nBbv\nS\nQmhroracn\nSoqpuqock\nIlrkrptx\nHjeqxvq\nKqxmtwmsf\nIze\nDlpr\nK\nMnmg\nSmzyx\nMmersfs\nMpmb\nWmgu\nCerelmhoxi\nXrdgmmtcpq\nXj\nCeh\nAknkwdkq\nLqo\nVej\nT\nNsdtpdg\nSr\nPibdq\nZaecfyqka\nVvvocqi\nA\nDylv\nOfccguo\nNevr\nBmo\nRlijc\nF"
},
{
"input": "2\nA captain\nB man",
"output": "B\nA"
},
{
"input": "2\nB woman\nA captain",
"output": "B\nA"
},
{
"input": "2\nX child\nY captain",
"output": "X\nY"
},
{
"input": "2\nX captain\nY rat",
"output": "Y\nX"
},
{
"input": "6\nAbdelrhman captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman",
"output": "Teddy\nAlice\nBob\nJulia\nCharlie\nAbdelrhman"
},
{
"input": "4\nAratos captain\nMind child\nAman woman\nCarrot man",
"output": "Mind\nAman\nCarrot\nAratos"
}
] | 1,614,265,301 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 124 | 409,600 | from collections import defaultdict
d=defaultdict(list)
for _ in range(int(input())):
a,b=input().split()
if b=="rat":
b=0
elif b=="woman" or b=="child":
b=1
elif b=="man":
b=2
else:b=3
d[b].append(a)
while d:
mi=min(d)
print(*d[mi],sep="\n")
del d[mi]
| Title: Sinking Ship
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The ship crashed into a reef and is sinking. Now the entire crew must be evacuated. All *n* crew members have already lined up in a row (for convenience let's label them all from left to right with positive integers from 1 to *n*) and await further instructions. However, one should evacuate the crew properly, in a strict order. Specifically:
The first crew members to leave the ship are rats. Then women and children (both groups have the same priority) leave the ship. After that all men are evacuated from the ship. The captain leaves the sinking ship last.
If we cannot determine exactly who should leave the ship first for any two members of the crew by the rules from the previous paragraph, then the one who stands to the left in the line leaves the ship first (or in other words, the one whose number in the line is less).
For each crew member we know his status as a crew member, and also his name. All crew members have different names. Determine the order in which to evacuate the crew.
Input Specification:
The first line contains an integer *n*, which is the number of people in the crew (1<=≤<=*n*<=≤<=100). Then follow *n* lines. The *i*-th of those lines contains two words — the name of the crew member who is *i*-th in line, and his status on the ship. The words are separated by exactly one space. There are no other spaces in the line. The names consist of Latin letters, the first letter is uppercase, the rest are lowercase. The length of any name is from 1 to 10 characters. The status can have the following values: rat for a rat, woman for a woman, child for a child, man for a man, captain for the captain. The crew contains exactly one captain.
Output Specification:
Print *n* lines. The *i*-th of them should contain the name of the crew member who must be the *i*-th one to leave the ship.
Demo Input:
['6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman\n']
Demo Output:
['Teddy\nAlice\nBob\nJulia\nCharlie\nJack\n']
Note:
none | ```python
from collections import defaultdict
d=defaultdict(list)
for _ in range(int(input())):
a,b=input().split()
if b=="rat":
b=0
elif b=="woman" or b=="child":
b=1
elif b=="man":
b=2
else:b=3
d[b].append(a)
while d:
mi=min(d)
print(*d[mi],sep="\n")
del d[mi]
``` | 3.968237 |
275 | A | Lights Out | PROGRAMMING | 900 | [
"implementation"
] | null | null | Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light. | The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed. | Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0". | [
"1 0 0\n0 0 0\n0 0 1\n",
"1 0 1\n8 8 8\n2 0 3\n"
] | [
"001\n010\n100\n",
"010\n011\n100\n"
] | none | 500 | [
{
"input": "1 0 0\n0 0 0\n0 0 1",
"output": "001\n010\n100"
},
{
"input": "1 0 1\n8 8 8\n2 0 3",
"output": "010\n011\n100"
},
{
"input": "13 85 77\n25 50 45\n65 79 9",
"output": "000\n010\n000"
},
{
"input": "96 95 5\n8 84 74\n67 31 61",
"output": "011\n011\n101"
},
{
"input": "24 54 37\n60 63 6\n1 84 26",
"output": "110\n101\n011"
},
{
"input": "23 10 40\n15 6 40\n92 80 77",
"output": "101\n100\n000"
},
{
"input": "62 74 80\n95 74 93\n2 47 95",
"output": "010\n001\n110"
},
{
"input": "80 83 48\n26 0 66\n47 76 37",
"output": "000\n000\n010"
},
{
"input": "32 15 65\n7 54 36\n5 51 3",
"output": "111\n101\n001"
},
{
"input": "22 97 12\n71 8 24\n100 21 64",
"output": "100\n001\n100"
},
{
"input": "46 37 13\n87 0 50\n90 8 55",
"output": "111\n011\n000"
},
{
"input": "57 43 58\n20 82 83\n66 16 52",
"output": "111\n010\n110"
},
{
"input": "45 56 93\n47 51 59\n18 51 63",
"output": "101\n011\n100"
},
{
"input": "47 66 67\n14 1 37\n27 81 69",
"output": "001\n001\n110"
},
{
"input": "26 69 69\n85 18 23\n14 22 74",
"output": "110\n001\n010"
},
{
"input": "10 70 65\n94 27 25\n74 66 30",
"output": "111\n010\n100"
},
{
"input": "97 1 74\n15 99 1\n88 68 86",
"output": "001\n011\n000"
},
{
"input": "36 48 42\n45 41 66\n26 64 1",
"output": "001\n111\n010"
},
{
"input": "52 81 97\n29 77 71\n66 11 2",
"output": "100\n100\n111"
},
{
"input": "18 66 33\n19 49 49\n48 46 26",
"output": "011\n100\n000"
},
{
"input": "68 79 52\n51 39 100\n29 14 26",
"output": "110\n000\n111"
},
{
"input": "91 69 77\n91 26 64\n91 88 57",
"output": "001\n011\n110"
},
{
"input": "16 69 64\n48 21 80\n81 51 51",
"output": "010\n101\n111"
},
{
"input": "96 14 2\n100 18 12\n65 34 89",
"output": "111\n010\n010"
},
{
"input": "93 95 90\n8 59 42\n53 13 19",
"output": "100\n001\n111"
},
{
"input": "71 84 18\n100 19 67\n9 76 15",
"output": "010\n010\n001"
},
{
"input": "38 93 85\n21 88 64\n4 96 25",
"output": "111\n011\n000"
},
{
"input": "75 20 20\n60 5 78\n77 4 69",
"output": "011\n001\n000"
},
{
"input": "65 70 96\n19 6 83\n33 37 82",
"output": "100\n000\n011"
},
{
"input": "11 13 60\n17 13 46\n42 21 39",
"output": "000\n011\n101"
},
{
"input": "0 0 0\n0 0 0\n0 0 0",
"output": "111\n111\n111"
},
{
"input": "0 0 0\n0 1 0\n0 0 0",
"output": "101\n000\n101"
},
{
"input": "0 0 0\n0 0 0\n0 0 1",
"output": "111\n110\n100"
}
] | 1,616,997,384 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 109 | 0 | a = []
b = []
for i in range(3):
a.append([int(i) for i in input().split()])
b.append([1]*3)
for i in range(3):
for j in range(3):
if a[i][j] % 2 == 0:
continue
b[i][j] = 1 - b[i][j]
for x in range(3):
if x == i or x == j:
continue
b[i][x] = 1 - b[i][x]
b[x][j] = 1 - b[x][j]
for i in b:
print(*i, sep='') | Title: Lights Out
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
Input Specification:
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
Output Specification:
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
Demo Input:
['1 0 0\n0 0 0\n0 0 1\n', '1 0 1\n8 8 8\n2 0 3\n']
Demo Output:
['001\n010\n100\n', '010\n011\n100\n']
Note:
none | ```python
a = []
b = []
for i in range(3):
a.append([int(i) for i in input().split()])
b.append([1]*3)
for i in range(3):
for j in range(3):
if a[i][j] % 2 == 0:
continue
b[i][j] = 1 - b[i][j]
for x in range(3):
if x == i or x == j:
continue
b[i][x] = 1 - b[i][x]
b[x][j] = 1 - b[x][j]
for i in b:
print(*i, sep='')
``` | 0 | |
202 | A | LLPS | PROGRAMMING | 800 | [
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null | This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z". | The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10. | Print the lexicographically largest palindromic subsequence of string *s*. | [
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] | [
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] | Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | 500 | [
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
{
"input": "helloworld",
"output": "w"
},
{
"input": "zzzzzzzazz",
"output": "zzzzzzzzz"
},
{
"input": "testcase",
"output": "tt"
},
{
"input": "hahahahaha",
"output": "hhhhh"
},
{
"input": "abbbbbbbbb",
"output": "bbbbbbbbb"
},
{
"input": "zaz",
"output": "zz"
},
{
"input": "aza",
"output": "z"
},
{
"input": "dcbaedcba",
"output": "e"
},
{
"input": "abcdeabcd",
"output": "e"
},
{
"input": "edcbabcde",
"output": "ee"
},
{
"input": "aaaaaaaaab",
"output": "b"
},
{
"input": "testzzzzzz",
"output": "zzzzzz"
},
{
"input": "zzzzzzwait",
"output": "zzzzzz"
},
{
"input": "rrrrrqponm",
"output": "rrrrr"
},
{
"input": "zzyzyy",
"output": "zzz"
},
{
"input": "aababb",
"output": "bbb"
},
{
"input": "zanzibar",
"output": "zz"
},
{
"input": "hhgfedcbaa",
"output": "hh"
},
{
"input": "aabcdefghh",
"output": "hh"
},
{
"input": "aruaru",
"output": "uu"
},
{
"input": "uraura",
"output": "uu"
},
{
"input": "aru",
"output": "u"
},
{
"input": "aburvabur",
"output": "v"
},
{
"input": "ura",
"output": "u"
},
{
"input": "eurottat",
"output": "u"
},
{
"input": "referee",
"output": "rr"
},
{
"input": "joking",
"output": "o"
},
{
"input": "seriously",
"output": "y"
},
{
"input": "sets",
"output": "t"
},
{
"input": "test",
"output": "tt"
},
{
"input": "klmgameklm",
"output": "mmm"
},
{
"input": "dfkjafdkdd",
"output": "kk"
},
{
"input": "zzzzzzzzzz",
"output": "zzzzzzzzzz"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaaaa"
},
{
"input": "buzz",
"output": "zz"
},
{
"input": "b",
"output": "b"
},
{
"input": "y",
"output": "y"
},
{
"input": "yy",
"output": "yy"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "bb",
"output": "bb"
},
{
"input": "aa",
"output": "aa"
},
{
"input": "a",
"output": "a"
},
{
"input": "z",
"output": "z"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "zzazazzzaz",
"output": "zzzzzzz"
},
{
"input": "hellhellhe",
"output": "llll"
},
{
"input": "hellohello",
"output": "oo"
},
{
"input": "refer",
"output": "rr"
}
] | 1,696,826,414 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 92 | 0 | s=input()
max='A'
cnt=0;
for i in s:
if(i>max):
max=i
cnt=1
elif(i==max):
cnt+=1
sc=max*cnt
print(sc) | Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | ```python
s=input()
max='A'
cnt=0;
for i in s:
if(i>max):
max=i
cnt=1
elif(i==max):
cnt+=1
sc=max*cnt
print(sc)
``` | 3 | |
508 | C | Anya and Ghosts | PROGRAMMING | 1,600 | [
"constructive algorithms",
"greedy"
] | null | null | Anya loves to watch horror movies. In the best traditions of horror, she will be visited by *m* ghosts tonight. Anya has lots of candles prepared for the visits, each candle can produce light for exactly *t* seconds. It takes the girl one second to light one candle. More formally, Anya can spend one second to light one candle, then this candle burns for exactly *t* seconds and then goes out and can no longer be used.
For each of the *m* ghosts Anya knows the time at which it comes: the *i*-th visit will happen *w**i* seconds after midnight, all *w**i*'s are distinct. Each visit lasts exactly one second.
What is the minimum number of candles Anya should use so that during each visit, at least *r* candles are burning? Anya can start to light a candle at any time that is integer number of seconds from midnight, possibly, at the time before midnight. That means, she can start to light a candle integer number of seconds before midnight or integer number of seconds after a midnight, or in other words in any integer moment of time. | The first line contains three integers *m*, *t*, *r* (1<=≤<=*m*,<=*t*,<=*r*<=≤<=300), representing the number of ghosts to visit Anya, the duration of a candle's burning and the minimum number of candles that should burn during each visit.
The next line contains *m* space-separated numbers *w**i* (1<=≤<=*i*<=≤<=*m*, 1<=≤<=*w**i*<=≤<=300), the *i*-th of them repesents at what second after the midnight the *i*-th ghost will come. All *w**i*'s are distinct, they follow in the strictly increasing order. | If it is possible to make at least *r* candles burn during each visit, then print the minimum number of candles that Anya needs to light for that.
If that is impossible, print <=-<=1. | [
"1 8 3\n10\n",
"2 10 1\n5 8\n",
"1 1 3\n10\n"
] | [
"3\n",
"1\n",
"-1\n"
] | Anya can start lighting a candle in the same second with ghost visit. But this candle isn't counted as burning at this visit.
It takes exactly one second to light up a candle and only after that second this candle is considered burning; it means that if Anya starts lighting candle at moment x, candle is buring from second x + 1 to second x + t inclusively.
In the first sample test three candles are enough. For example, Anya can start lighting them at the 3-rd, 5-th and 7-th seconds after the midnight.
In the second sample test one candle is enough. For example, Anya can start lighting it one second before the midnight.
In the third sample test the answer is - 1, since during each second at most one candle can burn but Anya needs three candles to light up the room at the moment when the ghost comes. | 1,500 | [
{
"input": "1 8 3\n10",
"output": "3"
},
{
"input": "2 10 1\n5 8",
"output": "1"
},
{
"input": "1 1 3\n10",
"output": "-1"
},
{
"input": "21 79 1\n13 42 51 60 69 77 94 103 144 189 196 203 210 215 217 222 224 234 240 260 282",
"output": "4"
},
{
"input": "125 92 2\n1 2 3 4 5 7 8 9 10 12 17 18 20 21 22 23 24 25 26 28 30 32 33 34 35 36 37 40 41 42 43 44 45 46 50 51 53 54 55 57 60 61 62 63 69 70 74 75 77 79 80 81 82 83 84 85 86 88 89 90 95 96 98 99 101 103 105 106 107 108 109 110 111 112 113 114 118 119 120 121 122 123 124 126 127 128 129 130 133 134 135 137 139 141 143 145 146 147 148 149 150 151 155 157 161 162 163 165 166 167 172 173 174 176 177 179 181 183 184 185 187 188 189 191 194",
"output": "6"
},
{
"input": "42 100 2\n55 56 57 58 60 61 63 66 71 73 75 76 77 79 82 86 87 91 93 96 97 98 99 100 101 103 108 109 111 113 114 117 119 122 128 129 134 135 137 141 142 149",
"output": "2"
},
{
"input": "31 23 2\n42 43 44 47 48 49 50 51 52 56 57 59 60 61 64 106 108 109 110 111 114 115 116 117 118 119 120 123 126 127 128",
"output": "6"
},
{
"input": "9 12 4\n1 2 3 4 5 7 8 9 10",
"output": "5"
},
{
"input": "9 16 2\n1 2 3 4 6 7 8 9 10",
"output": "2"
},
{
"input": "7 17 3\n1 3 4 5 7 9 10",
"output": "3"
},
{
"input": "1 1 1\n4",
"output": "1"
},
{
"input": "9 1 3\n1 2 4 5 6 7 8 9 10",
"output": "-1"
},
{
"input": "9 10 4\n1 2 3 4 5 6 8 9 10",
"output": "7"
},
{
"input": "7 2 2\n1 2 3 4 6 7 9",
"output": "10"
},
{
"input": "5 3 3\n1 4 5 6 10",
"output": "11"
},
{
"input": "9 7 1\n2 3 4 5 6 7 8 9 10",
"output": "2"
},
{
"input": "8 18 3\n2 3 4 5 6 7 8 9",
"output": "3"
},
{
"input": "88 82 36\n16 17 36 40 49 52 57 59 64 66 79 80 81 82 87 91 94 99 103 104 105 112 115 117 119 122 123 128 129 134 135 140 146 148 150 159 162 163 164 165 166 168 171 175 177 179 181 190 192 194 196 197 198 202 203 209 211 215 216 223 224 226 227 228 230 231 232 234 235 242 245 257 260 262 263 266 271 274 277 278 280 281 282 284 287 290 296 297",
"output": "144"
},
{
"input": "131 205 23\n1 3 8 9 10 11 12 13 14 17 18 19 23 25 26 27 31 32 33 36 37 39 40 41 43 44 51 58 61 65 68 69 71 72 73 75 79 80 82 87 88 89 90 91 92 93 96 99 100 103 107 109 113 114 119 121 122 123 124 127 135 136 137 139 141 142 143 144 148 149 151 152 153 154 155 157 160 162 168 169 170 171 172 174 176 177 179 182 183 185 186 187 190 193 194 196 197 200 206 209 215 220 224 226 230 232 233 235 237 240 242 243 244 247 251 252 260 264 265 269 272 278 279 280 281 288 290 292 294 296 300",
"output": "46"
},
{
"input": "45 131 15\n14 17 26 31 32 43 45 56 64 73 75 88 89 93 98 103 116 117 119 123 130 131 135 139 140 153 156 161 163 172 197 212 217 230 232 234 239 240 252 256 265 266 272 275 290",
"output": "45"
},
{
"input": "63 205 38\n47 50 51 54 56 64 67 69 70 72 73 75 78 81 83 88 91 99 109 114 118 122 136 137 138 143 146 147 149 150 158 159 160 168 171 172 174 176 181 189 192 195 198 201 204 205 226 232 235 238 247 248 253 254 258 260 270 276 278 280 282 284 298",
"output": "76"
},
{
"input": "44 258 19\n3 9 10 19 23 32 42 45 52 54 65 66 69 72 73 93 108 116 119 122 141 150 160 162 185 187 199 205 206 219 225 229 234 235 240 242 253 261 264 268 275 277 286 295",
"output": "38"
},
{
"input": "138 245 30\n3 5 6 8 9 13 15 16 19 20 24 25 27 29 30 32 33 34 35 36 37 38 40 42 47 51 52 53 55 56 58 59 63 66 67 68 69 72 73 74 75 77 78 80 81 82 85 86 87 89 91 93 95 96 99 100 102 104 105 108 110 111 112 117 122 124 125 128 129 131 133 136 139 144 145 146 147 148 149 151 153 155 156 159 162 163 164 165 168 174 175 176 183 191 193 194 195 203 204 205 206 211 216 217 218 219 228 229 230 235 237 238 239 242 244 248 249 250 252 253 255 257 258 260 264 265 266 268 270 271 272 277 278 280 285 288 290 291",
"output": "60"
},
{
"input": "21 140 28\n40 46 58 67 71 86 104 125 129 141 163 184 193 215 219 222 234 237 241 246 263",
"output": "56"
},
{
"input": "77 268 24\n2 6 15 18 24 32 35 39 41 44 49 54 59 63 70 73 74 85 90 91 95 98 100 104 105 108 114 119 120 125 126 128 131 137 139 142 148 150 151 153 155 158 160 163 168 171 175 183 195 198 202 204 205 207 208 213 220 224 230 239 240 244 256 258 260 262 264 265 266 272 274 277 280 284 291 299 300",
"output": "48"
},
{
"input": "115 37 25\n1 3 6 8 10 13 14 15 16 17 20 24 28 32 34 36 38 40 41 45 49 58 59 60 62 63 64 77 79 80 85 88 90 91 97 98 100 101 105 109 111 112 114 120 122 123 124 128 132 133 139 144 145 150 151 152 154 155 158 159 160 162 164 171 178 181 182 187 190 191 192 193 194 196 197 198 206 207 213 216 219 223 224 233 235 238 240 243 244 248 249 250 251 252 254 260 261 262 267 268 270 272 273 275 276 278 279 280 283 286 288 289 292 293 300",
"output": "224"
},
{
"input": "100 257 21\n50 56 57 58 59 60 62 66 71 75 81 84 86 90 91 92 94 95 96 97 100 107 110 111 112 114 115 121 123 125 126 127 129 130 133 134 136 137 147 151 152 156 162 167 168 172 176 177 178 179 181 182 185 186 188 189 190 191 193 199 200 201 202 205 209 213 216 218 220 222 226 231 232 235 240 241 244 248 249 250 252 253 254 256 257 258 260 261 263 264 268 270 274 276 278 279 282 294 297 300",
"output": "35"
},
{
"input": "84 55 48\n8 9 10 12 14 17 22 28 31 33 36 37 38 40 45 46 48 50 51 58 60 71 73 74 76 77 78 82 83 87 88 90 92 96 98 99 103 104 105 108 109 111 113 117 124 125 147 148 149 152 156 159 161 163 169 170 171 177 179 180 185 186 190 198 199 201 254 256 259 260 261 262 264 267 273 275 280 282 283 286 288 289 292 298",
"output": "296"
},
{
"input": "11 1 37\n18 48 50 133 141 167 168 173 188 262 267",
"output": "-1"
},
{
"input": "48 295 12\n203 205 207 208 213 214 218 219 222 223 224 225 228 229 230 234 239 241 243 245 246 247 248 251 252 253 254 255 259 260 261 262 264 266 272 277 278 280 282 285 286 287 289 292 293 296 299 300",
"output": "12"
},
{
"input": "2 3 1\n2 4",
"output": "1"
},
{
"input": "2 3 1\n2 5",
"output": "2"
},
{
"input": "2 2 2\n1 3",
"output": "4"
},
{
"input": "2 2 2\n1 2",
"output": "3"
},
{
"input": "2 1 2\n1 2",
"output": "-1"
},
{
"input": "1 300 300\n1",
"output": "300"
},
{
"input": "1 299 300\n300",
"output": "-1"
}
] | 1,690,174,715 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 3 | 30 | 0 | import sys
#import threading
#sys.setrecursionlimit(10**8)
#threading.stack_size(10**8)
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
############ ---- Input Functions ---- ############
def Anya_and_Ghosts():
m,t,r = invr()
sequence = inlt()
minimum_candels_to_light = 0
for index,time in enumerate(sequence):
if index == 0:
candles_stopping_time = [(time-i+t) for i in range(1,r+1)]
minimum_candels_to_light += r
if candles_stopping_time[-1] < time:
print(-1)
return
elif time <= max(candles_stopping_time):
continue
else:
candles_stopped = 0
max_stopping_time = -1
for index,st in candles_stopping_time:
if st < time:
candles_stopped += 1
candles_stopping_time[index] = time - candles_stopped + t
minimum_candels_to_light += 1
if candles_stopping_time[index] > max_stopping_time:
max_stopping_time = candles_stopping_time[index]
if max_stopping_time < time:
print(-1)
return
print(minimum_candels_to_light)
return
Anya_and_Ghosts() | Title: Anya and Ghosts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anya loves to watch horror movies. In the best traditions of horror, she will be visited by *m* ghosts tonight. Anya has lots of candles prepared for the visits, each candle can produce light for exactly *t* seconds. It takes the girl one second to light one candle. More formally, Anya can spend one second to light one candle, then this candle burns for exactly *t* seconds and then goes out and can no longer be used.
For each of the *m* ghosts Anya knows the time at which it comes: the *i*-th visit will happen *w**i* seconds after midnight, all *w**i*'s are distinct. Each visit lasts exactly one second.
What is the minimum number of candles Anya should use so that during each visit, at least *r* candles are burning? Anya can start to light a candle at any time that is integer number of seconds from midnight, possibly, at the time before midnight. That means, she can start to light a candle integer number of seconds before midnight or integer number of seconds after a midnight, or in other words in any integer moment of time.
Input Specification:
The first line contains three integers *m*, *t*, *r* (1<=≤<=*m*,<=*t*,<=*r*<=≤<=300), representing the number of ghosts to visit Anya, the duration of a candle's burning and the minimum number of candles that should burn during each visit.
The next line contains *m* space-separated numbers *w**i* (1<=≤<=*i*<=≤<=*m*, 1<=≤<=*w**i*<=≤<=300), the *i*-th of them repesents at what second after the midnight the *i*-th ghost will come. All *w**i*'s are distinct, they follow in the strictly increasing order.
Output Specification:
If it is possible to make at least *r* candles burn during each visit, then print the minimum number of candles that Anya needs to light for that.
If that is impossible, print <=-<=1.
Demo Input:
['1 8 3\n10\n', '2 10 1\n5 8\n', '1 1 3\n10\n']
Demo Output:
['3\n', '1\n', '-1\n']
Note:
Anya can start lighting a candle in the same second with ghost visit. But this candle isn't counted as burning at this visit.
It takes exactly one second to light up a candle and only after that second this candle is considered burning; it means that if Anya starts lighting candle at moment x, candle is buring from second x + 1 to second x + t inclusively.
In the first sample test three candles are enough. For example, Anya can start lighting them at the 3-rd, 5-th and 7-th seconds after the midnight.
In the second sample test one candle is enough. For example, Anya can start lighting it one second before the midnight.
In the third sample test the answer is - 1, since during each second at most one candle can burn but Anya needs three candles to light up the room at the moment when the ghost comes. | ```python
import sys
#import threading
#sys.setrecursionlimit(10**8)
#threading.stack_size(10**8)
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
############ ---- Input Functions ---- ############
def Anya_and_Ghosts():
m,t,r = invr()
sequence = inlt()
minimum_candels_to_light = 0
for index,time in enumerate(sequence):
if index == 0:
candles_stopping_time = [(time-i+t) for i in range(1,r+1)]
minimum_candels_to_light += r
if candles_stopping_time[-1] < time:
print(-1)
return
elif time <= max(candles_stopping_time):
continue
else:
candles_stopped = 0
max_stopping_time = -1
for index,st in candles_stopping_time:
if st < time:
candles_stopped += 1
candles_stopping_time[index] = time - candles_stopped + t
minimum_candels_to_light += 1
if candles_stopping_time[index] > max_stopping_time:
max_stopping_time = candles_stopping_time[index]
if max_stopping_time < time:
print(-1)
return
print(minimum_candels_to_light)
return
Anya_and_Ghosts()
``` | -1 | |
893 | E | Counting Arrays | PROGRAMMING | 2,000 | [
"combinatorics",
"dp",
"math",
"number theory"
] | null | null | You are given two positive integer numbers *x* and *y*. An array *F* is called an *y*-factorization of *x* iff the following conditions are met:
- There are *y* elements in *F*, and all of them are integer numbers; - .
You have to count the number of pairwise distinct arrays that are *y*-factorizations of *x*. Two arrays *A* and *B* are considered different iff there exists at least one index *i* (1<=≤<=*i*<=≤<=*y*) such that *A**i*<=≠<=*B**i*. Since the answer can be very large, print it modulo 109<=+<=7. | The first line contains one integer *q* (1<=≤<=*q*<=≤<=105) — the number of testcases to solve.
Then *q* lines follow, each containing two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=106). Each of these lines represents a testcase. | Print *q* integers. *i*-th integer has to be equal to the number of *y**i*-factorizations of *x**i* modulo 109<=+<=7. | [
"2\n6 3\n4 2\n"
] | [
"36\n6\n"
] | In the second testcase of the example there are six *y*-factorizations:
- { - 4, - 1}; - { - 2, - 2}; - { - 1, - 4}; - {1, 4}; - {2, 2}; - {4, 1}. | 0 | [
{
"input": "2\n6 3\n4 2",
"output": "36\n6"
},
{
"input": "1\n524288 1000000",
"output": "645043186"
},
{
"input": "1\n65536 1000000",
"output": "928522471"
},
{
"input": "1\n5612 11399",
"output": "215664246"
}
] | 1,630,953,697 | 4,297 | PyPy 3 | OK | TESTS | 45 | 1,497 | 181,555,200 | from collections import defaultdict
import sys
input = sys.stdin.readline
def fast_prime_factorization(n):
d = [(i + 1) % 2 * 2 for i in range(n + 1)]
d[0], d[1] = 0, 1
for i in range(3, n + 1):
if d[i]:
continue
for j in range(i, n + 1, 2 * i):
if not d[j]:
d[j] = i
return d
l = pow(10, 6) + 100
mod = pow(10, 9) + 7
fact = [1] * (l + 1)
for i in range(1, l + 1):
fact[i] = i * fact[i - 1] % mod
inv = [1] * (l + 1)
inv[l] = pow(fact[l], mod - 2, mod)
for i in range(l - 1, -1, -1):
inv[i] = (i + 1) * inv[i + 1] % mod
def comb(n, r):
return fact[n] * inv[r] * inv[n - r] % mod if n >= r >= 0 else 0
d = fast_prime_factorization(l)
pow2 = [1]
for _ in range(l):
pow2.append(2 * pow2[-1] % mod)
q = int(input())
for _ in range(q):
x, y = map(int, input().split())
cnt = defaultdict(lambda : 0)
while x ^ d[x]:
cnt[d[x]] += 1
x //= d[x]
if x > 1:
cnt[x] += 1
ans = 1
for i in cnt.values():
ans *= comb(y + i - 1, i)
ans %= mod
ans *= pow2[y - 1]
ans %= mod
print(ans) | Title: Counting Arrays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two positive integer numbers *x* and *y*. An array *F* is called an *y*-factorization of *x* iff the following conditions are met:
- There are *y* elements in *F*, and all of them are integer numbers; - .
You have to count the number of pairwise distinct arrays that are *y*-factorizations of *x*. Two arrays *A* and *B* are considered different iff there exists at least one index *i* (1<=≤<=*i*<=≤<=*y*) such that *A**i*<=≠<=*B**i*. Since the answer can be very large, print it modulo 109<=+<=7.
Input Specification:
The first line contains one integer *q* (1<=≤<=*q*<=≤<=105) — the number of testcases to solve.
Then *q* lines follow, each containing two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=106). Each of these lines represents a testcase.
Output Specification:
Print *q* integers. *i*-th integer has to be equal to the number of *y**i*-factorizations of *x**i* modulo 109<=+<=7.
Demo Input:
['2\n6 3\n4 2\n']
Demo Output:
['36\n6\n']
Note:
In the second testcase of the example there are six *y*-factorizations:
- { - 4, - 1}; - { - 2, - 2}; - { - 1, - 4}; - {1, 4}; - {2, 2}; - {4, 1}. | ```python
from collections import defaultdict
import sys
input = sys.stdin.readline
def fast_prime_factorization(n):
d = [(i + 1) % 2 * 2 for i in range(n + 1)]
d[0], d[1] = 0, 1
for i in range(3, n + 1):
if d[i]:
continue
for j in range(i, n + 1, 2 * i):
if not d[j]:
d[j] = i
return d
l = pow(10, 6) + 100
mod = pow(10, 9) + 7
fact = [1] * (l + 1)
for i in range(1, l + 1):
fact[i] = i * fact[i - 1] % mod
inv = [1] * (l + 1)
inv[l] = pow(fact[l], mod - 2, mod)
for i in range(l - 1, -1, -1):
inv[i] = (i + 1) * inv[i + 1] % mod
def comb(n, r):
return fact[n] * inv[r] * inv[n - r] % mod if n >= r >= 0 else 0
d = fast_prime_factorization(l)
pow2 = [1]
for _ in range(l):
pow2.append(2 * pow2[-1] % mod)
q = int(input())
for _ in range(q):
x, y = map(int, input().split())
cnt = defaultdict(lambda : 0)
while x ^ d[x]:
cnt[d[x]] += 1
x //= d[x]
if x > 1:
cnt[x] += 1
ans = 1
for i in cnt.values():
ans *= comb(y + i - 1, i)
ans %= mod
ans *= pow2[y - 1]
ans %= mod
print(ans)
``` | 3 | |
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"
}
] | 1,603,529,564 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 216 | 307,200 | p=int(input())
lst=str()
for x in range(p):
lst=lst+input()+' '
lst=lst.split()
ci=dict()
n=-2
while n<2*p:
n=n+2
try:
if lst[n]in ci:
ci[lst[n]]=ci[lst[n]]+int(lst[n+1])
else:
ci[lst[n]]=int(lst[n+1])
except:break
lstnm=[]
for va in ci.values():
lstnm.append(va)
lar=max(lstnm)
di=dict()
n=-2
while n<2*p:
n=n+2
try:
if lst[n]in di:
di[lst[n]]=di[lst[n]]+int(lst[n+1])
else:
di[lst[n]]=int(lst[n+1])
if di[lst[n]]is lar:
print(lst[n])
break
except:break
| 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
p=int(input())
lst=str()
for x in range(p):
lst=lst+input()+' '
lst=lst.split()
ci=dict()
n=-2
while n<2*p:
n=n+2
try:
if lst[n]in ci:
ci[lst[n]]=ci[lst[n]]+int(lst[n+1])
else:
ci[lst[n]]=int(lst[n+1])
except:break
lstnm=[]
for va in ci.values():
lstnm.append(va)
lar=max(lstnm)
di=dict()
n=-2
while n<2*p:
n=n+2
try:
if lst[n]in di:
di[lst[n]]=di[lst[n]]+int(lst[n+1])
else:
di[lst[n]]=int(lst[n+1])
if di[lst[n]]is lar:
print(lst[n])
break
except:break
``` | 0 |
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"
}
] | 1,568,040,968 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 39 | 296 | 11,059,200 | na,nb = map(int,input().split())
k,m = map(int,input().split())
a = input().split(" ")
b = input().split(" ")
if(b[nb-m]<=a[k-1]):
print("NO")
else:
print("YES")
| 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 = input().split(" ")
b = input().split(" ")
if(b[nb-m]<=a[k-1]):
print("NO")
else:
print("YES")
``` | 0 | |
445 | A | DZY Loves Chessboard | PROGRAMMING | 1,200 | [
"dfs and similar",
"implementation"
] | null | null | DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad. | Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists. | [
"1 1\n.\n",
"2 2\n..\n..\n",
"3 3\n.-.\n---\n--."
] | [
"B\n",
"BW\nWB\n",
"B-B\n---\n--B"
] | In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | 500 | [
{
"input": "1 1\n.",
"output": "B"
},
{
"input": "2 2\n..\n..",
"output": "BW\nWB"
},
{
"input": "3 3\n.-.\n---\n--.",
"output": "B-B\n---\n--B"
},
{
"input": "3 1\n-\n.\n.",
"output": "-\nW\nB"
},
{
"input": "11 11\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------",
"output": "-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------"
},
{
"input": "1 1\n-",
"output": "-"
},
{
"input": "4 2\n..\n..\n..\n..",
"output": "BW\nWB\nBW\nWB"
},
{
"input": "5 1\n.\n.\n.\n.\n.",
"output": "B\nW\nB\nW\nB"
}
] | 1,617,283,794 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 46 | 0 | n,m=input().split(" ")
n,m=int(n),int(m)
def split(word):
return [char for char in word]
a=[0]*n
for i in range(n):
a[i]=split(input())
count=0
while count<n*m:
i=count//n
j=count%n
if a[i][j]!="-":
if a[i][j]=='W':
a[i][j]='B'
else:
a[i][j]='W'
temp=a[i][j]
backtrack=0
try:
if a[i-1][j]==temp:
backtrack=1
except :
pass
try:
if a[i][j-1]==temp:
backtrack=1
except :
pass
try:
if a[i+1][j]==temp:
backtrack=1
except :
pass
try:
if a[i][j+1]==temp:
backtrack=1
except :
pass
if backtrack==1:
while(a[count//n][count%n]!='W'):
count-=1
count-=1
count+=1
count=0
for count in range(m*n):
print(a[count//n][count%n],end="")
if(count%n==n-1):
print() | Title: DZY Loves Chessboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output Specification:
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
Demo Input:
['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.']
Demo Output:
['B\n', 'BW\nWB\n', 'B-B\n---\n--B']
Note:
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | ```python
n,m=input().split(" ")
n,m=int(n),int(m)
def split(word):
return [char for char in word]
a=[0]*n
for i in range(n):
a[i]=split(input())
count=0
while count<n*m:
i=count//n
j=count%n
if a[i][j]!="-":
if a[i][j]=='W':
a[i][j]='B'
else:
a[i][j]='W'
temp=a[i][j]
backtrack=0
try:
if a[i-1][j]==temp:
backtrack=1
except :
pass
try:
if a[i][j-1]==temp:
backtrack=1
except :
pass
try:
if a[i+1][j]==temp:
backtrack=1
except :
pass
try:
if a[i][j+1]==temp:
backtrack=1
except :
pass
if backtrack==1:
while(a[count//n][count%n]!='W'):
count-=1
count-=1
count+=1
count=0
for count in range(m*n):
print(a[count//n][count%n],end="")
if(count%n==n-1):
print()
``` | -1 | |
0 | none | none | none | 0 | [
"none"
] | null | null | We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109<=+<=7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string. | The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106. | Print the minimum number of steps modulo 109<=+<=7. | [
"ab\n",
"aab\n"
] | [
"1\n",
"3\n"
] | The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa". | 0 | [
{
"input": "ab",
"output": "1"
},
{
"input": "aab",
"output": "3"
},
{
"input": "aaaaabaabababaaaaaba",
"output": "17307"
},
{
"input": "abaabaaabbabaabab",
"output": "1795"
},
{
"input": "abbaa",
"output": "2"
},
{
"input": "abbaaabaabaaaaabbbbaababaaaaabaabbaaaaabbaabbaaaabbbabbbabb",
"output": "690283580"
},
{
"input": "aababbaaaabbaabbbbbbbbabbababbbaaabbaaabbabbba",
"output": "2183418"
},
{
"input": "aabbaababbabbbaabbaababaaaabbaaaabaaaaaababbaaaabaababbabbbb",
"output": "436420225"
},
{
"input": "aaabaaaabbababbaabbababbbbaaaaaaabbabbba",
"output": "8431094"
},
{
"input": "abbbbababbabbbbbabaabbbaabbbbbbbaaab",
"output": "8180"
},
{
"input": "bbababbbaabaaaaaaaabbabbbb",
"output": "40979"
},
{
"input": "abbbaaabbbbbabaabbaaabbbababbbaabaabababababa",
"output": "2065758"
},
{
"input": "abaaaaaabaaaabbabbaaabbbbabababaaaaabbaabbaaaaabbbaababaaaaaaabbbbbaaaaabaababbabababbabbbbaabbaabbabbbabaabbaabbaaaaaab",
"output": "235606597"
},
{
"input": "abbbbbbbbbbbbbbbbbbbbbbbbbbaababaaaaaaabaabaaababaabaababaaabababaababab",
"output": "7"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbaaaaaaaaabaabaaababaabaababaaabababaabbbbbbb",
"output": "557763786"
}
] | 1,589,896,271 | 3,311 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 140 | 0 | s=input()
ans=0
j=0
for i in range(0,len(s)):
if i>=j:
cnt1=0
while i<len(s) and s[i]=='a':
cnt1=cnt1+1
i=i+1
cnt2=0
while i<len(s) and s[i]=='b':
cnt2=cnt2+1
i=i+1
i=i
j=i;
if cnt1>0 and cnt2>0:
ans=ans + cnt2
bb= cnt2*2
cnt1=cnt1-1
ans = ans + bb*(pow(2,cnt1)-1)
#print(ans)
print(ans%1000000007)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109<=+<=7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input Specification:
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output Specification:
Print the minimum number of steps modulo 109<=+<=7.
Demo Input:
['ab\n', 'aab\n']
Demo Output:
['1\n', '3\n']
Note:
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa". | ```python
s=input()
ans=0
j=0
for i in range(0,len(s)):
if i>=j:
cnt1=0
while i<len(s) and s[i]=='a':
cnt1=cnt1+1
i=i+1
cnt2=0
while i<len(s) and s[i]=='b':
cnt2=cnt2+1
i=i+1
i=i
j=i;
if cnt1>0 and cnt2>0:
ans=ans + cnt2
bb= cnt2*2
cnt1=cnt1-1
ans = ans + bb*(pow(2,cnt1)-1)
#print(ans)
print(ans%1000000007)
``` | 0 | |
13 | C | Sequence | PROGRAMMING | 2,200 | [
"dp",
"sortings"
] | C. Sequence | 1 | 64 | Little Petya likes to play very much. And most of all he likes to play the following game:
He is given a sequence of *N* integer numbers. At each step it is allowed to increase the value of any number by 1 or to decrease it by 1. The goal of the game is to make the sequence non-decreasing with the smallest number of steps. Petya is not good at math, so he asks for your help.
The sequence *a* is called non-decreasing if *a*1<=≤<=*a*2<=≤<=...<=≤<=*a**N* holds, where *N* is the length of the sequence. | The first line of the input contains single integer *N* (1<=≤<=*N*<=≤<=5000) — the length of the initial sequence. The following *N* lines contain one integer each — elements of the sequence. These numbers do not exceed 109 by absolute value. | Output one integer — minimum number of steps required to achieve the goal. | [
"5\n3 2 -1 2 11\n",
"5\n2 1 1 1 1\n"
] | [
"4\n",
"1\n"
] | none | 0 | [
{
"input": "5\n3 2 -1 2 11",
"output": "4"
},
{
"input": "5\n2 1 1 1 1",
"output": "1"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "1\n11",
"output": "0"
},
{
"input": "2\n10 2",
"output": "8"
},
{
"input": "6\n1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000",
"output": "6000000000"
},
{
"input": "7\n1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000",
"output": "6000000000"
},
{
"input": "10\n3 0 5 10 10 9 10 8 8 1",
"output": "16"
},
{
"input": "20\n5 5 6 5 10 9 4 0 0 0 7 7 7 5 6 7 5 6 5 6",
"output": "33"
},
{
"input": "20\n10 6 1 9 10 10 7 0 10 5 9 1 5 7 4 9 3 9 7 2",
"output": "55"
},
{
"input": "20\n0 0 0 1 5 10 7 7 8 7 66 64 62 65 39 38 72 69 71 72",
"output": "63"
},
{
"input": "5\n1 2 3 1 1",
"output": "3"
}
] | 1,697,696,838 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 34 | 248 | 8,499,200 | import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
q = []
ans = 0
import heapq
for i in range(n):
heapq.heappush(q,-a[i])
if -q[0] > a[i]:
T = heapq.heappop(q)
heapq.heappush(q,-a[i])
ans += -T - a[i]
print(ans) | Title: Sequence
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Little Petya likes to play very much. And most of all he likes to play the following game:
He is given a sequence of *N* integer numbers. At each step it is allowed to increase the value of any number by 1 or to decrease it by 1. The goal of the game is to make the sequence non-decreasing with the smallest number of steps. Petya is not good at math, so he asks for your help.
The sequence *a* is called non-decreasing if *a*1<=≤<=*a*2<=≤<=...<=≤<=*a**N* holds, where *N* is the length of the sequence.
Input Specification:
The first line of the input contains single integer *N* (1<=≤<=*N*<=≤<=5000) — the length of the initial sequence. The following *N* lines contain one integer each — elements of the sequence. These numbers do not exceed 109 by absolute value.
Output Specification:
Output one integer — minimum number of steps required to achieve the goal.
Demo Input:
['5\n3 2 -1 2 11\n', '5\n2 1 1 1 1\n']
Demo Output:
['4\n', '1\n']
Note:
none | ```python
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
q = []
ans = 0
import heapq
for i in range(n):
heapq.heappush(q,-a[i])
if -q[0] > a[i]:
T = heapq.heappop(q)
heapq.heappush(q,-a[i])
ans += -T - a[i]
print(ans)
``` | 3.812676 |
3 | A | Shortest path of the king | PROGRAMMING | 1,000 | [
"greedy",
"shortest paths"
] | A. Shortest path of the king | 1 | 64 | The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square *t*. As the king is not in habit of wasting his time, he wants to get from his current position *s* to square *t* in the least number of moves. Help him to do this.
In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to). | The first line contains the chessboard coordinates of square *s*, the second line — of square *t*.
Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8. | In the first line print *n* — minimum number of the king's moves. Then in *n* lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.
L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them. | [
"a8\nh1\n"
] | [
"7\nRD\nRD\nRD\nRD\nRD\nRD\nRD\n"
] | none | 0 | [
{
"input": "a8\nh1",
"output": "7\nRD\nRD\nRD\nRD\nRD\nRD\nRD"
},
{
"input": "b2\nb4",
"output": "2\nU\nU"
},
{
"input": "a5\na5",
"output": "0"
},
{
"input": "h1\nb2",
"output": "6\nLU\nL\nL\nL\nL\nL"
},
{
"input": "c5\nh2",
"output": "5\nRD\nRD\nRD\nR\nR"
},
{
"input": "e1\nf2",
"output": "1\nRU"
},
{
"input": "g4\nd2",
"output": "3\nLD\nLD\nL"
},
{
"input": "a8\nb2",
"output": "6\nRD\nD\nD\nD\nD\nD"
},
{
"input": "d4\nh2",
"output": "4\nRD\nRD\nR\nR"
},
{
"input": "c5\na2",
"output": "3\nLD\nLD\nD"
},
{
"input": "h5\nf8",
"output": "3\nLU\nLU\nU"
},
{
"input": "e6\nb6",
"output": "3\nL\nL\nL"
},
{
"input": "a6\ng4",
"output": "6\nRD\nRD\nR\nR\nR\nR"
},
{
"input": "f7\nc2",
"output": "5\nLD\nLD\nLD\nD\nD"
},
{
"input": "b7\nh8",
"output": "6\nRU\nR\nR\nR\nR\nR"
},
{
"input": "g7\nd6",
"output": "3\nLD\nL\nL"
},
{
"input": "c8\na3",
"output": "5\nLD\nLD\nD\nD\nD"
},
{
"input": "h8\nf1",
"output": "7\nLD\nLD\nD\nD\nD\nD\nD"
},
{
"input": "d1\nb7",
"output": "6\nLU\nLU\nU\nU\nU\nU"
},
{
"input": "a7\ne5",
"output": "4\nRD\nRD\nR\nR"
},
{
"input": "d6\nb1",
"output": "5\nLD\nLD\nD\nD\nD"
},
{
"input": "f5\ng5",
"output": "1\nR"
},
{
"input": "h4\nd1",
"output": "4\nLD\nLD\nLD\nL"
},
{
"input": "b3\na5",
"output": "2\nLU\nU"
},
{
"input": "d2\nf1",
"output": "2\nRD\nR"
},
{
"input": "f1\nc5",
"output": "4\nLU\nLU\nLU\nU"
},
{
"input": "a8\nh1",
"output": "7\nRD\nRD\nRD\nRD\nRD\nRD\nRD"
},
{
"input": "c7\ne5",
"output": "2\nRD\nRD"
},
{
"input": "e7\nb1",
"output": "6\nLD\nLD\nLD\nD\nD\nD"
},
{
"input": "g8\na8",
"output": "6\nL\nL\nL\nL\nL\nL"
},
{
"input": "g6\nf2",
"output": "4\nLD\nD\nD\nD"
},
{
"input": "g4\nc4",
"output": "4\nL\nL\nL\nL"
},
{
"input": "g2\na6",
"output": "6\nLU\nLU\nLU\nLU\nL\nL"
},
{
"input": "f8\nf8",
"output": "0"
},
{
"input": "f5\nd2",
"output": "3\nLD\nLD\nD"
}
] | 1,480,000,031 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 124 | 0 | king = input()
destiny = input()
xk = ord(king[0])
xd = ord(destiny[0])
yk = int(king[1])
yd = int(destiny[1])
h = ''
v = ''
moves = []
while xk != xd or yk != yd:
if xk < xd:
h = 'R'
xk += 1
elif xk > xd:
h = 'L'
xk -= 1
if yk < yd:
v = 'U'
yk += 1
elif yk > yd:
v = 'D'
yk -= 1
moves.append(h + v)
print(len(moves))
for i in moves:
print(i)
| Title: Shortest path of the king
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square *t*. As the king is not in habit of wasting his time, he wants to get from his current position *s* to square *t* in the least number of moves. Help him to do this.
In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).
Input Specification:
The first line contains the chessboard coordinates of square *s*, the second line — of square *t*.
Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8.
Output Specification:
In the first line print *n* — minimum number of the king's moves. Then in *n* lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.
L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.
Demo Input:
['a8\nh1\n']
Demo Output:
['7\nRD\nRD\nRD\nRD\nRD\nRD\nRD\n']
Note:
none | ```python
king = input()
destiny = input()
xk = ord(king[0])
xd = ord(destiny[0])
yk = int(king[1])
yd = int(destiny[1])
h = ''
v = ''
moves = []
while xk != xd or yk != yd:
if xk < xd:
h = 'R'
xk += 1
elif xk > xd:
h = 'L'
xk -= 1
if yk < yd:
v = 'U'
yk += 1
elif yk > yd:
v = 'D'
yk -= 1
moves.append(h + v)
print(len(moves))
for i in moves:
print(i)
``` | 0 |
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"
}
] | 1,585,557,380 | 2,147,483,647 | PyPy 3 | OK | TESTS | 40 | 140 | 0 | s = str(input())
res = 0
h = 'hello'
i = 0
for c in s:
if res == 5:
break
if i < 5 and c == h[i]:
res += 1
i += 1
if res == 5:
print("YES")
else:
print("NO") | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
s = str(input())
res = 0
h = 'hello'
i = 0
for c in s:
if res == 5:
break
if i < 5 and c == h[i]:
res += 1
i += 1
if res == 5:
print("YES")
else:
print("NO")
``` | 3.93 |
337 | A | Puzzles | PROGRAMMING | 900 | [
"greedy"
] | null | null | The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*. | The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop. | Print a single integer — the least possible difference the teacher can obtain. | [
"4 6\n10 12 10 7 5 22\n"
] | [
"5\n"
] | Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5. | 500 | [
{
"input": "4 6\n10 12 10 7 5 22",
"output": "5"
},
{
"input": "2 2\n4 4",
"output": "0"
},
{
"input": "2 10\n4 5 6 7 8 9 10 11 12 12",
"output": "0"
},
{
"input": "4 5\n818 136 713 59 946",
"output": "759"
},
{
"input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903",
"output": "13"
},
{
"input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94",
"output": "3"
},
{
"input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190",
"output": "31"
},
{
"input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997",
"output": "0"
},
{
"input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176",
"output": "412"
},
{
"input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886",
"output": "938"
},
{
"input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993",
"output": "937"
},
{
"input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7",
"output": "0"
},
{
"input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19",
"output": "12"
},
{
"input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17",
"output": "31"
},
{
"input": "2 2\n1000 4",
"output": "996"
},
{
"input": "2 3\n4 502 1000",
"output": "498"
},
{
"input": "3 3\n4 1000 4",
"output": "996"
}
] | 1,698,253,597 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 60 | 0 | n,m = map(int,input().split())
l = list(map(int,input().split()))
while n<len(l):
l.remove(max(l))
print(max(l)-min(l)) | Title: Puzzles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
Input Specification:
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Output Specification:
Print a single integer — the least possible difference the teacher can obtain.
Demo Input:
['4 6\n10 12 10 7 5 22\n']
Demo Output:
['5\n']
Note:
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5. | ```python
n,m = map(int,input().split())
l = list(map(int,input().split()))
while n<len(l):
l.remove(max(l))
print(max(l)-min(l))
``` | 0 | |
139 | A | Petr and Book | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book. | The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero. | Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday. | [
"100\n15 20 20 15 10 30 45\n",
"2\n1 0 0 0 0 0 0\n"
] | [
"6\n",
"1\n"
] | Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book. | 500 | [
{
"input": "100\n15 20 20 15 10 30 45",
"output": "6"
},
{
"input": "2\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100\n100 200 100 200 300 400 500",
"output": "1"
},
{
"input": "3\n1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "1\n1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "20\n5 3 7 2 1 6 4",
"output": "6"
},
{
"input": "10\n5 1 1 1 1 1 5",
"output": "6"
},
{
"input": "50\n10 1 10 1 10 1 10",
"output": "1"
},
{
"input": "77\n11 11 11 11 11 11 10",
"output": "1"
},
{
"input": "1\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n100 100 100 100 100 100 100",
"output": "3"
},
{
"input": "999\n10 20 10 20 30 20 10",
"output": "3"
},
{
"input": "433\n109 58 77 10 39 125 15",
"output": "7"
},
{
"input": "1\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n1 0 1 0 1 0 1",
"output": "1"
},
{
"input": "997\n1 1 0 0 1 0 1",
"output": "1"
},
{
"input": "1000\n1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1000\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "1000\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "1000\n1 0 0 1 0 0 1",
"output": "1"
},
{
"input": "509\n105 23 98 0 7 0 155",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2\n1 1 0 0 0 0 0",
"output": "2"
},
{
"input": "1\n0 0 0 0 0 1 0",
"output": "6"
},
{
"input": "10\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 6 0",
"output": "6"
},
{
"input": "3\n0 1 0 0 0 0 0",
"output": "2"
},
{
"input": "10\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "28\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "100\n5 5 5 5 5 5 5",
"output": "6"
},
{
"input": "4\n1 0 0 0 0 0 1",
"output": "7"
},
{
"input": "2\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "7\n0 0 0 0 0 0 7",
"output": "7"
},
{
"input": "7\n2 1 1 1 1 1 0",
"output": "6"
},
{
"input": "2\n0 0 1 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 1 1 1 1 1 0",
"output": "6"
},
{
"input": "5\n1 1 1 0 0 1 1",
"output": "7"
},
{
"input": "100\n10 20 30 10 10 10 10",
"output": "7"
},
{
"input": "1\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "70\n10 10 10 10 10 10 10",
"output": "7"
},
{
"input": "22\n1 2 3 4 5 6 10",
"output": "7"
},
{
"input": "5\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "2\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 0 0 0 0 0 2",
"output": "7"
},
{
"input": "10\n1 2 2 1 2 1 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "4\n0 1 1 0 0 0 0",
"output": "3"
},
{
"input": "100\n0 0 0 0 0 1 0",
"output": "6"
}
] | 1,687,467,751 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 92 | 0 | pages=int(input())
schedule=list(map(int,input().split()))
while pages>0:
for x in range (0,7):
if pages<=0:
break
day=x
pages-=schedule[x]
print(day+1) | Title: Petr and Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
Input Specification:
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Output Specification:
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Demo Input:
['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n']
Demo Output:
['6\n', '1\n']
Note:
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book. | ```python
pages=int(input())
schedule=list(map(int,input().split()))
while pages>0:
for x in range (0,7):
if pages<=0:
break
day=x
pages-=schedule[x]
print(day+1)
``` | 3 | |
771 | A | Bear and Friendship Condition | PROGRAMMING | 1,500 | [
"dfs and similar",
"dsu",
"graphs"
] | null | null | Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes. | The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input. | If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes). | [
"4 3\n1 3\n3 4\n1 4\n",
"4 4\n3 1\n2 3\n3 4\n1 2\n",
"10 4\n4 3\n5 10\n8 9\n1 2\n",
"3 2\n1 2\n2 3\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"NO\n"
] | The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not. | 250 | [
{
"input": "4 3\n1 3\n3 4\n1 4",
"output": "YES"
},
{
"input": "4 4\n3 1\n2 3\n3 4\n1 2",
"output": "NO"
},
{
"input": "10 4\n4 3\n5 10\n8 9\n1 2",
"output": "YES"
},
{
"input": "3 2\n1 2\n2 3",
"output": "NO"
},
{
"input": "3 0",
"output": "YES"
},
{
"input": "15 42\n8 1\n3 14\n7 14\n12 3\n7 9\n6 7\n6 12\n14 12\n3 10\n10 14\n6 3\n3 13\n13 10\n7 12\n7 2\n6 10\n11 4\n9 3\n8 4\n7 3\n2 3\n2 10\n9 13\n2 14\n6 14\n13 2\n1 4\n13 6\n7 10\n13 14\n12 10\n13 7\n12 2\n9 10\n13 12\n2 6\n9 14\n6 9\n12 9\n11 1\n2 9\n11 8",
"output": "YES"
},
{
"input": "20 80\n17 4\n10 1\n11 10\n17 7\n15 10\n14 15\n13 1\n18 13\n3 13\n12 7\n9 13\n10 12\n14 12\n18 11\n4 7\n10 13\n11 3\n19 8\n14 7\n10 17\n14 3\n7 11\n11 14\n19 5\n10 14\n15 17\n3 1\n9 10\n11 1\n4 1\n11 4\n9 1\n12 3\n13 7\n1 14\n11 12\n7 1\n9 12\n18 15\n17 3\n7 15\n4 10\n7 18\n7 9\n12 17\n14 18\n3 18\n18 17\n9 15\n14 4\n14 9\n9 18\n12 4\n7 10\n15 4\n4 18\n15 13\n1 12\n7 3\n13 11\n4 13\n5 8\n12 18\n12 15\n17 9\n11 15\n3 10\n18 10\n4 3\n15 3\n13 12\n9 4\n9 11\n14 17\n13 17\n3 9\n13 14\n1 17\n15 1\n17 11",
"output": "NO"
},
{
"input": "99 26\n64 17\n48 70\n71 50\n3 50\n9 60\n61 64\n53 50\n25 12\n3 71\n71 53\n3 53\n65 70\n9 25\n9 12\n59 56\n39 60\n64 69\n65 94\n70 94\n25 60\n60 12\n94 48\n17 69\n61 17\n65 48\n61 69",
"output": "NO"
},
{
"input": "3 1\n1 2",
"output": "YES"
},
{
"input": "3 2\n3 2\n1 3",
"output": "NO"
},
{
"input": "3 3\n2 3\n1 2\n1 3",
"output": "YES"
},
{
"input": "4 2\n4 1\n2 1",
"output": "NO"
},
{
"input": "4 3\n3 1\n2 1\n3 2",
"output": "YES"
},
{
"input": "5 9\n1 2\n5 1\n3 1\n1 4\n2 4\n5 3\n5 4\n2 3\n5 2",
"output": "NO"
},
{
"input": "10 5\n9 5\n1 2\n6 8\n6 3\n10 6",
"output": "NO"
},
{
"input": "10 8\n10 7\n9 7\n5 7\n6 8\n3 5\n8 10\n3 4\n7 8",
"output": "NO"
},
{
"input": "10 20\n8 2\n8 3\n1 8\n9 5\n2 4\n10 1\n10 5\n7 5\n7 8\n10 7\n6 5\n3 7\n1 9\n9 8\n7 2\n2 10\n2 1\n6 4\n9 7\n4 3",
"output": "NO"
},
{
"input": "150000 10\n62562 50190\n48849 60549\n139470 18456\n21436 25159\n66845 120884\n99972 114453\n11631 99153\n62951 134848\n78114 146050\n136760 131762",
"output": "YES"
},
{
"input": "150000 0",
"output": "YES"
},
{
"input": "4 4\n1 2\n2 3\n3 4\n1 4",
"output": "NO"
},
{
"input": "30 73\n25 2\n2 16\n20 12\n16 20\n7 18\n11 15\n13 11\n30 29\n16 12\n12 25\n2 1\n18 14\n9 8\n28 16\n2 9\n22 21\n1 25\n12 28\n14 7\n4 9\n26 7\n14 27\n12 2\n29 22\n1 9\n13 15\n3 10\n1 12\n8 20\n30 24\n25 20\n4 1\n4 12\n20 1\n8 4\n2 28\n25 16\n16 8\n20 4\n9 12\n21 30\n23 11\n19 6\n28 4\n29 21\n9 28\n30 10\n22 24\n25 8\n27 26\n25 4\n28 20\n9 25\n24 29\n20 9\n18 26\n1 28\n30 22\n23 15\n28 27\n8 2\n23 13\n12 8\n14 26\n16 4\n28 25\n8 1\n4 2\n9 16\n20 2\n18 27\n28 8\n27 7",
"output": "NO"
},
{
"input": "5 4\n1 2\n2 5\n3 4\n4 5",
"output": "NO"
},
{
"input": "4 4\n1 2\n2 3\n3 4\n4 1",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 4\n4 3\n1 5\n5 6\n6 3",
"output": "NO"
},
{
"input": "3 2\n1 2\n1 3",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 4\n4 5\n5 6\n1 6",
"output": "NO"
},
{
"input": "4 4\n1 2\n1 3\n2 4\n3 4",
"output": "NO"
},
{
"input": "6 9\n1 4\n1 5\n1 6\n2 4\n2 5\n2 6\n3 4\n3 5\n3 6",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n3 4",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n2 4",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 4\n4 5\n5 6\n6 1",
"output": "NO"
},
{
"input": "4 3\n1 2\n1 3\n1 4",
"output": "NO"
},
{
"input": "150000 3\n150000 149999\n149998 149999\n149998 150000",
"output": "YES"
},
{
"input": "10 15\n1 2\n1 3\n2 4\n2 5\n3 4\n3 5\n4 5\n1 6\n6 7\n6 8\n7 9\n7 10\n8 9\n8 10\n9 10",
"output": "NO"
},
{
"input": "5 5\n1 2\n2 3\n3 4\n4 5\n5 1",
"output": "NO"
},
{
"input": "5 4\n1 2\n1 3\n1 4\n1 5",
"output": "NO"
},
{
"input": "9 9\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 1",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n1 4\n2 3\n3 4",
"output": "NO"
},
{
"input": "4 4\n1 2\n1 3\n4 2\n4 3",
"output": "NO"
},
{
"input": "5 5\n1 4\n5 3\n5 1\n2 3\n4 5",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n2 3\n1 4\n2 4",
"output": "NO"
},
{
"input": "4 5\n1 2\n1 3\n1 4\n2 3\n2 4",
"output": "NO"
},
{
"input": "6 12\n1 2\n1 4\n1 5\n1 6\n2 3\n2 5\n2 6\n3 4\n3 5\n3 6\n4 5\n4 6",
"output": "NO"
},
{
"input": "1000 4\n100 132\n100 164\n196 132\n196 164",
"output": "NO"
}
] | 1,509,957,416 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 1,000 | 3,481,600 | n,m=map(int,input().split())
g=[]
for i in range(n):
g.append([])
for i in range(m):
o=list(map(int,input().split()))
o[0]-=1
o[1]-=1
g[o[0]].append(o[1])
g[o[1]].append(o[0])
for i in g:
l=len(i)
for j in range(l):
for k in range(j+1,l):
if not i[k] in g[i[j]]:
print('NO')
exit()
print('YES')
| Title: Bear and Friendship Condition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.
Input Specification:
The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input.
Output Specification:
If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).
Demo Input:
['4 3\n1 3\n3 4\n1 4\n', '4 4\n3 1\n2 3\n3 4\n1 2\n', '10 4\n4 3\n5 10\n8 9\n1 2\n', '3 2\n1 2\n2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n']
Note:
The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not. | ```python
n,m=map(int,input().split())
g=[]
for i in range(n):
g.append([])
for i in range(m):
o=list(map(int,input().split()))
o[0]-=1
o[1]-=1
g[o[0]].append(o[1])
g[o[1]].append(o[0])
for i in g:
l=len(i)
for j in range(l):
for k in range(j+1,l):
if not i[k] in g[i[j]]:
print('NO')
exit()
print('YES')
``` | 0 | |
96 | A | Football | PROGRAMMING | 900 | [
"implementation",
"strings"
] | A. Football | 2 | 256 | Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. | The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. | Print "YES" if the situation is dangerous. Otherwise, print "NO". | [
"001001\n",
"1000000001\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"output": "NO"
},
{
"input": "1010010100000000010",
"output": "YES"
},
{
"input": "101010101",
"output": "NO"
},
{
"input": "000000000100000000000110101100000",
"output": "YES"
},
{
"input": "100001000000110101100000",
"output": "NO"
},
{
"input": "100001000011010110000",
"output": "NO"
},
{
"input": "010",
"output": "NO"
},
{
"input": "10101011111111111111111111111100",
"output": "YES"
},
{
"input": "1001101100",
"output": "NO"
},
{
"input": "1001101010",
"output": "NO"
},
{
"input": "1111100111",
"output": "NO"
},
{
"input": "00110110001110001111",
"output": "NO"
},
{
"input": "11110001001111110001",
"output": "NO"
},
{
"input": "10001111001011111101",
"output": "NO"
},
{
"input": "10000010100000001000110001010100001001001010011",
"output": "YES"
},
{
"input": "01111011111010111100101100001011001010111110000010",
"output": "NO"
},
{
"input": "00100000100100101110011001011011101110110110010100",
"output": "NO"
},
{
"input": "10110100110001001011110101110010100010000000000100101010111110111110100011",
"output": "YES"
},
{
"input": "00011101010101111001011011001101101011111101000010100000111000011100101011",
"output": "NO"
},
{
"input": "01110000110100110101110100111000101101011101011110110100100111100001110111",
"output": "NO"
},
{
"input": "11110110011000100111100111101101011111110100010101011011111101110110110111",
"output": "YES"
},
{
"input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100",
"output": "NO"
},
{
"input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010",
"output": "NO"
},
{
"input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110",
"output": "NO"
},
{
"input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000",
"output": "NO"
},
{
"input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110",
"output": "YES"
},
{
"input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000",
"output": "NO"
},
{
"input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101",
"output": "YES"
},
{
"input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101",
"output": "YES"
},
{
"input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000",
"output": "NO"
},
{
"input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000",
"output": "YES"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "YES"
},
{
"input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111",
"output": "YES"
},
{
"input": "10100101000",
"output": "NO"
},
{
"input": "11110111011101",
"output": "NO"
},
{
"input": "10000000",
"output": "YES"
},
{
"input": "00000001",
"output": "YES"
},
{
"input": "01111111",
"output": "YES"
},
{
"input": "11111110",
"output": "YES"
}
] | 1,675,573,150 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 92 | 0 | string = input()
count = 0
tempLetter = "."
for letter in string:
if tempLetter != letter:
count = 1
tempLetter = letter
else:
count += 1
if count == 7:
print("YES")
break
if count < 7:
print("NO") | Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
string = input()
count = 0
tempLetter = "."
for letter in string:
if tempLetter != letter:
count = 1
tempLetter = letter
else:
count += 1
if count == 7:
print("YES")
break
if count < 7:
print("NO")
``` | 3.977 |
1,004 | A | Sonya and Hotels | PROGRAMMING | 900 | [
"implementation"
] | null | null | Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has $n$ hotels, where the $i$-th hotel is located in the city with coordinate $x_i$. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to $d$. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original $n$ hotels to the new one is equal to $d$. | The first line contains two integers $n$ and $d$ ($1\leq n\leq 100$, $1\leq d\leq 10^9$) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains $n$ different integers in strictly increasing order $x_1, x_2, \ldots, x_n$ ($-10^9\leq x_i\leq 10^9$) — coordinates of Sonya's hotels. | Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to $d$. | [
"4 3\n-3 2 9 16\n",
"5 2\n4 8 11 18 19\n"
] | [
"6\n",
"5\n"
] | In the first example, there are $6$ possible cities where Sonya can build a hotel. These cities have coordinates $-6$, $5$, $6$, $12$, $13$, and $19$.
In the second example, there are $5$ possible cities where Sonya can build a hotel. These cities have coordinates $2$, $6$, $13$, $16$, and $21$. | 500 | [
{
"input": "4 3\n-3 2 9 16",
"output": "6"
},
{
"input": "5 2\n4 8 11 18 19",
"output": "5"
},
{
"input": "10 10\n-67 -59 -49 -38 -8 20 41 59 74 83",
"output": "8"
},
{
"input": "10 10\n0 20 48 58 81 95 111 137 147 159",
"output": "9"
},
{
"input": "100 1\n0 1 2 3 4 5 7 8 10 11 12 13 14 15 16 17 19 21 22 23 24 25 26 27 28 30 32 33 36 39 40 41 42 46 48 53 54 55 59 60 61 63 65 68 70 71 74 75 76 79 80 81 82 84 88 89 90 91 93 94 96 97 98 100 101 102 105 106 107 108 109 110 111 113 114 115 116 117 118 120 121 122 125 126 128 131 132 133 134 135 137 138 139 140 143 144 146 147 148 149",
"output": "47"
},
{
"input": "1 1000000000\n-1000000000",
"output": "2"
},
{
"input": "2 1000000000\n-1000000000 1000000000",
"output": "3"
},
{
"input": "100 2\n1 3 5 6 8 9 12 13 14 17 18 21 22 23 24 25 26 27 29 30 34 35 36 39 41 44 46 48 52 53 55 56 57 59 61 63 64 66 68 69 70 71 72 73 75 76 77 79 80 81 82 87 88 91 92 93 94 95 96 97 99 100 102 103 104 106 109 110 111 112 113 114 115 117 118 119 120 122 124 125 127 128 129 130 131 132 133 134 136 137 139 140 141 142 143 145 146 148 149 150",
"output": "6"
},
{
"input": "100 3\n0 1 3 6 7 8 9 10 13 14 16 17 18 20 21 22 24 26 27 30 33 34 35 36 37 39 42 43 44 45 46 48 53 54 55 56 57 58 61 63 64 65 67 69 70 72 73 76 77 78 79 81 82 83 85 86 87 88 90 92 93 95 96 97 98 99 100 101 104 105 108 109 110 113 114 115 116 118 120 121 123 124 125 128 130 131 132 133 134 135 136 137 139 140 141 142 146 147 148 150",
"output": "2"
},
{
"input": "1 1000000000\n1000000000",
"output": "2"
},
{
"input": "10 2\n-93 -62 -53 -42 -38 11 57 58 87 94",
"output": "17"
},
{
"input": "2 500000000\n-1000000000 1000000000",
"output": "4"
},
{
"input": "100 10\n-489 -476 -445 -432 -430 -421 -420 -418 -412 -411 -404 -383 -356 -300 -295 -293 -287 -276 -265 -263 -258 -251 -249 -246 -220 -219 -205 -186 -166 -157 -143 -137 -136 -130 -103 -86 -80 -69 -67 -55 -43 -41 -40 -26 -19 -9 16 29 41 42 54 76 84 97 98 99 101 115 134 151 157 167 169 185 197 204 208 226 227 232 234 249 259 266 281 282 293 298 300 306 308 313 319 328 331 340 341 344 356 362 366 380 390 399 409 411 419 444 455 498",
"output": "23"
},
{
"input": "1 1000000000\n999999999",
"output": "2"
},
{
"input": "1 1\n-5",
"output": "2"
},
{
"input": "2 1\n-1000000000 1000000000",
"output": "4"
},
{
"input": "1 2\n1",
"output": "2"
},
{
"input": "4 5\n0 20 40 60",
"output": "8"
},
{
"input": "1 1\n-10",
"output": "2"
},
{
"input": "1 1\n-1000000000",
"output": "2"
},
{
"input": "1 1000000000\n0",
"output": "2"
},
{
"input": "1 2\n-10",
"output": "2"
},
{
"input": "1 1\n2",
"output": "2"
},
{
"input": "1 5\n-100",
"output": "2"
},
{
"input": "2 1000000000\n999999999 1000000000",
"output": "2"
},
{
"input": "2 2\n0 5",
"output": "4"
},
{
"input": "1 2\n-5",
"output": "2"
},
{
"input": "1 1\n1",
"output": "2"
},
{
"input": "1 5\n-20",
"output": "2"
},
{
"input": "2 100000000\n-1000000000 1000000000",
"output": "4"
},
{
"input": "1 1000000000\n147483641",
"output": "2"
},
{
"input": "3 3\n1 8 9",
"output": "4"
},
{
"input": "2 1\n1 1000000000",
"output": "4"
},
{
"input": "1 2\n-4",
"output": "2"
},
{
"input": "2 1\n-1000000000 -1",
"output": "4"
}
] | 1,530,850,433 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 78 | 0 | noh, dist = [int(x) for x in input().split()]
hotels = [int(x) for x in input().split()]
print( 2 + sum([ (b - a >= dist) + (b - a > dist) for a, b in zip(hotels, hotels[1:])]) ) | Title: Sonya and Hotels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has $n$ hotels, where the $i$-th hotel is located in the city with coordinate $x_i$. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to $d$. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original $n$ hotels to the new one is equal to $d$.
Input Specification:
The first line contains two integers $n$ and $d$ ($1\leq n\leq 100$, $1\leq d\leq 10^9$) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains $n$ different integers in strictly increasing order $x_1, x_2, \ldots, x_n$ ($-10^9\leq x_i\leq 10^9$) — coordinates of Sonya's hotels.
Output Specification:
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to $d$.
Demo Input:
['4 3\n-3 2 9 16\n', '5 2\n4 8 11 18 19\n']
Demo Output:
['6\n', '5\n']
Note:
In the first example, there are $6$ possible cities where Sonya can build a hotel. These cities have coordinates $-6$, $5$, $6$, $12$, $13$, and $19$.
In the second example, there are $5$ possible cities where Sonya can build a hotel. These cities have coordinates $2$, $6$, $13$, $16$, and $21$. | ```python
noh, dist = [int(x) for x in input().split()]
hotels = [int(x) for x in input().split()]
print( 2 + sum([ (b - a >= dist) + (b - a > dist) for a, b in zip(hotels, hotels[1:])]) )
``` | 0 | |
811 | B | Vladik and Complicated Book | PROGRAMMING | 1,200 | [
"implementation",
"sortings"
] | null | null | Vladik had started reading a complicated book about algorithms containing *n* pages. To improve understanding of what is written, his friends advised him to read pages in some order given by permutation *P*<==<=[*p*1,<=*p*2,<=...,<=*p**n*], where *p**i* denotes the number of page that should be read *i*-th in turn.
Sometimes Vladik’s mom sorted some subsegment of permutation *P* from position *l* to position *r* inclusive, because she loves the order. For every of such sorting Vladik knows number *x* — what index of page in permutation he should read. He is wondered if the page, which he will read after sorting, has changed. In other words, has *p**x* changed? After every sorting Vladik return permutation to initial state, so you can assume that each sorting is independent from each other. | First line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=104) — length of permutation and number of times Vladik's mom sorted some subsegment of the book.
Second line contains *n* space-separated integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — permutation *P*. Note that elements in permutation are distinct.
Each of the next *m* lines contains three space-separated integers *l**i*, *r**i*, *x**i* (1<=≤<=*l**i*<=≤<=*x**i*<=≤<=*r**i*<=≤<=*n*) — left and right borders of sorted subsegment in *i*-th sorting and position that is interesting to Vladik. | For each mom’s sorting on it’s own line print "Yes", if page which is interesting to Vladik hasn't changed, or "No" otherwise. | [
"5 5\n5 4 3 2 1\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3\n",
"6 5\n1 4 3 2 5 6\n2 4 3\n1 6 2\n4 5 4\n1 3 3\n2 6 3\n"
] | [
"Yes\nNo\nYes\nYes\nNo\n",
"Yes\nNo\nYes\nNo\nYes\n"
] | Explanation of first test case:
1. [1, 2, 3, 4, 5] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [3, 4, 5, 2, 1] — permutation after sorting, 1-st element has changed, so answer is "No". 1. [5, 2, 3, 4, 1] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [5, 4, 3, 2, 1] — permutation after sorting, 4-th element hasn’t changed, so answer is "Yes". 1. [5, 1, 2, 3, 4] — permutation after sorting, 3-rd element has changed, so answer is "No". | 1,000 | [
{
"input": "5 5\n5 4 3 2 1\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3",
"output": "Yes\nNo\nYes\nYes\nNo"
},
{
"input": "6 5\n1 4 3 2 5 6\n2 4 3\n1 6 2\n4 5 4\n1 3 3\n2 6 3",
"output": "Yes\nNo\nYes\nNo\nYes"
},
{
"input": "10 10\n10 1 6 7 9 8 4 3 5 2\n1 1 1\n4 4 4\n7 7 7\n3 3 3\n1 6 5\n2 6 2\n6 8 7\n1 1 1\n7 9 9\n2 9 4",
"output": "Yes\nYes\nYes\nYes\nYes\nYes\nYes\nYes\nYes\nNo"
},
{
"input": "20 20\n18 17 2 3 16 15 1 9 12 8 20 11 13 14 4 5 19 7 10 6\n13 15 15\n1 1 1\n2 2 2\n11 14 13\n10 11 10\n2 8 6\n12 18 16\n4 8 8\n2 2 2\n5 11 11\n4 9 9\n5 6 6\n3 20 12\n8 8 8\n6 16 11\n9 18 18\n8 18 17\n1 1 1\n2 6 5\n1 4 3",
"output": "No\nYes\nYes\nNo\nYes\nYes\nNo\nNo\nYes\nYes\nNo\nNo\nNo\nYes\nNo\nNo\nYes\nYes\nYes\nNo"
},
{
"input": "5 10\n5 4 3 2 1\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3",
"output": "Yes\nNo\nYes\nYes\nNo\nYes\nNo\nYes\nYes\nNo"
}
] | 1,498,159,650 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 2,000 | 6,041,600 | x,y=map(int,input().split())
t=0
awan=list(map(int,input().split()))
for v in range(0,y) :
l,r,x=map(int,input().split())
l-=1
x-=1
t=0
for j in range(l,r):
if awan[x]>awan[j] :
t+=1
if (l+t)==x :
print("Yes")
else :
print("No")
| Title: Vladik and Complicated Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vladik had started reading a complicated book about algorithms containing *n* pages. To improve understanding of what is written, his friends advised him to read pages in some order given by permutation *P*<==<=[*p*1,<=*p*2,<=...,<=*p**n*], where *p**i* denotes the number of page that should be read *i*-th in turn.
Sometimes Vladik’s mom sorted some subsegment of permutation *P* from position *l* to position *r* inclusive, because she loves the order. For every of such sorting Vladik knows number *x* — what index of page in permutation he should read. He is wondered if the page, which he will read after sorting, has changed. In other words, has *p**x* changed? After every sorting Vladik return permutation to initial state, so you can assume that each sorting is independent from each other.
Input Specification:
First line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=104) — length of permutation and number of times Vladik's mom sorted some subsegment of the book.
Second line contains *n* space-separated integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — permutation *P*. Note that elements in permutation are distinct.
Each of the next *m* lines contains three space-separated integers *l**i*, *r**i*, *x**i* (1<=≤<=*l**i*<=≤<=*x**i*<=≤<=*r**i*<=≤<=*n*) — left and right borders of sorted subsegment in *i*-th sorting and position that is interesting to Vladik.
Output Specification:
For each mom’s sorting on it’s own line print "Yes", if page which is interesting to Vladik hasn't changed, or "No" otherwise.
Demo Input:
['5 5\n5 4 3 2 1\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3\n', '6 5\n1 4 3 2 5 6\n2 4 3\n1 6 2\n4 5 4\n1 3 3\n2 6 3\n']
Demo Output:
['Yes\nNo\nYes\nYes\nNo\n', 'Yes\nNo\nYes\nNo\nYes\n']
Note:
Explanation of first test case:
1. [1, 2, 3, 4, 5] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [3, 4, 5, 2, 1] — permutation after sorting, 1-st element has changed, so answer is "No". 1. [5, 2, 3, 4, 1] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [5, 4, 3, 2, 1] — permutation after sorting, 4-th element hasn’t changed, so answer is "Yes". 1. [5, 1, 2, 3, 4] — permutation after sorting, 3-rd element has changed, so answer is "No". | ```python
x,y=map(int,input().split())
t=0
awan=list(map(int,input().split()))
for v in range(0,y) :
l,r,x=map(int,input().split())
l-=1
x-=1
t=0
for j in range(l,r):
if awan[x]>awan[j] :
t+=1
if (l+t)==x :
print("Yes")
else :
print("No")
``` | 0 | |
961 | A | Tetris | PROGRAMMING | 900 | [
"implementation"
] | null | null | You are given a following process.
There is a platform with $n$ columns. $1 \times 1$ squares are appearing one after another in some columns on this platform. If there are no squares in the column, a square will occupy the bottom row. Otherwise a square will appear at the top of the highest square of this column.
When all of the $n$ columns have at least one square in them, the bottom row is being removed. You will receive $1$ point for this, and all the squares left will fall down one row.
You task is to calculate the amount of points you will receive. | The first line of input contain 2 integer numbers $n$ and $m$ ($1 \le n, m \le 1000$) — the length of the platform and the number of the squares.
The next line contain $m$ integer numbers $c_1, c_2, \dots, c_m$ ($1 \le c_i \le n$) — column in which $i$-th square will appear. | Print one integer — the amount of points you will receive. | [
"3 9\n1 1 2 2 2 3 1 2 3\n"
] | [
"2\n"
] | In the sample case the answer will be equal to $2$ because after the appearing of $6$-th square will be removed one row (counts of the squares on the platform will look like $[2~ 3~ 1]$, and after removing one row will be $[1~ 2~ 0]$).
After the appearing of $9$-th square counts will be $[2~ 3~ 1]$, and after removing one row it will look like $[1~ 2~ 0]$.
So the answer will be equal to $2$. | 0 | [
{
"input": "3 9\n1 1 2 2 2 3 1 2 3",
"output": "2"
},
{
"input": "1 7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "3 5\n1 1 1 2 3",
"output": "1"
},
{
"input": "4 6\n4 4 4 4 4 4",
"output": "0"
},
{
"input": "4 6\n2 3 4 4 4 4",
"output": "0"
},
{
"input": "3 12\n1 1 1 1 2 2 2 2 3 3 3 3",
"output": "4"
},
{
"input": "8 8\n2 2 3 4 5 6 7 8",
"output": "0"
},
{
"input": "100 1\n50",
"output": "0"
},
{
"input": "2 1\n2",
"output": "0"
},
{
"input": "2 1\n1",
"output": "0"
},
{
"input": "2 4\n1 2 1 1",
"output": "1"
},
{
"input": "3 4\n3 2 2 2",
"output": "0"
},
{
"input": "2 2\n2 2",
"output": "0"
},
{
"input": "2 5\n2 1 1 2 1",
"output": "2"
},
{
"input": "15 3\n13 14 15",
"output": "0"
},
{
"input": "4 9\n1 2 3 1 2 3 1 2 3",
"output": "0"
},
{
"input": "100 3\n1 2 3",
"output": "0"
},
{
"input": "1000 10\n999 999 998 34 454 546 343 35 34 1000",
"output": "0"
},
{
"input": "4 2\n1 2",
"output": "0"
}
] | 1,669,184,160 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 29 | 62 | 0 | import sys
input=sys.stdin.readline
n, m = map(int, input().split())
vals = [0] * n
x = list(map(int, input().split()))
for i in x: vals[i - 1] += 1
print(min(vals))
| Title: Tetris
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a following process.
There is a platform with $n$ columns. $1 \times 1$ squares are appearing one after another in some columns on this platform. If there are no squares in the column, a square will occupy the bottom row. Otherwise a square will appear at the top of the highest square of this column.
When all of the $n$ columns have at least one square in them, the bottom row is being removed. You will receive $1$ point for this, and all the squares left will fall down one row.
You task is to calculate the amount of points you will receive.
Input Specification:
The first line of input contain 2 integer numbers $n$ and $m$ ($1 \le n, m \le 1000$) — the length of the platform and the number of the squares.
The next line contain $m$ integer numbers $c_1, c_2, \dots, c_m$ ($1 \le c_i \le n$) — column in which $i$-th square will appear.
Output Specification:
Print one integer — the amount of points you will receive.
Demo Input:
['3 9\n1 1 2 2 2 3 1 2 3\n']
Demo Output:
['2\n']
Note:
In the sample case the answer will be equal to $2$ because after the appearing of $6$-th square will be removed one row (counts of the squares on the platform will look like $[2~ 3~ 1]$, and after removing one row will be $[1~ 2~ 0]$).
After the appearing of $9$-th square counts will be $[2~ 3~ 1]$, and after removing one row it will look like $[1~ 2~ 0]$.
So the answer will be equal to $2$. | ```python
import sys
input=sys.stdin.readline
n, m = map(int, input().split())
vals = [0] * n
x = list(map(int, input().split()))
for i in x: vals[i - 1] += 1
print(min(vals))
``` | 3 | |
4 | A | Watermelon | PROGRAMMING | 800 | [
"brute force",
"math"
] | A. Watermelon | 1 | 64 | One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. | The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. | Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. | [
"8\n"
] | [
"YES\n"
] | For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | 0 | [
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "10",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "53",
"output": "NO"
},
{
"input": "77",
"output": "NO"
},
{
"input": "32",
"output": "YES"
},
{
"input": "44",
"output": "YES"
},
{
"input": "98",
"output": "YES"
},
{
"input": "99",
"output": "NO"
},
{
"input": "90",
"output": "YES"
},
{
"input": "67",
"output": "NO"
},
{
"input": "100",
"output": "YES"
},
{
"input": "88",
"output": "YES"
}
] | 1,688,932,073 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | w=int(input("enter w"))
def divid(w):
if w%2==0:
return "TRUE"
else:
return "FALSE"
print(divid(w)) | Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | ```python
w=int(input("enter w"))
def divid(w):
if w%2==0:
return "TRUE"
else:
return "FALSE"
print(divid(w))
``` | 0 |
631 | B | Print Check | PROGRAMMING | 1,200 | [
"constructive algorithms",
"implementation"
] | null | null | Kris works in a large company "Blake Technologies". As a best engineer of the company he was assigned a task to develop a printer that will be able to print horizontal and vertical strips. First prototype is already built and Kris wants to tests it. He wants you to implement the program that checks the result of the printing.
Printer works with a rectangular sheet of paper of size *n*<=×<=*m*. Consider the list as a table consisting of *n* rows and *m* columns. Rows are numbered from top to bottom with integers from 1 to *n*, while columns are numbered from left to right with integers from 1 to *m*. Initially, all cells are painted in color 0.
Your program has to support two operations:
1. Paint all cells in row *r**i* in color *a**i*; 1. Paint all cells in column *c**i* in color *a**i*.
If during some operation *i* there is a cell that have already been painted, the color of this cell also changes to *a**i*.
Your program has to print the resulting table after *k* operation. | The first line of the input contains three integers *n*, *m* and *k* (1<=<=≤<=<=*n*,<=<=*m*<=<=≤<=5000, *n*·*m*<=≤<=100<=000, 1<=≤<=*k*<=≤<=100<=000) — the dimensions of the sheet and the number of operations, respectively.
Each of the next *k* lines contains the description of exactly one query:
- 1 *r**i* *a**i* (1<=≤<=*r**i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=109), means that row *r**i* is painted in color *a**i*; - 2 *c**i* *a**i* (1<=≤<=*c**i*<=≤<=*m*, 1<=≤<=*a**i*<=≤<=109), means that column *c**i* is painted in color *a**i*. | Print *n* lines containing *m* integers each — the resulting table after all operations are applied. | [
"3 3 3\n1 1 3\n2 2 1\n1 2 2\n",
"5 3 5\n1 1 1\n1 3 1\n1 5 1\n2 1 1\n2 3 1\n"
] | [
"3 1 3 \n2 2 2 \n0 1 0 \n",
"1 1 1 \n1 0 1 \n1 1 1 \n1 0 1 \n1 1 1 \n"
] | The figure below shows all three operations for the first sample step by step. The cells that were painted on the corresponding step are marked gray. | 1,000 | [
{
"input": "3 3 3\n1 1 3\n2 2 1\n1 2 2",
"output": "3 1 3 \n2 2 2 \n0 1 0 "
},
{
"input": "5 3 5\n1 1 1\n1 3 1\n1 5 1\n2 1 1\n2 3 1",
"output": "1 1 1 \n1 0 1 \n1 1 1 \n1 0 1 \n1 1 1 "
},
{
"input": "5 5 4\n1 2 1\n1 4 1\n2 2 1\n2 4 1",
"output": "0 1 0 1 0 \n1 1 1 1 1 \n0 1 0 1 0 \n1 1 1 1 1 \n0 1 0 1 0 "
},
{
"input": "4 6 8\n1 2 1\n2 2 2\n2 5 2\n1 1 1\n1 4 1\n1 3 2\n2 1 1\n2 6 1",
"output": "1 1 1 1 1 1 \n1 2 1 1 2 1 \n1 2 2 2 2 1 \n1 1 1 1 1 1 "
},
{
"input": "2 2 3\n1 1 1\n1 2 1\n2 1 2",
"output": "2 1 \n2 1 "
},
{
"input": "1 2 4\n1 1 1\n2 1 2\n2 2 3\n1 1 4",
"output": "4 4 "
},
{
"input": "2 1 5\n1 1 7\n1 2 77\n2 1 777\n1 1 77\n1 2 7",
"output": "77 \n7 "
},
{
"input": "2 1 1\n1 2 1000000000",
"output": "0 \n1000000000 "
},
{
"input": "1 2 1\n2 2 1000000000",
"output": "0 1000000000 "
},
{
"input": "160 600 1\n1 132 589472344",
"output": "0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "600 160 1\n1 124 542622711",
"output": "0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 \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..."
},
{
"input": "10 1 1\n2 1 1000000000",
"output": "1000000000 \n1000000000 \n1000000000 \n1000000000 \n1000000000 \n1000000000 \n1000000000 \n1000000000 \n1000000000 \n1000000000 "
},
{
"input": "1 10 1\n1 1 1000000000",
"output": "1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 "
},
{
"input": "5000 20 15\n2 13 447246914\n2 10 89345638\n2 6 393683717\n2 1 62225152\n2 12 990340161\n2 4 227462932\n1 4011 327145900\n1 1915 981331082\n1 802 437883065\n2 11 205232924\n2 15 303578182\n1 2667 835309719\n2 3 550440583\n1 3852 766318960\n2 5 204700467",
"output": "62225152 0 550440583 227462932 204700467 393683717 0 0 0 89345638 205232924 990340161 447246914 0 303578182 0 0 0 0 0 \n62225152 0 550440583 227462932 204700467 393683717 0 0 0 89345638 205232924 990340161 447246914 0 303578182 0 0 0 0 0 \n62225152 0 550440583 227462932 204700467 393683717 0 0 0 89345638 205232924 990340161 447246914 0 303578182 0 0 0 0 0 \n62225152 0 550440583 227462932 204700467 393683717 0 0 0 89345638 205232924 990340161 447246914 0 303578182 0 0 0 0 0 \n62225152 0 550440583 227462932 ..."
},
{
"input": "20 5000 15\n1 14 94104058\n1 3 170575572\n1 12 613859505\n2 4797 363375206\n1 10 554079405\n2 4481 223609505\n2 4974 385117513\n1 7 363246561\n2 2503 128936152\n1 2 152944073\n1 14 25797327\n1 20 258831369\n1 10 453164078\n2 2159 753235825\n2 1823 827844660",
"output": "0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "1 1 1\n1 1 1",
"output": "1 "
},
{
"input": "1 1 1\n1 1 1000000000",
"output": "1000000000 "
},
{
"input": "1 1 2\n1 1 123\n1 1 321",
"output": "321 "
},
{
"input": "1 1 2\n1 1 321\n1 1 123",
"output": "123 "
}
] | 1,574,288,221 | 2,147,483,647 | Python 3 | OK | TESTS | 56 | 546 | 7,475,200 | n, m, k = [ int(x) for x in input().split() ]
instante_linhas = [0] * n
instante_colunas = [0] * m
cores = [0] * (k + 1)
for i in range(k):
op, lc, cor = [ int(x) for x in input().split() ]
cores[i + 1] = cor
if op == 1:
instante_linhas[lc - 1] = i + 1
elif op == 2:
instante_colunas[lc - 1] = i + 1
resp = ""
for i in range(n):
for j in range(m):
resp += str(cores[max(instante_linhas[i], instante_colunas[j])])
if j != m - 1:
resp += " "
if i != n - 1:
resp += "\n"
print(resp)
| Title: Print Check
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kris works in a large company "Blake Technologies". As a best engineer of the company he was assigned a task to develop a printer that will be able to print horizontal and vertical strips. First prototype is already built and Kris wants to tests it. He wants you to implement the program that checks the result of the printing.
Printer works with a rectangular sheet of paper of size *n*<=×<=*m*. Consider the list as a table consisting of *n* rows and *m* columns. Rows are numbered from top to bottom with integers from 1 to *n*, while columns are numbered from left to right with integers from 1 to *m*. Initially, all cells are painted in color 0.
Your program has to support two operations:
1. Paint all cells in row *r**i* in color *a**i*; 1. Paint all cells in column *c**i* in color *a**i*.
If during some operation *i* there is a cell that have already been painted, the color of this cell also changes to *a**i*.
Your program has to print the resulting table after *k* operation.
Input Specification:
The first line of the input contains three integers *n*, *m* and *k* (1<=<=≤<=<=*n*,<=<=*m*<=<=≤<=5000, *n*·*m*<=≤<=100<=000, 1<=≤<=*k*<=≤<=100<=000) — the dimensions of the sheet and the number of operations, respectively.
Each of the next *k* lines contains the description of exactly one query:
- 1 *r**i* *a**i* (1<=≤<=*r**i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=109), means that row *r**i* is painted in color *a**i*; - 2 *c**i* *a**i* (1<=≤<=*c**i*<=≤<=*m*, 1<=≤<=*a**i*<=≤<=109), means that column *c**i* is painted in color *a**i*.
Output Specification:
Print *n* lines containing *m* integers each — the resulting table after all operations are applied.
Demo Input:
['3 3 3\n1 1 3\n2 2 1\n1 2 2\n', '5 3 5\n1 1 1\n1 3 1\n1 5 1\n2 1 1\n2 3 1\n']
Demo Output:
['3 1 3 \n2 2 2 \n0 1 0 \n', '1 1 1 \n1 0 1 \n1 1 1 \n1 0 1 \n1 1 1 \n']
Note:
The figure below shows all three operations for the first sample step by step. The cells that were painted on the corresponding step are marked gray. | ```python
n, m, k = [ int(x) for x in input().split() ]
instante_linhas = [0] * n
instante_colunas = [0] * m
cores = [0] * (k + 1)
for i in range(k):
op, lc, cor = [ int(x) for x in input().split() ]
cores[i + 1] = cor
if op == 1:
instante_linhas[lc - 1] = i + 1
elif op == 2:
instante_colunas[lc - 1] = i + 1
resp = ""
for i in range(n):
for j in range(m):
resp += str(cores[max(instante_linhas[i], instante_colunas[j])])
if j != m - 1:
resp += " "
if i != n - 1:
resp += "\n"
print(resp)
``` | 3 |
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