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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
266 | B | Queue at the School | PROGRAMMING | 800 | [
"constructive algorithms",
"graph matchings",
"implementation",
"shortest paths"
] | null | null | During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds. | The first line contains two integers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G". | Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G". | [
"5 1\nBGGBG\n",
"5 2\nBGGBG\n",
"4 1\nGGGB\n"
] | [
"GBGGB\n",
"GGBGB\n",
"GGGB\n"
] | none | 500 | [
{
"input": "5 1\nBGGBG",
"output": "GBGGB"
},
{
"input": "5 2\nBGGBG",
"output": "GGBGB"
},
{
"input": "4 1\nGGGB",
"output": "GGGB"
},
{
"input": "2 1\nBB",
"output": "BB"
},
{
"input": "2 1\nBG",
"output": "GB"
},
{
"input": "6 2\nBBGBBG",
"output": "GBBGBB"
},
{
"input": "8 3\nBBGBGBGB",
"output": "GGBGBBBB"
},
{
"input": "10 3\nBBGBBBBBBG",
"output": "GBBBBBGBBB"
},
{
"input": "22 7\nGBGGBGGGGGBBBGGBGBGBBB",
"output": "GGGGGGGGBGGBGGBBBBBBBB"
},
{
"input": "50 4\nGBBGBBBGGGGGBBGGBBBBGGGBBBGBBBGGBGGBGBBBGGBGGBGGBG",
"output": "GGBGBGBGBGBGGGBBGBGBGBGBBBGBGBGBGBGBGBGBGBGBGGBGBB"
},
{
"input": "50 8\nGGGGBGGBGGGBGBBBGGGGGGGGBBGBGBGBBGGBGGBGGGGGGGGBBG",
"output": "GGGGGGGGGGGGBGGBGBGBGBGBGGGGGGBGBGBGBGBGBGGBGGBGBB"
},
{
"input": "50 30\nBGGGGGGBGGBGBGGGGBGBBGBBBGGBBBGBGBGGGGGBGBBGBGBGGG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "20 20\nBBGGBGGGGBBBGBBGGGBB",
"output": "GGGGGGGGGGBBBBBBBBBB"
},
{
"input": "27 6\nGBGBGBGGGGGGBGGBGGBBGBBBGBB",
"output": "GGGGGGGBGBGBGGGGGBGBBBBBBBB"
},
{
"input": "46 11\nBGGGGGBGBGGBGGGBBGBBGBBGGBBGBBGBGGGGGGGBGBGBGB",
"output": "GGGGGGGGGGGBGGGGGBBGBGBGBGBGBGBGBGBGBGBGBBBBBB"
},
{
"input": "50 6\nBGGBBBBGGBBBBBBGGBGBGBBBBGBBBBBBGBBBBBBBBBBBBBBBBB",
"output": "GGGGBBBBBGBGBGBGBBBGBBBBBBGBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 10\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 8\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "50 10\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBGB",
"output": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBGBBBBBBBBBBB"
},
{
"input": "50 13\nGGGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "GGGGGGGGGGGGGGGGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "1 1\nB",
"output": "B"
},
{
"input": "1 1\nG",
"output": "G"
},
{
"input": "1 50\nB",
"output": "B"
},
{
"input": "1 50\nG",
"output": "G"
},
{
"input": "50 50\nBBBBBBBBGGBBBBBBGBBBBBBBBBBBGBBBBBBBBBBBBBBGBBBBBB",
"output": "GGGGGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 50\nGGBBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBGGGGGGBG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBBBB"
},
{
"input": "6 3\nGGBBBG",
"output": "GGGBBB"
},
{
"input": "26 3\nGBBGBBBBBGGGBGBGGGBGBGGBBG",
"output": "GGBBBBGBGBGBGGGBGBGGGBGBBB"
},
{
"input": "46 3\nGGBBGGGGBBGBGBBBBBGGGBGGGBBGGGBBBGGBGGBBBGBGBB",
"output": "GGGGBGBGGGBBBBBGBGBGBGGGBGGBGBGBGBGBGBGBGBBBBB"
},
{
"input": "44 8\nBGBBBBBBBBBGGBBGBGBGGBBBBBGBBGBBBBBBBBBGBBGB",
"output": "GBBGBGBGBGBGBGBBBBGBBGBBBBBBBBBGBBGBBBBBBBBB"
},
{
"input": "20 20\nBBGGBGGGGBBBGBBGGGBB",
"output": "GGGGGGGGGGBBBBBBBBBB"
},
{
"input": "30 25\nBGGBBGBGGBGBGBBGBGGGGBGBGGBBBB",
"output": "GGGGGGGGGGGGGGGBBBBBBBBBBBBBBB"
},
{
"input": "17 42\nBBGBGBGGGGGGBBGGG",
"output": "GGGGGGGGGGGBBBBBB"
},
{
"input": "30 50\nBGGBBGGGGGGGGBBGGGBBGGBBBGBBGG",
"output": "GGGGGGGGGGGGGGGGGGBBBBBBBBBBBB"
},
{
"input": "31 33\nBBGGBBGGBGBBBBGGBGBBBGBGGBBGBBB",
"output": "GGGGGGGGGGGGBBBBBBBBBBBBBBBBBBB"
},
{
"input": "2 2\nBG",
"output": "GB"
},
{
"input": "8 8\nGGGBGGBB",
"output": "GGGGGBBB"
},
{
"input": "34 34\nBGGBBBBGGBBGBGGGGGGBBGGGGGBGGBGGGB",
"output": "GGGGGGGGGGGGGGGGGGGGGBBBBBBBBBBBBB"
},
{
"input": "34 20\nBBBBGBGGGBGGGBGBGGBBBBGGBBGGGBGBBG",
"output": "GGGGGGGGGGGGGGGGGBBBBBBBBBBBBBBBBB"
},
{
"input": "50 50\nBBGBBBBBBBBBBBBGBBBGBBBBBBBGBBBBBBGBBBGBBGBBGBBBGG",
"output": "GGGGGGGGGGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "10 10\nGGGGGGGGGG",
"output": "GGGGGGGGGG"
},
{
"input": "10 10\nBBBBBBBBBB",
"output": "BBBBBBBBBB"
},
{
"input": "10 10\nBGBGBGBGBG",
"output": "GGGGGBBBBB"
},
{
"input": "1 1\nB",
"output": "B"
}
] | 1,699,017,342 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 92 | 0 | n,move=map(int,input().split())
a=input()
l=[]
for i in a:
l.append(i)
for j in range(move):
i=0
while(i<len(l)-1):
if(l[i]=='B'and l[i+1]=='G'):
temp=l[i]
l[i]=l[i+1]
l[i+1]=temp
i=i+2
else:
i=i+1
a=""
for k in l:
a=a+k
print(a)
| Title: Queue at the School
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G".
Output Specification:
Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G".
Demo Input:
['5 1\nBGGBG\n', '5 2\nBGGBG\n', '4 1\nGGGB\n']
Demo Output:
['GBGGB\n', 'GGBGB\n', 'GGGB\n']
Note:
none | ```python
n,move=map(int,input().split())
a=input()
l=[]
for i in a:
l.append(i)
for j in range(move):
i=0
while(i<len(l)-1):
if(l[i]=='B'and l[i+1]=='G'):
temp=l[i]
l[i]=l[i+1]
l[i+1]=temp
i=i+2
else:
i=i+1
a=""
for k in l:
a=a+k
print(a)
``` | 3 | |
673 | A | Bear and Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Bear Limak likes watching sports on TV. He is going to watch a game today. The game lasts 90 minutes and there are no breaks.
Each minute can be either interesting or boring. If 15 consecutive minutes are boring then Limak immediately turns TV off.
You know that there will be *n* interesting minutes *t*1,<=*t*2,<=...,<=*t**n*. Your task is to calculate for how many minutes Limak will watch the game. | The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=90) — the number of interesting minutes.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=... *t**n*<=≤<=90), given in the increasing order. | Print the number of minutes Limak will watch the game. | [
"3\n7 20 88\n",
"9\n16 20 30 40 50 60 70 80 90\n",
"9\n15 20 30 40 50 60 70 80 90\n"
] | [
"35\n",
"15\n",
"90\n"
] | In the first sample, minutes 21, 22, ..., 35 are all boring and thus Limak will turn TV off immediately after the 35-th minute. So, he would watch the game for 35 minutes.
In the second sample, the first 15 minutes are boring.
In the third sample, there are no consecutive 15 boring minutes. So, Limak will watch the whole game. | 500 | [
{
"input": "3\n7 20 88",
"output": "35"
},
{
"input": "9\n16 20 30 40 50 60 70 80 90",
"output": "15"
},
{
"input": "9\n15 20 30 40 50 60 70 80 90",
"output": "90"
},
{
"input": "30\n6 11 12 15 22 24 30 31 32 33 34 35 40 42 44 45 47 50 53 54 57 58 63 67 75 77 79 81 83 88",
"output": "90"
},
{
"input": "60\n1 2 4 5 6 7 11 14 16 18 20 21 22 23 24 25 26 33 34 35 36 37 38 39 41 42 43 44 46 47 48 49 52 55 56 57 58 59 60 61 63 64 65 67 68 70 71 72 73 74 75 77 78 80 82 83 84 85 86 88",
"output": "90"
},
{
"input": "90\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90",
"output": "90"
},
{
"input": "1\n1",
"output": "16"
},
{
"input": "5\n15 30 45 60 75",
"output": "90"
},
{
"input": "6\n14 29 43 59 70 74",
"output": "58"
},
{
"input": "1\n15",
"output": "30"
},
{
"input": "1\n16",
"output": "15"
},
{
"input": "14\n14 22 27 31 35 44 46 61 62 69 74 79 88 89",
"output": "90"
},
{
"input": "76\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 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",
"output": "90"
},
{
"input": "1\n90",
"output": "15"
},
{
"input": "6\n13 17 32 47 60 66",
"output": "81"
},
{
"input": "84\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84",
"output": "90"
},
{
"input": "9\n6 20 27 28 40 53 59 70 85",
"output": "90"
},
{
"input": "12\n14 22 27 31 35 44 62 69 74 79 88 89",
"output": "59"
},
{
"input": "5\n15 30 45 60 74",
"output": "89"
},
{
"input": "72\n3 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 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",
"output": "54"
},
{
"input": "8\n1 16 30 31 32 33 34 50",
"output": "49"
},
{
"input": "12\n1 3 6 10 15 21 28 36 45 55 66 78",
"output": "90"
},
{
"input": "25\n1 2 3 4 5 6 7 8 9 10 11 23 36 50 65 81 82 83 84 85 86 87 88 89 90",
"output": "80"
},
{
"input": "8\n5 17 20 35 42 53 67 76",
"output": "90"
},
{
"input": "9\n15 28 39 48 55 60 63 64 74",
"output": "89"
},
{
"input": "10\n15 28 39 48 55 60 63 64 74 82",
"output": "90"
},
{
"input": "2\n1 18",
"output": "16"
},
{
"input": "9\n10 20 30 40 50 60 70 80 84",
"output": "90"
},
{
"input": "2\n16 50",
"output": "15"
},
{
"input": "6\n15 30 45 60 75 84",
"output": "90"
},
{
"input": "8\n15 20 30 40 50 60 73 83",
"output": "90"
},
{
"input": "8\n10 20 30 40 50 60 70 80",
"output": "90"
},
{
"input": "3\n1 20 90",
"output": "16"
},
{
"input": "6\n15 30 45 60 74 89",
"output": "90"
}
] | 1,553,330,340 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 93 | 0 | n=int(input())
a=[int(x) for x in input().split()]
rez=0
for e in range(n):
if a[e]-rez<=15:
if e==0:
rez+=a[e]
else:
rez+=a[e]-rez
else:
rez+=15
break
print(rez) | Title: Bear and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak likes watching sports on TV. He is going to watch a game today. The game lasts 90 minutes and there are no breaks.
Each minute can be either interesting or boring. If 15 consecutive minutes are boring then Limak immediately turns TV off.
You know that there will be *n* interesting minutes *t*1,<=*t*2,<=...,<=*t**n*. Your task is to calculate for how many minutes Limak will watch the game.
Input Specification:
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=90) — the number of interesting minutes.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=... *t**n*<=≤<=90), given in the increasing order.
Output Specification:
Print the number of minutes Limak will watch the game.
Demo Input:
['3\n7 20 88\n', '9\n16 20 30 40 50 60 70 80 90\n', '9\n15 20 30 40 50 60 70 80 90\n']
Demo Output:
['35\n', '15\n', '90\n']
Note:
In the first sample, minutes 21, 22, ..., 35 are all boring and thus Limak will turn TV off immediately after the 35-th minute. So, he would watch the game for 35 minutes.
In the second sample, the first 15 minutes are boring.
In the third sample, there are no consecutive 15 boring minutes. So, Limak will watch the whole game. | ```python
n=int(input())
a=[int(x) for x in input().split()]
rez=0
for e in range(n):
if a[e]-rez<=15:
if e==0:
rez+=a[e]
else:
rez+=a[e]-rez
else:
rez+=15
break
print(rez)
``` | 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"
}
] | 1,521,019,497 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | def inputArray2D(arr2d):
#n, m = [int (x) for x in raw_input().split()]
n = input()
m = 2
for i in range(n):
arr2d.append([])
arr2d[i] = map(int, raw_input().split())
return n, m
A = []
inputArray2D(A)
segmentCoverAll = -1
aMax = A[0][0]
aMin = A[0][0]
for i in range(len(A)):
for j in range(0,2):
if(A[i][j] > aMax):
aMax = A[i][j]
if(A[i][j] < aMin):
aMin = A[i][j]
for i in range(len(A)):
if(A[i][0] == aMin and A[i][1] == aMax):
segmentCoverAll = i+1
break
print segmentCoverAll | 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 inputArray2D(arr2d):
#n, m = [int (x) for x in raw_input().split()]
n = input()
m = 2
for i in range(n):
arr2d.append([])
arr2d[i] = map(int, raw_input().split())
return n, m
A = []
inputArray2D(A)
segmentCoverAll = -1
aMax = A[0][0]
aMin = A[0][0]
for i in range(len(A)):
for j in range(0,2):
if(A[i][j] > aMax):
aMax = A[i][j]
if(A[i][j] < aMin):
aMin = A[i][j]
for i in range(len(A)):
if(A[i][0] == aMin and A[i][1] == aMax):
segmentCoverAll = i+1
break
print segmentCoverAll
``` | -1 | |
306 | A | Candies | PROGRAMMING | 800 | [
"implementation"
] | null | null | Polycarpus has got *n* candies and *m* friends (*n*<=≥<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one. | The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100;*n*<=≥<=*m*) — the number of candies and the number of Polycarpus's friends. | Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value. | [
"12 3\n",
"15 4\n",
"18 7\n"
] | [
"4 4 4 ",
"3 4 4 4 ",
"2 2 2 3 3 3 3 "
] | Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces. | 500 | [
{
"input": "12 4",
"output": "3 3 3 3 "
},
{
"input": "15 6",
"output": "2 2 2 3 3 3 "
},
{
"input": "18 8",
"output": "2 2 2 2 2 2 3 3 "
},
{
"input": "1 1",
"output": "1 "
},
{
"input": "2 1",
"output": "2 "
},
{
"input": "100 1",
"output": "100 "
},
{
"input": "100 100",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "100 10",
"output": "10 10 10 10 10 10 10 10 10 10 "
},
{
"input": "64 8",
"output": "8 8 8 8 8 8 8 8 "
},
{
"input": "81 27",
"output": "3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "100 99",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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 "
},
{
"input": "100 51",
"output": "1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100 49",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 "
},
{
"input": "100 13",
"output": "7 7 7 7 8 8 8 8 8 8 8 8 8 "
},
{
"input": "97 11",
"output": "8 8 9 9 9 9 9 9 9 9 9 "
},
{
"input": "19 3",
"output": "6 6 7 "
},
{
"input": "99 10",
"output": "9 10 10 10 10 10 10 10 10 10 "
},
{
"input": "88 5",
"output": "17 17 18 18 18 "
},
{
"input": "87 34",
"output": "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 "
},
{
"input": "97 55",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
}
] | 1,675,833,942 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | n,f=map(int,input().split())
for t in range(f,0,-1):
print(n//i,end=' ')
n-=n//i | Title: Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has got *n* candies and *m* friends (*n*<=≥<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one.
Input Specification:
The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100;*n*<=≥<=*m*) — the number of candies and the number of Polycarpus's friends.
Output Specification:
Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value.
Demo Input:
['12 3\n', '15 4\n', '18 7\n']
Demo Output:
['4 4 4 ', '3 4 4 4 ', '2 2 2 3 3 3 3 ']
Note:
Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces. | ```python
n,f=map(int,input().split())
for t in range(f,0,-1):
print(n//i,end=' ')
n-=n//i
``` | -1 | |
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"
}
] | 1,636,277,537 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 264 | 27,340,800 | #!/usr/bin/env python
# coding=utf-8
'''
Author: Deean
Date: 2021-11-07 17:25:33
LastEditTime: 2021-11-07 17:31:42
Description: Optimal Point on a Line
FilePath: CF710B.py
'''
def func():
n = int(input())
position = sorted(map(int, input().strip().split()))
if n % 2 == 0:
print(position[n // 2 - 1])
else:
print(position[n // 2])
if __name__ == '__main__':
func()
| 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
#!/usr/bin/env python
# coding=utf-8
'''
Author: Deean
Date: 2021-11-07 17:25:33
LastEditTime: 2021-11-07 17:31:42
Description: Optimal Point on a Line
FilePath: CF710B.py
'''
def func():
n = int(input())
position = sorted(map(int, input().strip().split()))
if n % 2 == 0:
print(position[n // 2 - 1])
else:
print(position[n // 2])
if __name__ == '__main__':
func()
``` | 3 | |
427 | A | Police Recruits | PROGRAMMING | 800 | [
"implementation"
] | null | null | The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated. | The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time. | Print a single integer, the number of crimes which will go untreated. | [
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] | [
"2\n",
"1\n",
"8\n"
] | Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated. | 500 | [
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 -1 2 2 1 6 5 -1 5",
"output": "5"
},
{
"input": "98\n-1 -1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -1 1 1 1 -1 1 1 -1 -1 -1 1 1 1 -1 -1 -1 1 -1 1 1",
"output": "13"
},
{
"input": "3\n-1 5 4",
"output": "1"
},
{
"input": "146\n4 -1 -1 -1 -1 -1 -1 -1 -1 -1 4 -1 3 -1 3 -1 -1 1 4 -1 2 -1 -1 3 -1 -1 -1 4 1 -1 -1 -1 4 -1 -1 -1 -1 -1 -1 3 2 -1 3 2 4 5 2 4 1 5 -1 -1 2 -1 -1 1 -1 5 3 -1 1 2 2 3 5 3 -1 -1 3 -1 -1 3 5 5 -1 -1 5 -1 4 4 1 -1 -1 -1 2 1 -1 -1 -1 2 5 3 -1 -1 -1 3 -1 5 4 -1 1 -1 -1 3 -1 -1 3 1 1 2 -1 -1 -1 1 3 1 -1 2 -1 -1 5 5 -1 -1 3 4 5 1 -1 2 -1 -1 -1 3 -1 5 3 2 -1 2 -1 -1 5 -1 3 -1",
"output": "5"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n1 -1",
"output": "0"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n-1 -1",
"output": "2"
},
{
"input": "4\n10 -1 -1 -1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "4\n-1 -1 1 1",
"output": "2"
}
] | 1,690,274,999 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 77 | 9,420,800 | n = int(input())
nums = [int(v) for v in input().split()]
police = untreated = 0
for v in nums:
untreated += 1 if v == -1 and not police else 0
police += v if v>0 else -1 if police else 0
print(untreated) | Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated. | ```python
n = int(input())
nums = [int(v) for v in input().split()]
police = untreated = 0
for v in nums:
untreated += 1 if v == -1 and not police else 0
police += v if v>0 else -1 if police else 0
print(untreated)
``` | 3 | |
166 | A | Rank List | PROGRAMMING | 1,100 | [
"binary search",
"implementation",
"sortings"
] | null | null | Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place.
You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*.
It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place.
Your task is to count what number of teams from the given list shared the *k*-th place. | The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=≤<=*p**i*,<=*t**i*<=≤<=50) — the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces. | In the only line print the sought number of teams that got the *k*-th place in the final results' table. | [
"7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n",
"5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n"
] | [
"3\n",
"4\n"
] | The final results' table for the first sample is:
- 1-3 places — 4 solved problems, the penalty time equals 10 - 4 place — 3 solved problems, the penalty time equals 20 - 5-6 places — 2 solved problems, the penalty time equals 1 - 7 place — 1 solved problem, the penalty time equals 10
The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams.
The final table for the second sample is:
- 1 place — 5 solved problems, the penalty time equals 3 - 2-5 places — 3 solved problems, the penalty time equals 1
The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams. | 500 | [
{
"input": "7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10",
"output": "3"
},
{
"input": "5 4\n3 1\n3 1\n5 3\n3 1\n3 1",
"output": "4"
},
{
"input": "5 1\n2 2\n1 1\n1 1\n1 1\n2 2",
"output": "2"
},
{
"input": "6 3\n2 2\n3 1\n2 2\n4 5\n2 2\n4 5",
"output": "1"
},
{
"input": "5 5\n3 1\n10 2\n2 2\n1 10\n10 2",
"output": "1"
},
{
"input": "3 2\n3 3\n3 3\n3 3",
"output": "3"
},
{
"input": "4 3\n10 3\n6 10\n5 2\n5 2",
"output": "2"
},
{
"input": "5 3\n10 10\n10 10\n1 1\n10 10\n4 3",
"output": "3"
},
{
"input": "3 1\n2 1\n1 1\n1 2",
"output": "1"
},
{
"input": "1 1\n28 28",
"output": "1"
},
{
"input": "2 2\n1 2\n1 2",
"output": "2"
},
{
"input": "5 3\n2 3\n4 2\n5 3\n2 4\n3 5",
"output": "1"
},
{
"input": "50 22\n4 9\n8 1\n3 7\n1 2\n3 8\n9 8\n8 5\n2 10\n5 8\n1 3\n1 8\n2 3\n7 9\n10 2\n9 9\n7 3\n8 6\n10 6\n5 4\n8 1\n1 5\n6 8\n9 5\n9 5\n3 2\n3 3\n3 8\n7 5\n4 5\n8 10\n8 2\n3 5\n3 2\n1 1\n7 2\n2 7\n6 8\n10 4\n7 5\n1 7\n6 5\n3 1\n4 9\n2 3\n3 6\n5 8\n4 10\n10 7\n7 10\n9 8",
"output": "1"
},
{
"input": "50 6\n11 20\n18 13\n1 13\n3 11\n4 17\n15 10\n15 8\n9 16\n11 17\n16 3\n3 20\n14 13\n12 15\n9 10\n14 2\n12 12\n13 17\n6 10\n20 9\n2 8\n13 7\n7 20\n15 3\n1 20\n2 13\n2 5\n14 7\n10 13\n15 12\n15 5\n17 6\n9 11\n18 5\n10 1\n15 14\n3 16\n6 12\n4 1\n14 9\n7 14\n8 17\n17 13\n4 6\n19 16\n5 6\n3 15\n4 19\n15 20\n2 10\n20 10",
"output": "1"
},
{
"input": "50 12\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "50"
},
{
"input": "50 28\n2 2\n1 1\n2 1\n1 2\n1 1\n1 1\n1 1\n2 2\n2 2\n2 2\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 1\n2 2\n1 2\n2 2\n2 2\n2 1\n1 1\n1 2\n1 2\n1 1\n1 1\n1 1\n2 2\n2 1\n2 1\n2 2\n1 2\n1 2\n1 2\n1 1\n2 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n2 1\n1 1\n2 2\n2 2\n2 2\n2 2",
"output": "13"
},
{
"input": "50 40\n2 3\n3 1\n2 1\n2 1\n2 1\n3 1\n1 1\n1 2\n2 3\n1 3\n1 3\n2 1\n3 1\n1 1\n3 1\n3 1\n2 2\n1 1\n3 3\n3 1\n3 2\n2 3\n3 3\n3 1\n1 3\n2 3\n2 1\n3 2\n3 3\n3 1\n2 1\n2 2\n1 3\n3 3\n1 1\n3 2\n1 2\n2 3\n2 1\n2 2\n3 2\n1 3\n3 1\n1 1\n3 3\n2 3\n2 1\n2 3\n2 3\n1 2",
"output": "5"
},
{
"input": "50 16\n2 1\n3 2\n5 2\n2 2\n3 4\n4 4\n3 3\n4 1\n2 3\n1 5\n4 1\n2 2\n1 5\n3 2\n2 1\n5 4\n5 2\n5 4\n1 1\n3 5\n2 1\n4 5\n5 1\n5 5\n5 4\n2 4\n1 2\n5 5\n4 4\n1 5\n4 2\n5 1\n2 4\n2 5\n2 2\n3 4\n3 1\n1 1\n5 5\n2 2\n3 4\n2 4\n5 2\n4 1\n3 1\n1 1\n4 1\n4 4\n1 4\n1 3",
"output": "1"
},
{
"input": "50 32\n6 6\n4 2\n5 5\n1 1\n2 4\n6 5\n2 3\n6 5\n2 3\n6 3\n1 4\n1 6\n3 3\n2 4\n3 2\n6 2\n4 1\n3 3\n3 1\n5 5\n1 2\n2 1\n5 4\n3 1\n4 4\n5 6\n4 1\n2 5\n3 1\n4 6\n2 3\n1 1\n6 5\n2 6\n3 3\n2 6\n2 3\n2 6\n3 4\n2 6\n4 5\n5 4\n1 6\n3 2\n5 1\n4 1\n4 6\n4 2\n1 2\n5 2",
"output": "1"
},
{
"input": "50 48\n5 1\n6 4\n3 2\n2 1\n4 7\n3 6\n7 1\n7 5\n6 5\n5 6\n4 7\n5 7\n5 7\n5 5\n7 3\n3 5\n4 3\n5 4\n6 2\n1 6\n6 3\n6 5\n5 2\n4 2\n3 1\n1 1\n5 6\n1 3\n6 5\n3 7\n1 5\n7 5\n6 5\n3 6\n2 7\n5 3\n5 3\n4 7\n5 2\n6 5\n5 7\n7 1\n2 3\n5 5\n2 6\n4 1\n6 2\n6 5\n3 3\n1 6",
"output": "1"
},
{
"input": "50 8\n5 3\n7 3\n4 3\n7 4\n2 2\n4 4\n5 4\n1 1\n7 7\n4 8\n1 1\n6 3\n1 5\n7 3\n6 5\n4 5\n8 6\n3 6\n2 1\n3 2\n2 5\n7 6\n5 8\n1 3\n5 5\n8 4\n4 5\n4 4\n8 8\n7 2\n7 2\n3 6\n2 8\n8 3\n3 2\n4 5\n8 1\n3 2\n8 7\n6 3\n2 3\n5 1\n3 4\n7 2\n6 3\n7 3\n3 3\n6 4\n2 2\n5 1",
"output": "3"
},
{
"input": "20 16\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "20"
},
{
"input": "20 20\n1 2\n2 2\n1 1\n2 1\n2 2\n1 1\n1 1\n2 1\n1 1\n1 2\n2 2\n1 2\n1 2\n2 2\n2 2\n1 2\n2 1\n2 1\n1 2\n2 2",
"output": "6"
},
{
"input": "30 16\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "30"
},
{
"input": "30 22\n2 1\n1 2\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n2 2\n1 2\n2 2\n1 2\n1 2\n2 1\n1 2\n2 2\n2 2\n1 2\n2 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 2\n2 2\n1 2\n2 2\n2 1\n1 1",
"output": "13"
},
{
"input": "30 22\n1 1\n1 3\n2 3\n3 1\n2 3\n3 1\n1 2\n3 3\n2 1\n2 1\n2 2\n3 1\n3 2\n2 3\n3 1\n1 3\n2 3\n3 1\n1 2\n1 2\n2 3\n2 1\n3 3\n3 2\n1 3\n3 3\n3 3\n3 3\n3 3\n3 1",
"output": "5"
},
{
"input": "50 16\n2 1\n3 2\n5 2\n2 2\n3 4\n4 4\n3 3\n4 1\n2 3\n1 5\n4 1\n2 2\n1 5\n3 2\n2 1\n5 4\n5 2\n5 4\n1 1\n3 5\n2 1\n4 5\n5 1\n5 5\n5 4\n2 4\n1 2\n5 5\n4 4\n1 5\n4 2\n5 1\n2 4\n2 5\n2 2\n3 4\n3 1\n1 1\n5 5\n2 2\n3 4\n2 4\n5 2\n4 1\n3 1\n1 1\n4 1\n4 4\n1 4\n1 3",
"output": "1"
},
{
"input": "50 22\n4 9\n8 1\n3 7\n1 2\n3 8\n9 8\n8 5\n2 10\n5 8\n1 3\n1 8\n2 3\n7 9\n10 2\n9 9\n7 3\n8 6\n10 6\n5 4\n8 1\n1 5\n6 8\n9 5\n9 5\n3 2\n3 3\n3 8\n7 5\n4 5\n8 10\n8 2\n3 5\n3 2\n1 1\n7 2\n2 7\n6 8\n10 4\n7 5\n1 7\n6 5\n3 1\n4 9\n2 3\n3 6\n5 8\n4 10\n10 7\n7 10\n9 8",
"output": "1"
},
{
"input": "50 22\n29 15\n18 10\n6 23\n38 28\n34 40\n40 1\n16 26\n22 33\n14 30\n26 7\n15 16\n22 40\n14 15\n6 28\n32 27\n33 3\n38 22\n40 17\n16 27\n21 27\n34 26\n5 15\n34 9\n38 23\n7 36\n17 6\n19 37\n40 1\n10 28\n9 14\n8 31\n40 8\n14 2\n24 16\n38 33\n3 37\n2 9\n21 21\n40 26\n28 33\n24 31\n10 12\n27 27\n17 4\n38 5\n21 31\n5 12\n29 7\n39 12\n26 14",
"output": "1"
},
{
"input": "50 14\n4 20\n37 50\n46 19\n20 25\n47 10\n6 34\n12 41\n47 9\n22 28\n41 34\n47 40\n12 42\n9 4\n15 15\n27 8\n38 9\n4 17\n8 13\n47 7\n9 38\n30 48\n50 7\n41 34\n23 11\n16 37\n2 32\n18 46\n37 48\n47 41\n13 9\n24 50\n46 14\n33 49\n9 50\n35 30\n49 44\n42 49\n39 15\n33 42\n3 18\n44 15\n44 28\n9 17\n16 4\n10 36\n4 22\n47 17\n24 12\n2 31\n6 30",
"output": "2"
},
{
"input": "2 1\n50 50\n50 50",
"output": "2"
},
{
"input": "2 2\n50 50\n50 50",
"output": "2"
},
{
"input": "2 1\n50 50\n50 49",
"output": "1"
},
{
"input": "2 2\n50 50\n50 49",
"output": "1"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50",
"output": "50"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n49 50",
"output": "1"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 49",
"output": "49"
},
{
"input": "10 1\n3 1\n3 2\n3 2\n2 3\n2 3\n2 3\n2 4\n2 4\n2 4\n2 4",
"output": "1"
},
{
"input": "4 3\n1 1\n1 2\n2 1\n2 2",
"output": "1"
},
{
"input": "4 2\n1 2\n1 3\n1 5\n1 2",
"output": "2"
},
{
"input": "6 1\n4 3\n4 2\n5 1\n5 2\n5 3\n5 3",
"output": "1"
},
{
"input": "5 3\n2 1\n1 3\n1 2\n1 1\n1 1",
"output": "2"
},
{
"input": "3 1\n1 1\n1 2\n1 3",
"output": "1"
},
{
"input": "10 10\n3 1\n3 2\n3 2\n2 3\n2 3\n2 3\n2 4\n2 4\n2 4\n2 4",
"output": "4"
},
{
"input": "5 1\n5 5\n5 6\n5 7\n5 8\n5 5",
"output": "2"
},
{
"input": "4 1\n1 1\n1 1\n1 2\n1 3",
"output": "2"
}
] | 1,645,806,222 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 92 | 0 | n,k=list(map(int,input().split()))
li=[]
for t in range(n):
li.append(list(map(int,input().split())))
li = sorted(li,key=lambda x: (-x[0],x[1]))
print(li.count(li[k-1])) | Title: Rank List
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place.
You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*.
It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place.
Your task is to count what number of teams from the given list shared the *k*-th place.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=≤<=*p**i*,<=*t**i*<=≤<=50) — the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces.
Output Specification:
In the only line print the sought number of teams that got the *k*-th place in the final results' table.
Demo Input:
['7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n', '5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n']
Demo Output:
['3\n', '4\n']
Note:
The final results' table for the first sample is:
- 1-3 places — 4 solved problems, the penalty time equals 10 - 4 place — 3 solved problems, the penalty time equals 20 - 5-6 places — 2 solved problems, the penalty time equals 1 - 7 place — 1 solved problem, the penalty time equals 10
The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams.
The final table for the second sample is:
- 1 place — 5 solved problems, the penalty time equals 3 - 2-5 places — 3 solved problems, the penalty time equals 1
The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams. | ```python
n,k=list(map(int,input().split()))
li=[]
for t in range(n):
li.append(list(map(int,input().split())))
li = sorted(li,key=lambda x: (-x[0],x[1]))
print(li.count(li[k-1]))
``` | 3 | |
25 | E | Test | PROGRAMMING | 2,200 | [
"hashing",
"strings"
] | E. Test | 2 | 256 | Sometimes it is hard to prepare tests for programming problems. Now Bob is preparing tests to new problem about strings — input data to his problem is one string. Bob has 3 wrong solutions to this problem. The first gives the wrong answer if the input data contains the substring *s*1, the second enters an infinite loop if the input data contains the substring *s*2, and the third requires too much memory if the input data contains the substring *s*3. Bob wants these solutions to fail single test. What is the minimal length of test, which couldn't be passed by all three Bob's solutions? | There are exactly 3 lines in the input data. The *i*-th line contains string *s**i*. All the strings are non-empty, consists of lowercase Latin letters, the length of each string doesn't exceed 105. | Output one number — what is minimal length of the string, containing *s*1, *s*2 and *s*3 as substrings. | [
"ab\nbc\ncd\n",
"abacaba\nabaaba\nx\n"
] | [
"4\n",
"11\n"
] | none | 0 | [
{
"input": "ab\nbc\ncd",
"output": "4"
},
{
"input": "abacaba\nabaaba\nx",
"output": "11"
},
{
"input": "syvncqmfhautvxudqdhggz\nhrpxzeghsocjpicuixskfuzupytsgjsdiyb\nybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehls",
"output": "100"
},
{
"input": "jwdezvgfm\nmdoqvylpuvyk\nqylldbziva",
"output": "30"
},
{
"input": "ujgquqxdlowuwnqkmbd\nwdwkhkdgsujgqu\njlxqvcuivagmw",
"output": "40"
},
{
"input": "rdtevvmiqmfgvafkdypxjthzhfsbavmhgkavkfonscaokdxoscenpxrc\nijbvueenzsmgkmkrskjspvfchwkqdglkxnrdtevvmiqmfgvafkdypxjthz\nkqdglkxnrdtevvmiqmfgvafkdypxjthzhfsbavmhgkavkfonscaokdxoscenpxrcivydtkrxjy",
"output": "100"
},
{
"input": "xufuzdlsjxmevrtessfbwlnzzclcqwevnnucxyvhngnxhcbdfwq\nwlwobhnmmgtfolfaeckbrnnglylydxtgtvrlmeeszoiuatzzzxufuzdlsjxmevrt\nbrnnglylydxtgtvrlmeeszoiuatzzzx",
"output": "100"
},
{
"input": "iefouqzxoyuieqdzalfktehtvdbvjmeubju\nocotspetkkhvwfgaqynhovjwjhciefouqzxoyuieqdzalfktehtvdbvjmeubjubcmnvpwgdpnchqhzjrchyrfpvigubp\nycnhjwgbocotspetkkhvwfgaqynhovjwjhcief",
"output": "100"
}
] | 1,591,283,001 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 38 | 2,000 | 10,240,000 | from itertools import permutations
from sys import stdin
input = stdin.readline
s1, s2, s3 = input().strip(), input().strip(), input().strip()
def findPattern(s):
slen = len(s)
p = [0]*slen
ci = 0
for i in range(1, slen):
if(s[ci] == s[i]):
ci+=1
p[i] = ci
else:
while(ci):
ci = p[ci-1]
if(s[ci]==s[i]):
ci+=1
p[i] = ci
break
return p
def searchPattern(s, ps, p):
slen = len(s)
plen = len(p)
smatch = [0]*slen
pi = 0
for i in range(slen):
if(s[i] == ps[pi]):
pi+=1
smatch[i] = pi
else:
while(pi):
pi = p[pi-1]
if(s[i] == ps[pi]):
pi+=1
smatch[i] = pi
break
if(pi==plen):
pi = p[pi-1]
return smatch
def calcLength(a, b, c):
pb = findPattern(b)
amatchb = searchPattern(a, b, pb)
if(len(b) not in amatchb): ab = a[:max(0, len(a) - amatchb[-1])] + b
else: ab = a
pc = findPattern(c)
abmatchc = searchPattern(ab, c, pc)
return max(0, len(ab) - abmatchc[-1]) + len(c) if len(c) not in abmatchc else len(ab)
ans = float('inf')
s = [s1, s2, s3]
for ai, bi, ci in permutations(list(range(3)), 3):
ans = min(ans, calcLength(s[ai], s[bi], s[ci]))
print(ans)
| Title: Test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes it is hard to prepare tests for programming problems. Now Bob is preparing tests to new problem about strings — input data to his problem is one string. Bob has 3 wrong solutions to this problem. The first gives the wrong answer if the input data contains the substring *s*1, the second enters an infinite loop if the input data contains the substring *s*2, and the third requires too much memory if the input data contains the substring *s*3. Bob wants these solutions to fail single test. What is the minimal length of test, which couldn't be passed by all three Bob's solutions?
Input Specification:
There are exactly 3 lines in the input data. The *i*-th line contains string *s**i*. All the strings are non-empty, consists of lowercase Latin letters, the length of each string doesn't exceed 105.
Output Specification:
Output one number — what is minimal length of the string, containing *s*1, *s*2 and *s*3 as substrings.
Demo Input:
['ab\nbc\ncd\n', 'abacaba\nabaaba\nx\n']
Demo Output:
['4\n', '11\n']
Note:
none | ```python
from itertools import permutations
from sys import stdin
input = stdin.readline
s1, s2, s3 = input().strip(), input().strip(), input().strip()
def findPattern(s):
slen = len(s)
p = [0]*slen
ci = 0
for i in range(1, slen):
if(s[ci] == s[i]):
ci+=1
p[i] = ci
else:
while(ci):
ci = p[ci-1]
if(s[ci]==s[i]):
ci+=1
p[i] = ci
break
return p
def searchPattern(s, ps, p):
slen = len(s)
plen = len(p)
smatch = [0]*slen
pi = 0
for i in range(slen):
if(s[i] == ps[pi]):
pi+=1
smatch[i] = pi
else:
while(pi):
pi = p[pi-1]
if(s[i] == ps[pi]):
pi+=1
smatch[i] = pi
break
if(pi==plen):
pi = p[pi-1]
return smatch
def calcLength(a, b, c):
pb = findPattern(b)
amatchb = searchPattern(a, b, pb)
if(len(b) not in amatchb): ab = a[:max(0, len(a) - amatchb[-1])] + b
else: ab = a
pc = findPattern(c)
abmatchc = searchPattern(ab, c, pc)
return max(0, len(ab) - abmatchc[-1]) + len(c) if len(c) not in abmatchc else len(ab)
ans = float('inf')
s = [s1, s2, s3]
for ai, bi, ci in permutations(list(range(3)), 3):
ans = min(ans, calcLength(s[ai], s[bi], s[ci]))
print(ans)
``` | 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"
}
] | 1,693,209,344 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 62 | 0 | a, b = map(int, input().split())
listodd = []
listeven = []
for x in range(a):
if x %2 != 0:
listodd.append(x)
else:
listeven.append(x)
listodd = listodd + listeven
print(listodd[b-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
a, b = map(int, input().split())
listodd = []
listeven = []
for x in range(a):
if x %2 != 0:
listodd.append(x)
else:
listeven.append(x)
listodd = listodd + listeven
print(listodd[b-1])
``` | 0 | |
265 | A | Colorful Stones (Simplified Edition) | PROGRAMMING | 800 | [
"implementation"
] | null | null | There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence. | The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence. | Print the final 1-based position of Liss in a single line. | [
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] | [
"2\n",
"3\n",
"15\n"
] | none | 500 | [
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
{
"input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB",
"output": "1"
},
{
"input": "RRGGBRGRBG\nBRRGGBBGGR",
"output": "7"
},
{
"input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG",
"output": "15"
},
{
"input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR",
"output": "8"
},
{
"input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR",
"output": "16"
},
{
"input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR",
"output": "18"
},
{
"input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG",
"output": "23"
},
{
"input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR",
"output": "14"
},
{
"input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG",
"output": "12"
},
{
"input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB",
"output": "12"
},
{
"input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR",
"output": "2"
},
{
"input": "R\nB",
"output": "1"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "50"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
},
{
"input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
}
] | 1,660,650,099 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | path = list(input())
steps = list(input())
currentBlock = path[0]
currentIndex = 0
for step in steps:
if(step == path[currentIndex+1]):
currentIndex+=1
print(currentIndex+1) | Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none | ```python
path = list(input())
steps = list(input())
currentBlock = path[0]
currentIndex = 0
for step in steps:
if(step == path[currentIndex+1]):
currentIndex+=1
print(currentIndex+1)
``` | 0 | |
779 | C | Dishonest Sellers | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy",
"sortings"
] | null | null | Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*.
Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.
Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items. | In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now.
The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now).
The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week). | Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now. | [
"3 1\n5 4 6\n3 1 5\n",
"5 3\n3 4 7 10 3\n4 5 5 12 5\n"
] | [
"10\n",
"25\n"
] | In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.
In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25. | 1,000 | [
{
"input": "3 1\n5 4 6\n3 1 5",
"output": "10"
},
{
"input": "5 3\n3 4 7 10 3\n4 5 5 12 5",
"output": "25"
},
{
"input": "1 0\n9\n8",
"output": "8"
},
{
"input": "2 0\n4 10\n1 2",
"output": "3"
},
{
"input": "4 2\n19 5 17 13\n3 18 8 10",
"output": "29"
},
{
"input": "5 3\n28 17 20 45 45\n39 12 41 27 9",
"output": "101"
},
{
"input": "10 5\n87 96 19 81 10 88 7 49 36 21\n11 75 28 28 74 17 64 19 81 31",
"output": "243"
},
{
"input": "50 45\n80 125 152 122 85 62 123 195 157 73 8 127 47 193 70 132 131 38 98 97 185 98 68 166 114 53 144 106 141 106 191 123 44 199 35 18 84 169 46 52 129 102 155 130 63 169 116 145 154 41\n155 187 79 106 136 173 17 146 9 47 173 21 37 183 110 74 185 14 82 156 147 25 162 195 26 25 22 80 171 180 7 92 53 18 100 17 159 184 37 127 116 192 163 49 94 95 118 102 8 30",
"output": "4680"
},
{
"input": "5 2\n1 1 1 1 1\n2 2 2 2 2",
"output": "5"
},
{
"input": "5 2\n2 4 4 4 4\n3 1 1 1 1",
"output": "9"
},
{
"input": "6 3\n9 9 9 2 2 2\n6 6 6 1 1 1",
"output": "24"
},
{
"input": "5 3\n10 17 23 26 35\n7 18 18 29 32",
"output": "103"
},
{
"input": "2 1\n10 12\n5 6",
"output": "16"
}
] | 1,552,187,384 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 109 | 0 | n,k=map(int,input().split())
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
s=sum(a[i]<=b[i] for i in range(n))
if s>=k:
sm=0
for i in range(n):
sm+=min(a[i],b[i])
print(sm)
exit()
sm=0
for i in range(n):
sm+=min(a[i],b[i])
delta=[a[i]-b[i] for i in range(n) if a[i]>=b[i]]
delta.sort()
sm+=sum(delta[0:k])
print(sm) | Title: Dishonest Sellers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*.
Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.
Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items.
Input Specification:
In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now.
The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now).
The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week).
Output Specification:
Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now.
Demo Input:
['3 1\n5 4 6\n3 1 5\n', '5 3\n3 4 7 10 3\n4 5 5 12 5\n']
Demo Output:
['10\n', '25\n']
Note:
In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.
In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25. | ```python
n,k=map(int,input().split())
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
s=sum(a[i]<=b[i] for i in range(n))
if s>=k:
sm=0
for i in range(n):
sm+=min(a[i],b[i])
print(sm)
exit()
sm=0
for i in range(n):
sm+=min(a[i],b[i])
delta=[a[i]-b[i] for i in range(n) if a[i]>=b[i]]
delta.sort()
sm+=sum(delta[0:k])
print(sm)
``` | 0 | |
22 | A | Second Order Statistics | PROGRAMMING | 800 | [
"brute force"
] | A. Second Order Statistics | 2 | 256 | Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem. | The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value. | If the given sequence has the second order statistics, output this order statistics, otherwise output NO. | [
"4\n1 2 2 -4\n",
"5\n1 2 3 1 1\n"
] | [
"1\n",
"2\n"
] | none | 0 | [
{
"input": "4\n1 2 2 -4",
"output": "1"
},
{
"input": "5\n1 2 3 1 1",
"output": "2"
},
{
"input": "1\n28",
"output": "NO"
},
{
"input": "2\n-28 12",
"output": "12"
},
{
"input": "3\n-83 40 -80",
"output": "-80"
},
{
"input": "8\n93 77 -92 26 21 -48 53 91",
"output": "-48"
},
{
"input": "20\n-72 -9 -86 80 7 -10 40 -27 -94 92 96 56 28 -19 79 36 -3 -73 -63 -49",
"output": "-86"
},
{
"input": "49\n-74 -100 -80 23 -8 -83 -41 -20 48 17 46 -73 -55 67 85 4 40 -60 -69 -75 56 -74 -42 93 74 -95 64 -46 97 -47 55 0 -78 -34 -31 40 -63 -49 -76 48 21 -1 -49 -29 -98 -11 76 26 94",
"output": "-98"
},
{
"input": "88\n63 48 1 -53 -89 -49 64 -70 -49 71 -17 -16 76 81 -26 -50 67 -59 -56 97 2 100 14 18 -91 -80 42 92 -25 -88 59 8 -56 38 48 -71 -78 24 -14 48 -1 69 73 -76 54 16 -92 44 47 33 -34 -17 -81 21 -59 -61 53 26 10 -76 67 35 -29 70 65 -13 -29 81 80 32 74 -6 34 46 57 1 -45 -55 69 79 -58 11 -2 22 -18 -16 -89 -46",
"output": "-91"
},
{
"input": "100\n34 32 88 20 76 53 -71 -39 -98 -10 57 37 63 -3 -54 -64 -78 -82 73 20 -30 -4 22 75 51 -64 -91 29 -52 -48 83 19 18 -47 46 57 -44 95 89 89 -30 84 -83 67 58 -99 -90 -53 92 -60 -5 -56 -61 27 68 -48 52 -95 64 -48 -30 -67 66 89 14 -33 -31 -91 39 7 -94 -54 92 -96 -99 -83 -16 91 -28 -66 81 44 14 -85 -21 18 40 16 -13 -82 -33 47 -10 -40 -19 10 25 60 -34 -89",
"output": "-98"
},
{
"input": "2\n-1 -1",
"output": "NO"
},
{
"input": "3\n-2 -2 -2",
"output": "NO"
},
{
"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": "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": "100"
},
{
"input": "10\n40 71 -85 -85 40 -85 -85 64 -85 47",
"output": "40"
},
{
"input": "23\n-90 -90 -41 -64 -64 -90 -15 10 -43 -90 -64 -64 89 -64 36 47 38 -90 -64 -90 -90 68 -90",
"output": "-64"
},
{
"input": "39\n-97 -93 -42 -93 -97 -93 56 -97 -97 -97 76 -33 -60 91 7 82 17 47 -97 -97 -93 73 -97 12 -97 -97 -97 -97 56 -92 -83 -93 -93 49 -93 -97 -97 -17 -93",
"output": "-93"
},
{
"input": "51\n-21 6 -35 -98 -86 -98 -86 -43 -65 32 -98 -40 96 -98 -98 -98 -98 -86 -86 -98 56 -86 -98 -98 -30 -98 -86 -31 -98 -86 -86 -86 -86 -30 96 -86 -86 -86 -60 25 88 -86 -86 58 31 -47 57 -86 37 44 -83",
"output": "-86"
},
{
"input": "66\n-14 -95 65 -95 -95 -97 -90 -71 -97 -97 70 -95 -95 -97 -95 -27 35 -87 -95 -5 -97 -97 87 34 -49 -95 -97 -95 -97 -95 -30 -95 -97 47 -95 -17 -97 -95 -97 -69 51 -97 -97 -95 -75 87 59 21 63 56 76 -91 98 -97 6 -97 -95 -95 -97 -73 11 -97 -35 -95 -95 -43",
"output": "-95"
},
{
"input": "77\n-67 -93 -93 -92 97 29 93 -93 -93 -5 -93 -7 60 -92 -93 44 -84 68 -92 -93 69 -92 -37 56 43 -93 35 -92 -93 19 -79 18 -92 -93 -93 -37 -93 -47 -93 -92 -92 74 67 19 40 -92 -92 -92 -92 -93 -93 -41 -93 -92 -93 -93 -92 -93 51 -80 6 -42 -92 -92 -66 -12 -92 -92 -3 93 -92 -49 -93 40 62 -92 -92",
"output": "-92"
},
{
"input": "89\n-98 40 16 -87 -98 63 -100 55 -96 -98 -21 -100 -93 26 -98 -98 -100 -89 -98 -5 -65 -28 -100 -6 -66 67 -100 -98 -98 10 -98 -98 -70 7 -98 2 -100 -100 -98 25 -100 -100 -98 23 -68 -100 -98 3 98 -100 -98 -98 -98 -98 -24 -100 -100 -9 -98 35 -100 99 -5 -98 -100 -100 37 -100 -84 57 -98 40 -47 -100 -1 -92 -76 -98 -98 -100 -100 -100 -63 30 21 -100 -100 -100 -12",
"output": "-98"
},
{
"input": "99\n10 -84 -100 -100 73 -64 -100 -94 33 -100 -100 -100 -100 71 64 24 7 -100 -32 -100 -100 77 -100 62 -12 55 45 -100 -100 -80 -100 -100 -100 -100 -100 -100 -100 -100 -100 -39 -48 -100 -34 47 -100 -100 -100 -100 -100 -77 -100 -100 -100 -100 -100 -100 -52 40 -55 -100 -44 -100 72 33 70 -100 -100 -78 -100 -3 100 -77 22 -100 95 -30 -100 10 -69 -100 -100 -100 -100 52 -39 -100 -100 -100 7 -100 -98 -66 95 -17 -100 52 -100 68 -100",
"output": "-98"
},
{
"input": "100\n-99 -98 -64 89 53 57 -99 29 -78 18 -3 -54 76 -98 -99 -98 37 -98 19 -47 89 73 -98 -91 -99 -99 -98 -48 -99 22 -99 -98 -99 -99 -98 -60 84 67 -99 -98 20 -98 88 -98 46 -98 -99 -98 -99 -71 -99 -98 -98 -39 83 95 -98 63 -98 -99 32 -98 -99 -64 57 -30 -53 -83 -4 -99 58 20 -98 -10 -99 -44 -99 -99 -99 -99 75 34 -98 -52 -98 -30 -98 -99 -98 -98 51 -99 -99 -99 -98 -99 -99 -82 -90 92",
"output": "-98"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "3\n1 3 2",
"output": "2"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "3\n3 1 2",
"output": "2"
},
{
"input": "3\n3 2 1",
"output": "2"
},
{
"input": "9\n99 99 99 99 100 100 100 100 100",
"output": "100"
},
{
"input": "5\n-100 -100 -100 -100 -100",
"output": "NO"
}
] | 1,530,200,677 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 184 | 0 | def second_minimum(lst):
first, second = float("inf")
for i in range(len(lst) - 1):
if lst[i] < first:
second = first
first = lst[i]
elif lst[i] < second and lst[i] != first:
second = lst[i]
if second == float("inf") :
return "NO"
return second
n = int(input())
a = [int(i) for i in input().split()]
print(second_minimum(a))
| Title: Second Order Statistics
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
Output Specification:
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
Demo Input:
['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n']
Demo Output:
['1\n', '2\n']
Note:
none | ```python
def second_minimum(lst):
first, second = float("inf")
for i in range(len(lst) - 1):
if lst[i] < first:
second = first
first = lst[i]
elif lst[i] < second and lst[i] != first:
second = lst[i]
if second == float("inf") :
return "NO"
return second
n = int(input())
a = [int(i) for i in input().split()]
print(second_minimum(a))
``` | -1 |
359 | A | Table | PROGRAMMING | 1,000 | [
"constructive algorithms",
"greedy",
"implementation"
] | null | null | Simon has a rectangular table consisting of *n* rows and *m* columns. Simon numbered the rows of the table from top to bottom starting from one and the columns — from left to right starting from one. We'll represent the cell on the *x*-th row and the *y*-th column as a pair of numbers (*x*,<=*y*). The table corners are cells: (1,<=1), (*n*,<=1), (1,<=*m*), (*n*,<=*m*).
Simon thinks that some cells in this table are good. Besides, it's known that no good cell is the corner of the table.
Initially, all cells of the table are colorless. Simon wants to color all cells of his table. In one move, he can choose any good cell of table (*x*1,<=*y*1), an arbitrary corner of the table (*x*2,<=*y*2) and color all cells of the table (*p*,<=*q*), which meet both inequations: *min*(*x*1,<=*x*2)<=≤<=*p*<=≤<=*max*(*x*1,<=*x*2), *min*(*y*1,<=*y*2)<=≤<=*q*<=≤<=*max*(*y*1,<=*y*2).
Help Simon! Find the minimum number of operations needed to color all cells of the table. Note that you can color one cell multiple times. | The first line contains exactly two integers *n*, *m* (3<=≤<=*n*,<=*m*<=≤<=50).
Next *n* lines contain the description of the table cells. Specifically, the *i*-th line contains *m* space-separated integers *a**i*1,<=*a**i*2,<=...,<=*a**im*. If *a**ij* equals zero, then cell (*i*,<=*j*) isn't good. Otherwise *a**ij* equals one. It is guaranteed that at least one cell is good. It is guaranteed that no good cell is a corner. | Print a single number — the minimum number of operations Simon needs to carry out his idea. | [
"3 3\n0 0 0\n0 1 0\n0 0 0\n",
"4 3\n0 0 0\n0 0 1\n1 0 0\n0 0 0\n"
] | [
"4\n",
"2\n"
] | In the first sample, the sequence of operations can be like this:
- For the first time you need to choose cell (2, 2) and corner (1, 1). - For the second time you need to choose cell (2, 2) and corner (3, 3). - For the third time you need to choose cell (2, 2) and corner (3, 1). - For the fourth time you need to choose cell (2, 2) and corner (1, 3).
In the second sample the sequence of operations can be like this:
- For the first time you need to choose cell (3, 1) and corner (4, 3). - For the second time you need to choose cell (2, 3) and corner (1, 1). | 500 | [
{
"input": "3 3\n0 0 0\n0 1 0\n0 0 0",
"output": "4"
},
{
"input": "4 3\n0 0 0\n0 0 1\n1 0 0\n0 0 0",
"output": "2"
},
{
"input": "50 4\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 1 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "4"
},
{
"input": "5 50\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\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\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\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 1\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",
"output": "2"
},
{
"input": "4 32\n0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\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\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\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",
"output": "2"
},
{
"input": "7 4\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0\n0 1 0 0",
"output": "2"
},
{
"input": "13 15\n0 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\n0 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\n0 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\n1 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\n0 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\n0 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\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "2"
},
{
"input": "3 3\n0 1 0\n0 0 0\n0 0 0",
"output": "2"
},
{
"input": "3 3\n0 0 0\n0 0 0\n0 1 0",
"output": "2"
},
{
"input": "3 3\n0 0 0\n1 0 0\n0 0 0",
"output": "2"
},
{
"input": "3 3\n0 0 0\n0 0 1\n0 0 0",
"output": "2"
},
{
"input": "3 4\n0 1 0 0\n0 0 0 0\n0 0 0 0",
"output": "2"
},
{
"input": "3 5\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "2"
},
{
"input": "3 5\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "3 5\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "2"
},
{
"input": "3 5\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "4 3\n0 1 0\n0 0 0\n0 0 0\n0 0 0",
"output": "2"
},
{
"input": "4 3\n0 0 0\n0 0 0\n0 0 0\n0 1 0",
"output": "2"
},
{
"input": "5 3\n0 0 0\n0 0 0\n1 0 0\n0 0 0\n0 0 0",
"output": "2"
},
{
"input": "5 3\n0 0 0\n0 0 1\n0 0 0\n0 0 0\n0 0 0",
"output": "2"
},
{
"input": "5 3\n0 0 0\n0 1 0\n0 0 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "4 4\n0 0 0 0\n0 1 1 0\n0 1 1 0\n0 0 0 0",
"output": "4"
},
{
"input": "5 3\n0 0 0\n0 0 1\n0 0 0\n0 1 0\n0 0 0",
"output": "2"
},
{
"input": "3 3\n0 0 0\n0 1 1\n0 0 0",
"output": "2"
},
{
"input": "4 3\n0 0 0\n0 0 0\n0 1 0\n0 0 0",
"output": "4"
},
{
"input": "5 5\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "5 3\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n0 0 0",
"output": "2"
}
] | 1,651,745,839 | 2,147,483,647 | PyPy 3 | OK | TESTS | 37 | 77 | 0 | import sys
input = sys.stdin.readline
n, m = map(int, input().split())
ans = 4
for i in range(n):
s = input()[:-1].replace(' ','')
if i == 0 or i == n-1:
if '1' in s:
ans = 2
break
if s[0] == '1' or s[-1] == '1':
ans = 2
break
print(ans)
| Title: Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon has a rectangular table consisting of *n* rows and *m* columns. Simon numbered the rows of the table from top to bottom starting from one and the columns — from left to right starting from one. We'll represent the cell on the *x*-th row and the *y*-th column as a pair of numbers (*x*,<=*y*). The table corners are cells: (1,<=1), (*n*,<=1), (1,<=*m*), (*n*,<=*m*).
Simon thinks that some cells in this table are good. Besides, it's known that no good cell is the corner of the table.
Initially, all cells of the table are colorless. Simon wants to color all cells of his table. In one move, he can choose any good cell of table (*x*1,<=*y*1), an arbitrary corner of the table (*x*2,<=*y*2) and color all cells of the table (*p*,<=*q*), which meet both inequations: *min*(*x*1,<=*x*2)<=≤<=*p*<=≤<=*max*(*x*1,<=*x*2), *min*(*y*1,<=*y*2)<=≤<=*q*<=≤<=*max*(*y*1,<=*y*2).
Help Simon! Find the minimum number of operations needed to color all cells of the table. Note that you can color one cell multiple times.
Input Specification:
The first line contains exactly two integers *n*, *m* (3<=≤<=*n*,<=*m*<=≤<=50).
Next *n* lines contain the description of the table cells. Specifically, the *i*-th line contains *m* space-separated integers *a**i*1,<=*a**i*2,<=...,<=*a**im*. If *a**ij* equals zero, then cell (*i*,<=*j*) isn't good. Otherwise *a**ij* equals one. It is guaranteed that at least one cell is good. It is guaranteed that no good cell is a corner.
Output Specification:
Print a single number — the minimum number of operations Simon needs to carry out his idea.
Demo Input:
['3 3\n0 0 0\n0 1 0\n0 0 0\n', '4 3\n0 0 0\n0 0 1\n1 0 0\n0 0 0\n']
Demo Output:
['4\n', '2\n']
Note:
In the first sample, the sequence of operations can be like this:
- For the first time you need to choose cell (2, 2) and corner (1, 1). - For the second time you need to choose cell (2, 2) and corner (3, 3). - For the third time you need to choose cell (2, 2) and corner (3, 1). - For the fourth time you need to choose cell (2, 2) and corner (1, 3).
In the second sample the sequence of operations can be like this:
- For the first time you need to choose cell (3, 1) and corner (4, 3). - For the second time you need to choose cell (2, 3) and corner (1, 1). | ```python
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
ans = 4
for i in range(n):
s = input()[:-1].replace(' ','')
if i == 0 or i == n-1:
if '1' in s:
ans = 2
break
if s[0] == '1' or s[-1] == '1':
ans = 2
break
print(ans)
``` | 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"
}
] | 1,554,270,609 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 342 | 0 | s = input()
small = 0
capital = 0
sign = 0
for letter in s:
if ord(letter) < 97:
sign -= 1
else:
sign += 1
if sign < 0:
sign = -1
else:
sign = 1
for letter in s:
if sign == 1:
if ord(letter) < 97:
print(chr(ord(letter) + (32 * sign)), end = '')
else:
print(letter, end = '')
else:
if ord(letter) > 96:
print(chr(ord(letter) + (32 * sign)), end = '')
else:
print(letter, end = '') | 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
s = input()
small = 0
capital = 0
sign = 0
for letter in s:
if ord(letter) < 97:
sign -= 1
else:
sign += 1
if sign < 0:
sign = -1
else:
sign = 1
for letter in s:
if sign == 1:
if ord(letter) < 97:
print(chr(ord(letter) + (32 * sign)), end = '')
else:
print(letter, end = '')
else:
if ord(letter) > 96:
print(chr(ord(letter) + (32 * sign)), end = '')
else:
print(letter, end = '')
``` | 3.9145 |
682 | A | Alyona and Numbers | PROGRAMMING | 1,100 | [
"constructive algorithms",
"math",
"number theory"
] | null | null | After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0.
As usual, Alyona has some troubles and asks you to help. | The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000). | Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5. | [
"6 12\n",
"11 14\n",
"1 5\n",
"3 8\n",
"5 7\n",
"21 21\n"
] | [
"14\n",
"31\n",
"1\n",
"5\n",
"7\n",
"88\n"
] | Following pairs are suitable in the first sample case:
- for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | 500 | [
{
"input": "6 12",
"output": "14"
},
{
"input": "11 14",
"output": "31"
},
{
"input": "1 5",
"output": "1"
},
{
"input": "3 8",
"output": "5"
},
{
"input": "5 7",
"output": "7"
},
{
"input": "21 21",
"output": "88"
},
{
"input": "10 15",
"output": "30"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 1000000",
"output": "200000"
},
{
"input": "1000000 1",
"output": "200000"
},
{
"input": "1000000 1000000",
"output": "200000000000"
},
{
"input": "944 844",
"output": "159348"
},
{
"input": "368 984",
"output": "72423"
},
{
"input": "792 828",
"output": "131155"
},
{
"input": "920 969",
"output": "178296"
},
{
"input": "640 325",
"output": "41600"
},
{
"input": "768 170",
"output": "26112"
},
{
"input": "896 310",
"output": "55552"
},
{
"input": "320 154",
"output": "9856"
},
{
"input": "744 999",
"output": "148652"
},
{
"input": "630 843",
"output": "106218"
},
{
"input": "54 688",
"output": "7431"
},
{
"input": "478 828",
"output": "79157"
},
{
"input": "902 184",
"output": "33194"
},
{
"input": "31 29",
"output": "180"
},
{
"input": "751 169",
"output": "25384"
},
{
"input": "879 14",
"output": "2462"
},
{
"input": "7 858",
"output": "1201"
},
{
"input": "431 702",
"output": "60512"
},
{
"input": "855 355",
"output": "60705"
},
{
"input": "553 29",
"output": "3208"
},
{
"input": "721767 525996",
"output": "75929310986"
},
{
"input": "805191 74841",
"output": "12052259926"
},
{
"input": "888615 590981",
"output": "105030916263"
},
{
"input": "4743 139826",
"output": "132638943"
},
{
"input": "88167 721374",
"output": "12720276292"
},
{
"input": "171591 13322",
"output": "457187060"
},
{
"input": "287719 562167",
"output": "32349225415"
},
{
"input": "371143 78307",
"output": "5812618980"
},
{
"input": "487271 627151",
"output": "61118498984"
},
{
"input": "261436 930642",
"output": "48660664382"
},
{
"input": "377564 446782",
"output": "33737759810"
},
{
"input": "460988 28330",
"output": "2611958008"
},
{
"input": "544412 352983",
"output": "38433636199"
},
{
"input": "660540 869123",
"output": "114818101284"
},
{
"input": "743964 417967",
"output": "62190480238"
},
{
"input": "827388 966812",
"output": "159985729411"
},
{
"input": "910812 515656",
"output": "93933134534"
},
{
"input": "26940 64501",
"output": "347531388"
},
{
"input": "110364 356449",
"output": "7867827488"
},
{
"input": "636358 355531",
"output": "45248999219"
},
{
"input": "752486 871672",
"output": "131184195318"
},
{
"input": "803206 420516",
"output": "67552194859"
},
{
"input": "919334 969361",
"output": "178233305115"
},
{
"input": "35462 261309",
"output": "1853307952"
},
{
"input": "118887 842857",
"output": "20040948031"
},
{
"input": "202311 358998",
"output": "14525848875"
},
{
"input": "285735 907842",
"output": "51880446774"
},
{
"input": "401863 456686",
"output": "36705041203"
},
{
"input": "452583 972827",
"output": "88056992428"
},
{
"input": "235473 715013",
"output": "33673251230"
},
{
"input": "318897 263858",
"output": "16828704925"
},
{
"input": "402321 812702",
"output": "65393416268"
},
{
"input": "518449 361546",
"output": "37488632431"
},
{
"input": "634577 910391",
"output": "115542637921"
},
{
"input": "685297 235043",
"output": "32214852554"
},
{
"input": "801425 751183",
"output": "120403367155"
},
{
"input": "884849 300028",
"output": "53095895155"
},
{
"input": "977 848872",
"output": "165869588"
},
{
"input": "51697 397716",
"output": "4112144810"
},
{
"input": "834588 107199",
"output": "17893399803"
},
{
"input": "918012 688747",
"output": "126455602192"
},
{
"input": "1436 237592",
"output": "68236422"
},
{
"input": "117564 753732",
"output": "17722349770"
},
{
"input": "200988 302576",
"output": "12162829017"
},
{
"input": "284412 818717",
"output": "46570587880"
},
{
"input": "400540 176073",
"output": "14104855884"
},
{
"input": "483964 724917",
"output": "70166746198"
},
{
"input": "567388 241058",
"output": "27354683301"
},
{
"input": "650812 789902",
"output": "102815540084"
},
{
"input": "400999 756281",
"output": "60653584944"
},
{
"input": "100 101",
"output": "2020"
},
{
"input": "100 102",
"output": "2040"
},
{
"input": "103 100",
"output": "2060"
},
{
"input": "100 104",
"output": "2080"
},
{
"input": "3 4",
"output": "3"
},
{
"input": "11 23",
"output": "50"
},
{
"input": "8 14",
"output": "23"
},
{
"input": "23423 34234",
"output": "160372597"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "999999 999999",
"output": "199999600001"
},
{
"input": "82 99",
"output": "1624"
},
{
"input": "21 18",
"output": "75"
},
{
"input": "234 234",
"output": "10952"
},
{
"input": "4 4",
"output": "4"
},
{
"input": "6 13",
"output": "15"
},
{
"input": "3 9",
"output": "6"
},
{
"input": "99999 99999",
"output": "1999960001"
},
{
"input": "34 33",
"output": "225"
},
{
"input": "2 2",
"output": "0"
},
{
"input": "333 1",
"output": "66"
},
{
"input": "3 3",
"output": "2"
},
{
"input": "8 2",
"output": "3"
},
{
"input": "2179 2218",
"output": "966605"
},
{
"input": "1000000 999999",
"output": "199999800000"
},
{
"input": "873828 774207",
"output": "135304750879"
},
{
"input": "13 19",
"output": "50"
},
{
"input": "1648 576469",
"output": "190004183"
},
{
"input": "11 13",
"output": "28"
},
{
"input": "5 8",
"output": "8"
},
{
"input": "650074 943659",
"output": "122689636154"
},
{
"input": "1 3",
"output": "0"
},
{
"input": "54 43",
"output": "465"
},
{
"input": "14 9",
"output": "26"
},
{
"input": "2 3",
"output": "1"
},
{
"input": "543 534",
"output": "57993"
},
{
"input": "321 123",
"output": "7896"
},
{
"input": "21 3",
"output": "12"
},
{
"input": "2 1",
"output": "0"
},
{
"input": "4 3",
"output": "3"
},
{
"input": "47474 74747",
"output": "709707816"
},
{
"input": "4 9",
"output": "8"
},
{
"input": "7 4",
"output": "6"
},
{
"input": "9 4",
"output": "8"
},
{
"input": "12414 4214",
"output": "10462520"
},
{
"input": "2 9",
"output": "4"
},
{
"input": "253 821",
"output": "41542"
},
{
"input": "2 4",
"output": "2"
}
] | 1,658,840,580 | 2,147,483,647 | PyPy 3 | OK | TESTS | 128 | 77 | 0 | # -*- coding: utf-8 -*-
"""
Created on Tue Jul 26 13:57:33 2022
@author: Conor
CFSheet A Problem 72 - CF682-DIV2A
"""
n,m = map(int,input().split())
fullN = n//5
fullM = m//5
haveN = [fullN+1]*(n%5) + [fullN]*(5-n%5)
haveM = [fullM+1]*(m%5) + [fullM]*(5-m%5)
ans = haveN[4]*haveM[4] + haveN[0]*haveM[3] + haveN[1]*haveM[2] + haveN[2]*haveM[1]+haveN[3]*haveM[0]
print(ans) | Title: Alyona and Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0.
As usual, Alyona has some troubles and asks you to help.
Input Specification:
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000).
Output Specification:
Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5.
Demo Input:
['6 12\n', '11 14\n', '1 5\n', '3 8\n', '5 7\n', '21 21\n']
Demo Output:
['14\n', '31\n', '1\n', '5\n', '7\n', '88\n']
Note:
Following pairs are suitable in the first sample case:
- for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | ```python
# -*- coding: utf-8 -*-
"""
Created on Tue Jul 26 13:57:33 2022
@author: Conor
CFSheet A Problem 72 - CF682-DIV2A
"""
n,m = map(int,input().split())
fullN = n//5
fullM = m//5
haveN = [fullN+1]*(n%5) + [fullN]*(5-n%5)
haveM = [fullM+1]*(m%5) + [fullM]*(5-m%5)
ans = haveN[4]*haveM[4] + haveN[0]*haveM[3] + haveN[1]*haveM[2] + haveN[2]*haveM[1]+haveN[3]*haveM[0]
print(ans)
``` | 3 | |
600 | B | Queries about less or equal elements | PROGRAMMING | 1,300 | [
"binary search",
"data structures",
"sortings",
"two pointers"
] | null | null | You are given two arrays of integers *a* and *b*. For each element of the second array *b**j* you should find the number of elements in array *a* that are less than or equal to the value *b**j*. | The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=2·105) — the sizes of arrays *a* and *b*.
The second line contains *n* integers — the elements of array *a* (<=-<=109<=≤<=*a**i*<=≤<=109).
The third line contains *m* integers — the elements of array *b* (<=-<=109<=≤<=*b**j*<=≤<=109). | Print *m* integers, separated by spaces: the *j*-th of which is equal to the number of such elements in array *a* that are less than or equal to the value *b**j*. | [
"5 4\n1 3 5 7 9\n6 4 2 8\n",
"5 5\n1 2 1 2 5\n3 1 4 1 5\n"
] | [
"3 2 1 4\n",
"4 2 4 2 5\n"
] | none | 0 | [
{
"input": "5 4\n1 3 5 7 9\n6 4 2 8",
"output": "3 2 1 4"
},
{
"input": "5 5\n1 2 1 2 5\n3 1 4 1 5",
"output": "4 2 4 2 5"
},
{
"input": "1 1\n-1\n-2",
"output": "0"
},
{
"input": "1 1\n-80890826\n686519510",
"output": "1"
},
{
"input": "11 11\n237468511 -779187544 -174606592 193890085 404563196 -71722998 -617934776 170102710 -442808289 109833389 953091341\n994454001 322957429 216874735 -606986750 -455806318 -663190696 3793295 41395397 -929612742 -787653860 -684738874",
"output": "11 9 8 2 2 1 5 5 0 0 1"
},
{
"input": "20 22\n858276994 -568758442 -918490847 -983345984 -172435358 389604931 200224783 486556113 413281867 -258259500 -627945379 -584563643 444685477 -602481243 -370745158 965672503 630955806 -626138773 -997221880 633102929\n-61330638 -977252080 -212144219 385501731 669589742 954357160 563935906 584468977 -895883477 405774444 853372186 186056475 -964575261 -952431965 632332084 -388829939 -23011650 310957048 -770695392 977376693 321435214 199223897",
"output": "11 2 10 12 18 19 16 16 3 13 18 11 2 2 17 8 11 12 3 20 12 11"
},
{
"input": "5 9\n1 3 5 7 9\n1 2 3 4 5 6 7 8 9",
"output": "1 1 2 2 3 3 4 4 5"
},
{
"input": "22 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22\n1",
"output": "1"
},
{
"input": "5 1\n1 3 3 3 5\n3",
"output": "4"
},
{
"input": "4 5\n1 1 1 4\n1 5 5 4 3",
"output": "3 4 4 4 3"
},
{
"input": "5 4\n0 5 5 5 6\n5 1 6 3",
"output": "4 1 5 1"
},
{
"input": "1 3\n0\n-1 0 1",
"output": "0 1 1"
},
{
"input": "96 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1",
"output": "96"
},
{
"input": "7 1\n1 2 3 4 5 6 7\n1",
"output": "1"
},
{
"input": "13 13\n-1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000\n-1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000",
"output": "7 13 7 13 7 13 7 13 7 13 7 13 7"
},
{
"input": "9 5\n1 2 3 4 5 6 7 8 9\n1 2 3 4 5",
"output": "1 2 3 4 5"
},
{
"input": "3 8\n1 1 1\n1 1 1 1 1 1 1 1",
"output": "3 3 3 3 3 3 3 3"
},
{
"input": "1 1\n-11111\n-5938",
"output": "1"
},
{
"input": "1 1\n1\n400000009",
"output": "1"
},
{
"input": "1 1\n1\n300000009",
"output": "1"
},
{
"input": "1 1\n1\n200000009",
"output": "1"
},
{
"input": "1 1\n1\n200000003",
"output": "1"
}
] | 1,683,309,012 | 2,147,483,647 | Python 3 | OK | TESTS | 60 | 607 | 30,720,000 | from bisect import bisect_right
input()
l = sorted(map(int, input().split()))
[print(bisect_right(l, x),end=' ') for x in map(int, input().split())]
print()
| Title: Queries about less or equal elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two arrays of integers *a* and *b*. For each element of the second array *b**j* you should find the number of elements in array *a* that are less than or equal to the value *b**j*.
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=2·105) — the sizes of arrays *a* and *b*.
The second line contains *n* integers — the elements of array *a* (<=-<=109<=≤<=*a**i*<=≤<=109).
The third line contains *m* integers — the elements of array *b* (<=-<=109<=≤<=*b**j*<=≤<=109).
Output Specification:
Print *m* integers, separated by spaces: the *j*-th of which is equal to the number of such elements in array *a* that are less than or equal to the value *b**j*.
Demo Input:
['5 4\n1 3 5 7 9\n6 4 2 8\n', '5 5\n1 2 1 2 5\n3 1 4 1 5\n']
Demo Output:
['3 2 1 4\n', '4 2 4 2 5\n']
Note:
none | ```python
from bisect import bisect_right
input()
l = sorted(map(int, input().split()))
[print(bisect_right(l, x),end=' ') for x in map(int, input().split())]
print()
``` | 3 | |
485 | A | Factory | PROGRAMMING | 1,400 | [
"implementation",
"math",
"matrices"
] | null | null | One industrial factory is reforming working plan. The director suggested to set a mythical detail production norm. If at the beginning of the day there were *x* details in the factory storage, then by the end of the day the factory has to produce (remainder after dividing *x* by *m*) more details. Unfortunately, no customer has ever bought any mythical detail, so all the details produced stay on the factory.
The board of directors are worried that the production by the given plan may eventually stop (that means that there will be а moment when the current number of details on the factory is divisible by *m*).
Given the number of details *a* on the first day and number *m* check if the production stops at some moment. | The first line contains two integers *a* and *m* (1<=≤<=*a*,<=*m*<=≤<=105). | Print "Yes" (without quotes) if the production will eventually stop, otherwise print "No". | [
"1 5\n",
"3 6\n"
] | [
"No\n",
"Yes\n"
] | none | 1,000 | [
{
"input": "1 5",
"output": "No"
},
{
"input": "3 6",
"output": "Yes"
},
{
"input": "1 8",
"output": "Yes"
},
{
"input": "2 3",
"output": "No"
},
{
"input": "3 24",
"output": "Yes"
},
{
"input": "1 1",
"output": "Yes"
},
{
"input": "100000 100000",
"output": "Yes"
},
{
"input": "1 99989",
"output": "No"
},
{
"input": "512 2",
"output": "Yes"
},
{
"input": "100 24",
"output": "No"
},
{
"input": "1 100000",
"output": "No"
},
{
"input": "100000 1",
"output": "Yes"
},
{
"input": "3 99929",
"output": "No"
},
{
"input": "99961 99971",
"output": "No"
},
{
"input": "1 65536",
"output": "Yes"
},
{
"input": "4 65536",
"output": "Yes"
},
{
"input": "3 65536",
"output": "Yes"
},
{
"input": "32768 65536",
"output": "Yes"
},
{
"input": "65535 65536",
"output": "Yes"
},
{
"input": "1 65535",
"output": "No"
},
{
"input": "98812 100000",
"output": "No"
},
{
"input": "10 5",
"output": "Yes"
},
{
"input": "6 8",
"output": "Yes"
}
] | 1,500,969,359 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 61 | 4,608,000 | string = input()
numbers = string.split()
a, b = int(numbers[0]), int(numbers[1])
n = a % b
while b % 2 == 0:
b //= 2
if b == n:
print("Yes")
else:
print("No") | Title: Factory
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One industrial factory is reforming working plan. The director suggested to set a mythical detail production norm. If at the beginning of the day there were *x* details in the factory storage, then by the end of the day the factory has to produce (remainder after dividing *x* by *m*) more details. Unfortunately, no customer has ever bought any mythical detail, so all the details produced stay on the factory.
The board of directors are worried that the production by the given plan may eventually stop (that means that there will be а moment when the current number of details on the factory is divisible by *m*).
Given the number of details *a* on the first day and number *m* check if the production stops at some moment.
Input Specification:
The first line contains two integers *a* and *m* (1<=≤<=*a*,<=*m*<=≤<=105).
Output Specification:
Print "Yes" (without quotes) if the production will eventually stop, otherwise print "No".
Demo Input:
['1 5\n', '3 6\n']
Demo Output:
['No\n', 'Yes\n']
Note:
none | ```python
string = input()
numbers = string.split()
a, b = int(numbers[0]), int(numbers[1])
n = a % b
while b % 2 == 0:
b //= 2
if b == n:
print("Yes")
else:
print("No")
``` | 0 | |
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,681,154,280 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 92 | 0 | str1=input()
str2=input()
if("".join(reversed(str1))==str2):
print("YES")
else:
print("NO") | Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
str1=input()
str2=input()
if("".join(reversed(str1))==str2):
print("YES")
else:
print("NO")
``` | 3.977 |
868 | D | Huge Strings | PROGRAMMING | 2,200 | [
"bitmasks",
"brute force",
"dp",
"implementation",
"strings"
] | null | null | You are given *n* strings *s*1,<=*s*2,<=...,<=*s**n* consisting of characters 0 and 1. *m* operations are performed, on each of them you concatenate two existing strings into a new one. On the *i*-th operation the concatenation *s**a**i**s**b**i* is saved into a new string *s**n*<=+<=*i* (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer *k* such that all possible strings consisting of 0 and 1 of length *k* (there are 2*k* such strings) are substrings of the new string. If there is no such *k*, print 0. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of strings. The next *n* lines contain strings *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=|*s**i*|<=≤<=100), one per line. The total length of strings is not greater than 100.
The next line contains single integer *m* (1<=≤<=*m*<=≤<=100) — the number of operations. *m* lines follow, each of them contains two integers *a**i* abd *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*<=+<=*i*<=-<=1) — the number of strings that are concatenated to form *s**n*<=+<=*i*. | Print *m* lines, each should contain one integer — the answer to the question after the corresponding operation. | [
"5\n01\n10\n101\n11111\n0\n3\n1 2\n6 5\n4 4\n"
] | [
"1\n2\n0\n"
] | On the first operation, a new string "0110" is created. For *k* = 1 the two possible binary strings of length *k* are "0" and "1", they are substrings of the new string. For *k* = 2 and greater there exist strings of length *k* that do not appear in this string (for *k* = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length *k* = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0. | 1,500 | [
{
"input": "5\n01\n10\n101\n11111\n0\n3\n1 2\n6 5\n4 4",
"output": "1\n2\n0"
},
{
"input": "5\n01\n1\n0011\n0\n01\n6\n5 5\n3 2\n4 2\n6 7\n5 1\n9 7",
"output": "1\n1\n1\n2\n1\n2"
},
{
"input": "5\n111101000111100011100110000100\n000111001\n01101000\n0000110100100010011001000000010100100111110110\n0110001\n10\n5 5\n2 2\n5 6\n1 1\n1 7\n10 6\n6 2\n11 1\n3 6\n8 2",
"output": "2\n2\n2\n3\n3\n4\n3\n4\n2\n3"
},
{
"input": "1\n1\n1\n1 1",
"output": "0"
},
{
"input": "5\n110101010101010110000011011\n111111\n1000100011100111100101101010011111100000001001\n00\n1111101100001110000\n10\n4 3\n6 6\n7 5\n8 8\n8 7\n10 8\n11 9\n10 12\n13 13\n12 13",
"output": "4\n4\n4\n4\n4\n4\n4\n4\n4\n4"
},
{
"input": "5\n100010010\n0\n1001100110010111\n0001000011000111000011011000110000010010010001110001000011011\n0100000100100\n10\n5 5\n6 6\n6 7\n7 8\n8 9\n10 8\n11 9\n10 9\n12 13\n12 13",
"output": "1\n1\n1\n1\n1\n1\n1\n1\n1\n1"
},
{
"input": "5\n0\n1\n11\n110000010001100101001\n1101011011111\n10\n5 3\n6 4\n7 6\n8 7\n9 8\n10 9\n11 10\n12 11\n13 12\n14 13",
"output": "1\n4\n5\n5\n5\n5\n5\n5\n5\n5"
},
{
"input": "10\n0\n1\n1111100000\n0\n1\n0000\n11000\n1010001110010010110\n01101001111\n010101110110111111\n20\n10 3\n11 4\n12 5\n13 6\n14 7\n15 8\n16 9\n17 16\n18 17\n19 18\n20 19\n21 20\n22 21\n23 22\n24 23\n25 24\n26 25\n27 26\n28 27\n29 28",
"output": "2\n2\n3\n3\n3\n4\n5\n6\n6\n6\n6\n6\n6\n6\n6\n6\n6\n6\n6\n6"
},
{
"input": "10\n0\n1\n1111\n110000000\n100000\n1\n1\n000010100001110001\n00100010110001101000111100100110010101001011\n100110110011101\n50\n10 3\n11 4\n12 5\n13 6\n14 7\n15 8\n16 9\n17 1\n18 1\n19 2\n20 2\n21 2\n22 2\n23 2\n24 1\n25 2\n26 1\n27 2\n28 1\n29 2\n30 2\n31 1\n32 2\n33 1\n34 2\n35 2\n36 2\n37 2\n38 1\n39 2\n40 2\n41 1\n42 2\n43 2\n44 2\n45 1\n46 2\n47 2\n48 2\n49 2\n50 2\n51 2\n52 2\n53 52\n54 53\n55 54\n56 55\n57 56\n58 57\n59 58",
"output": "2\n2\n3\n3\n3\n4\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n5\n6\n6\n6\n6\n6\n6\n6\n6\n6\n6\n6\n6\n6\n6\n6"
},
{
"input": "2\n001010011100101110111\n001100110011001100110011001100110011001100110011001100111001\n14\n1 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15",
"output": "2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2"
},
{
"input": "2\n1\n0\n40\n1 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 20\n21 21\n22 22\n23 23\n24 24\n25 25\n26 26\n27 27\n28 28\n29 29\n30 30\n31 31\n32 32\n33 33\n34 34\n35 35\n36 36\n37 37\n38 38\n39 39\n40 40\n41 41",
"output": "1\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"
},
{
"input": "2\n011\n100\n63\n1 1\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 20\n21 21\n22 22\n23 23\n24 24\n25 25\n26 26\n27 27\n28 28\n29 29\n30 30\n31 31\n32 32\n2 2\n34 34\n35 35\n36 36\n37 37\n38 38\n39 39\n40 40\n41 41\n42 42\n43 43\n44 44\n45 45\n46 46\n47 47\n48 48\n49 49\n50 50\n51 51\n52 52\n53 53\n54 54\n55 55\n56 56\n57 57\n58 58\n59 59\n60 60\n61 61\n62 62\n63 63\n33 64",
"output": "1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n2"
},
{
"input": "1\n0000000000000000000000000000000000000000000000000000000000000000\n25\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 20\n21 21\n22 22\n23 23\n24 24\n25 25",
"output": "0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0"
}
] | 1,511,006,490 | 6,990 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 2,000 | 14,233,600 | cr=""
def check(l):
i=0;
dick={}
while(i+l<=len(cr)):
q=cr[i:i+l]
if(q in dick):
dick[q]+=1
else:
dick[q]=1
i+=1
if(len(dick)==2**l):
return 1
return 0
def bs(f,l):
if(f==l-1):
return f
mid=(f+l)//2
if(check(mid)):
return bs(mid,l)
return bs(f,mid)
n=int(input())
s=[input().strip() for i in range(0,n)]
m=int(input())
for i in range(0,m):
t1,t2=map(int,input().split())
cr=s[t1-1]+s[t2-1]
s.append(cr)
print(bs(0,len(cr)))
| Title: Huge Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* strings *s*1,<=*s*2,<=...,<=*s**n* consisting of characters 0 and 1. *m* operations are performed, on each of them you concatenate two existing strings into a new one. On the *i*-th operation the concatenation *s**a**i**s**b**i* is saved into a new string *s**n*<=+<=*i* (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer *k* such that all possible strings consisting of 0 and 1 of length *k* (there are 2*k* such strings) are substrings of the new string. If there is no such *k*, print 0.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of strings. The next *n* lines contain strings *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=|*s**i*|<=≤<=100), one per line. The total length of strings is not greater than 100.
The next line contains single integer *m* (1<=≤<=*m*<=≤<=100) — the number of operations. *m* lines follow, each of them contains two integers *a**i* abd *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*<=+<=*i*<=-<=1) — the number of strings that are concatenated to form *s**n*<=+<=*i*.
Output Specification:
Print *m* lines, each should contain one integer — the answer to the question after the corresponding operation.
Demo Input:
['5\n01\n10\n101\n11111\n0\n3\n1 2\n6 5\n4 4\n']
Demo Output:
['1\n2\n0\n']
Note:
On the first operation, a new string "0110" is created. For *k* = 1 the two possible binary strings of length *k* are "0" and "1", they are substrings of the new string. For *k* = 2 and greater there exist strings of length *k* that do not appear in this string (for *k* = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length *k* = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0. | ```python
cr=""
def check(l):
i=0;
dick={}
while(i+l<=len(cr)):
q=cr[i:i+l]
if(q in dick):
dick[q]+=1
else:
dick[q]=1
i+=1
if(len(dick)==2**l):
return 1
return 0
def bs(f,l):
if(f==l-1):
return f
mid=(f+l)//2
if(check(mid)):
return bs(mid,l)
return bs(f,mid)
n=int(input())
s=[input().strip() for i in range(0,n)]
m=int(input())
for i in range(0,m):
t1,t2=map(int,input().split())
cr=s[t1-1]+s[t2-1]
s.append(cr)
print(bs(0,len(cr)))
``` | 0 | |
278 | A | Circle Line | PROGRAMMING | 800 | [
"implementation"
] | null | null | The circle line of the Berland subway has *n* stations. We know the distances between all pairs of neighboring stations:
- *d*1 is the distance between the 1-st and the 2-nd station;- *d*2 is the distance between the 2-nd and the 3-rd station;...- *d**n*<=-<=1 is the distance between the *n*<=-<=1-th and the *n*-th station;- *d**n* is the distance between the *n*-th and the 1-st station.
The trains go along the circle line in both directions. Find the shortest distance between stations with numbers *s* and *t*. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — the number of stations on the circle line. The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (1<=≤<=*d**i*<=≤<=100) — the distances between pairs of neighboring stations. The third line contains two integers *s* and *t* (1<=≤<=*s*,<=*t*<=≤<=*n*) — the numbers of stations, between which you need to find the shortest distance. These numbers can be the same.
The numbers in the lines are separated by single spaces. | Print a single number — the length of the shortest path between stations number *s* and *t*. | [
"4\n2 3 4 9\n1 3\n",
"4\n5 8 2 100\n4 1\n",
"3\n1 1 1\n3 1\n",
"3\n31 41 59\n1 1\n"
] | [
"5\n",
"15\n",
"1\n",
"0\n"
] | In the first sample the length of path 1 → 2 → 3 equals 5, the length of path 1 → 4 → 3 equals 13.
In the second sample the length of path 4 → 1 is 100, the length of path 4 → 3 → 2 → 1 is 15.
In the third sample the length of path 3 → 1 is 1, the length of path 3 → 2 → 1 is 2.
In the fourth sample the numbers of stations are the same, so the shortest distance equals 0. | 500 | [
{
"input": "4\n2 3 4 9\n1 3",
"output": "5"
},
{
"input": "4\n5 8 2 100\n4 1",
"output": "15"
},
{
"input": "3\n1 1 1\n3 1",
"output": "1"
},
{
"input": "3\n31 41 59\n1 1",
"output": "0"
},
{
"input": "5\n16 13 10 30 15\n4 2",
"output": "23"
},
{
"input": "6\n89 82 87 32 67 33\n4 4",
"output": "0"
},
{
"input": "7\n2 3 17 10 2 2 2\n4 2",
"output": "18"
},
{
"input": "3\n4 37 33\n3 3",
"output": "0"
},
{
"input": "8\n87 40 96 7 86 86 72 97\n6 8",
"output": "158"
},
{
"input": "10\n91 94 75 99 100 91 79 86 79 92\n2 8",
"output": "348"
},
{
"input": "19\n1 1 1 1 2 1 1 1 1 1 2 1 3 2 2 1 1 1 2\n7 7",
"output": "0"
},
{
"input": "34\n96 65 24 99 74 76 97 93 99 69 94 82 92 91 98 83 95 97 96 81 90 95 86 87 43 78 88 86 82 62 76 99 83 96\n21 16",
"output": "452"
},
{
"input": "50\n75 98 65 75 99 89 84 65 9 53 62 61 61 53 80 7 6 47 86 1 89 27 67 1 31 39 53 92 19 20 76 41 60 15 29 94 76 82 87 89 93 38 42 6 87 36 100 97 93 71\n2 6",
"output": "337"
},
{
"input": "99\n1 15 72 78 23 22 26 98 7 2 75 58 100 98 45 79 92 69 79 72 33 88 62 9 15 87 17 73 68 54 34 89 51 91 28 44 20 11 74 7 85 61 30 46 95 72 36 18 48 22 42 46 29 46 86 53 96 55 98 34 60 37 75 54 1 81 20 68 84 19 18 18 75 84 86 57 73 34 23 43 81 87 47 96 57 41 69 1 52 44 54 7 85 35 5 1 19 26 7\n4 64",
"output": "1740"
},
{
"input": "100\n33 63 21 27 49 82 86 93 43 55 4 72 89 85 5 34 80 7 23 13 21 49 22 73 89 65 81 25 6 92 82 66 58 88 48 96 1 1 16 48 67 96 84 63 87 76 20 100 36 4 31 41 35 62 55 76 74 70 68 41 4 16 39 81 2 41 34 73 66 57 41 89 78 93 68 96 87 47 92 60 40 58 81 12 19 74 56 83 56 61 83 97 26 92 62 52 39 57 89 95\n71 5",
"output": "2127"
},
{
"input": "100\n95 98 99 81 98 96 100 92 96 90 99 91 98 98 91 78 97 100 96 98 87 93 96 99 91 92 96 92 90 97 85 83 99 95 66 91 87 89 100 95 100 88 99 84 96 79 99 100 94 100 99 99 92 89 99 91 100 94 98 97 91 92 90 87 84 99 97 98 93 100 90 85 75 95 86 71 98 93 91 87 92 95 98 94 95 94 100 98 96 100 97 96 95 95 86 86 94 97 98 96\n67 57",
"output": "932"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 97 100 100 100 100 100 99 100 100 99 99 100 99 100 100 100 100 100 100 100 100 100 97 99 98 98 100 98 98 100 99 100 100 100 100 99 100 98 100 99 98 99 98 98 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 98 100 99 99 100 96 100 96 100 99 100 100 99 100 99 100 100 100 99 100 100 100 100 98 98 97 100 100 99 98\n16 6",
"output": "997"
},
{
"input": "100\n3 6 23 4 23 1 2 14 2 3 3 9 17 8 10 5 1 14 8 5 7 4 13 8 5 6 24 3 12 3 4 9 2 8 2 1 2 1 3 2 1 6 14 23 8 6 3 5 7 8 18 9 2 5 22 6 13 16 2 4 31 20 4 3 3 6 6 1 1 18 5 11 1 14 4 16 6 37 11 1 8 3 7 11 21 14 3 3 12 2 5 1 9 16 3 1 3 4 4 2\n98 24",
"output": "195"
},
{
"input": "100\n1 1 3 1 1 2 1 2 1 1 2 2 2 1 1 1 1 1 1 3 1 1 1 3 1 3 3 1 1 2 1 1 1 1 1 2 1 1 1 4 1 1 3 3 2 1 1 1 1 1 2 2 1 3 1 1 1 2 4 1 1 2 5 2 1 1 2 1 1 1 2 3 1 1 1 1 2 1 1 1 1 2 1 1 1 1 1 2 2 3 1 7 3 1 3 1 2 1 2 1\n49 10",
"output": "60"
},
{
"input": "100\n75 62 31 96 62 76 93 96 72 67 88 35 67 34 60 56 95 86 82 48 64 61 74 100 56 98 76 98 78 55 53 10 12 78 58 45 86 90 93 77 69 73 88 66 92 88 33 50 95 69 89 12 93 57 93 89 59 53 71 86 15 13 61 93 24 100 58 76 46 95 76 82 50 20 79 38 5 72 99 81 55 90 90 65 85 44 63 39 6 34 98 72 88 30 59 73 84 61 25 67\n86 25",
"output": "2523"
},
{
"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\n1 51",
"output": "5000"
},
{
"input": "4\n1 1 1 1\n2 4",
"output": "2"
},
{
"input": "4\n1 1 2 1\n2 4",
"output": "2"
}
] | 1,426,418,742 | 2,147,483,647 | Python 3 | OK | TESTS | 23 | 124 | 0 | # coding: utf-8
n = int(input())
d = [int(i) for i in input().split()]
s, t = [int(i) for i in input().split()]
if s > t:
s, t = t, s
circu = sum(d)
len1 = sum(d[s-1:t-1])
len2 = circu-len1
print(min(len1,len2))
| Title: Circle Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The circle line of the Berland subway has *n* stations. We know the distances between all pairs of neighboring stations:
- *d*1 is the distance between the 1-st and the 2-nd station;- *d*2 is the distance between the 2-nd and the 3-rd station;...- *d**n*<=-<=1 is the distance between the *n*<=-<=1-th and the *n*-th station;- *d**n* is the distance between the *n*-th and the 1-st station.
The trains go along the circle line in both directions. Find the shortest distance between stations with numbers *s* and *t*.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — the number of stations on the circle line. The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (1<=≤<=*d**i*<=≤<=100) — the distances between pairs of neighboring stations. The third line contains two integers *s* and *t* (1<=≤<=*s*,<=*t*<=≤<=*n*) — the numbers of stations, between which you need to find the shortest distance. These numbers can be the same.
The numbers in the lines are separated by single spaces.
Output Specification:
Print a single number — the length of the shortest path between stations number *s* and *t*.
Demo Input:
['4\n2 3 4 9\n1 3\n', '4\n5 8 2 100\n4 1\n', '3\n1 1 1\n3 1\n', '3\n31 41 59\n1 1\n']
Demo Output:
['5\n', '15\n', '1\n', '0\n']
Note:
In the first sample the length of path 1 → 2 → 3 equals 5, the length of path 1 → 4 → 3 equals 13.
In the second sample the length of path 4 → 1 is 100, the length of path 4 → 3 → 2 → 1 is 15.
In the third sample the length of path 3 → 1 is 1, the length of path 3 → 2 → 1 is 2.
In the fourth sample the numbers of stations are the same, so the shortest distance equals 0. | ```python
# coding: utf-8
n = int(input())
d = [int(i) for i in input().split()]
s, t = [int(i) for i in input().split()]
if s > t:
s, t = t, s
circu = sum(d)
len1 = sum(d[s-1:t-1])
len2 = circu-len1
print(min(len1,len2))
``` | 3 | |
723 | A | The New Year: Meeting Friends | PROGRAMMING | 800 | [
"implementation",
"math",
"sortings"
] | null | null | There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer. | The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively. | Print one integer — the minimum total distance the friends need to travel in order to meet together. | [
"7 1 4\n",
"30 20 10\n"
] | [
"6\n",
"20\n"
] | In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4. | 500 | [
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,682,010,252 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 77 | 0 | a, b, c = map(int, input().split())
m = max(a, b, c)
n = min(a, b, c)
x = (m + n) // 2
s = abs(a - x)
f = abs(b - x)
d = abs(c - x)
print(s + f + d)
| Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4. | ```python
a, b, c = map(int, input().split())
m = max(a, b, c)
n = min(a, b, c)
x = (m + n) // 2
s = abs(a - x)
f = abs(b - x)
d = abs(c - x)
print(s + f + d)
``` | 0 | |
22 | A | Second Order Statistics | PROGRAMMING | 800 | [
"brute force"
] | A. Second Order Statistics | 2 | 256 | Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem. | The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value. | If the given sequence has the second order statistics, output this order statistics, otherwise output NO. | [
"4\n1 2 2 -4\n",
"5\n1 2 3 1 1\n"
] | [
"1\n",
"2\n"
] | none | 0 | [
{
"input": "4\n1 2 2 -4",
"output": "1"
},
{
"input": "5\n1 2 3 1 1",
"output": "2"
},
{
"input": "1\n28",
"output": "NO"
},
{
"input": "2\n-28 12",
"output": "12"
},
{
"input": "3\n-83 40 -80",
"output": "-80"
},
{
"input": "8\n93 77 -92 26 21 -48 53 91",
"output": "-48"
},
{
"input": "20\n-72 -9 -86 80 7 -10 40 -27 -94 92 96 56 28 -19 79 36 -3 -73 -63 -49",
"output": "-86"
},
{
"input": "49\n-74 -100 -80 23 -8 -83 -41 -20 48 17 46 -73 -55 67 85 4 40 -60 -69 -75 56 -74 -42 93 74 -95 64 -46 97 -47 55 0 -78 -34 -31 40 -63 -49 -76 48 21 -1 -49 -29 -98 -11 76 26 94",
"output": "-98"
},
{
"input": "88\n63 48 1 -53 -89 -49 64 -70 -49 71 -17 -16 76 81 -26 -50 67 -59 -56 97 2 100 14 18 -91 -80 42 92 -25 -88 59 8 -56 38 48 -71 -78 24 -14 48 -1 69 73 -76 54 16 -92 44 47 33 -34 -17 -81 21 -59 -61 53 26 10 -76 67 35 -29 70 65 -13 -29 81 80 32 74 -6 34 46 57 1 -45 -55 69 79 -58 11 -2 22 -18 -16 -89 -46",
"output": "-91"
},
{
"input": "100\n34 32 88 20 76 53 -71 -39 -98 -10 57 37 63 -3 -54 -64 -78 -82 73 20 -30 -4 22 75 51 -64 -91 29 -52 -48 83 19 18 -47 46 57 -44 95 89 89 -30 84 -83 67 58 -99 -90 -53 92 -60 -5 -56 -61 27 68 -48 52 -95 64 -48 -30 -67 66 89 14 -33 -31 -91 39 7 -94 -54 92 -96 -99 -83 -16 91 -28 -66 81 44 14 -85 -21 18 40 16 -13 -82 -33 47 -10 -40 -19 10 25 60 -34 -89",
"output": "-98"
},
{
"input": "2\n-1 -1",
"output": "NO"
},
{
"input": "3\n-2 -2 -2",
"output": "NO"
},
{
"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": "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": "100"
},
{
"input": "10\n40 71 -85 -85 40 -85 -85 64 -85 47",
"output": "40"
},
{
"input": "23\n-90 -90 -41 -64 -64 -90 -15 10 -43 -90 -64 -64 89 -64 36 47 38 -90 -64 -90 -90 68 -90",
"output": "-64"
},
{
"input": "39\n-97 -93 -42 -93 -97 -93 56 -97 -97 -97 76 -33 -60 91 7 82 17 47 -97 -97 -93 73 -97 12 -97 -97 -97 -97 56 -92 -83 -93 -93 49 -93 -97 -97 -17 -93",
"output": "-93"
},
{
"input": "51\n-21 6 -35 -98 -86 -98 -86 -43 -65 32 -98 -40 96 -98 -98 -98 -98 -86 -86 -98 56 -86 -98 -98 -30 -98 -86 -31 -98 -86 -86 -86 -86 -30 96 -86 -86 -86 -60 25 88 -86 -86 58 31 -47 57 -86 37 44 -83",
"output": "-86"
},
{
"input": "66\n-14 -95 65 -95 -95 -97 -90 -71 -97 -97 70 -95 -95 -97 -95 -27 35 -87 -95 -5 -97 -97 87 34 -49 -95 -97 -95 -97 -95 -30 -95 -97 47 -95 -17 -97 -95 -97 -69 51 -97 -97 -95 -75 87 59 21 63 56 76 -91 98 -97 6 -97 -95 -95 -97 -73 11 -97 -35 -95 -95 -43",
"output": "-95"
},
{
"input": "77\n-67 -93 -93 -92 97 29 93 -93 -93 -5 -93 -7 60 -92 -93 44 -84 68 -92 -93 69 -92 -37 56 43 -93 35 -92 -93 19 -79 18 -92 -93 -93 -37 -93 -47 -93 -92 -92 74 67 19 40 -92 -92 -92 -92 -93 -93 -41 -93 -92 -93 -93 -92 -93 51 -80 6 -42 -92 -92 -66 -12 -92 -92 -3 93 -92 -49 -93 40 62 -92 -92",
"output": "-92"
},
{
"input": "89\n-98 40 16 -87 -98 63 -100 55 -96 -98 -21 -100 -93 26 -98 -98 -100 -89 -98 -5 -65 -28 -100 -6 -66 67 -100 -98 -98 10 -98 -98 -70 7 -98 2 -100 -100 -98 25 -100 -100 -98 23 -68 -100 -98 3 98 -100 -98 -98 -98 -98 -24 -100 -100 -9 -98 35 -100 99 -5 -98 -100 -100 37 -100 -84 57 -98 40 -47 -100 -1 -92 -76 -98 -98 -100 -100 -100 -63 30 21 -100 -100 -100 -12",
"output": "-98"
},
{
"input": "99\n10 -84 -100 -100 73 -64 -100 -94 33 -100 -100 -100 -100 71 64 24 7 -100 -32 -100 -100 77 -100 62 -12 55 45 -100 -100 -80 -100 -100 -100 -100 -100 -100 -100 -100 -100 -39 -48 -100 -34 47 -100 -100 -100 -100 -100 -77 -100 -100 -100 -100 -100 -100 -52 40 -55 -100 -44 -100 72 33 70 -100 -100 -78 -100 -3 100 -77 22 -100 95 -30 -100 10 -69 -100 -100 -100 -100 52 -39 -100 -100 -100 7 -100 -98 -66 95 -17 -100 52 -100 68 -100",
"output": "-98"
},
{
"input": "100\n-99 -98 -64 89 53 57 -99 29 -78 18 -3 -54 76 -98 -99 -98 37 -98 19 -47 89 73 -98 -91 -99 -99 -98 -48 -99 22 -99 -98 -99 -99 -98 -60 84 67 -99 -98 20 -98 88 -98 46 -98 -99 -98 -99 -71 -99 -98 -98 -39 83 95 -98 63 -98 -99 32 -98 -99 -64 57 -30 -53 -83 -4 -99 58 20 -98 -10 -99 -44 -99 -99 -99 -99 75 34 -98 -52 -98 -30 -98 -99 -98 -98 51 -99 -99 -99 -98 -99 -99 -82 -90 92",
"output": "-98"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "3\n1 3 2",
"output": "2"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "3\n3 1 2",
"output": "2"
},
{
"input": "3\n3 2 1",
"output": "2"
},
{
"input": "9\n99 99 99 99 100 100 100 100 100",
"output": "100"
},
{
"input": "5\n-100 -100 -100 -100 -100",
"output": "NO"
}
] | 1,597,134,670 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 216 | 6,758,400 | n=int(input())
l=list(map(int,input().split()))
if(n==1 or len(set(l))==1):
print('NO')
else:
l.sort()
prev=l[0]
for i in range(1,n):
if(l[i]!=prev):
print(l[i])
break
| Title: Second Order Statistics
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
Output Specification:
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
Demo Input:
['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n']
Demo Output:
['1\n', '2\n']
Note:
none | ```python
n=int(input())
l=list(map(int,input().split()))
if(n==1 or len(set(l))==1):
print('NO')
else:
l.sort()
prev=l[0]
for i in range(1,n):
if(l[i]!=prev):
print(l[i])
break
``` | 3.933411 |
165 | A | Supercentral Point | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | [
"8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n",
"5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n"
] | [
"2\n",
"1\n"
] | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | 500 | [
{
"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3",
"output": "2"
},
{
"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0",
"output": "1"
},
{
"input": "9\n-565 -752\n-184 723\n-184 -752\n-184 1\n950 723\n-565 723\n950 -752\n950 1\n-565 1",
"output": "1"
},
{
"input": "25\n-651 897\n916 897\n-651 -808\n-748 301\n-734 414\n-651 -973\n-734 897\n916 -550\n-758 414\n916 180\n-758 -808\n-758 -973\n125 -550\n125 -973\n125 301\n916 414\n-748 -808\n-651 301\n-734 301\n-307 897\n-651 -550\n-651 414\n125 -808\n-748 -550\n916 -808",
"output": "7"
},
{
"input": "1\n487 550",
"output": "0"
},
{
"input": "10\n990 -396\n990 736\n990 646\n990 -102\n990 -570\n990 155\n990 528\n990 489\n990 268\n990 676",
"output": "0"
},
{
"input": "30\n507 836\n525 836\n-779 196\n507 -814\n525 -814\n525 42\n525 196\n525 -136\n-779 311\n507 -360\n525 300\n507 578\n507 311\n-779 836\n507 300\n525 -360\n525 311\n-779 -360\n-779 578\n-779 300\n507 42\n525 578\n-779 379\n507 196\n525 379\n507 379\n-779 -814\n-779 42\n-779 -136\n507 -136",
"output": "8"
},
{
"input": "25\n890 -756\n890 -188\n-37 -756\n-37 853\n523 998\n-261 853\n-351 853\n-351 -188\n523 -756\n-261 -188\n-37 998\n523 -212\n-351 998\n-37 -188\n-351 -756\n-37 -212\n890 998\n890 -212\n523 853\n-351 -212\n-261 -212\n-261 998\n-261 -756\n890 853\n523 -188",
"output": "9"
},
{
"input": "21\n-813 -11\n486 254\n685 254\n-708 254\n-55 -11\n-671 -191\n486 -11\n-671 -11\n685 -11\n685 -191\n486 -191\n-55 254\n-708 -11\n-813 254\n-708 -191\n41 -11\n-671 254\n-813 -191\n41 254\n-55 -191\n41 -191",
"output": "5"
},
{
"input": "4\n1 0\n2 0\n1 1\n1 -1",
"output": "0"
}
] | 1,625,896,422 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 186 | 0 | l=[]
for i in range(int(input())):
l.append(list(map(int,input().split())))
c=0
for i in range(len(l)):
p1=p2=p3=p4=0
for j in range(len(l)):
if l[j][0]>l[i][0] and l[j][1]==l[i][1]:
p1+=1
elif l[j][1]>l[i][1] and l[j][0]==l[i][0]:
p2+=1
elif l[j][0]<l[i][0] and l[j][1]==l[i][1]:
p3+=1
elif l[j][1]<l[i][1] and l[j][0]==l[i][0]:
p4+=1
if p1 and p2 and p3 and p4:
c+=1
print(c) | Title: Supercentral Point
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
Input Specification:
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
Output Specification:
Print the only number — the number of supercentral points of the given set.
Demo Input:
['8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n', '5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | ```python
l=[]
for i in range(int(input())):
l.append(list(map(int,input().split())))
c=0
for i in range(len(l)):
p1=p2=p3=p4=0
for j in range(len(l)):
if l[j][0]>l[i][0] and l[j][1]==l[i][1]:
p1+=1
elif l[j][1]>l[i][1] and l[j][0]==l[i][0]:
p2+=1
elif l[j][0]<l[i][0] and l[j][1]==l[i][1]:
p3+=1
elif l[j][1]<l[i][1] and l[j][0]==l[i][0]:
p4+=1
if p1 and p2 and p3 and p4:
c+=1
print(c)
``` | 3 | |
908 | A | New Year and Counting Cards | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true. | The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit. | Print a single integer, the minimum number of cards you must turn over to verify your claim. | [
"ee\n",
"z\n",
"0ay1\n"
] | [
"2\n",
"0\n",
"2\n"
] | In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | 500 | [
{
"input": "ee",
"output": "2"
},
{
"input": "z",
"output": "0"
},
{
"input": "0ay1",
"output": "2"
},
{
"input": "0abcdefghijklmnopqrstuvwxyz1234567896",
"output": "10"
},
{
"input": "0a0a9e9e2i2i9o9o6u6u9z9z4x4x9b9b",
"output": "18"
},
{
"input": "01234567890123456789012345678901234567890123456789",
"output": "25"
},
{
"input": "qwertyuioplkjhgfdsazxcvbnmqwertyuioplkjhgfdsazxcvb",
"output": "10"
},
{
"input": "cjw2dwmr10pku4yxohe0wglktd",
"output": "4"
},
{
"input": "6z2tx805jie8cfybwtfqvmlveec3iak5z5u3lu62vbxyqht6",
"output": "13"
},
{
"input": "kaq7jyialrfp4ilkni90eq8v3amcbygon7py0hb8z26fbl8ss1",
"output": "13"
},
{
"input": "hpwn50zgbmct80k9rizjqg40nycgs0acwikjqt11nr6m61krfs",
"output": "8"
},
{
"input": "l3rw91a4m25l8iytxyeuixsegzcbm4h41ornf3pixkrmwznrzc",
"output": "14"
},
{
"input": "2222",
"output": "0"
},
{
"input": "13579",
"output": "5"
},
{
"input": "1",
"output": "1"
},
{
"input": "0",
"output": "0"
},
{
"input": "a",
"output": "1"
},
{
"input": "y",
"output": "0"
},
{
"input": "w",
"output": "0"
},
{
"input": "oo",
"output": "2"
},
{
"input": "oy",
"output": "1"
},
{
"input": "yo",
"output": "1"
},
{
"input": "yy",
"output": "0"
},
{
"input": "a0",
"output": "1"
},
{
"input": "a9",
"output": "2"
},
{
"input": "y0",
"output": "0"
},
{
"input": "y7",
"output": "1"
},
{
"input": "0a",
"output": "1"
},
{
"input": "3a",
"output": "2"
},
{
"input": "06",
"output": "0"
},
{
"input": "07",
"output": "1"
},
{
"input": "70",
"output": "1"
},
{
"input": "77",
"output": "2"
},
{
"input": "13570",
"output": "4"
},
{
"input": "0000000000011111",
"output": "5"
},
{
"input": "1357",
"output": "4"
},
{
"input": "uuuuuuuuuuuuuuuuuu",
"output": "18"
},
{
"input": "gabieurat",
"output": "5"
}
] | 1,514,940,011 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 5,529,600 | import sys
s = sys.stdin.readline()
vowel = consonant = odd_digit = even_digit = 0
i = 0
while i < len(s):
if s[i] >= 'a' and s[i] <= 'z':
if s[i] == 'a' or s[i] == 'e' or s[i] == 'i' or s[i] == 'o' or s[i] == 'u':
vowel += 1
else:
consonant += 1
elif s[i] >= '0' and s[i] <= '9':
if s[i] == '0' or s[i] == '2' or s[i] == '4' or s[i] == '6' or s[i] == '8':
even_digit += 1
else:
odd_digit += 1
i += 1
if len(s) == consonant + odd_digit:
print(0)
else:
print(vowel + even_digit)
| Title: New Year and Counting Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input Specification:
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
Output Specification:
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Demo Input:
['ee\n', 'z\n', '0ay1\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | ```python
import sys
s = sys.stdin.readline()
vowel = consonant = odd_digit = even_digit = 0
i = 0
while i < len(s):
if s[i] >= 'a' and s[i] <= 'z':
if s[i] == 'a' or s[i] == 'e' or s[i] == 'i' or s[i] == 'o' or s[i] == 'u':
vowel += 1
else:
consonant += 1
elif s[i] >= '0' and s[i] <= '9':
if s[i] == '0' or s[i] == '2' or s[i] == '4' or s[i] == '6' or s[i] == '8':
even_digit += 1
else:
odd_digit += 1
i += 1
if len(s) == consonant + odd_digit:
print(0)
else:
print(vowel + even_digit)
``` | 0 | |
939 | A | Love Triangle | PROGRAMMING | 800 | [
"graphs"
] | null | null | As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th. | Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case. | [
"5\n2 4 5 1 3\n",
"5\n5 5 5 5 1\n"
] | [
"YES\n",
"NO\n"
] | In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | 500 | [
{
"input": "5\n2 4 5 1 3",
"output": "YES"
},
{
"input": "5\n5 5 5 5 1",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "10\n4 10 9 5 3 1 5 10 6 4",
"output": "NO"
},
{
"input": "10\n5 5 4 9 10 9 9 5 3 1",
"output": "YES"
},
{
"input": "100\n50 40 60 87 39 58 44 84 46 68 16 57 77 87 92 95 42 31 74 15 36 84 30 3 47 15 87 90 76 66 6 63 74 19 40 49 6 84 41 9 77 34 7 12 11 73 58 24 81 14 81 29 65 100 1 85 64 32 38 4 54 67 32 81 80 7 100 71 29 80 4 52 47 7 78 56 52 75 81 37 16 41 27 28 58 60 62 47 29 40 37 14 59 91 12 54 25 58 12 43",
"output": "NO"
},
{
"input": "100\n25 6 46 37 87 99 70 31 46 12 94 40 87 56 28 8 94 39 13 12 67 13 71 39 83 48 40 14 62 41 16 71 20 41 83 41 68 98 23 82 62 83 62 35 49 22 31 21 66 98 54 39 34 52 11 28 47 89 25 44 68 36 91 46 82 86 88 48 27 93 7 9 53 36 16 100 84 84 44 25 58 66 16 46 72 21 91 78 4 17 44 17 47 67 93 89 75 44 56 50",
"output": "YES"
},
{
"input": "2\n2 1",
"output": "NO"
},
{
"input": "3\n2 3 1",
"output": "YES"
},
{
"input": "5\n2 1 4 5 3",
"output": "YES"
},
{
"input": "5\n5 4 5 5 2",
"output": "YES"
}
] | 1,622,024,202 | 2,147,483,647 | Python 3 | OK | TESTS | 16 | 62 | 102,400 | import sys
n=int(input())
l=list(map(int,input().split()))
for i in range(n):
a=l[i]
b=l[a-1]
if a==b:
continue
else:
c=l[b-1]
if i==c-1:
print("YES")
sys.exit()
else:
continue
else:
print("NO") | Title: Love Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th.
Output Specification:
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Demo Input:
['5\n2 4 5 1 3\n', '5\n5 5 5 5 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | ```python
import sys
n=int(input())
l=list(map(int,input().split()))
for i in range(n):
a=l[i]
b=l[a-1]
if a==b:
continue
else:
c=l[b-1]
if i==c-1:
print("YES")
sys.exit()
else:
continue
else:
print("NO")
``` | 3 | |
689 | E | Mike and Geometry Problem | PROGRAMMING | 2,000 | [
"combinatorics",
"data structures",
"dp",
"geometry",
"implementation"
] | null | null | Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him an interesting geometry problem. Let's define *f*([*l*,<=*r*])<==<=*r*<=-<=*l*<=+<=1 to be the number of integer points in the segment [*l*,<=*r*] with *l*<=≤<=*r* (say that ). You are given two integers *n* and *k* and *n* closed intervals [*l**i*,<=*r**i*] on *OX* axis and you have to find:
In other words, you should find the sum of the number of integer points in the intersection of any *k* of the segments.
As the answer may be very large, output it modulo 1000000007 (109<=+<=7).
Mike can't solve this problem so he needs your help. You will help him, won't you? | The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200<=000) — the number of segments and the number of segments in intersection groups respectively.
Then *n* lines follow, the *i*-th line contains two integers *l**i*,<=*r**i* (<=-<=109<=≤<=*l**i*<=≤<=*r**i*<=≤<=109), describing *i*-th segment bounds. | Print one integer number — the answer to Mike's problem modulo 1000000007 (109<=+<=7) in the only line. | [
"3 2\n1 2\n1 3\n2 3\n",
"3 3\n1 3\n1 3\n1 3\n",
"3 1\n1 2\n2 3\n3 4\n"
] | [
"5\n",
"3\n",
"6\n"
] | In the first example:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/099f833590960ffc5dafcbc207172a93605c44a8.png" style="max-width: 100.0%;max-height: 100.0%;"/>;
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/6f56e6e99db45efff1c9be404aeb569c7e2bbb1d.png" style="max-width: 100.0%;max-height: 100.0%;"/>;
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eaff785f85562a9ff263ebbbfafe78a6b2581a57.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
So the answer is 2 + 1 + 2 = 5. | 2,500 | [
{
"input": "3 2\n1 2\n1 3\n2 3",
"output": "5"
},
{
"input": "3 3\n1 3\n1 3\n1 3",
"output": "3"
},
{
"input": "3 1\n1 2\n2 3\n3 4",
"output": "6"
},
{
"input": "1 1\n45 70",
"output": "26"
},
{
"input": "1 1\n-35 -8",
"output": "28"
},
{
"input": "1 1\n-79 -51",
"output": "29"
},
{
"input": "2 2\n26 99\n-56 40",
"output": "15"
},
{
"input": "9 6\n-44 -29\n-11 85\n11 84\n-63 1\n75 89\n-37 61\n14 73\n78 88\n-22 -18",
"output": "0"
},
{
"input": "2 2\n-93 -22\n12 72",
"output": "0"
}
] | 1,697,698,434 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 69 | 1,513 | 48,128,000 | from collections import defaultdict
mod=10**9+7
MAXN=300001
def power(a,b):
res=1
while b>0:
if b&1:
res=(res*a)%mod
b>>=1
a=(a*a)%mod
return res
fact=[1 for _ in range(MAXN)]
inv_fact=[1 for _ in range(MAXN)]
for i in range(2,MAXN):
fact[i]=(fact[i-1]*i)%mod
inv_fact[i]=power(fact[i],mod-2)
def ncr(n,r):
if n<r:
return 0
return (((fact[n]*inv_fact[r])%mod)*inv_fact[n-r])%mod
n,k=map(int,input().split())
mp=defaultdict(int)
end_points=[]
for _ in range(n):
l,r=map(int,input().split())
mp[l]+=1
mp[r+1]-=1
end_points=list(mp.keys())
end_points.sort()
ans=0
curr=0
for i in range(len(end_points)-1):
curr+=mp[end_points[i]]
ans+=(end_points[i+1]-end_points[i])*ncr(curr,k)
if ans>=mod:
ans%=mod
print(ans)
| Title: Mike and Geometry Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him an interesting geometry problem. Let's define *f*([*l*,<=*r*])<==<=*r*<=-<=*l*<=+<=1 to be the number of integer points in the segment [*l*,<=*r*] with *l*<=≤<=*r* (say that ). You are given two integers *n* and *k* and *n* closed intervals [*l**i*,<=*r**i*] on *OX* axis and you have to find:
In other words, you should find the sum of the number of integer points in the intersection of any *k* of the segments.
As the answer may be very large, output it modulo 1000000007 (109<=+<=7).
Mike can't solve this problem so he needs your help. You will help him, won't you?
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200<=000) — the number of segments and the number of segments in intersection groups respectively.
Then *n* lines follow, the *i*-th line contains two integers *l**i*,<=*r**i* (<=-<=109<=≤<=*l**i*<=≤<=*r**i*<=≤<=109), describing *i*-th segment bounds.
Output Specification:
Print one integer number — the answer to Mike's problem modulo 1000000007 (109<=+<=7) in the only line.
Demo Input:
['3 2\n1 2\n1 3\n2 3\n', '3 3\n1 3\n1 3\n1 3\n', '3 1\n1 2\n2 3\n3 4\n']
Demo Output:
['5\n', '3\n', '6\n']
Note:
In the first example:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/099f833590960ffc5dafcbc207172a93605c44a8.png" style="max-width: 100.0%;max-height: 100.0%;"/>;
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/6f56e6e99db45efff1c9be404aeb569c7e2bbb1d.png" style="max-width: 100.0%;max-height: 100.0%;"/>;
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eaff785f85562a9ff263ebbbfafe78a6b2581a57.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
So the answer is 2 + 1 + 2 = 5. | ```python
from collections import defaultdict
mod=10**9+7
MAXN=300001
def power(a,b):
res=1
while b>0:
if b&1:
res=(res*a)%mod
b>>=1
a=(a*a)%mod
return res
fact=[1 for _ in range(MAXN)]
inv_fact=[1 for _ in range(MAXN)]
for i in range(2,MAXN):
fact[i]=(fact[i-1]*i)%mod
inv_fact[i]=power(fact[i],mod-2)
def ncr(n,r):
if n<r:
return 0
return (((fact[n]*inv_fact[r])%mod)*inv_fact[n-r])%mod
n,k=map(int,input().split())
mp=defaultdict(int)
end_points=[]
for _ in range(n):
l,r=map(int,input().split())
mp[l]+=1
mp[r+1]-=1
end_points=list(mp.keys())
end_points.sort()
ans=0
curr=0
for i in range(len(end_points)-1):
curr+=mp[end_points[i]]
ans+=(end_points[i+1]-end_points[i])*ncr(curr,k)
if ans>=mod:
ans%=mod
print(ans)
``` | 3 | |
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"
}
] | 1,690,546,093 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 60 | 0 | n = input()
x = [0]*n
y = [0]*n
z = [0]*n
for i in range(n):
x1 = input()
y1 = input()
z1 = input()
x[i] = x1
y[i] = y1
z[i] = z1
for i in range(n):
if x[i]+y[i]+z[i] != 0:
print("NO")
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
n = input()
x = [0]*n
y = [0]*n
z = [0]*n
for i in range(n):
x1 = input()
y1 = input()
z1 = input()
x[i] = x1
y[i] = y1
z[i] = z1
for i in range(n):
if x[i]+y[i]+z[i] != 0:
print("NO")
print("YES")
``` | -1 |
248 | A | Cupboards | PROGRAMMING | 800 | [
"implementation"
] | null | null | One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*. | The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces. | In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs. | [
"5\n0 1\n1 0\n0 1\n1 1\n0 1\n"
] | [
"3\n"
] | none | 500 | [
{
"input": "5\n0 1\n1 0\n0 1\n1 1\n0 1",
"output": "3"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1\n1 1\n1 1",
"output": "1"
},
{
"input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0",
"output": "7"
},
{
"input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 1\n1 1\n0 1",
"output": "6"
},
{
"input": "15\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "5\n1 0\n1 0\n1 0\n0 1\n0 1",
"output": "4"
}
] | 1,663,664,222 | 2,147,483,647 | Python 3 | OK | TESTS | 57 | 154 | 0 | a=int(input())
sayacm=0
sayacn=0
hamle=0
for i in range(a):
m,n=input().split()
if m == "0":
sayacm=sayacm+1
if n == "0":
sayacn=sayacn+1
if sayacm>=(a/2):
hamle = hamle+(a-sayacm)
if sayacm<(a/2):
hamle = hamle +(sayacm)
if sayacn>=(a/2):
hamle = hamle+(a-sayacn)
if sayacn<(a/2):
hamle = hamle +(sayacn)
print(hamle) | Title: Cupboards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
Input Specification:
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
Output Specification:
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
Demo Input:
['5\n0 1\n1 0\n0 1\n1 1\n0 1\n']
Demo Output:
['3\n']
Note:
none | ```python
a=int(input())
sayacm=0
sayacn=0
hamle=0
for i in range(a):
m,n=input().split()
if m == "0":
sayacm=sayacm+1
if n == "0":
sayacn=sayacn+1
if sayacm>=(a/2):
hamle = hamle+(a-sayacm)
if sayacm<(a/2):
hamle = hamle +(sayacm)
if sayacn>=(a/2):
hamle = hamle+(a-sayacn)
if sayacn<(a/2):
hamle = hamle +(sayacn)
print(hamle)
``` | 3 | |
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"
}
] | 1,464,620,627 | 2,147,483,647 | Python 3 | OK | TESTS | 16 | 234 | 6,656,000 | from functools import cmp_to_key as ctk
n = int(input())
c = []
for i in range(n):
c.append(input())
def sort(a,b):
if a + b < b + a:
return -1
else:
return 1
c.sort(key = ctk(sort))
print(''.join(c))
| 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
from functools import cmp_to_key as ctk
n = int(input())
c = []
for i in range(n):
c.append(input())
def sort(a,b):
if a + b < b + a:
return -1
else:
return 1
c.sort(key = ctk(sort))
print(''.join(c))
``` | 3 | |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,683,178,766 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 62 | 0 | n, m, a = input().split(' ')
count_n = 0
count_m = 0
count_n = -(-int(n)//int(a))
count_m = -(-int(m)//int(a))
print(count_n * count_m) | Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
n, m, a = input().split(' ')
count_n = 0
count_m = 0
count_n = -(-int(n)//int(a))
count_m = -(-int(m)//int(a))
print(count_n * count_m)
``` | 3.969 |
818 | A | Diplomas and Certificates | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners. | The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas. | Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners. | [
"18 2\n",
"9 10\n",
"1000000000000 5\n",
"1000000000000 499999999999\n"
] | [
"3 6 9\n",
"0 0 9\n",
"83333333333 416666666665 500000000002\n",
"1 499999999999 500000000000\n"
] | none | 0 | [
{
"input": "18 2",
"output": "3 6 9"
},
{
"input": "9 10",
"output": "0 0 9"
},
{
"input": "1000000000000 5",
"output": "83333333333 416666666665 500000000002"
},
{
"input": "1000000000000 499999999999",
"output": "1 499999999999 500000000000"
},
{
"input": "1 1",
"output": "0 0 1"
},
{
"input": "5 3",
"output": "0 0 5"
},
{
"input": "42 6",
"output": "3 18 21"
},
{
"input": "1000000000000 1000",
"output": "499500499 499500499000 500000000501"
},
{
"input": "999999999999 999999",
"output": "499999 499998500001 500000999999"
},
{
"input": "732577309725 132613",
"output": "2762066 366285858458 366288689201"
},
{
"input": "152326362626 15",
"output": "4760198832 71402982480 76163181314"
},
{
"input": "2 1",
"output": "0 0 2"
},
{
"input": "1000000000000 500000000000",
"output": "0 0 1000000000000"
},
{
"input": "100000000000 50000000011",
"output": "0 0 100000000000"
},
{
"input": "1000000000000 32416187567",
"output": "15 486242813505 513757186480"
},
{
"input": "1000000000000 7777777777",
"output": "64 497777777728 502222222208"
},
{
"input": "1000000000000 77777777777",
"output": "6 466666666662 533333333332"
},
{
"input": "100000000000 578485652",
"output": "86 49749766072 50250233842"
},
{
"input": "999999999999 10000000000",
"output": "49 490000000000 509999999950"
},
{
"input": "7 2",
"output": "1 2 4"
},
{
"input": "420506530901 752346673804",
"output": "0 0 420506530901"
},
{
"input": "960375521135 321688347872",
"output": "1 321688347872 638687173262"
},
{
"input": "1000000000000 1000000000000",
"output": "0 0 1000000000000"
},
{
"input": "99999999999 15253636363",
"output": "3 45760909089 54239090907"
},
{
"input": "19 2",
"output": "3 6 10"
},
{
"input": "999999999999 1000000000000",
"output": "0 0 999999999999"
},
{
"input": "1000000000000 5915587276",
"output": "84 496909331184 503090668732"
},
{
"input": "1000000000000 1000000006",
"output": "499 499000002994 500999996507"
},
{
"input": "549755813888 134217728",
"output": "2047 274743689216 275012122625"
},
{
"input": "99999999999 3333333",
"output": "14999 49996661667 50003323333"
},
{
"input": "9 1",
"output": "2 2 5"
},
{
"input": "1000000000000 250000000001",
"output": "1 250000000001 749999999998"
},
{
"input": "5 1",
"output": "1 1 3"
},
{
"input": "3107038133 596040207",
"output": "2 1192080414 1914957717"
},
{
"input": "1000000000000 73786977",
"output": "6776 499980556152 500019437072"
},
{
"input": "1000000000000 73786976",
"output": "6776 499980549376 500019443848"
},
{
"input": "1000000000000 25000000000",
"output": "19 475000000000 524999999981"
},
{
"input": "216929598879 768233755932",
"output": "0 0 216929598879"
},
{
"input": "1000000000000 250000000000",
"output": "1 250000000000 749999999999"
},
{
"input": "1000000000000 100000000001",
"output": "4 400000000004 599999999992"
},
{
"input": "100000000000 100000000001",
"output": "0 0 100000000000"
},
{
"input": "900000000000 100281800001",
"output": "4 401127200004 498872799992"
},
{
"input": "906028900004 109123020071",
"output": "4 436492080284 469536819716"
},
{
"input": "1000000000000 1",
"output": "250000000000 250000000000 500000000000"
}
] | 1,625,211,439 | 2,147,483,647 | PyPy 3 | OK | TESTS | 44 | 124 | 0 | n, k = map(int, input().split())
h = n//2
a = h //(k+1)
print(a,k * a,n-a*(k + 1),end=" ")
| Title: Diplomas and Certificates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners.
Input Specification:
The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas.
Output Specification:
Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners.
Demo Input:
['18 2\n', '9 10\n', '1000000000000 5\n', '1000000000000 499999999999\n']
Demo Output:
['3 6 9\n', '0 0 9\n', '83333333333 416666666665 500000000002\n', '1 499999999999 500000000000\n']
Note:
none | ```python
n, k = map(int, input().split())
h = n//2
a = h //(k+1)
print(a,k * a,n-a*(k + 1),end=" ")
``` | 3 | |
75 | C | Modified GCD | PROGRAMMING | 1,600 | [
"binary search",
"number theory"
] | C. Modified GCD | 2 | 256 | Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor *d* between two integers *a* and *b* that is in a given range from *low* to *high* (inclusive), i.e. *low*<=≤<=*d*<=≤<=*high*. It is possible that there is no common divisor in the given range.
You will be given the two integers *a* and *b*, then *n* queries. Each query is a range from *low* to *high* and you have to answer each query. | The first line contains two integers *a* and *b*, the two integers as described above (1<=≤<=*a*,<=*b*<=≤<=109). The second line contains one integer *n*, the number of queries (1<=≤<=*n*<=≤<=104). Then *n* lines follow, each line contains one query consisting of two integers, *low* and *high* (1<=≤<=*low*<=≤<=*high*<=≤<=109). | Print *n* lines. The *i*-th of them should contain the result of the *i*-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query. | [
"9 27\n3\n1 5\n10 11\n9 11\n"
] | [
"3\n-1\n9\n"
] | none | 1,500 | [
{
"input": "9 27\n3\n1 5\n10 11\n9 11",
"output": "3\n-1\n9"
},
{
"input": "48 72\n2\n8 29\n29 37",
"output": "24\n-1"
},
{
"input": "90 100\n10\n51 61\n6 72\n1 84\n33 63\n37 69\n18 21\n9 54\n49 90\n14 87\n37 90",
"output": "-1\n10\n10\n-1\n-1\n-1\n10\n-1\n-1\n-1"
},
{
"input": "84 36\n1\n18 32",
"output": "-1"
},
{
"input": "90 36\n16\n13 15\n5 28\n11 30\n26 35\n2 8\n19 36\n3 17\n5 14\n4 26\n22 33\n16 33\n18 27\n4 17\n1 2\n29 31\n18 36",
"output": "-1\n18\n18\n-1\n6\n-1\n9\n9\n18\n-1\n18\n18\n9\n2\n-1\n18"
},
{
"input": "84 90\n18\n10 75\n2 40\n30 56\n49 62\n19 33\n5 79\n61 83\n13 56\n73 78\n1 18\n23 35\n14 72\n22 33\n1 21\n8 38\n54 82\n6 80\n57 75",
"output": "-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n6\n-1"
},
{
"input": "84 100\n16\n10 64\n3 61\n19 51\n42 67\n51 68\n12 40\n10 47\n52 53\n37 67\n2 26\n23 47\n17 75\n49 52\n3 83\n63 81\n8 43",
"output": "-1\n4\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n4\n-1\n-1\n-1\n4\n-1\n-1"
},
{
"input": "36 60\n2\n17 25\n16 20",
"output": "-1\n-1"
},
{
"input": "90 100\n8\n55 75\n46 68\n44 60\n32 71\n43 75\n23 79\n47 86\n11 57",
"output": "-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1"
},
{
"input": "90 36\n8\n1 19\n10 12\n14 28\n21 24\n8 8\n33 34\n10 26\n15 21",
"output": "18\n-1\n18\n-1\n-1\n-1\n18\n18"
},
{
"input": "48 80\n19\n1 1\n16 16\n1 16\n16 48\n16 80\n16 1000000000\n1000000000 1000000000\n1 1000000000\n500000000 1000000000\n15 17\n17 17\n15 15\n8 8\n8 15\n8 16\n8 17\n7 17\n7 15\n9 15",
"output": "1\n16\n16\n16\n16\n16\n-1\n16\n-1\n16\n-1\n-1\n8\n8\n16\n16\n16\n8\n-1"
},
{
"input": "31607 999002449\n18\n31607 31607\n31606 31608\n31607 31608\n31606 31607\n31606 31606\n31608 31608\n1 31607\n1 31606\n1 31608\n1 1000000000\n31607 1000000000\n31606 1000000000\n31608 1000000000\n1000000000 1000000000\n1 1\n2 31606\n2 31607\n2 31608",
"output": "31607\n31607\n31607\n31607\n-1\n-1\n31607\n1\n31607\n31607\n31607\n31607\n-1\n-1\n1\n-1\n31607\n31607"
},
{
"input": "999999937 999999929\n12\n999999929 999999937\n1 1\n1 1000000000\n2 1000000000\n1 2\n999999937 999999937\n999999929 999999929\n2 2\n3 3\n1 100\n1 999999937\n1 999999929",
"output": "-1\n1\n1\n-1\n1\n-1\n-1\n-1\n-1\n1\n1\n1"
}
] | 1,593,866,323 | 2,147,483,647 | PyPy 3 | OK | TESTS | 60 | 686 | 5,222,400 | # import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
def gcd(num1, num2):
if num2 == 0:
return num1
return gcd(num2, num1 % num2)
# def binary_s(left, right):
# while left <= right:
# mid = left + (right - left)//2
# if res1[mid] > right:
# right = mid - 1
# else
a, b = map(int, input().split())
n = int(input())
k = gcd(a, b)
res1 = []
i = 1
while i * i <= k:
if k % i == 0:
res1.append(i)
if i != k // i:
res1.append(k // i)
i += 1
for _ in range(n):
l, r = map(int, input().split())
ans = -1
for d in res1:
if l <= d <= r:
ans = max(ans, d)
print(ans) | Title: Modified GCD
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor *d* between two integers *a* and *b* that is in a given range from *low* to *high* (inclusive), i.e. *low*<=≤<=*d*<=≤<=*high*. It is possible that there is no common divisor in the given range.
You will be given the two integers *a* and *b*, then *n* queries. Each query is a range from *low* to *high* and you have to answer each query.
Input Specification:
The first line contains two integers *a* and *b*, the two integers as described above (1<=≤<=*a*,<=*b*<=≤<=109). The second line contains one integer *n*, the number of queries (1<=≤<=*n*<=≤<=104). Then *n* lines follow, each line contains one query consisting of two integers, *low* and *high* (1<=≤<=*low*<=≤<=*high*<=≤<=109).
Output Specification:
Print *n* lines. The *i*-th of them should contain the result of the *i*-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.
Demo Input:
['9 27\n3\n1 5\n10 11\n9 11\n']
Demo Output:
['3\n-1\n9\n']
Note:
none | ```python
# import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
def gcd(num1, num2):
if num2 == 0:
return num1
return gcd(num2, num1 % num2)
# def binary_s(left, right):
# while left <= right:
# mid = left + (right - left)//2
# if res1[mid] > right:
# right = mid - 1
# else
a, b = map(int, input().split())
n = int(input())
k = gcd(a, b)
res1 = []
i = 1
while i * i <= k:
if k % i == 0:
res1.append(i)
if i != k // i:
res1.append(k // i)
i += 1
for _ in range(n):
l, r = map(int, input().split())
ans = -1
for d in res1:
if l <= d <= r:
ans = max(ans, d)
print(ans)
``` | 3.818773 |
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"
}
] | 1,693,990,252 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | # LUOGU_RID: 123939062
a=int(input(a))
if a==0:
print('NO')
elif a%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
# LUOGU_RID: 123939062
a=int(input(a))
if a==0:
print('NO')
elif a%2=0:
print('YES')
else:
print('NO')
``` | -1 |
347 | B | Fixed Points | PROGRAMMING | 1,100 | [
"brute force",
"implementation",
"math"
] | null | null | A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, sequence [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] are not.
A fixed point of a function is a point that is mapped to itself by the function. A permutation can be regarded as a bijective function. We'll get a definition of a fixed point in a permutation. An integer *i* is a fixed point of permutation *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 if and only if *a**i*<==<=*i*. For example, permutation [0,<=2,<=1] has 1 fixed point and permutation [0,<=1,<=2] has 3 fixed points.
You are given permutation *a*. You are allowed to swap two elements of the permutation at most once. Your task is to maximize the number of fixed points in the resulting permutation. Note that you are allowed to make at most one swap operation. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 — the given permutation. | Print a single integer — the maximum possible number of fixed points in the permutation after at most one swap operation. | [
"5\n0 1 3 4 2\n"
] | [
"3\n"
] | none | 1,000 | [
{
"input": "5\n0 1 3 4 2",
"output": "3"
},
{
"input": "10\n6 9 4 7 8 2 3 5 0 1",
"output": "2"
},
{
"input": "100\n99 5 40 32 4 31 38 57 94 47 26 16 89 72 9 80 55 86 78 90 42 41 46 74 56 97 21 48 66 27 93 85 88 59 64 95 10 45 12 22 84 60 8 98 62 51 14 65 39 30 11 71 92 19 76 43 87 54 15 53 37 6 25 18 96 35 13 91 2 3 0 23 1 7 49 75 81 33 50 52 63 44 69 36 17 61 24 20 68 34 73 29 70 83 58 79 82 28 77 67",
"output": "3"
},
{
"input": "3\n0 1 2",
"output": "3"
},
{
"input": "3\n2 1 0",
"output": "3"
},
{
"input": "3\n1 2 0",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "5\n0 1 2 3 4",
"output": "5"
},
{
"input": "4\n0 1 2 3",
"output": "4"
},
{
"input": "7\n0 1 2 4 3 6 5",
"output": "5"
},
{
"input": "6\n0 1 2 3 5 4",
"output": "6"
}
] | 1,653,043,755 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 60 | 0 | import sys
input = sys.stdin.readline
n = int(input())
w = list(map(int, input().split()))
s = sorted(w)
c, d = 0, 0
for i in range(n):
if w[i] == s[i]:
c += 1
else:
if d == 0:
if i == w[i]:
d = 1
print(c+1+d) | Title: Fixed Points
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, sequence [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] are not.
A fixed point of a function is a point that is mapped to itself by the function. A permutation can be regarded as a bijective function. We'll get a definition of a fixed point in a permutation. An integer *i* is a fixed point of permutation *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 if and only if *a**i*<==<=*i*. For example, permutation [0,<=2,<=1] has 1 fixed point and permutation [0,<=1,<=2] has 3 fixed points.
You are given permutation *a*. You are allowed to swap two elements of the permutation at most once. Your task is to maximize the number of fixed points in the resulting permutation. Note that you are allowed to make at most one swap operation.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 — the given permutation.
Output Specification:
Print a single integer — the maximum possible number of fixed points in the permutation after at most one swap operation.
Demo Input:
['5\n0 1 3 4 2\n']
Demo Output:
['3\n']
Note:
none | ```python
import sys
input = sys.stdin.readline
n = int(input())
w = list(map(int, input().split()))
s = sorted(w)
c, d = 0, 0
for i in range(n):
if w[i] == s[i]:
c += 1
else:
if d == 0:
if i == w[i]:
d = 1
print(c+1+d)
``` | 0 | |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,694,769,895 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 46 | 0 | s1chi = str(input())
s2chi = str(input())
stringchi = ''
l1chi = []
l2chi = []
for i1chi in s1chi:
l1chi.append(i1chi)
for i2chi in s2chi:
l2chi.append(i2chi)
for ichi in range(len(l1chi)):
if l1chi[ichi] == l2chi[ichi]:
stringchi += '0'
else:
stringchi += '1'
print(stringchi)
| Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
s1chi = str(input())
s2chi = str(input())
stringchi = ''
l1chi = []
l2chi = []
for i1chi in s1chi:
l1chi.append(i1chi)
for i2chi in s2chi:
l2chi.append(i2chi)
for ichi in range(len(l1chi)):
if l1chi[ichi] == l2chi[ichi]:
stringchi += '0'
else:
stringchi += '1'
print(stringchi)
``` | 3.9885 |
816 | B | Karen and Coffee | PROGRAMMING | 1,400 | [
"binary search",
"data structures",
"implementation"
] | null | null | To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range? | The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive. | For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive. | [
"3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n",
"2 1 1\n1 1\n200000 200000\n90 100\n"
] | [
"3\n3\n0\n4\n",
"0\n"
] | In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none. | 1,000 | [
{
"input": "3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100",
"output": "3\n3\n0\n4"
},
{
"input": "2 1 1\n1 1\n200000 200000\n90 100",
"output": "0"
},
{
"input": "1 1 1\n1 1\n1 1",
"output": "1"
},
{
"input": "1 1 1\n200000 200000\n200000 200000",
"output": "1"
}
] | 1,564,920,970 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 2,152 | 73,318,400 |
def computeAdmissableTempNo(n, k):
freqOfTemp = [0] * 2000007
admTempsUntilTemp = [0] * 2000007
for i in range(0, n):
lr = [int(i) for i in input().split()]
freqOfTemp[lr[0]] += 1
freqOfTemp[lr[1] + 1] -= 1
for i in range(0, 2000007):
freqOfTemp[i] += freqOfTemp[i - 1]
admTempsUntilTemp[i] += admTempsUntilTemp[i - 1] + (freqOfTemp[i] >= k)
return admTempsUntilTemp
def computeNoOfAdmTemp(q, admTempsUntilTemp):
ans = []
for i in range(0, q):
ab = [int(i) for i in input().split()]
ans.append(admTempsUntilTemp[ab[1]] - admTempsUntilTemp[ab[0] - 1])
return '\n'.join([str(i) for i in ans])
def start():
nkq = [int(i) for i in input().split()]
return computeNoOfAdmTemp(nkq[2], computeAdmissableTempNo(nkq[0], nkq[1]))
print(start())
| Title: Karen and Coffee
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
Input Specification:
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Output Specification:
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Demo Input:
['3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n', '2 1 1\n1 1\n200000 200000\n90 100\n']
Demo Output:
['3\n3\n0\n4\n', '0\n']
Note:
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none. | ```python
def computeAdmissableTempNo(n, k):
freqOfTemp = [0] * 2000007
admTempsUntilTemp = [0] * 2000007
for i in range(0, n):
lr = [int(i) for i in input().split()]
freqOfTemp[lr[0]] += 1
freqOfTemp[lr[1] + 1] -= 1
for i in range(0, 2000007):
freqOfTemp[i] += freqOfTemp[i - 1]
admTempsUntilTemp[i] += admTempsUntilTemp[i - 1] + (freqOfTemp[i] >= k)
return admTempsUntilTemp
def computeNoOfAdmTemp(q, admTempsUntilTemp):
ans = []
for i in range(0, q):
ab = [int(i) for i in input().split()]
ans.append(admTempsUntilTemp[ab[1]] - admTempsUntilTemp[ab[0] - 1])
return '\n'.join([str(i) for i in ans])
def start():
nkq = [int(i) for i in input().split()]
return computeNoOfAdmTemp(nkq[2], computeAdmissableTempNo(nkq[0], nkq[1]))
print(start())
``` | 3 | |
270 | A | Fancy Fence | PROGRAMMING | 1,100 | [
"geometry",
"implementation",
"math"
] | null | null | Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*? | The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees. | For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible. | [
"3\n30\n60\n90\n"
] | [
"NO\nYES\nYES\n"
] | In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | 500 | [
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,644,107,439 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 3 | 156 | 0 | lst = []
for sides in range(3, 361):
temp = ((sides - 2) * 180) / sides
if temp == int(round(temp)):
lst.append(int(temp))
amt = int(input())
for x in range(amt):
if int(input()) in lst:
print('YES')
else:
print('NO') | Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | ```python
lst = []
for sides in range(3, 361):
temp = ((sides - 2) * 180) / sides
if temp == int(round(temp)):
lst.append(int(temp))
amt = int(input())
for x in range(amt):
if int(input()) in lst:
print('YES')
else:
print('NO')
``` | 3 | |
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"
}
] | 1,619,027,254 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 82 | 2,000 | 1,433,600 | def main():
valueA = input()
valueB = input()
if check(valueA, valueB) :
print("YES")
else:
print("NO")
def check(a, b):
n = len(a)
if(n%2 ==1):
return a==b
else:
a1 = a[0:int(n/2)]
a2 = a[int(n/2): n]
b1 = b[0:int(n/2)]
b2 = b[int(n/2):n]
return ((check(a1,b1) and check(a2,b2)) or (check(a1,b2) and check(a2,b1)))
if __name__ == '__main__':
main()
| 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 main():
valueA = input()
valueB = input()
if check(valueA, valueB) :
print("YES")
else:
print("NO")
def check(a, b):
n = len(a)
if(n%2 ==1):
return a==b
else:
a1 = a[0:int(n/2)]
a2 = a[int(n/2): n]
b1 = b[0:int(n/2)]
b2 = b[int(n/2):n]
return ((check(a1,b1) and check(a2,b2)) or (check(a1,b2) and check(a2,b1)))
if __name__ == '__main__':
main()
``` | 0 | |
767 | A | Snacktower | PROGRAMMING | 1,100 | [
"data structures",
"implementation"
] | null | null | According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*. | Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty. | [
"3\n3 1 2\n",
"5\n4 5 1 2 3\n"
] | [
"3\n \n2 1",
"5 4\n \n \n3 2 1\n"
] | In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | 500 | [
{
"input": "3\n3 1 2",
"output": "3 \n\n2 1 "
},
{
"input": "5\n4 5 1 2 3",
"output": "5 4 \n\n\n3 2 1 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "10\n5 1 6 2 8 3 4 10 9 7",
"output": "10 \n9 8 \n7 6 5 4 3 2 1 "
},
{
"input": "30\n16 10 4 29 5 28 12 21 11 30 18 6 14 3 17 22 20 15 9 1 27 19 24 26 13 25 2 23 8 7",
"output": "30 29 28 \n\n\n\n\n\n\n\n\n\n\n27 \n\n\n26 \n\n25 24 \n\n23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 \n8 \n7 6 5 4 3 2 1 "
},
{
"input": "100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 4 80 72 39",
"output": "100 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n99 98 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 \n\n\n\n\n\n\n\n\n80 79 78 77 76 75 74 73 \n72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 \n39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "2\n2 1",
"output": "2 \n1 "
}
] | 1,687,592,850 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 2,000 | 12,390,400 | n = int(input())
y = list(map(int, input().split()))
for l in range(0, y.index(n)):
print()
count = y.index(n)
print(n, end=" ")
for i in reversed(range(1, n)):
if y.index(i) < count:
print(i, end=" ")
else:
count2 = 0
for r in range(0, y.index(i)):
if y[r] > i:
count2 += 1
if count2 == y.index(i):
print(i, end=" ")
else:
count2 = 0
for u in range(count, y.index(i)):
print()
print(i, end=" ")
count2 = 0
count = y.index(i)
| Title: Snacktower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
Output Specification:
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
Demo Input:
['3\n3 1 2\n', '5\n4 5 1 2 3\n']
Demo Output:
['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n']
Note:
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | ```python
n = int(input())
y = list(map(int, input().split()))
for l in range(0, y.index(n)):
print()
count = y.index(n)
print(n, end=" ")
for i in reversed(range(1, n)):
if y.index(i) < count:
print(i, end=" ")
else:
count2 = 0
for r in range(0, y.index(i)):
if y[r] > i:
count2 += 1
if count2 == y.index(i):
print(i, end=" ")
else:
count2 = 0
for u in range(count, y.index(i)):
print()
print(i, end=" ")
count2 = 0
count = y.index(i)
``` | 0 | |
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..."
}
] | 1,581,505,544 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 8 | 109 | 307,200 | def fibo(n):
f = [0, 1]
for i in range(2, n + 1):
f.append(f[i-1] + f[i-2])
return f
n=int(input())
t=fibo(n)
r=''
if n!=2:
for i in range(1,n+1):
if i not in t:
r+='o'
else:
r+='O'
print(r)
else:
print('OO')
| 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
def fibo(n):
f = [0, 1]
for i in range(2, n + 1):
f.append(f[i-1] + f[i-2])
return f
n=int(input())
t=fibo(n)
r=''
if n!=2:
for i in range(1,n+1):
if i not in t:
r+='o'
else:
r+='O'
print(r)
else:
print('OO')
``` | 0 | |
1,005 | A | Tanya and Stairways | PROGRAMMING | 800 | [
"implementation"
] | null | null | Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$.
You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways. | The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya.
The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways. | In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways. | [
"7\n1 2 3 1 2 3 4\n",
"4\n1 1 1 1\n",
"5\n1 2 3 4 5\n",
"5\n1 2 1 2 1\n"
] | [
"2\n3 4 ",
"4\n1 1 1 1 ",
"1\n5 ",
"3\n2 2 1 "
] | none | 0 | [
{
"input": "7\n1 2 3 1 2 3 4",
"output": "2\n3 4 "
},
{
"input": "4\n1 1 1 1",
"output": "4\n1 1 1 1 "
},
{
"input": "5\n1 2 3 4 5",
"output": "1\n5 "
},
{
"input": "5\n1 2 1 2 1",
"output": "3\n2 2 1 "
},
{
"input": "1\n1",
"output": "1\n1 "
},
{
"input": "48\n1 2 3 4 1 2 3 1 1 2 3 1 2 3 4 1 1 2 3 4 1 2 3 4 1 2 3 4 1 1 2 1 2 1 2 1 1 2 1 2 1 2 3 1 2 1 2 1",
"output": "20\n4 3 1 3 4 1 4 4 4 1 2 2 2 1 2 2 3 2 2 1 "
},
{
"input": "2\n1 2",
"output": "1\n2 "
},
{
"input": "3\n1 1 2",
"output": "2\n1 2 "
},
{
"input": "4\n1 1 2 3",
"output": "2\n1 3 "
},
{
"input": "8\n1 2 3 1 2 3 4 5",
"output": "2\n3 5 "
},
{
"input": "5\n1 1 1 2 3",
"output": "3\n1 1 3 "
}
] | 1,645,882,841 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 16 | 62 | 2,355,200 | n = int(input())
a = list(map(int,input().split()))
ans =[]
cur = 0
for i in a:
if i<=cur:
ans.append(cur)
cur = i
ans.append(cur)
print(len(ans))
print(*ans)
| Title: Tanya and Stairways
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$.
You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
Input Specification:
The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya.
The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
Output Specification:
In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.
Demo Input:
['7\n1 2 3 1 2 3 4\n', '4\n1 1 1 1\n', '5\n1 2 3 4 5\n', '5\n1 2 1 2 1\n']
Demo Output:
['2\n3 4 ', '4\n1 1 1 1 ', '1\n5 ', '3\n2 2 1 ']
Note:
none | ```python
n = int(input())
a = list(map(int,input().split()))
ans =[]
cur = 0
for i in a:
if i<=cur:
ans.append(cur)
cur = i
ans.append(cur)
print(len(ans))
print(*ans)
``` | 3 | |
676 | C | Vasya and String | PROGRAMMING | 1,500 | [
"binary search",
"dp",
"strings",
"two pointers"
] | null | null | High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve? | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=0<=≤<=*k*<=≤<=*n*) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only. | Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than *k* characters. | [
"4 2\nabba\n",
"8 1\naabaabaa\n"
] | [
"4\n",
"5\n"
] | In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". | 1,500 | [
{
"input": "4 2\nabba",
"output": "4"
},
{
"input": "8 1\naabaabaa",
"output": "5"
},
{
"input": "1 0\na",
"output": "1"
},
{
"input": "1 1\nb",
"output": "1"
},
{
"input": "1 0\nb",
"output": "1"
},
{
"input": "1 1\na",
"output": "1"
},
{
"input": "10 10\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 2\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 1\nbbabbabbba",
"output": "6"
},
{
"input": "10 10\nbbabbbaabb",
"output": "10"
},
{
"input": "10 9\nbabababbba",
"output": "10"
},
{
"input": "10 4\nbababbaaab",
"output": "9"
},
{
"input": "10 10\naabaaabaaa",
"output": "10"
},
{
"input": "10 10\naaaabbbaaa",
"output": "10"
},
{
"input": "10 1\nbaaaaaaaab",
"output": "9"
},
{
"input": "10 5\naaaaabaaaa",
"output": "10"
},
{
"input": "10 4\naaaaaaaaaa",
"output": "10"
},
{
"input": "100 10\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "100"
},
{
"input": "100 7\nbbbbabbbbbaabbbabbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbabbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbab",
"output": "93"
},
{
"input": "100 30\nbbaabaaabbbbbbbbbbaababababbbbbbaabaabbbbbbbbabbbbbabbbbabbbbbbbbaabbbbbbbbbabbbbbabbbbbbbbbaaaaabba",
"output": "100"
},
{
"input": "100 6\nbaababbbaabbabbaaabbabbaabbbbbbbbaabbbabbbbaabbabbbbbabababbbbabbbbbbabbbbbbbbbaaaabbabbbbaabbabaabb",
"output": "34"
},
{
"input": "100 45\naabababbabbbaaabbbbbbaabbbabbaabbbbbabbbbbbbbabbbbbbabbaababbaabbababbbbbbababbbbbaabbbbbbbaaaababab",
"output": "100"
},
{
"input": "100 2\nababaabababaaababbaaaabbaabbbababbbaaabbbbabababbbabababaababaaabaabbbbaaabbbabbbbbabbbbbbbaabbabbba",
"output": "17"
},
{
"input": "100 25\nbabbbaaababaaabbbaabaabaabbbabbabbbbaaaaaaabaaabaaaaaaaaaabaaaabaaabbbaaabaaababaaabaabbbbaaaaaaaaaa",
"output": "80"
},
{
"input": "100 14\naabaaaaabababbabbabaaaabbaaaabaaabbbaaabaaaaaaaabaaaaabbaaaaaaaaabaaaaaaabbaababaaaababbbbbabaaaabaa",
"output": "61"
},
{
"input": "100 8\naaaaabaaaabaabaaaaaaaabaaaabaaaaaaaaaaaaaabaaaaabaaaaaaaaaaaaaaaaabaaaababaabaaaaaaaaaaaaabbabaaaaaa",
"output": "76"
},
{
"input": "100 12\naaaaaaaaaaaaaaaabaaabaaaaaaaaaabbaaaabbabaaaaaaaaaaaaaaaaaaaaabbaaabaaaaaaaaaaaabaaaaaaaabaaaaaaaaaa",
"output": "100"
},
{
"input": "100 65\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "100"
},
{
"input": "10 0\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 0\nbbbbabbbbb",
"output": "5"
},
{
"input": "10 0\nbbabbbabba",
"output": "3"
},
{
"input": "10 0\nbaabbbbaba",
"output": "4"
},
{
"input": "10 0\naababbbbaa",
"output": "4"
},
{
"input": "10 2\nabbbbbaaba",
"output": "8"
},
{
"input": "10 0\nabbaaabaaa",
"output": "3"
},
{
"input": "10 0\naabbaaabaa",
"output": "3"
},
{
"input": "10 1\naaaaaababa",
"output": "8"
},
{
"input": "10 0\nbaaaaaaaaa",
"output": "9"
},
{
"input": "10 0\naaaaaaaaaa",
"output": "10"
},
{
"input": "100 0\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "100"
},
{
"input": "100 0\nbbbbbbbbbbabbbbaaabbbbbbbbbbbabbbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbabbbbbbbbbbbbbab",
"output": "40"
},
{
"input": "100 11\nbaabbbbbababbbbabbbbbbbabbbbbbbbbbbbbbabbbbbbababbbbababbbbaaabbbbabbbbbabbbbbbbbabababbbabbbbbbbabb",
"output": "65"
},
{
"input": "100 8\nbbababbbbbaabbbaaababbbbababababbbbababbabbbabbbbbaabbbabbbababbabbbbabbbabbbbaabbbbabbbaabbbbaaaabb",
"output": "33"
},
{
"input": "100 21\nabbaaaabbbababaabbbababbbbbbbbabbaababababbbabbbaaabbaaabbbbabbabbbabbbabaababbbabbbbbabbbbbbabbbbab",
"output": "65"
},
{
"input": "100 9\nabbbaabaabaaaaaaabbabbbababbaaabbbaaabbaabaaaaabbbbbabbaabaabbbbbaaaaababbaaabbabaabaaabababbaababbb",
"output": "26"
},
{
"input": "100 5\naababababbaaaaaaaabbbabaaaabbabaaaabbaabaaaaabababbabaabaaabaaaaaaaabaababbabbaaabaabbabbaaaaabbabba",
"output": "22"
},
{
"input": "100 9\naababaabaaaaaaaaabbbaabaaaaaaabaaaaaaaaaaaaabaaabaabaabbbbabbaababbabbaaaabbababaabaababaabaaaaaaaaa",
"output": "49"
},
{
"input": "100 6\naaaaabbaaaaaaaaaaabaaaabaaaaaaaaabaaabaaaaaabaaaaaaaaaaabaabaaaabaaaaaaaaaaaaaaabaabbaaaaaaaaaaaaaaa",
"output": "56"
},
{
"input": "100 7\nabaaabaabaabaaaaaabaaaaaaaaaaaaaaaaaaaaaaaaaaaaabaaabaaaaaaabbabaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaba",
"output": "86"
},
{
"input": "100 0\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "100"
}
] | 1,697,903,822 | 2,147,483,647 | Python 3 | OK | TESTS | 117 | 93 | 614,400 | from collections import defaultdict
MOD = 10000007
def max_substring_length(n, k, s):
mp = defaultdict(int)
ans = 0
x = 0
for i in range(n):
mp[s[i]] += 1
if min(mp['a'], mp['b']) > k:
mp[s[x]] -= 1
x += 1
ans = max(ans, mp['a'] + mp['b'])
return ans
if __name__ == "__main__":
n, k = map(int, input().split())
s = input()
result = max_substring_length(n, k, s)
print(result)
| Title: Vasya and String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve?
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=0<=≤<=*k*<=≤<=*n*) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only.
Output Specification:
Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than *k* characters.
Demo Input:
['4 2\nabba\n', '8 1\naabaabaa\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". | ```python
from collections import defaultdict
MOD = 10000007
def max_substring_length(n, k, s):
mp = defaultdict(int)
ans = 0
x = 0
for i in range(n):
mp[s[i]] += 1
if min(mp['a'], mp['b']) > k:
mp[s[x]] -= 1
x += 1
ans = max(ans, mp['a'] + mp['b'])
return ans
if __name__ == "__main__":
n, k = map(int, input().split())
s = input()
result = max_substring_length(n, k, s)
print(result)
``` | 3 | |
620 | A | Professor GukiZ's Robot | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Professor GukiZ makes a new robot. The robot are in the point with coordinates (*x*1,<=*y*1) and should go to the point (*x*2,<=*y*2). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrease or increase). So the robot can move in one of the 8 directions. Find the minimal number of steps the robot should make to get the finish position. | The first line contains two integers *x*1,<=*y*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109) — the start position of the robot.
The second line contains two integers *x*2,<=*y*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109) — the finish position of the robot. | Print the only integer *d* — the minimal number of steps to get the finish position. | [
"0 0\n4 5\n",
"3 4\n6 1\n"
] | [
"5\n",
"3\n"
] | In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its *y* coordinate and get the finish position.
In the second example robot should simultaneously increase *x* coordinate and decrease *y* coordinate by one three times. | 0 | [
{
"input": "0 0\n4 5",
"output": "5"
},
{
"input": "3 4\n6 1",
"output": "3"
},
{
"input": "0 0\n4 6",
"output": "6"
},
{
"input": "1 1\n-3 -5",
"output": "6"
},
{
"input": "-1 -1\n-10 100",
"output": "101"
},
{
"input": "1 -1\n100 -100",
"output": "99"
},
{
"input": "-1000000000 -1000000000\n1000000000 1000000000",
"output": "2000000000"
},
{
"input": "-1000000000 -1000000000\n0 999999999",
"output": "1999999999"
},
{
"input": "0 0\n2 1",
"output": "2"
},
{
"input": "10 0\n100 0",
"output": "90"
},
{
"input": "1 5\n6 4",
"output": "5"
},
{
"input": "0 0\n5 4",
"output": "5"
},
{
"input": "10 1\n20 1",
"output": "10"
},
{
"input": "1 1\n-3 4",
"output": "4"
},
{
"input": "-863407280 504312726\n786535210 -661703810",
"output": "1649942490"
},
{
"input": "-588306085 -741137832\n341385643 152943311",
"output": "929691728"
},
{
"input": "0 0\n4 0",
"output": "4"
},
{
"input": "93097194 -48405232\n-716984003 -428596062",
"output": "810081197"
},
{
"input": "9 1\n1 1",
"output": "8"
},
{
"input": "4 6\n0 4",
"output": "4"
},
{
"input": "2 4\n5 2",
"output": "3"
},
{
"input": "-100000000 -100000000\n100000000 100000123",
"output": "200000123"
},
{
"input": "5 6\n5 7",
"output": "1"
},
{
"input": "12 16\n12 1",
"output": "15"
},
{
"input": "0 0\n5 1",
"output": "5"
},
{
"input": "0 1\n1 1",
"output": "1"
},
{
"input": "-44602634 913365223\n-572368780 933284951",
"output": "527766146"
},
{
"input": "-2 0\n2 -2",
"output": "4"
},
{
"input": "0 0\n3 1",
"output": "3"
},
{
"input": "-458 2\n1255 4548",
"output": "4546"
},
{
"input": "-5 -4\n-3 -3",
"output": "2"
},
{
"input": "4 5\n7 3",
"output": "3"
},
{
"input": "-1000000000 -999999999\n1000000000 999999998",
"output": "2000000000"
},
{
"input": "-1000000000 -1000000000\n1000000000 -1000000000",
"output": "2000000000"
},
{
"input": "-464122675 -898521847\n656107323 -625340409",
"output": "1120229998"
},
{
"input": "-463154699 -654742385\n-699179052 -789004997",
"output": "236024353"
},
{
"input": "982747270 -593488945\n342286841 -593604186",
"output": "640460429"
},
{
"input": "-80625246 708958515\n468950878 574646184",
"output": "549576124"
},
{
"input": "0 0\n1 0",
"output": "1"
},
{
"input": "109810 1\n2 3",
"output": "109808"
},
{
"input": "-9 0\n9 9",
"output": "18"
},
{
"input": "9 9\n9 9",
"output": "0"
},
{
"input": "1 1\n4 3",
"output": "3"
},
{
"input": "1 2\n45 1",
"output": "44"
},
{
"input": "207558188 -313753260\n-211535387 -721675423",
"output": "419093575"
},
{
"input": "-11 0\n0 0",
"output": "11"
},
{
"input": "-1000000000 1000000000\n1000000000 -1000000000",
"output": "2000000000"
},
{
"input": "0 0\n1 1",
"output": "1"
},
{
"input": "0 0\n0 1",
"output": "1"
},
{
"input": "0 0\n-1 1",
"output": "1"
},
{
"input": "0 0\n-1 0",
"output": "1"
},
{
"input": "0 0\n-1 -1",
"output": "1"
},
{
"input": "0 0\n0 -1",
"output": "1"
},
{
"input": "0 0\n1 -1",
"output": "1"
},
{
"input": "10 90\n90 10",
"output": "80"
},
{
"input": "851016864 573579544\n-761410925 -380746263",
"output": "1612427789"
},
{
"input": "1 9\n9 9",
"output": "8"
},
{
"input": "1000 1000\n1000 1000",
"output": "0"
},
{
"input": "1 9\n9 1",
"output": "8"
},
{
"input": "1 90\n90 90",
"output": "89"
},
{
"input": "100 100\n1000 1000",
"output": "900"
},
{
"input": "-1 0\n0 0",
"output": "1"
},
{
"input": "-750595959 -2984043\n649569876 -749608783",
"output": "1400165835"
},
{
"input": "958048496 712083589\n423286949 810566863",
"output": "534761547"
},
{
"input": "146316710 53945094\n-523054748 147499505",
"output": "669371458"
},
{
"input": "50383856 -596516251\n-802950224 -557916272",
"output": "853334080"
},
{
"input": "-637204864 -280290367\n-119020929 153679771",
"output": "518183935"
},
{
"input": "-100 -100\n-60 -91",
"output": "40"
},
{
"input": "337537326 74909428\n-765558776 167951547",
"output": "1103096102"
},
{
"input": "0 81\n18 90",
"output": "18"
},
{
"input": "283722202 -902633305\n-831696497 -160868946",
"output": "1115418699"
},
{
"input": "1000 1000\n-1000 1000",
"output": "2000"
},
{
"input": "5 6\n4 8",
"output": "2"
},
{
"input": "40572000 597493595\n-935051731 368493185",
"output": "975623731"
},
{
"input": "-5 5\n5 5",
"output": "10"
}
] | 1,626,792,812 | 2,147,483,647 | Python 3 | OK | TESTS | 75 | 77 | 6,758,400 | x1,y1 = map(int,input().split())
x2,y2 = map(int,input().split())
if abs(x1-x2)>abs(y1-y2):
print(abs(x1-x2))
else:
print(abs(y1-y2))
| Title: Professor GukiZ's Robot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ makes a new robot. The robot are in the point with coordinates (*x*1,<=*y*1) and should go to the point (*x*2,<=*y*2). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrease or increase). So the robot can move in one of the 8 directions. Find the minimal number of steps the robot should make to get the finish position.
Input Specification:
The first line contains two integers *x*1,<=*y*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109) — the start position of the robot.
The second line contains two integers *x*2,<=*y*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109) — the finish position of the robot.
Output Specification:
Print the only integer *d* — the minimal number of steps to get the finish position.
Demo Input:
['0 0\n4 5\n', '3 4\n6 1\n']
Demo Output:
['5\n', '3\n']
Note:
In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its *y* coordinate and get the finish position.
In the second example robot should simultaneously increase *x* coordinate and decrease *y* coordinate by one three times. | ```python
x1,y1 = map(int,input().split())
x2,y2 = map(int,input().split())
if abs(x1-x2)>abs(y1-y2):
print(abs(x1-x2))
else:
print(abs(y1-y2))
``` | 3 | |
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"
}
] | 1,696,769,436 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 31 | 0 | rows_columns = input()
place_list = []
colours = ["B","W"]
for i in range(int(rows_columns[0])):
places = input()
place_list.append(places)
output = ""
for i in place_list:
count = 0
for j in i:
if j == "-":
output += "-"
else:
if count > 0:
if output[(i.index(j))-1] == "B":
output += "W"
else:
output += "B"
else:
output += "B"
count += 1
print(output)
output = ""
| 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
rows_columns = input()
place_list = []
colours = ["B","W"]
for i in range(int(rows_columns[0])):
places = input()
place_list.append(places)
output = ""
for i in place_list:
count = 0
for j in i:
if j == "-":
output += "-"
else:
if count > 0:
if output[(i.index(j))-1] == "B":
output += "W"
else:
output += "B"
else:
output += "B"
count += 1
print(output)
output = ""
``` | 0 | |
215 | A | Bicycle Chain | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed. | The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces. | Print the number of "integer" gears with the maximum ratio among all "integer" gears. | [
"2\n4 5\n3\n12 13 15\n",
"4\n1 2 3 4\n5\n10 11 12 13 14\n"
] | [
"2\n",
"1\n"
] | In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15. | 500 | [
{
"input": "2\n4 5\n3\n12 13 15",
"output": "2"
},
{
"input": "4\n1 2 3 4\n5\n10 11 12 13 14",
"output": "1"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1 2\n1\n1",
"output": "1"
},
{
"input": "1\n1\n2\n1 2",
"output": "1"
},
{
"input": "4\n3 7 11 13\n4\n51 119 187 221",
"output": "4"
},
{
"input": "4\n2 3 4 5\n3\n1 2 3",
"output": "2"
},
{
"input": "10\n6 12 13 20 48 53 74 92 96 97\n10\n1 21 32 36 47 54 69 75 95 97",
"output": "1"
},
{
"input": "10\n5 9 10 14 15 17 19 22 24 26\n10\n2 11 17 19 21 22 24 25 27 28",
"output": "1"
},
{
"input": "10\n24 53 56 126 354 432 442 740 795 856\n10\n273 438 494 619 689 711 894 947 954 958",
"output": "1"
},
{
"input": "10\n3 4 6 7 8 10 14 16 19 20\n10\n3 4 5 7 8 10 15 16 18 20",
"output": "1"
},
{
"input": "10\n1 6 8 14 15 17 25 27 34 39\n10\n1 8 16 17 19 22 32 39 44 50",
"output": "1"
},
{
"input": "10\n5 21 22 23 25 32 35 36 38 39\n10\n3 7 8 9 18 21 23 24 36 38",
"output": "4"
},
{
"input": "50\n5 8 13 16 19 20 21 22 24 27 28 29 30 32 33 34 35 43 45 48 50 51 54 55 58 59 60 61 62 65 70 71 72 76 78 79 80 81 83 84 85 87 89 91 92 94 97 98 99 100\n50\n2 3 5 6 7 10 15 16 17 20 23 28 29 30 31 34 36 37 40 42 45 46 48 54 55 56 58 59 61 62 69 70 71 72 75 76 78 82 84 85 86 87 88 89 90 91 92 97 99 100",
"output": "1"
},
{
"input": "50\n3 5 6 8 9 11 13 19 21 23 24 32 34 35 42 50 51 52 56 58 59 69 70 72 73 75 76 77 78 80 83 88 90 95 96 100 101 102 108 109 113 119 124 135 138 141 142 143 145 150\n50\n5 8 10 11 18 19 23 30 35 43 51 53 55 58 63 68 69 71 77 78 79 82 83 86 88 89 91 92 93 94 96 102 103 105 109 110 113 114 116 123 124 126 127 132 133 135 136 137 142 149",
"output": "1"
},
{
"input": "50\n6 16 24 25 27 33 36 40 51 60 62 65 71 72 75 77 85 87 91 93 98 102 103 106 117 118 120 121 122 123 125 131 134 136 143 148 155 157 160 161 164 166 170 178 184 187 188 192 194 197\n50\n5 9 17 23 27 34 40 44 47 59 62 70 81 82 87 88 89 90 98 101 102 110 113 114 115 116 119 122 124 128 130 137 138 140 144 150 152 155 159 164 166 169 171 175 185 186 187 189 190 193",
"output": "1"
},
{
"input": "50\n14 22 23 31 32 35 48 63 76 79 88 97 101 102 103 104 106 113 114 115 116 126 136 138 145 152 155 156 162 170 172 173 179 180 182 203 208 210 212 222 226 229 231 232 235 237 245 246 247 248\n50\n2 5 6 16 28 44 45 46 54 55 56 63 72 80 87 93 94 96 97 100 101 103 132 135 140 160 164 165 167 168 173 180 182 185 186 192 194 198 199 202 203 211 213 216 217 227 232 233 236 245",
"output": "1"
},
{
"input": "50\n14 19 33 35 38 41 51 54 69 70 71 73 76 80 84 94 102 104 105 106 107 113 121 128 131 168 180 181 187 191 195 201 205 207 210 216 220 238 249 251 263 271 272 275 281 283 285 286 291 294\n50\n2 3 5 20 21 35 38 40 43 48 49 52 55 64 73 77 82 97 109 113 119 121 125 132 137 139 145 146 149 180 182 197 203 229 234 241 244 251 264 271 274 281 284 285 287 291 292 293 294 298",
"output": "1"
},
{
"input": "50\n2 4 5 16 18 19 22 23 25 26 34 44 48 54 67 79 80 84 92 110 116 133 138 154 163 171 174 202 205 218 228 229 234 245 247 249 250 263 270 272 274 275 277 283 289 310 312 334 339 342\n50\n1 5 17 18 25 37 46 47 48 59 67 75 80 83 84 107 115 122 137 141 159 162 175 180 184 204 221 224 240 243 247 248 249 258 259 260 264 266 269 271 274 293 294 306 329 330 334 335 342 350",
"output": "1"
},
{
"input": "50\n6 9 11 21 28 39 42 56 60 63 81 88 91 95 105 110 117 125 149 165 174 176 185 189 193 196 205 231 233 268 278 279 281 286 289 292 298 303 305 306 334 342 350 353 361 371 372 375 376 378\n50\n6 17 20 43 45 52 58 59 82 83 88 102 111 118 121 131 145 173 190 191 200 216 224 225 232 235 243 256 260 271 290 291 321 322 323 329 331 333 334 341 343 348 351 354 356 360 366 379 387 388",
"output": "1"
},
{
"input": "10\n17 239 443 467 661 1069 1823 2333 3767 4201\n20\n51 83 97 457 593 717 997 1329 1401 1459 1471 1983 2371 2539 3207 3251 3329 5469 6637 6999",
"output": "8"
},
{
"input": "20\n179 359 401 467 521 601 919 941 1103 1279 1709 1913 1949 2003 2099 2143 2179 2213 2399 4673\n20\n151 181 191 251 421 967 1109 1181 1249 1447 1471 1553 1619 2327 2551 2791 3049 3727 6071 7813",
"output": "3"
},
{
"input": "20\n79 113 151 709 809 983 1291 1399 1409 1429 2377 2659 2671 2897 3217 3511 3557 3797 3823 4363\n10\n19 101 659 797 1027 1963 2129 2971 3299 9217",
"output": "3"
},
{
"input": "30\n19 47 109 179 307 331 389 401 461 509 547 569 617 853 883 1249 1361 1381 1511 1723 1741 1783 2459 2531 2621 3533 3821 4091 5557 6217\n20\n401 443 563 941 967 997 1535 1567 1655 1747 1787 1945 1999 2251 2305 2543 2735 4415 6245 7555",
"output": "8"
},
{
"input": "30\n3 43 97 179 257 313 353 359 367 389 397 457 547 599 601 647 1013 1021 1063 1433 1481 1531 1669 3181 3373 3559 3769 4157 4549 5197\n50\n13 15 17 19 29 79 113 193 197 199 215 223 271 293 359 485 487 569 601 683 895 919 941 967 1283 1285 1289 1549 1565 1765 1795 1835 1907 1931 1945 1985 1993 2285 2731 2735 2995 3257 4049 4139 5105 5315 7165 7405 7655 8345",
"output": "20"
},
{
"input": "50\n11 17 23 53 59 109 137 149 173 251 353 379 419 421 439 503 593 607 661 773 821 877 941 997 1061 1117 1153 1229 1289 1297 1321 1609 1747 2311 2389 2543 2693 3041 3083 3137 3181 3209 3331 3373 3617 3767 4201 4409 4931 6379\n50\n55 59 67 73 85 89 101 115 211 263 295 353 545 599 607 685 739 745 997 1031 1255 1493 1523 1667 1709 1895 1949 2161 2195 2965 3019 3035 3305 3361 3373 3673 3739 3865 3881 4231 4253 4385 4985 5305 5585 5765 6145 6445 8045 8735",
"output": "23"
},
{
"input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782",
"output": "3"
},
{
"input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428",
"output": "4"
},
{
"input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959",
"output": "5"
},
{
"input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664",
"output": "24"
},
{
"input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782",
"output": "3"
},
{
"input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428",
"output": "4"
},
{
"input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959",
"output": "5"
},
{
"input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664",
"output": "24"
},
{
"input": "47\n66 262 357 457 513 530 538 540 592 691 707 979 1015 1242 1246 1667 1823 1886 1963 2133 2649 2679 2916 2949 3413 3523 3699 3958 4393 4922 5233 5306 5799 6036 6302 6629 7208 7282 7315 7822 7833 7927 8068 8150 8870 8962 9987\n39\n167 199 360 528 1515 1643 1986 1988 2154 2397 2856 3552 3656 3784 3980 4096 4104 4240 4320 4736 4951 5266 5656 5849 5850 6169 6517 6875 7244 7339 7689 7832 8120 8716 9503 9509 9933 9936 9968",
"output": "12"
},
{
"input": "1\n94\n50\n423 446 485 1214 1468 1507 1853 1930 1999 2258 2271 2285 2425 2543 2715 2743 2992 3196 4074 4108 4448 4475 4652 5057 5250 5312 5356 5375 5731 5986 6298 6501 6521 7146 7255 7276 7332 7481 7998 8141 8413 8665 8908 9221 9336 9491 9504 9677 9693 9706",
"output": "1"
},
{
"input": "50\n51 67 75 186 194 355 512 561 720 876 1077 1221 1503 1820 2153 2385 2568 2608 2937 2969 3271 3311 3481 4081 4093 4171 4255 4256 4829 5020 5192 5636 5817 6156 6712 6717 7153 7436 7608 7612 7866 7988 8264 8293 8867 9311 9879 9882 9889 9908\n1\n5394",
"output": "1"
},
{
"input": "50\n26 367 495 585 675 789 855 1185 1312 1606 2037 2241 2587 2612 2628 2807 2873 2924 3774 4067 4376 4668 4902 5001 5082 5100 5104 5209 5345 5515 5661 5777 5902 5907 6155 6323 6675 6791 7503 8159 8207 8254 8740 8848 8855 8933 9069 9164 9171 9586\n5\n1557 6246 7545 8074 8284",
"output": "1"
},
{
"input": "5\n25 58 91 110 2658\n50\n21 372 909 1172 1517 1554 1797 1802 1843 1977 2006 2025 2137 2225 2317 2507 2645 2754 2919 3024 3202 3212 3267 3852 4374 4487 4553 4668 4883 4911 4916 5016 5021 5068 5104 5162 5683 5856 6374 6871 7333 7531 8099 8135 8173 8215 8462 8776 9433 9790",
"output": "4"
},
{
"input": "45\n37 48 56 59 69 70 79 83 85 86 99 114 131 134 135 145 156 250 1739 1947 2116 2315 2449 3104 3666 4008 4406 4723 4829 5345 5836 6262 6296 6870 7065 7110 7130 7510 7595 8092 8442 8574 9032 9091 9355\n50\n343 846 893 1110 1651 1837 2162 2331 2596 3012 3024 3131 3294 3394 3528 3717 3997 4125 4347 4410 4581 4977 5030 5070 5119 5229 5355 5413 5418 5474 5763 5940 6151 6161 6164 6237 6506 6519 6783 7182 7413 7534 8069 8253 8442 8505 9135 9308 9828 9902",
"output": "17"
},
{
"input": "50\n17 20 22 28 36 38 46 47 48 50 52 57 58 62 63 69 70 74 75 78 79 81 82 86 87 90 93 95 103 202 292 442 1756 1769 2208 2311 2799 2957 3483 4280 4324 4932 5109 5204 6225 6354 6561 7136 8754 9670\n40\n68 214 957 1649 1940 2078 2134 2716 3492 3686 4462 4559 4656 4756 4850 5044 5490 5529 5592 5626 6014 6111 6693 6790 7178 7275 7566 7663 7702 7857 7954 8342 8511 8730 8957 9021 9215 9377 9445 9991",
"output": "28"
},
{
"input": "39\n10 13 21 25 36 38 47 48 58 64 68 69 73 79 86 972 2012 2215 2267 2503 3717 3945 4197 4800 5266 6169 6612 6824 7023 7322 7582 7766 8381 8626 8879 9079 9088 9838 9968\n50\n432 877 970 1152 1202 1223 1261 1435 1454 1578 1843 1907 2003 2037 2183 2195 2215 2425 3065 3492 3615 3637 3686 3946 4189 4415 4559 4656 4665 4707 4886 4887 5626 5703 5955 6208 6521 6581 6596 6693 6985 7013 7081 7343 7663 8332 8342 8637 9207 9862",
"output": "15"
},
{
"input": "50\n7 144 269 339 395 505 625 688 709 950 1102 1152 1350 1381 1641 1830 1977 1999 2093 2180 2718 3308 3574 4168 4232 4259 4393 4689 4982 5154 5476 5581 5635 5721 6159 6302 6741 7010 7152 7315 7417 7482 8116 8239 8640 9347 9395 9614 9661 9822\n20\n84 162 292 1728 1866 2088 3228 3470 4068 5318 5470 6060 6380 6929 7500 8256 8399 8467 8508 9691",
"output": "8"
},
{
"input": "50\n159 880 1070 1139 1358 1608 1691 1841 2073 2171 2213 2597 2692 2759 2879 2931 3173 3217 3441 4201 4878 5106 5129 5253 5395 5647 5968 6019 6130 6276 6286 6330 6409 6728 7488 7713 7765 7828 7899 8064 8264 8457 8483 8685 8900 8946 8965 9133 9187 9638\n45\n57 159 1070 1139 1391 1608 1691 1841 2171 2213 2692 2759 2931 3173 3217 3441 4201 4878 5106 5129 5253 5647 5968 6130 6276 6286 6409 7488 7694 7713 7765 7828 7899 8003 8064 8081 8244 8264 8685 8900 8946 8965 9133 9638 9673",
"output": "38"
},
{
"input": "3\n3 4 5\n3\n6 20 25",
"output": "2"
},
{
"input": "4\n2 3 5 8\n4\n2 6 8 10",
"output": "1"
},
{
"input": "4\n3 5 7 11\n4\n3 5 7 22",
"output": "1"
},
{
"input": "2\n2 3\n3\n20 30 50",
"output": "1"
},
{
"input": "3\n1 2 3\n4\n2 4 6 49",
"output": "1"
},
{
"input": "2\n4 5\n3\n12 15 20",
"output": "1"
},
{
"input": "3\n2 5 7\n3\n4 5 7",
"output": "1"
},
{
"input": "3\n3 5 8\n3\n6 8 10",
"output": "2"
},
{
"input": "2\n2 3\n4\n4 6 9 33",
"output": "1"
},
{
"input": "2\n2 3\n4\n4 6 21 40",
"output": "1"
},
{
"input": "3\n4 9 10\n3\n8 9 10",
"output": "1"
},
{
"input": "5\n1 5 6 9 51\n5\n5 12 18 27 10000",
"output": "1"
},
{
"input": "13\n1 2 3 4 5 6 7 8 9 10 11 12 13\n1\n14",
"output": "1"
}
] | 1,631,603,444 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 122 | 6,758,400 | # A. Bicycle Chain
n = int(input())
pedal = list(map(int,input().split(' ')))
m = int(input())
gear = list(map(int,input().split(' ')))
count = 0
for i in gear:
for j in pedal:
if i%j == 0:
count+=1
print(count)
| Title: Bicycle Chain
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces.
Output Specification:
Print the number of "integer" gears with the maximum ratio among all "integer" gears.
Demo Input:
['2\n4 5\n3\n12 13 15\n', '4\n1 2 3 4\n5\n10 11 12 13 14\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15. | ```python
# A. Bicycle Chain
n = int(input())
pedal = list(map(int,input().split(' ')))
m = int(input())
gear = list(map(int,input().split(' ')))
count = 0
for i in gear:
for j in pedal:
if i%j == 0:
count+=1
print(count)
``` | 0 | |
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"
}
] | 1,582,049,230 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 2 | 109 | 0 | n,k = map(int,input().split())
s = input()
a = [0]*k
for i in s:
c = ord(i) - 97
a[c] += 1
if max(a) > k:
print('NO')
else:
print('YES') | 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
n,k = map(int,input().split())
s = input()
a = [0]*k
for i in s:
c = ord(i) - 97
a[c] += 1
if max(a) > k:
print('NO')
else:
print('YES')
``` | -1 | |
939 | A | Love Triangle | PROGRAMMING | 800 | [
"graphs"
] | null | null | As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th. | Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case. | [
"5\n2 4 5 1 3\n",
"5\n5 5 5 5 1\n"
] | [
"YES\n",
"NO\n"
] | In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | 500 | [
{
"input": "5\n2 4 5 1 3",
"output": "YES"
},
{
"input": "5\n5 5 5 5 1",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "10\n4 10 9 5 3 1 5 10 6 4",
"output": "NO"
},
{
"input": "10\n5 5 4 9 10 9 9 5 3 1",
"output": "YES"
},
{
"input": "100\n50 40 60 87 39 58 44 84 46 68 16 57 77 87 92 95 42 31 74 15 36 84 30 3 47 15 87 90 76 66 6 63 74 19 40 49 6 84 41 9 77 34 7 12 11 73 58 24 81 14 81 29 65 100 1 85 64 32 38 4 54 67 32 81 80 7 100 71 29 80 4 52 47 7 78 56 52 75 81 37 16 41 27 28 58 60 62 47 29 40 37 14 59 91 12 54 25 58 12 43",
"output": "NO"
},
{
"input": "100\n25 6 46 37 87 99 70 31 46 12 94 40 87 56 28 8 94 39 13 12 67 13 71 39 83 48 40 14 62 41 16 71 20 41 83 41 68 98 23 82 62 83 62 35 49 22 31 21 66 98 54 39 34 52 11 28 47 89 25 44 68 36 91 46 82 86 88 48 27 93 7 9 53 36 16 100 84 84 44 25 58 66 16 46 72 21 91 78 4 17 44 17 47 67 93 89 75 44 56 50",
"output": "YES"
},
{
"input": "2\n2 1",
"output": "NO"
},
{
"input": "3\n2 3 1",
"output": "YES"
},
{
"input": "5\n2 1 4 5 3",
"output": "YES"
},
{
"input": "5\n5 4 5 5 2",
"output": "YES"
}
] | 1,694,596,608 | 2,147,483,647 | Python 3 | OK | TESTS | 16 | 46 | 512,000 | def has_love_triangle(n, likes):
for i in range(n):
a = i + 1 # Plane numbers are 1-indexed
b = likes[i]
c = likes[b - 1]
if likes[c - 1] == a:
return "YES" # Found a love triangle
return "NO" # No love triangle found
# Read input
n = int(input())
likes = list(map(int, input().split()))
# Check if there is a love triangle
result = has_love_triangle(n, likes)
print(result)
| Title: Love Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th.
Output Specification:
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Demo Input:
['5\n2 4 5 1 3\n', '5\n5 5 5 5 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | ```python
def has_love_triangle(n, likes):
for i in range(n):
a = i + 1 # Plane numbers are 1-indexed
b = likes[i]
c = likes[b - 1]
if likes[c - 1] == a:
return "YES" # Found a love triangle
return "NO" # No love triangle found
# Read input
n = int(input())
likes = list(map(int, input().split()))
# Check if there is a love triangle
result = has_love_triangle(n, likes)
print(result)
``` | 3 | |
770 | A | New Password | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions. | The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists. | Print any password which satisfies all conditions given by Innokentiy. | [
"4 3\n",
"6 6\n",
"5 2\n"
] | [
"java\n",
"python\n",
"phphp\n"
] | In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests. | 500 | [
{
"input": "4 3",
"output": "abca"
},
{
"input": "6 6",
"output": "abcdef"
},
{
"input": "5 2",
"output": "ababa"
},
{
"input": "3 2",
"output": "aba"
},
{
"input": "10 2",
"output": "ababababab"
},
{
"input": "26 13",
"output": "abcdefghijklmabcdefghijklm"
},
{
"input": "100 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "100 10",
"output": "abcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij"
},
{
"input": "3 3",
"output": "abc"
},
{
"input": "6 3",
"output": "abcabc"
},
{
"input": "10 3",
"output": "abcabcabca"
},
{
"input": "50 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab"
},
{
"input": "90 2",
"output": "ababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "6 2",
"output": "ababab"
},
{
"input": "99 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc"
},
{
"input": "4 2",
"output": "abab"
},
{
"input": "100 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "40 22",
"output": "abcdefghijklmnopqrstuvabcdefghijklmnopqr"
},
{
"input": "13 8",
"output": "abcdefghabcde"
},
{
"input": "16 15",
"output": "abcdefghijklmnoa"
},
{
"input": "17 17",
"output": "abcdefghijklmnopq"
},
{
"input": "19 4",
"output": "abcdabcdabcdabcdabc"
},
{
"input": "100 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv"
},
{
"input": "100 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxy"
},
{
"input": "26 26",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "27 26",
"output": "abcdefghijklmnopqrstuvwxyza"
},
{
"input": "2 2",
"output": "ab"
},
{
"input": "26 25",
"output": "abcdefghijklmnopqrstuvwxya"
},
{
"input": "99 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababa"
},
{
"input": "99 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstu"
},
{
"input": "4 4",
"output": "abcd"
},
{
"input": "5 3",
"output": "abcab"
},
{
"input": "5 4",
"output": "abcda"
},
{
"input": "5 5",
"output": "abcde"
},
{
"input": "24 22",
"output": "abcdefghijklmnopqrstuvab"
},
{
"input": "26 14",
"output": "abcdefghijklmnabcdefghijkl"
},
{
"input": "26 15",
"output": "abcdefghijklmnoabcdefghijk"
},
{
"input": "30 12",
"output": "abcdefghijklabcdefghijklabcdef"
},
{
"input": "35 4",
"output": "abcdabcdabcdabcdabcdabcdabcdabcdabc"
},
{
"input": "79 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "79 14",
"output": "abcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghi"
},
{
"input": "85 13",
"output": "abcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefg"
},
{
"input": "90 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmno"
},
{
"input": "90 19",
"output": "abcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmn"
},
{
"input": "26 24",
"output": "abcdefghijklmnopqrstuvwxab"
},
{
"input": "100 17",
"output": "abcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmno"
},
{
"input": "26 2",
"output": "ababababababababababababab"
}
] | 1,616,172,921 | 2,147,483,647 | Python 3 | OK | TESTS | 47 | 77 | 0 | n, k = map(int, input().split())
alpha = 'abcdefghijklmnopqrstuvwxyz'
idx = 0
for i in range(0, n):
if idx == k:
idx = 0
print(alpha[idx], end='')
idx += 1
| Title: New Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
Input Specification:
The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
Output Specification:
Print any password which satisfies all conditions given by Innokentiy.
Demo Input:
['4 3\n', '6 6\n', '5 2\n']
Demo Output:
['java\n', 'python\n', 'phphp\n']
Note:
In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests. | ```python
n, k = map(int, input().split())
alpha = 'abcdefghijklmnopqrstuvwxyz'
idx = 0
for i in range(0, n):
if idx == k:
idx = 0
print(alpha[idx], end='')
idx += 1
``` | 3 | |
733 | A | Grasshopper And the String | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump.
Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability.
The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'. | The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100. | Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels. | [
"ABABBBACFEYUKOTT\n",
"AAA\n"
] | [
"4",
"1"
] | none | 500 | [
{
"input": "ABABBBACFEYUKOTT",
"output": "4"
},
{
"input": "AAA",
"output": "1"
},
{
"input": "A",
"output": "1"
},
{
"input": "B",
"output": "2"
},
{
"input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIKLMJNHGTRWSDZXCVBNMHGFDSXVWRTPPPLKMNBXIUOIUOIUOIUOOIU",
"output": "39"
},
{
"input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIAEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOI",
"output": "1"
},
{
"input": "KMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVCKMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVC",
"output": "85"
},
{
"input": "QWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZ",
"output": "18"
},
{
"input": "PKLKBWTXVJ",
"output": "11"
},
{
"input": "CFHFPTGMOKXVLJJZJDQW",
"output": "12"
},
{
"input": "TXULTFSBUBFLRNQORMMULWNVLPWTYJXZBPBGAWNX",
"output": "9"
},
{
"input": "DAIUSEAUEUYUWEIOOEIOUYVYYOPEEWEBZOOOAOXUOIEUKYYOJOYAUYUUIYUXOUJLGIYEIIYUOCUAACRY",
"output": "4"
},
{
"input": "VRPHBNWNWVWBWMFJJDCTJQJDJBKSJRZLVQRVVFLTZFSGCGDXCWQVWWWMFVCQHPKXXVRKTGWGPSMQTPKNDQJHNSKLXPCXDJDQDZZD",
"output": "101"
},
{
"input": "SGDDFCDRDWGPNNFBBZZJSPXFYMZKPRXTCHVJSJJBWZXXQMDZBNKDHRGSRLGLRKPMWXNSXJPNJLDPXBSRCQMHJKPZNTPNTZXNPCJC",
"output": "76"
},
{
"input": "NVTQVNLGWFDBCBKSDLTBGWBMNQZWZQJWNGVCTCQBGWNTYJRDBPZJHXCXFMIXNRGSTXHQPCHNFQPCMDZWJGLJZWMRRFCVLBKDTDSC",
"output": "45"
},
{
"input": "SREZXQFVPQCLRCQGMKXCBRWKYZKWKRMZGXPMKWNMFZTRDPHJFCSXVPPXWKZMZTBFXGNLPLHZIPLFXNRRQFDTLFPKBGCXKTMCFKKT",
"output": "48"
},
{
"input": "ICKJKMVPDNZPLKDSLTPZNRLSQSGHQJQQPJJSNHNWVDLJRLZEJSXZDPHYXGGWXHLCTVQSKWNWGTLJMOZVJNZPVXGVPJKHFVZTGCCX",
"output": "47"
},
{
"input": "XXFPZDRPXLNHGDVCBDKJMKLGUQZXLLWYLOKFZVGXVNPJWZZZNRMQBRJCZTSDRHSNCVDMHKVXCXPCRBWSJCJWDRDPVZZLCZRTDRYA",
"output": "65"
},
{
"input": "HDDRZDKCHHHEDKHZMXQSNQGSGNNSCCPVJFGXGNCEKJMRKSGKAPQWPCWXXWHLSMRGSJWEHWQCSJJSGLQJXGVTBYALWMLKTTJMFPFS",
"output": "28"
},
{
"input": "PXVKJHXVDPWGLHWFWMJPMCCNHCKSHCPZXGIHHNMYNFQBUCKJJTXXJGKRNVRTQFDFMLLGPQKFOVNNLTNDIEXSARRJKGSCZKGGJCBW",
"output": "35"
},
{
"input": "EXNMTTFPJLDHXDQBJJRDRYBZVFFHUDCHCPNFZWXSMZXNFVJGHZWXVBRQFNUIDVLZOVPXQNVMFNBTJDSCKRLNGXPSADTGCAHCBJKL",
"output": "30"
},
{
"input": "NRNLSQQJGIJBCZFTNKJCXMGPARGWXPSHZXOBNSFOLDQVXTVAGJZNLXULHBRDGMNQKQGWMRRDPYCSNFVPUFTFBUBRXVJGNGSPJKLL",
"output": "19"
},
{
"input": "SRHOKCHQQMVZKTCVQXJJCFGYFXGMBZSZFNAFETXILZHPGHBWZRZQFMGSEYRUDVMCIQTXTBTSGFTHRRNGNTHHWWHCTDFHSVARMCMB",
"output": "30"
},
{
"input": "HBSVZHDKGNIRQUBYKYHUPJCEETGFMVBZJTHYHFQPFBVBSMQACYAVWZXSBGNKWXFNMQJFMSCHJVWBZXZGSNBRUHTHAJKVLEXFBOFB",
"output": "34"
},
{
"input": "NXKMUGOPTUQNSRYTKUKSCWCRQSZKKFPYUMDIBJAHJCEKZJVWZAWOLOEFBFXLQDDPNNZKCQHUPBFVDSXSUCVLMZXQROYQYIKPQPWR",
"output": "17"
},
{
"input": "TEHJDICFNOLQVQOAREVAGUAWODOCXJXIHYXFAEPEXRHPKEIIRCRIVASKNTVYUYDMUQKSTSSBYCDVZKDDHTSDWJWACPCLYYOXGCLT",
"output": "15"
},
{
"input": "LCJJUZZFEIUTMSEXEYNOOAIZMORQDOANAMUCYTFRARDCYHOYOPHGGYUNOGNXUAOYSEMXAZOOOFAVHQUBRNGORSPNQWZJYQQUNPEB",
"output": "9"
},
{
"input": "UUOKAOOJBXUTSMOLOOOOSUYYFTAVBNUXYFVOOGCGZYQEOYISIYOUULUAIJUYVVOENJDOCLHOSOHIHDEJOIGZNIXEMEGZACHUAQFW",
"output": "5"
},
{
"input": "OUUBEHXOOURMOAIAEHXCUOIYHUJEVAWYRCIIAGDRIPUIPAIUYAIWJEVYEYYUYBYOGVYESUJCFOJNUAHIOOKBUUHEJFEWPOEOUHYA",
"output": "4"
},
{
"input": "EMNOYEEUIOUHEWZITIAEZNCJUOUAOQEAUYEIHYUSUYUUUIAEDIOOERAEIRBOJIEVOMECOGAIAIUIYYUWYIHIOWVIJEYUEAFYULSE",
"output": "5"
},
{
"input": "BVOYEAYOIEYOREJUYEUOEOYIISYAEOUYAAOIOEOYOOOIEFUAEAAESUOOIIEUAAGAEISIAPYAHOOEYUJHUECGOYEIDAIRTBHOYOYA",
"output": "5"
},
{
"input": "GOIEOAYIEYYOOEOAIAEOOUWYEIOTNYAANAYOOXEEOEAVIOIAAIEOIAUIAIAAUEUAOIAEUOUUZYIYAIEUEGOOOOUEIYAEOSYAEYIO",
"output": "3"
},
{
"input": "AUEAOAYIAOYYIUIOAULIOEUEYAIEYYIUOEOEIEYRIYAYEYAEIIMMAAEAYAAAAEOUICAUAYOUIAOUIAIUOYEOEEYAEYEYAAEAOYIY",
"output": "3"
},
{
"input": "OAIIYEYYAOOEIUOEEIOUOIAEFIOAYETUYIOAAAEYYOYEYOEAUIIUEYAYYIIAOIEEYGYIEAAOOWYAIEYYYIAOUUOAIAYAYYOEUEOY",
"output": "2"
},
{
"input": "EEEAOEOEEIOUUUEUEAAOEOIUYJEYAIYIEIYYEAUOIIYIUOOEUCYEOOOYYYIUUAYIAOEUEIEAOUOIAACAOOUAUIYYEAAAOOUYIAAE",
"output": "2"
},
{
"input": "AYEYIIEUIYOYAYEUEIIIEUYUUAUEUIYAIAAUYONIEYIUIAEUUOUOYYOUUUIUIAEYEOUIIUOUUEOAIUUYAAEOAAEOYUUIYAYRAIII",
"output": "2"
},
{
"input": "YOOAAUUAAAYEUYIUIUYIUOUAEIEEIAUEOAUIIAAIUYEUUOYUIYEAYAAAYUEEOEEAEOEEYYOUAEUYEEAIIYEUEYJOIIYUIOIUOIEE",
"output": "2"
},
{
"input": "UYOIIIAYOOAIUUOOEEUYIOUAEOOEIOUIAIEYOAEAIOOEOOOIUYYUYIAAUIOUYYOOUAUIEYYUOAAUUEAAIEUIAUEUUIAUUOYOAYIU",
"output": "1"
},
{
"input": "ABBABBB",
"output": "4"
},
{
"input": "ABCD",
"output": "4"
},
{
"input": "XXYC",
"output": "3"
},
{
"input": "YYY",
"output": "1"
},
{
"input": "ABABBBBBBB",
"output": "8"
},
{
"input": "YYYY",
"output": "1"
},
{
"input": "YYYYY",
"output": "1"
},
{
"input": "AXXX",
"output": "4"
},
{
"input": "YYYYYYY",
"output": "1"
},
{
"input": "BYYBBB",
"output": "4"
},
{
"input": "YYYYYYYYY",
"output": "1"
},
{
"input": "CAAAAA",
"output": "2"
},
{
"input": "CCCACCCC",
"output": "5"
},
{
"input": "ABABBBACFEYUKOTTTT",
"output": "5"
},
{
"input": "AABBYYYYYYYY",
"output": "3"
},
{
"input": "BYBACYC",
"output": "2"
},
{
"input": "Y",
"output": "1"
},
{
"input": "ABBBBBB",
"output": "7"
},
{
"input": "BACDYDI",
"output": "3"
},
{
"input": "XEXXXXXXXXXXXXXXX",
"output": "16"
},
{
"input": "TTYTT",
"output": "3"
},
{
"input": "AAYBC",
"output": "3"
},
{
"input": "ABABBBACFEYUKOTTTTT",
"output": "6"
},
{
"input": "YYAYY",
"output": "1"
},
{
"input": "YZZY",
"output": "3"
},
{
"input": "YY",
"output": "1"
},
{
"input": "ZZYZZ",
"output": "3"
},
{
"input": "YBBBY",
"output": "4"
},
{
"input": "BBBACCCCCCC",
"output": "8"
},
{
"input": "YBBBBY",
"output": "5"
},
{
"input": "YYYYYYYYYY",
"output": "1"
},
{
"input": "ABABBBBBBBBBBBB",
"output": "13"
}
] | 1,608,948,077 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 109 | 307,200 | x = list(str(input()))
y = []
z = 0
for i in range(0,len(x)):
if x[i] == "A" or x[i] == "E" or x[i] == "I" or x[i] == "O" or x[i] =="U" or x[i]=="Y":
y.append(i)
for i in range(0, len(y)-2):
q = y[i+1] - y[i]
r = y[i+2] - y[i+1]
if r>z:
z = r
print(z)
| Title: Grasshopper And the String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump.
Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability.
The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'.
Input Specification:
The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100.
Output Specification:
Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels.
Demo Input:
['ABABBBACFEYUKOTT\n', 'AAA\n']
Demo Output:
['4', '1']
Note:
none | ```python
x = list(str(input()))
y = []
z = 0
for i in range(0,len(x)):
if x[i] == "A" or x[i] == "E" or x[i] == "I" or x[i] == "O" or x[i] =="U" or x[i]=="Y":
y.append(i)
for i in range(0, len(y)-2):
q = y[i+1] - y[i]
r = y[i+2] - y[i+1]
if r>z:
z = r
print(z)
``` | 0 | |
20 | C | Dijkstra? | PROGRAMMING | 1,900 | [
"graphs",
"shortest paths"
] | C. Dijkstra? | 1 | 64 | You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*. | The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | [
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n",
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n"
] | [
"1 4 3 5 ",
"1 4 3 5 "
] | none | 1,500 | [
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "2 1\n1 2 1",
"output": "1 2 "
},
{
"input": "3 1\n1 2 1",
"output": "-1"
},
{
"input": "3 3\n1 2 1\n1 3 2\n2 3 1",
"output": "1 3 "
},
{
"input": "10 10\n1 5 12\n2 4 140\n2 10 149\n3 6 154\n3 7 9\n3 8 226\n3 10 132\n4 10 55\n5 8 33\n7 8 173",
"output": "1 5 8 7 3 10 "
},
{
"input": "10 10\n1 5 178\n1 8 221\n2 7 92\n2 8 159\n3 5 55\n3 6 179\n3 10 237\n4 8 205\n5 6 191\n8 10 157",
"output": "1 8 10 "
},
{
"input": "10 10\n1 4 200\n1 9 197\n3 4 79\n3 5 213\n3 6 149\n5 8 3\n5 9 189\n6 7 130\n6 9 51\n8 10 135",
"output": "1 9 5 8 10 "
},
{
"input": "10 10\n1 4 201\n2 3 238\n3 4 40\n3 6 231\n3 8 45\n4 5 227\n4 6 58\n4 9 55\n5 7 14\n6 10 242",
"output": "1 4 6 10 "
}
] | 1,479,246,112 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 27 | 1,000 | 26,214,400 | from heapq import heappush, heappop
from itertools import count
counter = count()
task_finder = {}
pq = []
INVALID = -1
VALID = 0
def priority_pop():
while len(pq)>0:
priority, count, task = heappop(pq)
if count is not INVALID:
del task_finder[task]
return task
return -1
def priority_add(priority, task, count=None):
if(count==None):
count=next(counter)
entry = [priority, count, task]
task_finder[task] = entry
heappush(pq, entry)
def priority_delete(task):
entry = task_finder[task]
entry[1] = INVALID
def priority_change(priority, task):
if(task not in task_finder):
return
entry = task_finder[task]
priority_add(priority, task, entry[1])
entry[1] = INVALID
N, M = [int(x) for x in input().split(" ")]
neighbours = [{} for x in range(N)]
for i in range(M):
v1,v2,c = [int(x) for x in input().split(" ")]
if((v1-1) in neighbours[v2-1]):
c = min(neighbours[v2-1][v1-1],c)
neighbours[v2-1][v1-1] = c
neighbours[v1-1][v2-1] = c
prev=[-1 for i in range(N)]
cost=[0 if i==0 else float("Inf") for i in range(N)]
for i in range(N):
priority_add(cost[i],i)
while(len(pq)>0):
v=priority_pop()
if(v==-1):
break
for n in neighbours[v]:
altCost = cost[v]+neighbours[v][n]
if(altCost < cost[n]):
cost[n] = altCost
prev[n] = v
priority_change(cost[n],n)
if(prev[N-1] == -1):
print(-1)
else:
path = [N]
cur = N-1
while(cur > 0):
cur = prev[cur]
path.append(cur+1)
print(" ".join(str(x) for x in reversed(path))) | Title: Dijkstra?
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
Output Specification:
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
Demo Input:
['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n']
Demo Output:
['1 4 3 5 ', '1 4 3 5 ']
Note:
none | ```python
from heapq import heappush, heappop
from itertools import count
counter = count()
task_finder = {}
pq = []
INVALID = -1
VALID = 0
def priority_pop():
while len(pq)>0:
priority, count, task = heappop(pq)
if count is not INVALID:
del task_finder[task]
return task
return -1
def priority_add(priority, task, count=None):
if(count==None):
count=next(counter)
entry = [priority, count, task]
task_finder[task] = entry
heappush(pq, entry)
def priority_delete(task):
entry = task_finder[task]
entry[1] = INVALID
def priority_change(priority, task):
if(task not in task_finder):
return
entry = task_finder[task]
priority_add(priority, task, entry[1])
entry[1] = INVALID
N, M = [int(x) for x in input().split(" ")]
neighbours = [{} for x in range(N)]
for i in range(M):
v1,v2,c = [int(x) for x in input().split(" ")]
if((v1-1) in neighbours[v2-1]):
c = min(neighbours[v2-1][v1-1],c)
neighbours[v2-1][v1-1] = c
neighbours[v1-1][v2-1] = c
prev=[-1 for i in range(N)]
cost=[0 if i==0 else float("Inf") for i in range(N)]
for i in range(N):
priority_add(cost[i],i)
while(len(pq)>0):
v=priority_pop()
if(v==-1):
break
for n in neighbours[v]:
altCost = cost[v]+neighbours[v][n]
if(altCost < cost[n]):
cost[n] = altCost
prev[n] = v
priority_change(cost[n],n)
if(prev[N-1] == -1):
print(-1)
else:
path = [N]
cur = N-1
while(cur > 0):
cur = prev[cur]
path.append(cur+1)
print(" ".join(str(x) for x in reversed(path)))
``` | 0 |
978 | A | Remove Duplicates | PROGRAMMING | 800 | [
"implementation"
] | null | null | Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed. | The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array. | In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left. | [
"6\n1 5 5 1 6 1\n",
"5\n2 4 2 4 4\n",
"5\n6 6 6 6 6\n"
] | [
"3\n5 6 1 \n",
"2\n2 4 \n",
"1\n6 \n"
] | In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$. | 0 | [
{
"input": "6\n1 5 5 1 6 1",
"output": "3\n5 6 1 "
},
{
"input": "5\n2 4 2 4 4",
"output": "2\n2 4 "
},
{
"input": "5\n6 6 6 6 6",
"output": "1\n6 "
},
{
"input": "7\n1 2 3 4 2 2 3",
"output": "4\n1 4 2 3 "
},
{
"input": "9\n100 100 100 99 99 99 100 100 100",
"output": "2\n99 100 "
},
{
"input": "27\n489 489 487 488 750 230 43 645 42 42 489 42 973 42 973 750 645 355 868 112 868 489 750 489 887 489 868",
"output": "13\n487 488 230 43 42 973 645 355 112 750 887 489 868 "
},
{
"input": "40\n151 421 421 909 117 222 909 954 227 421 227 954 954 222 421 227 421 421 421 151 421 227 222 222 222 222 421 183 421 227 421 954 222 421 954 421 222 421 909 421",
"output": "8\n117 151 183 227 954 222 909 421 "
},
{
"input": "48\n2 2 2 903 903 2 726 2 2 2 2 2 2 2 2 2 2 726 2 2 2 2 2 2 2 726 2 2 2 2 62 2 2 2 2 2 2 2 2 726 62 726 2 2 2 903 903 2",
"output": "4\n62 726 903 2 "
},
{
"input": "1\n1",
"output": "1\n1 "
},
{
"input": "13\n5 37 375 5 37 33 37 375 37 2 3 3 2",
"output": "6\n5 33 375 37 3 2 "
},
{
"input": "50\n1 2 3 4 5 4 3 2 1 2 3 2 1 4 5 5 4 3 2 1 1 2 3 4 5 4 3 2 1 2 3 2 1 4 5 5 4 3 2 1 4 3 2 5 1 6 6 6 6 6",
"output": "6\n4 3 2 5 1 6 "
},
{
"input": "47\n233 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "2\n233 1 "
},
{
"input": "47\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",
"output": "1\n1 "
},
{
"input": "2\n964 964",
"output": "1\n964 "
},
{
"input": "2\n1000 1000",
"output": "1\n1000 "
},
{
"input": "1\n1000",
"output": "1\n1000 "
},
{
"input": "45\n991 991 996 996 992 992 999 1000 998 1000 992 999 996 999 991 991 999 993 992 999 1000 997 992 999 996 991 994 996 991 999 1000 993 999 997 999 992 991 997 991 998 998 995 998 994 993",
"output": "10\n996 1000 999 992 997 991 995 998 994 993 "
},
{
"input": "6\n994 993 1000 998 991 994",
"output": "5\n993 1000 998 991 994 "
},
{
"input": "48\n992 995 992 991 994 992 995 999 996 993 999 995 993 992 1000 992 997 996 991 993 992 998 998 998 999 995 992 992 993 992 992 995 996 995 997 991 997 991 999 994 994 997 1000 998 1000 992 1000 999",
"output": "10\n993 996 995 991 994 997 998 992 1000 999 "
},
{
"input": "3\n6 6 3",
"output": "2\n6 3 "
},
{
"input": "36\n999 1000 993 993 1000 999 996 997 998 995 995 997 999 995 1000 998 998 994 993 998 994 999 1000 995 996 994 991 991 999 996 993 999 996 998 991 997",
"output": "9\n1000 995 994 993 999 996 998 991 997 "
},
{
"input": "49\n991 997 995 991 995 998 994 996 997 991 998 993 992 999 996 995 991 991 999 997 995 993 997 1000 997 993 993 994 999 994 992 991 1000 996 994 996 996 999 992 992 1000 991 997 993 991 994 997 991 996",
"output": "10\n998 995 999 992 1000 993 994 997 991 996 "
},
{
"input": "2\n1000 999",
"output": "2\n1000 999 "
},
{
"input": "28\n997 994 991 994 994 995 1000 992 995 994 994 995 991 996 991 996 991 999 999 993 994 997 995 992 991 992 998 1000",
"output": "10\n996 999 993 994 997 995 991 992 998 1000 "
},
{
"input": "15\n991 995 995 1000 991 993 997 998 1000 994 1000 992 991 995 992",
"output": "8\n993 997 998 994 1000 991 995 992 "
},
{
"input": "6\n1000 1 1000 2 1000 1",
"output": "3\n2 1000 1 "
},
{
"input": "6\n1000 1 1000 2 1 1000",
"output": "3\n2 1 1000 "
},
{
"input": "7\n1 1000 1 999 1000 1 2",
"output": "4\n999 1000 1 2 "
}
] | 1,679,201,051 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | a = int(input())
b = input()
count = 0
c = 0
if a>=3 and a<=100:
if len(b)==a:
for i in b:
if i=='x':
count+=1
elif i!='x':
count = 0
if count>=3:
c+=1
print(c) | Title: Remove Duplicates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array.
Output Specification:
In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left.
Demo Input:
['6\n1 5 5 1 6 1\n', '5\n2 4 2 4 4\n', '5\n6 6 6 6 6\n']
Demo Output:
['3\n5 6 1 \n', '2\n2 4 \n', '1\n6 \n']
Note:
In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$. | ```python
a = int(input())
b = input()
count = 0
c = 0
if a>=3 and a<=100:
if len(b)==a:
for i in b:
if i=='x':
count+=1
elif i!='x':
count = 0
if count>=3:
c+=1
print(c)
``` | 0 | |
4 | B | Before an Exam | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy"
] | B. Before an Exam | 0 | 64 | Tomorrow Peter has a Biology exam. He does not like this subject much, but *d* days ago he learnt that he would have to take this exam. Peter's strict parents made him prepare for the exam immediately, for this purpose he has to study not less than *minTime**i* and not more than *maxTime**i* hours per each *i*-th day. Moreover, they warned Peter that a day before the exam they would check how he has followed their instructions.
So, today is the day when Peter's parents ask him to show the timetable of his preparatory studies. But the boy has counted only the sum of hours *sumTime* spent him on preparation, and now he wants to know if he can show his parents a timetable *sсhedule* with *d* numbers, where each number *sсhedule**i* stands for the time in hours spent by Peter each *i*-th day on biology studies, and satisfying the limitations imposed by his parents, and at the same time the sum total of all *schedule**i* should equal to *sumTime*. | The first input line contains two integer numbers *d*,<=*sumTime* (1<=≤<=*d*<=≤<=30,<=0<=≤<=*sumTime*<=≤<=240) — the amount of days, during which Peter studied, and the total amount of hours, spent on preparation. Each of the following *d* lines contains two integer numbers *minTime**i*,<=*maxTime**i* (0<=≤<=*minTime**i*<=≤<=*maxTime**i*<=≤<=8), separated by a space — minimum and maximum amount of hours that Peter could spent in the *i*-th day. | In the first line print YES, and in the second line print *d* numbers (separated by a space), each of the numbers — amount of hours, spent by Peter on preparation in the corresponding day, if he followed his parents' instructions; or print NO in the unique line. If there are many solutions, print any of them. | [
"1 48\n5 7\n",
"2 5\n0 1\n3 5\n"
] | [
"NO\n",
"YES\n1 4 "
] | none | 0 | [
{
"input": "1 48\n5 7",
"output": "NO"
},
{
"input": "2 5\n0 1\n3 5",
"output": "YES\n1 4 "
},
{
"input": "1 1\n5 6",
"output": "NO"
},
{
"input": "1 4\n2 4",
"output": "YES\n4 "
},
{
"input": "2 5\n4 6\n0 0",
"output": "YES\n5 0 "
},
{
"input": "27 97\n2 8\n0 5\n5 6\n3 6\n5 5\n1 2\n3 5\n1 8\n0 4\n3 3\n0 2\n0 0\n4 8\n5 6\n5 8\n0 7\n1 4\n0 4\n1 5\n3 7\n2 5\n5 6\n4 7\n3 8\n0 1\n3 4\n5 7",
"output": "YES\n8 5 6 6 5 2 5 8 4 3 2 0 6 5 5 0 1 0 1 3 2 5 4 3 0 3 5 "
},
{
"input": "30 92\n4 5\n4 7\n2 6\n8 8\n7 8\n4 5\n1 5\n7 8\n1 2\n6 8\n2 7\n2 4\n0 0\n1 3\n4 5\n1 1\n0 7\n2 5\n2 5\n3 3\n1 2\n1 7\n5 5\n5 8\n6 7\n0 3\n2 6\n0 7\n5 6\n2 5",
"output": "YES\n5 7 2 8 7 4 1 7 1 6 2 2 0 1 4 1 0 2 2 3 1 1 5 5 6 0 2 0 5 2 "
},
{
"input": "30 178\n1 6\n2 7\n2 5\n2 8\n1 6\n2 8\n3 4\n2 7\n0 2\n0 8\n0 3\n0 2\n2 4\n4 8\n6 8\n0 8\n0 6\n1 8\n0 3\n6 7\n4 8\n2 7\n1 1\n3 7\n3 6\n2 5\n4 7\n2 2\n1 8\n5 6",
"output": "NO"
},
{
"input": "30 71\n1 3\n0 6\n3 5\n3 6\n2 4\n2 8\n2 4\n3 8\n3 5\n2 4\n2 3\n3 7\n0 0\n5 7\n0 2\n5 8\n0 8\n4 7\n0 3\n3 7\n2 3\n4 5\n7 8\n7 7\n2 7\n1 3\n0 1\n1 5\n6 7\n5 8",
"output": "NO"
},
{
"input": "30 119\n2 7\n1 3\n0 3\n3 4\n7 7\n7 7\n0 5\n2 3\n0 8\n0 8\n0 5\n5 7\n2 2\n2 6\n2 5\n3 7\n0 8\n0 2\n1 3\n2 3\n1 4\n0 1\n3 7\n7 8\n1 2\n0 6\n1 8\n1 7\n4 8\n1 4",
"output": "YES\n7 3 3 4 7 7 5 3 8 8 5 7 2 6 5 7 8 2 1 2 1 0 3 7 1 0 1 1 4 1 "
},
{
"input": "1 0\n0 0",
"output": "YES\n0 "
},
{
"input": "30 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "YES\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 "
},
{
"input": "30 240\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8\n8 8",
"output": "YES\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 "
},
{
"input": "30 64\n7 7\n3 4\n0 4\n5 8\n1 3\n1 1\n0 2\n0 4\n7 7\n3 5\n4 4\n0 4\n4 5\n1 4\n4 6\n5 7\n1 4\n1 2\n3 3\n0 2\n3 7\n3 4\n0 3\n3 8\n0 8\n0 1\n0 3\n4 5\n1 5\n0 3",
"output": "YES\n7 3 0 5 1 1 0 0 7 3 4 0 4 1 4 5 1 1 3 0 3 3 0 3 0 0 0 4 1 0 "
},
{
"input": "30 189\n0 8\n3 8\n5 8\n2 2\n7 7\n2 8\n7 8\n2 5\n1 7\n0 1\n3 4\n1 7\n1 6\n5 6\n3 7\n4 7\n4 7\n4 7\n5 5\n0 5\n4 7\n3 5\n0 6\n4 8\n2 5\n5 6\n5 6\n2 8\n0 7\n7 8",
"output": "YES\n8 8 8 2 7 8 8 5 7 1 4 7 6 6 7 7 7 7 5 5 7 5 6 8 5 6 6 8 7 8 "
}
] | 1,691,661,168 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 15 | 61 | 0 | import sys
R = lambda:map(int, sys.stdin.readline().rstrip().split())
def solve():
d, sumtime = R()
arr = []
minsum = maxsum = 0
for i in range(d):
mint, maxt = R()
minsum += mint
maxsum += maxt
arr.append([mint, maxt])
if sumtime < minsum or sumtime > maxsum:
print('NO')
return
summ = 0
ans = []
for i in range(d):
now = arr[i][0]
summ += now
ans.append(now)
i = 0
while i < d and summ != sumtime:
dif = arr[i][1] - ans[i]
if dif + summ < sumtime:
summ += dif
ans[i] = arr[i][1]
else:
temp = sumtime - summ
ans[i] += temp
print('YES')
print(*ans)
return
i += 1
if sum(ans) == sumtime:
print('YES')
print(*ans)
solve()
| Title: Before an Exam
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
Tomorrow Peter has a Biology exam. He does not like this subject much, but *d* days ago he learnt that he would have to take this exam. Peter's strict parents made him prepare for the exam immediately, for this purpose he has to study not less than *minTime**i* and not more than *maxTime**i* hours per each *i*-th day. Moreover, they warned Peter that a day before the exam they would check how he has followed their instructions.
So, today is the day when Peter's parents ask him to show the timetable of his preparatory studies. But the boy has counted only the sum of hours *sumTime* spent him on preparation, and now he wants to know if he can show his parents a timetable *sсhedule* with *d* numbers, where each number *sсhedule**i* stands for the time in hours spent by Peter each *i*-th day on biology studies, and satisfying the limitations imposed by his parents, and at the same time the sum total of all *schedule**i* should equal to *sumTime*.
Input Specification:
The first input line contains two integer numbers *d*,<=*sumTime* (1<=≤<=*d*<=≤<=30,<=0<=≤<=*sumTime*<=≤<=240) — the amount of days, during which Peter studied, and the total amount of hours, spent on preparation. Each of the following *d* lines contains two integer numbers *minTime**i*,<=*maxTime**i* (0<=≤<=*minTime**i*<=≤<=*maxTime**i*<=≤<=8), separated by a space — minimum and maximum amount of hours that Peter could spent in the *i*-th day.
Output Specification:
In the first line print YES, and in the second line print *d* numbers (separated by a space), each of the numbers — amount of hours, spent by Peter on preparation in the corresponding day, if he followed his parents' instructions; or print NO in the unique line. If there are many solutions, print any of them.
Demo Input:
['1 48\n5 7\n', '2 5\n0 1\n3 5\n']
Demo Output:
['NO\n', 'YES\n1 4 ']
Note:
none | ```python
import sys
R = lambda:map(int, sys.stdin.readline().rstrip().split())
def solve():
d, sumtime = R()
arr = []
minsum = maxsum = 0
for i in range(d):
mint, maxt = R()
minsum += mint
maxsum += maxt
arr.append([mint, maxt])
if sumtime < minsum or sumtime > maxsum:
print('NO')
return
summ = 0
ans = []
for i in range(d):
now = arr[i][0]
summ += now
ans.append(now)
i = 0
while i < d and summ != sumtime:
dif = arr[i][1] - ans[i]
if dif + summ < sumtime:
summ += dif
ans[i] = arr[i][1]
else:
temp = sumtime - summ
ans[i] += temp
print('YES')
print(*ans)
return
i += 1
if sum(ans) == sumtime:
print('YES')
print(*ans)
solve()
``` | 3 |
599 | A | Patrick and Shopping | PROGRAMMING | 800 | [
"implementation"
] | null | null | Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled. | The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops. | Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house. | [
"10 20 30\n",
"1 1 5\n"
] | [
"60\n",
"4\n"
] | The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. | 500 | [
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
{
"input": "789 101112 131415",
"output": "203802"
},
{
"input": "27485716 99999999 35182",
"output": "55041796"
},
{
"input": "1 293548 5",
"output": "12"
},
{
"input": "12059 259855 5874875",
"output": "543828"
},
{
"input": "46981 105809 585858",
"output": "305580"
},
{
"input": "9889 1221 2442",
"output": "7326"
},
{
"input": "100500 200600 300700",
"output": "601800"
},
{
"input": "318476 318476 318476",
"output": "955428"
},
{
"input": "23985 3353 75633",
"output": "54676"
},
{
"input": "120 1298 2222",
"output": "2836"
},
{
"input": "98437 23487 666672",
"output": "243848"
},
{
"input": "100000000 100000000 100000000",
"output": "300000000"
},
{
"input": "2 5 2",
"output": "8"
},
{
"input": "1 1000 1",
"output": "4"
},
{
"input": "1 100000000 1",
"output": "4"
}
] | 1,623,656,876 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 62 | 0 | if __name__ == "__main__":
d1,d2,d3 = map(int,input().split())
(minm,dist1,dist2) = (0,0,0)
# case 1
dist1 += d1 + d2 + d3
if(minm == 0):
minm = dist1
else:
if(dist1 < minm):
minm = dist1
# case 2
dist2 += 2*d1 + 2*d2
if(dist2 < minm):
minm = dist2
print(minm) | Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. | ```python
if __name__ == "__main__":
d1,d2,d3 = map(int,input().split())
(minm,dist1,dist2) = (0,0,0)
# case 1
dist1 += d1 + d2 + d3
if(minm == 0):
minm = dist1
else:
if(dist1 < minm):
minm = dist1
# case 2
dist2 += 2*d1 + 2*d2
if(dist2 < minm):
minm = dist2
print(minm)
``` | 0 | |
331 | C1 | The Great Julya Calendar | PROGRAMMING | 1,100 | [
"dp"
] | null | null | Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero. | The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3). | Print a single integer — the minimum number of subtractions that turns the magic number to a zero. | [
"24\n"
] | [
"5"
] | In the first test sample the minimum number of operations can be reached by the following sequence of subtractions: | 20 | [
{
"input": "24",
"output": "5"
},
{
"input": "0",
"output": "0"
},
{
"input": "3",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "10",
"output": "2"
},
{
"input": "31",
"output": "6"
},
{
"input": "701",
"output": "116"
},
{
"input": "222",
"output": "39"
},
{
"input": "156",
"output": "28"
},
{
"input": "12343",
"output": "1778"
},
{
"input": "7984",
"output": "1178"
},
{
"input": "1000000",
"output": "128207"
},
{
"input": "1",
"output": "1"
},
{
"input": "3",
"output": "1"
},
{
"input": "5",
"output": "1"
},
{
"input": "11",
"output": "3"
},
{
"input": "77",
"output": "14"
},
{
"input": "216",
"output": "37"
},
{
"input": "1468",
"output": "233"
},
{
"input": "1995",
"output": "305"
},
{
"input": "11010",
"output": "1568"
},
{
"input": "47320",
"output": "6746"
},
{
"input": "258634",
"output": "35024"
}
] | 1,652,312,940 | 2,147,483,647 | Python 3 | OK | TESTS1 | 24 | 248 | 0 |
def solve (n):
count = 0
s = int(n)
while True:
if s == 0:
print(count)
break
s -= int(max(str(n)))
n = s
count += 1
n = input()
solve(n) | Title: The Great Julya Calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero.
Input Specification:
The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3).
Output Specification:
Print a single integer — the minimum number of subtractions that turns the magic number to a zero.
Demo Input:
['24\n']
Demo Output:
['5']
Note:
In the first test sample the minimum number of operations can be reached by the following sequence of subtractions: | ```python
def solve (n):
count = 0
s = int(n)
while True:
if s == 0:
print(count)
break
s -= int(max(str(n)))
n = s
count += 1
n = input()
solve(n)
``` | 3 | |
348 | A | Mafia | PROGRAMMING | 1,600 | [
"binary search",
"math",
"sortings"
] | null | null | One day *n* friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other *n*<=-<=1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the *i*-th person wants to play *a**i* rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want? | The first line contains integer *n* (3<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the *i*-th number in the list is the number of rounds the *i*-th person wants to play. | In a single line print a single integer — the minimum number of game rounds the friends need to let the *i*-th person play at least *a**i* rounds.
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. | [
"3\n3 2 2\n",
"4\n2 2 2 2\n"
] | [
"4\n",
"3\n"
] | You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game). | 500 | [
{
"input": "3\n3 2 2",
"output": "4"
},
{
"input": "4\n2 2 2 2",
"output": "3"
},
{
"input": "7\n9 7 7 8 8 7 8",
"output": "9"
},
{
"input": "10\n13 12 10 13 13 14 10 10 12 12",
"output": "14"
},
{
"input": "10\n94 96 91 95 99 94 96 92 95 99",
"output": "106"
},
{
"input": "100\n1 555 876 444 262 234 231 598 416 261 206 165 181 988 469 123 602 592 533 97 864 716 831 156 962 341 207 377 892 51 866 96 757 317 832 476 549 472 770 1000 887 145 956 515 992 653 972 677 973 527 984 559 280 346 580 30 372 547 209 929 492 520 446 726 47 170 699 560 814 206 688 955 308 287 26 102 77 430 262 71 415 586 532 562 419 615 732 658 108 315 268 574 86 12 23 429 640 995 342 305",
"output": "1000"
},
{
"input": "3\n1 1 1",
"output": "2"
},
{
"input": "30\n94 93 90 94 90 91 93 91 93 94 93 90 100 94 97 94 94 95 94 96 94 98 97 95 97 91 91 95 98 96",
"output": "100"
},
{
"input": "5\n1000000000 5 5 4 4",
"output": "1000000000"
},
{
"input": "3\n1 2 1",
"output": "2"
},
{
"input": "3\n2 1 1",
"output": "2"
},
{
"input": "4\n1 2 3 4",
"output": "4"
},
{
"input": "3\n1000000000 1000000000 10000000",
"output": "1005000000"
},
{
"input": "3\n677876423 834056477 553175531",
"output": "1032554216"
},
{
"input": "5\n1000000000 1 1 1 1",
"output": "1000000000"
},
{
"input": "4\n1000000000 1000000000 1000000000 1000000000",
"output": "1333333334"
},
{
"input": "3\n4 10 11",
"output": "13"
},
{
"input": "5\n1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1250000000"
}
] | 1,654,743,403 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 124 | 0 | import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
a.sort()
s = sum(a)
g = a[-1]
while((n * g - s) < a[-1]):
g += 1
print(g) | Title: Mafia
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day *n* friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other *n*<=-<=1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the *i*-th person wants to play *a**i* rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the *i*-th number in the list is the number of rounds the *i*-th person wants to play.
Output Specification:
In a single line print a single integer — the minimum number of game rounds the friends need to let the *i*-th person play at least *a**i* rounds.
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:
['3\n3 2 2\n', '4\n2 2 2 2\n']
Demo Output:
['4\n', '3\n']
Note:
You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game). | ```python
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
a.sort()
s = sum(a)
g = a[-1]
while((n * g - s) < a[-1]):
g += 1
print(g)
``` | 0 | |
611 | B | New Year and Old Property | PROGRAMMING | 1,300 | [
"bitmasks",
"brute force",
"implementation"
] | null | null | The year 2015 is almost over.
Limak is a little polar bear. He has recently learnt about the binary system. He noticed that the passing year has exactly one zero in its representation in the binary system — 201510<==<=111110111112. Note that he doesn't care about the number of zeros in the decimal representation.
Limak chose some interval of years. He is going to count all years from this interval that have exactly one zero in the binary representation. Can you do it faster?
Assume that all positive integers are always written without leading zeros. | The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=1018) — the first year and the last year in Limak's interval respectively. | Print one integer – the number of years Limak will count in his chosen interval. | [
"5 10\n",
"2015 2015\n",
"100 105\n",
"72057594000000000 72057595000000000\n"
] | [
"2\n",
"1\n",
"0\n",
"26\n"
] | In the first sample Limak's interval contains numbers 5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>, 6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>, 7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>, 8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>, 9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub> and 10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>. Two of them (101<sub class="lower-index">2</sub> and 110<sub class="lower-index">2</sub>) have the described property. | 750 | [
{
"input": "5 10",
"output": "2"
},
{
"input": "2015 2015",
"output": "1"
},
{
"input": "100 105",
"output": "0"
},
{
"input": "72057594000000000 72057595000000000",
"output": "26"
},
{
"input": "1 100",
"output": "16"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "0"
},
{
"input": "1 1000000000000000000",
"output": "1712"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 7",
"output": "3"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "2 3",
"output": "1"
},
{
"input": "2 4",
"output": "1"
},
{
"input": "2 5",
"output": "2"
},
{
"input": "2 6",
"output": "3"
},
{
"input": "2 7",
"output": "3"
},
{
"input": "3 3",
"output": "0"
},
{
"input": "3 4",
"output": "0"
},
{
"input": "3 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "3 7",
"output": "2"
},
{
"input": "4 4",
"output": "0"
},
{
"input": "4 5",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "4 7",
"output": "2"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "5 6",
"output": "2"
},
{
"input": "5 7",
"output": "2"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "6 7",
"output": "1"
},
{
"input": "7 7",
"output": "0"
},
{
"input": "1 8",
"output": "3"
},
{
"input": "6 8",
"output": "1"
},
{
"input": "7 8",
"output": "0"
},
{
"input": "8 8",
"output": "0"
},
{
"input": "1 1022",
"output": "45"
},
{
"input": "1 1023",
"output": "45"
},
{
"input": "1 1024",
"output": "45"
},
{
"input": "1 1025",
"output": "45"
},
{
"input": "1 1026",
"output": "45"
},
{
"input": "509 1022",
"output": "11"
},
{
"input": "510 1022",
"output": "10"
},
{
"input": "511 1022",
"output": "9"
},
{
"input": "512 1022",
"output": "9"
},
{
"input": "513 1022",
"output": "9"
},
{
"input": "509 1023",
"output": "11"
},
{
"input": "510 1023",
"output": "10"
},
{
"input": "511 1023",
"output": "9"
},
{
"input": "512 1023",
"output": "9"
},
{
"input": "513 1023",
"output": "9"
},
{
"input": "509 1024",
"output": "11"
},
{
"input": "510 1024",
"output": "10"
},
{
"input": "511 1024",
"output": "9"
},
{
"input": "512 1024",
"output": "9"
},
{
"input": "513 1024",
"output": "9"
},
{
"input": "509 1025",
"output": "11"
},
{
"input": "510 1025",
"output": "10"
},
{
"input": "511 1025",
"output": "9"
},
{
"input": "512 1025",
"output": "9"
},
{
"input": "513 1025",
"output": "9"
},
{
"input": "1 1000000000",
"output": "408"
},
{
"input": "10000000000 70000000000000000",
"output": "961"
},
{
"input": "1 935829385028502935",
"output": "1712"
},
{
"input": "500000000000000000 1000000000000000000",
"output": "58"
},
{
"input": "500000000000000000 576460752303423488",
"output": "57"
},
{
"input": "576460752303423488 1000000000000000000",
"output": "1"
},
{
"input": "999999999999999999 1000000000000000000",
"output": "0"
},
{
"input": "1124800395214847 36011204832919551",
"output": "257"
},
{
"input": "1124800395214847 36011204832919550",
"output": "256"
},
{
"input": "1124800395214847 36011204832919552",
"output": "257"
},
{
"input": "1124800395214846 36011204832919551",
"output": "257"
},
{
"input": "1124800395214848 36011204832919551",
"output": "256"
},
{
"input": "1 287104476244869119",
"output": "1603"
},
{
"input": "1 287104476244869118",
"output": "1602"
},
{
"input": "1 287104476244869120",
"output": "1603"
},
{
"input": "492581209243647 1000000000000000000",
"output": "583"
},
{
"input": "492581209243646 1000000000000000000",
"output": "583"
},
{
"input": "492581209243648 1000000000000000000",
"output": "582"
},
{
"input": "1099444518911 1099444518911",
"output": "1"
},
{
"input": "1099444518910 1099444518911",
"output": "1"
},
{
"input": "1099444518911 1099444518912",
"output": "1"
},
{
"input": "1099444518910 1099444518912",
"output": "1"
},
{
"input": "864691128455135231 864691128455135231",
"output": "1"
},
{
"input": "864691128455135231 864691128455135232",
"output": "1"
},
{
"input": "864691128455135230 864691128455135232",
"output": "1"
},
{
"input": "864691128455135230 864691128455135231",
"output": "1"
},
{
"input": "864691128455135231 1000000000000000000",
"output": "1"
},
{
"input": "864691128455135232 1000000000000000000",
"output": "0"
},
{
"input": "864691128455135230 1000000000000000000",
"output": "1"
},
{
"input": "576460752303423487 576460752303423487",
"output": "0"
},
{
"input": "1 576460752303423487",
"output": "1711"
},
{
"input": "1 576460752303423486",
"output": "1711"
},
{
"input": "2 1000000000000000000",
"output": "1712"
},
{
"input": "3 1000000000000000000",
"output": "1711"
},
{
"input": "4 1000000000000000000",
"output": "1711"
},
{
"input": "5 1000000000000000000",
"output": "1711"
},
{
"input": "6 1000000000000000000",
"output": "1710"
},
{
"input": "5 6",
"output": "2"
},
{
"input": "1 2",
"output": "1"
}
] | 1,683,179,566 | 2,147,483,647 | Python 3 | OK | TESTS | 103 | 46 | 0 | a,b=map(int,input().split())
print(len([(2**l-1)-2**j for l in range(65) for j in range(l-1) if a<=2**l-1-2**j<=b]))
| Title: New Year and Old Property
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The year 2015 is almost over.
Limak is a little polar bear. He has recently learnt about the binary system. He noticed that the passing year has exactly one zero in its representation in the binary system — 201510<==<=111110111112. Note that he doesn't care about the number of zeros in the decimal representation.
Limak chose some interval of years. He is going to count all years from this interval that have exactly one zero in the binary representation. Can you do it faster?
Assume that all positive integers are always written without leading zeros.
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=1018) — the first year and the last year in Limak's interval respectively.
Output Specification:
Print one integer – the number of years Limak will count in his chosen interval.
Demo Input:
['5 10\n', '2015 2015\n', '100 105\n', '72057594000000000 72057595000000000\n']
Demo Output:
['2\n', '1\n', '0\n', '26\n']
Note:
In the first sample Limak's interval contains numbers 5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>, 6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>, 7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>, 8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>, 9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub> and 10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>. Two of them (101<sub class="lower-index">2</sub> and 110<sub class="lower-index">2</sub>) have the described property. | ```python
a,b=map(int,input().split())
print(len([(2**l-1)-2**j for l in range(65) for j in range(l-1) if a<=2**l-1-2**j<=b]))
``` | 3 | |
884 | A | Book Reading | PROGRAMMING | 800 | [
"implementation"
] | null | null | Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed *n*.
Remember that there are 86400 seconds in a day. | The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day. | Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed *n*. | [
"2 2\n86400 86398\n",
"2 86400\n0 86400\n"
] | [
"2\n",
"1\n"
] | none | 0 | [
{
"input": "2 2\n86400 86398",
"output": "2"
},
{
"input": "2 86400\n0 86400",
"output": "1"
},
{
"input": "2 86400\n1 86399",
"output": "2"
},
{
"input": "100 1000000\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": "12"
},
{
"input": "1 1\n86399",
"output": "1"
},
{
"input": "6 1200\n86400 86400 86000 86000 86000 86400",
"output": "5"
},
{
"input": "6 1200\n86400 86400 86000 86000 86001 86399",
"output": "6"
},
{
"input": "4 172799\n1 1 86400 0",
"output": "4"
},
{
"input": "4 172799\n0 86400 86399 0",
"output": "4"
},
{
"input": "6 1\n1 1 86400 1 86399 1",
"output": "1"
},
{
"input": "4 1\n86400 86399 86400 86400",
"output": "2"
},
{
"input": "4 1\n86400 86400 0 86400",
"output": "3"
}
] | 1,654,264,561 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 16 | 46 | 0 | n,t = tuple(map(int, input().split()))
day_seconds = 86400
A = tuple(map(int, input().split()))
def calc(A):
free_time = 0
counter = 0
for a in A:
free_time += day_seconds - a
counter += 1
if free_time >= t:
return counter
print(calc(A))
| Title: Book Reading
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed *n*.
Remember that there are 86400 seconds in a day.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day.
Output Specification:
Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed *n*.
Demo Input:
['2 2\n86400 86398\n', '2 86400\n0 86400\n']
Demo Output:
['2\n', '1\n']
Note:
none | ```python
n,t = tuple(map(int, input().split()))
day_seconds = 86400
A = tuple(map(int, input().split()))
def calc(A):
free_time = 0
counter = 0
for a in A:
free_time += day_seconds - a
counter += 1
if free_time >= t:
return counter
print(calc(A))
``` | 3 | |
892 | A | Greed | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not! | The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans.
The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans. | Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower). | [
"2\n3 5\n3 6\n",
"3\n6 8 9\n6 10 12\n",
"5\n0 0 5 0 0\n1 1 8 10 5\n",
"4\n4 1 0 3\n5 2 2 3\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"YES\n"
] | In the first sample, there are already 2 cans, so the answer is "YES". | 500 | [
{
"input": "2\n3 5\n3 6",
"output": "YES"
},
{
"input": "3\n6 8 9\n6 10 12",
"output": "NO"
},
{
"input": "5\n0 0 5 0 0\n1 1 8 10 5",
"output": "YES"
},
{
"input": "4\n4 1 0 3\n5 2 2 3",
"output": "YES"
},
{
"input": "10\n9 10 24 11 1 7 8 3 28 14\n86 20 34 11 22 94 8 16 73 85",
"output": "YES"
},
{
"input": "4\n25 35 7 31\n70 37 43 35",
"output": "YES"
},
{
"input": "10\n15 26 15 14 14 39 40 4 25 39\n27 72 16 44 69 48 53 17 63 42",
"output": "NO"
},
{
"input": "5\n22 5 19 16 32\n26 10 43 38 37",
"output": "NO"
},
{
"input": "5\n32 4 22 40 26\n39 20 36 98 44",
"output": "YES"
},
{
"input": "6\n18 25 3 10 13 37\n38 73 19 35 24 37",
"output": "YES"
},
{
"input": "2\n2 2\n2 2",
"output": "YES"
},
{
"input": "2\n2 5\n2 5",
"output": "YES"
},
{
"input": "2\n1000 1008\n10000 2352",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5\n1 2 3 4 11",
"output": "YES"
},
{
"input": "4\n1 0 0 0\n2 0 0 0",
"output": "YES"
},
{
"input": "2\n0 0\n1 2",
"output": "YES"
},
{
"input": "3\n9 13 4\n10 14 5",
"output": "NO"
},
{
"input": "2\n0 0\n1 1",
"output": "YES"
},
{
"input": "5\n1 1 2 3 1\n1 1 2 3 4",
"output": "NO"
},
{
"input": "2\n0 0\n0 0",
"output": "YES"
},
{
"input": "3\n5 1 1\n5 5 5",
"output": "YES"
}
] | 1,592,629,669 | 2,147,483,647 | PyPy 3 | OK | TESTS | 39 | 327 | 13,312,000 | r=lambda :list(map(int,input().split()));n=r()[0];a=r();b=sorted(r())
print('YES') if b[-1]+b[-2]>=sum(a) else print('NO') | Title: Greed
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input Specification:
The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans.
The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans.
Output Specification:
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Demo Input:
['2\n3 5\n3 6\n', '3\n6 8 9\n6 10 12\n', '5\n0 0 5 0 0\n1 1 8 10 5\n', '4\n4 1 0 3\n5 2 2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'YES\n']
Note:
In the first sample, there are already 2 cans, so the answer is "YES". | ```python
r=lambda :list(map(int,input().split()));n=r()[0];a=r();b=sorted(r())
print('YES') if b[-1]+b[-2]>=sum(a) else print('NO')
``` | 3 | |
579 | A | Raising Bacteria | PROGRAMMING | 1,000 | [
"bitmasks"
] | null | null | You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days? | The only line containing one integer *x* (1<=≤<=*x*<=≤<=109). | The only line containing one integer: the answer. | [
"5\n",
"8\n"
] | [
"2\n",
"1\n"
] | For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1. | 250 | [
{
"input": "5",
"output": "2"
},
{
"input": "8",
"output": "1"
},
{
"input": "536870911",
"output": "29"
},
{
"input": "1",
"output": "1"
},
{
"input": "343000816",
"output": "14"
},
{
"input": "559980448",
"output": "12"
},
{
"input": "697681824",
"output": "14"
},
{
"input": "41313494",
"output": "14"
},
{
"input": "673935585",
"output": "13"
},
{
"input": "117422204",
"output": "19"
},
{
"input": "954746654",
"output": "15"
},
{
"input": "536838144",
"output": "14"
},
{
"input": "1000000000",
"output": "13"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "6",
"output": "2"
},
{
"input": "7",
"output": "3"
},
{
"input": "9",
"output": "2"
},
{
"input": "10",
"output": "2"
},
{
"input": "999999999",
"output": "21"
},
{
"input": "999999998",
"output": "20"
},
{
"input": "999999997",
"output": "20"
},
{
"input": "999999996",
"output": "19"
},
{
"input": "999999995",
"output": "20"
},
{
"input": "999999994",
"output": "19"
},
{
"input": "999999993",
"output": "19"
},
{
"input": "999999992",
"output": "18"
},
{
"input": "999999991",
"output": "20"
},
{
"input": "999999990",
"output": "19"
},
{
"input": "536870910",
"output": "28"
},
{
"input": "536870912",
"output": "1"
},
{
"input": "536870913",
"output": "2"
},
{
"input": "536870914",
"output": "2"
},
{
"input": "22",
"output": "3"
},
{
"input": "14",
"output": "3"
}
] | 1,697,474,014 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | l=[1]
n=int(input())
while l[-1]<=n:
l.append(2*l[-1])
a=n-l[-2]
l2=[1]
while l2[-1]<=a:
l2.append(2*l2[-1])
s=0
for i in l2[:-1][::-1]:
s+=1
print(s+1) | Title: Raising Bacteria
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input Specification:
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
Output Specification:
The only line containing one integer: the answer.
Demo Input:
['5\n', '8\n']
Demo Output:
['2\n', '1\n']
Note:
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1. | ```python
l=[1]
n=int(input())
while l[-1]<=n:
l.append(2*l[-1])
a=n-l[-2]
l2=[1]
while l2[-1]<=a:
l2.append(2*l2[-1])
s=0
for i in l2[:-1][::-1]:
s+=1
print(s+1)
``` | 0 | |
236 | A | Boy or Girl | PROGRAMMING | 800 | [
"brute force",
"implementation",
"strings"
] | null | null | Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method. | The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters. | If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes). | [
"wjmzbmr\n",
"xiaodao\n",
"sevenkplus\n"
] | [
"CHAT WITH HER!\n",
"IGNORE HIM!\n",
"CHAT WITH HER!\n"
] | For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!". | 500 | [
{
"input": "wjmzbmr",
"output": "CHAT WITH HER!"
},
{
"input": "xiaodao",
"output": "IGNORE HIM!"
},
{
"input": "sevenkplus",
"output": "CHAT WITH HER!"
},
{
"input": "pezu",
"output": "CHAT WITH HER!"
},
{
"input": "wnemlgppy",
"output": "CHAT WITH HER!"
},
{
"input": "zcinitufxoldnokacdvtmdohsfdjepyfioyvclhmujiqwvmudbfjzxjfqqxjmoiyxrfsbvseawwoyynn",
"output": "IGNORE HIM!"
},
{
"input": "qsxxuoynwtebujwpxwpajitiwxaxwgbcylxneqiebzfphugwkftpaikixmumkhfbjiswmvzbtiyifbx",
"output": "CHAT WITH HER!"
},
{
"input": "qwbdfzfylckctudyjlyrtmvbidfatdoqfmrfshsqqmhzohhsczscvwzpwyoyswhktjlykumhvaounpzwpxcspxwlgt",
"output": "IGNORE HIM!"
},
{
"input": "nuezoadauueermoeaabjrkxttkatspjsjegjcjcdmcxgodowzbwuqncfbeqlhkk",
"output": "IGNORE HIM!"
},
{
"input": "lggvdmulrsvtuagoavstuyufhypdxfomjlzpnduulukszqnnwfvxbvxyzmleocmofwclmzz",
"output": "IGNORE HIM!"
},
{
"input": "tgcdptnkc",
"output": "IGNORE HIM!"
},
{
"input": "wvfgnfrzabgibzxhzsojskmnlmrokydjoexnvi",
"output": "IGNORE HIM!"
},
{
"input": "sxtburpzskucowowebgrbovhadrrayamuwypmmxhscrujkmcgvyinp",
"output": "IGNORE HIM!"
},
{
"input": "pjqxhvxkyeqqvyuujxhmbspatvrckhhkfloottuybjivkkhpyivcighxumavrxzxslfpggnwbtalmhysyfllznphzia",
"output": "IGNORE HIM!"
},
{
"input": "fpellxwskyekoyvrfnuf",
"output": "CHAT WITH HER!"
},
{
"input": "xninyvkuvakfbs",
"output": "IGNORE HIM!"
},
{
"input": "vnxhrweyvhqufpfywdwftoyrfgrhxuamqhblkvdpxmgvphcbeeqbqssresjifwyzgfhurmamhkwupymuomak",
"output": "CHAT WITH HER!"
},
{
"input": "kmsk",
"output": "IGNORE HIM!"
},
{
"input": "lqonogasrkzhryjxppjyriyfxmdfubieglthyswz",
"output": "CHAT WITH HER!"
},
{
"input": "ndormkufcrkxlihdhmcehzoimcfhqsmombnfjrlcalffq",
"output": "CHAT WITH HER!"
},
{
"input": "zqzlnnuwcfufwujygtczfakhcpqbtxtejrbgoodychepzdphdahtxyfpmlrycyicqthsgm",
"output": "IGNORE HIM!"
},
{
"input": "ppcpbnhwoizajrl",
"output": "IGNORE HIM!"
},
{
"input": "sgubujztzwkzvztitssxxxwzanfmddfqvv",
"output": "CHAT WITH HER!"
},
{
"input": "ptkyaxycecpbrjnvxcjtbqiocqcswnmicxbvhdsptbxyxswbw",
"output": "IGNORE HIM!"
},
{
"input": "yhbtzfppwcycxqjpqdfmjnhwaogyuaxamwxpnrdrnqsgdyfvxu",
"output": "CHAT WITH HER!"
},
{
"input": "ojjvpnkrxibyevxk",
"output": "CHAT WITH HER!"
},
{
"input": "wjweqcrqfuollfvfbiyriijovweg",
"output": "IGNORE HIM!"
},
{
"input": "hkdbykboclchfdsuovvpknwqr",
"output": "IGNORE HIM!"
},
{
"input": "stjvyfrfowopwfjdveduedqylerqugykyu",
"output": "IGNORE HIM!"
},
{
"input": "rafcaanqytfclvfdegak",
"output": "CHAT WITH HER!"
},
{
"input": "xczn",
"output": "CHAT WITH HER!"
},
{
"input": "arcoaeozyeawbveoxpmafxxzdjldsielp",
"output": "IGNORE HIM!"
},
{
"input": "smdfafbyehdylhaleevhoggiurdgeleaxkeqdixyfztkuqsculgslheqfafxyghyuibdgiuwrdxfcitojxika",
"output": "CHAT WITH HER!"
},
{
"input": "vbpfgjqnhfazmvtkpjrdasfhsuxnpiepxfrzvoh",
"output": "CHAT WITH HER!"
},
{
"input": "dbdokywnpqnotfrhdbrzmuyoxfdtrgrzcccninbtmoqvxfatcqg",
"output": "CHAT WITH HER!"
},
{
"input": "udlpagtpq",
"output": "CHAT WITH HER!"
},
{
"input": "zjurevbytijifnpfuyswfchdzelxheboruwjqijxcucylysmwtiqsqqhktexcynquvcwhbjsipy",
"output": "CHAT WITH HER!"
},
{
"input": "qagzrqjomdwhagkhrjahhxkieijyten",
"output": "CHAT WITH HER!"
},
{
"input": "achhcfjnnfwgoufxamcqrsontgjjhgyfzuhklkmiwybnrlsvblnsrjqdytglipxsulpnphpjpoewvlusalsgovwnsngb",
"output": "CHAT WITH HER!"
},
{
"input": "qbkjsdwpahdbbohggbclfcufqelnojoehsxxkr",
"output": "CHAT WITH HER!"
},
{
"input": "cpvftiwgyvnlmbkadiafddpgfpvhqqvuehkypqjsoibpiudfvpkhzlfrykc",
"output": "IGNORE HIM!"
},
{
"input": "lnpdosnceumubvk",
"output": "IGNORE HIM!"
},
{
"input": "efrk",
"output": "CHAT WITH HER!"
},
{
"input": "temnownneghnrujforif",
"output": "IGNORE HIM!"
},
{
"input": "ottnneymszwbumgobazfjyxewkjakglbfflsajuzescplpcxqta",
"output": "IGNORE HIM!"
},
{
"input": "eswpaclodzcwhgixhpyzvhdwsgneqidanbzdzszquefh",
"output": "IGNORE HIM!"
},
{
"input": "gwntwbpj",
"output": "IGNORE HIM!"
},
{
"input": "wuqvlbblkddeindiiswsinkfrnkxghhwunzmmvyovpqapdfbolyim",
"output": "IGNORE HIM!"
},
{
"input": "swdqsnzmzmsyvktukaoyqsqzgfmbzhezbfaqeywgwizrwjyzquaahucjchegknqaioliqd",
"output": "CHAT WITH HER!"
},
{
"input": "vlhrpzezawyolhbmvxbwhtjustdbqggexmzxyieihjlelvwjosmkwesfjmramsikhkupzvfgezmrqzudjcalpjacmhykhgfhrjx",
"output": "IGNORE HIM!"
},
{
"input": "lxxwbkrjgnqjwsnflfnsdyxihmlspgivirazsbveztnkuzpaxtygidniflyjheejelnjyjvgkgvdqks",
"output": "CHAT WITH HER!"
},
{
"input": "wpxbxzfhtdecetpljcrvpjjnllosdqirnkzesiqeukbedkayqx",
"output": "CHAT WITH HER!"
},
{
"input": "vmzxgacicvweclaodrunmjnfwtimceetsaoickarqyrkdghcmyjgmtgsqastcktyrjgvjqimdc",
"output": "CHAT WITH HER!"
},
{
"input": "yzlzmesxdttfcztooypjztlgxwcr",
"output": "IGNORE HIM!"
},
{
"input": "qpbjwzwgdzmeluheirjrvzrhbmagfsjdgvzgwumjtjzecsfkrfqjasssrhhtgdqqfydlmrktlgfc",
"output": "IGNORE HIM!"
},
{
"input": "aqzftsvezdgouyrirsxpbuvdjupnzvbhguyayeqozfzymfnepvwgblqzvmxxkxcilmsjvcgyqykpoaktjvsxbygfgsalbjoq",
"output": "CHAT WITH HER!"
},
{
"input": "znicjjgijhrbdlnwmtjgtdgziollrfxroabfhadygnomodaembllreorlyhnehijfyjbfxucazellblegyfrzuraogadj",
"output": "IGNORE HIM!"
},
{
"input": "qordzrdiknsympdrkgapjxokbldorpnmnpucmwakklmqenpmkom",
"output": "CHAT WITH HER!"
},
{
"input": "wqfldgihuxfktzanyycluzhtewmwvnawqlfoavuguhygqrrxtstxwouuzzsryjqtfqo",
"output": "CHAT WITH HER!"
},
{
"input": "vujtrrpshinkskgyknlcfckmqdrwtklkzlyipmetjvaqxdsslkskschbalmdhzsdrrjmxdltbtnxbh",
"output": "IGNORE HIM!"
},
{
"input": "zioixjibuhrzyrbzqcdjbbhhdmpgmqykixcxoqupggaqajuzonrpzihbsogjfsrrypbiphehonyhohsbybnnukqebopppa",
"output": "CHAT WITH HER!"
},
{
"input": "oh",
"output": "CHAT WITH HER!"
},
{
"input": "kxqthadqesbpgpsvpbcbznxpecqrzjoilpauttzlnxvaczcqwuri",
"output": "IGNORE HIM!"
},
{
"input": "zwlunigqnhrwirkvufqwrnwcnkqqonebrwzcshcbqqwkjxhymjjeakuzjettebciadjlkbfp",
"output": "CHAT WITH HER!"
},
{
"input": "fjuldpuejgmggvvigkwdyzytfxzwdlofrpifqpdnhfyroginqaufwgjcbgshyyruwhofctsdaisqpjxqjmtpp",
"output": "CHAT WITH HER!"
},
{
"input": "xiwntnheuitbtqxrmzvxmieldudakogealwrpygbxsbluhsqhtwmdlpjwzyafckrqrdduonkgo",
"output": "CHAT WITH HER!"
},
{
"input": "mnmbupgo",
"output": "IGNORE HIM!"
},
{
"input": "mcjehdiygkbmrbfjqwpwxidbdfelifwhstaxdapigbymmsgrhnzsdjhsqchl",
"output": "IGNORE HIM!"
},
{
"input": "yocxrzspinchmhtmqo",
"output": "CHAT WITH HER!"
},
{
"input": "vasvvnpymtgjirnzuynluluvmgpquskuaafwogeztfnvybblajvuuvfomtifeuzpikjrolzeeoftv",
"output": "CHAT WITH HER!"
},
{
"input": "ecsdicrznvglwggrdbrvehwzaenzjutjydhvimtqegweurpxtjkmpcznshtrvotkvrghxhacjkedidqqzrduzad",
"output": "IGNORE HIM!"
},
{
"input": "ubvhyaebyxoghakajqrpqpctwbrfqzli",
"output": "CHAT WITH HER!"
},
{
"input": "gogbxfeqylxoummvgxpkoqzsmobasesxbqjjktqbwqxeiaagnnhbvepbpy",
"output": "IGNORE HIM!"
},
{
"input": "nheihhxkbbrmlpxpxbhnpofcjmxemyvqqdbanwd",
"output": "IGNORE HIM!"
},
{
"input": "acrzbavz",
"output": "CHAT WITH HER!"
},
{
"input": "drvzznznvrzskftnrhvvzxcalwutxmdza",
"output": "IGNORE HIM!"
},
{
"input": "oacwxipdfcoabhkwxqdbtowiekpnflnqhlrkustgzryvws",
"output": "CHAT WITH HER!"
},
{
"input": "tpnwfmfsibnccvdwjvzviyvjfljupinfigfunyff",
"output": "CHAT WITH HER!"
},
{
"input": "gavaihhamfolcndgytcsgucqdqngxkrlovpthvteacmmthoglxu",
"output": "CHAT WITH HER!"
},
{
"input": "hsfcfvameeupldgvchmogrvwxrvsmnwxxkxoawwodtsahqvehlcw",
"output": "IGNORE HIM!"
},
{
"input": "sbkydrscoojychxchqsuciperfroumenelgiyiwlqfwximrgdbyvkmacy",
"output": "CHAT WITH HER!"
},
{
"input": "rhh",
"output": "CHAT WITH HER!"
},
{
"input": "zhdouqfmlkenjzdijxdfxnlegxeuvhelo",
"output": "IGNORE HIM!"
},
{
"input": "yufkkfwyhhvcjntsgsvpzbhqtmtgyxifqoewmuplphykmptfdebjxuaxigomjtwgtljwdjhjernkitifbomifbhysnmadtnyn",
"output": "CHAT WITH HER!"
},
{
"input": "urigreuzpxnej",
"output": "CHAT WITH HER!"
}
] | 1,698,319,343 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | s=input()
x=[]
if 1<=len(s)<=100:
for i in s:
if i not in x:
x.append(i)
if len(x)%2==0:
print("CHAT WITH HER !")
else:
print("IGNORE HIM!") | Title: Boy or Girl
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
Input Specification:
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
Output Specification:
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
Demo Input:
['wjmzbmr\n', 'xiaodao\n', 'sevenkplus\n']
Demo Output:
['CHAT WITH HER!\n', 'IGNORE HIM!\n', 'CHAT WITH HER!\n']
Note:
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!". | ```python
s=input()
x=[]
if 1<=len(s)<=100:
for i in s:
if i not in x:
x.append(i)
if len(x)%2==0:
print("CHAT WITH HER !")
else:
print("IGNORE HIM!")
``` | 0 | |
727 | A | Transformation: from A to B | PROGRAMMING | 1,000 | [
"brute force",
"dfs and similar",
"math"
] | null | null | Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations:
- multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1).
You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible.
Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*. | The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have. | If there is no way to get *b* from *a*, print "NO" (without quotes).
Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where:
- *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*).
If there are multiple answers, print any of them. | [
"2 162\n",
"4 42\n",
"100 40021\n"
] | [
"YES\n5\n2 4 8 81 162 \n",
"NO\n",
"YES\n5\n100 200 2001 4002 40021 \n"
] | none | 1,000 | [
{
"input": "2 162",
"output": "YES\n5\n2 4 8 81 162 "
},
{
"input": "4 42",
"output": "NO"
},
{
"input": "100 40021",
"output": "YES\n5\n100 200 2001 4002 40021 "
},
{
"input": "1 111111111",
"output": "YES\n9\n1 11 111 1111 11111 111111 1111111 11111111 111111111 "
},
{
"input": "1 1000000000",
"output": "NO"
},
{
"input": "999999999 1000000000",
"output": "NO"
},
{
"input": "1 2",
"output": "YES\n2\n1 2 "
},
{
"input": "1 536870912",
"output": "YES\n30\n1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 "
},
{
"input": "11111 11111111",
"output": "YES\n4\n11111 111111 1111111 11111111 "
},
{
"input": "59139 946224",
"output": "YES\n5\n59139 118278 236556 473112 946224 "
},
{
"input": "9859 19718",
"output": "YES\n2\n9859 19718 "
},
{
"input": "25987 51974222",
"output": "YES\n5\n25987 259871 2598711 25987111 51974222 "
},
{
"input": "9411 188222222",
"output": "YES\n6\n9411 94111 941111 9411111 94111111 188222222 "
},
{
"input": "25539 510782222",
"output": "YES\n6\n25539 255391 2553911 25539111 255391111 510782222 "
},
{
"input": "76259 610072",
"output": "YES\n4\n76259 152518 305036 610072 "
},
{
"input": "92387 184774",
"output": "YES\n2\n92387 184774 "
},
{
"input": "8515 85151111",
"output": "YES\n5\n8515 85151 851511 8515111 85151111 "
},
{
"input": "91939 9193911",
"output": "YES\n3\n91939 919391 9193911 "
},
{
"input": "30518 610361",
"output": "YES\n3\n30518 61036 610361 "
},
{
"input": "46646 373168844",
"output": "YES\n7\n46646 466461 932922 9329221 93292211 186584422 373168844 "
},
{
"input": "30070 300701",
"output": "YES\n2\n30070 300701 "
},
{
"input": "13494 1079528",
"output": "YES\n5\n13494 134941 269882 539764 1079528 "
},
{
"input": "96918 775344422",
"output": "YES\n7\n96918 193836 1938361 3876722 38767221 387672211 775344422 "
},
{
"input": "13046 260921",
"output": "YES\n3\n13046 26092 260921 "
},
{
"input": "29174 5834811",
"output": "YES\n4\n29174 58348 583481 5834811 "
},
{
"input": "79894 319576421",
"output": "YES\n6\n79894 798941 1597882 15978821 31957642 319576421 "
},
{
"input": "96022 1920442",
"output": "YES\n3\n96022 960221 1920442 "
},
{
"input": "79446 6355681",
"output": "YES\n5\n79446 158892 317784 635568 6355681 "
},
{
"input": "5440 27853056",
"output": "YES\n11\n5440 10880 108801 217602 435204 870408 1740816 3481632 6963264 13926528 27853056 "
},
{
"input": "250000000 705032705",
"output": "NO"
},
{
"input": "17 35",
"output": "NO"
},
{
"input": "1 3",
"output": "NO"
},
{
"input": "2 11",
"output": "NO"
}
] | 1,664,812,225 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | import copy
ab = input()
ab = list(map(int, ab.split()))
a,b = ab[0],ab[1]
vector = [a]
answer = []
def a_to_b(a,b,vector,answer):
if a>b:
#print('No')
return False
elif a==b:
answer = copy.deepcopy(vector)
print('Yes')
for i in answer:
print(i, end=' ')
#print(answer)
#print('pahuch gye!')
return True
vector.append(2*a)
p1 = a_to_b(2*a,b,vector,answer)
vector.remove(vector[-1])
vector.append(10*a + 1)
p2 = a_to_b(10*a + 1,b,vector,answer)
vector.remove(vector[-1])
return p1 or p2
p = a_to_b(a,b,vector,answer)#==True:
#print()
if p==False:
print('No') | Title: Transformation: from A to B
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations:
- multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1).
You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible.
Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*.
Input Specification:
The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have.
Output Specification:
If there is no way to get *b* from *a*, print "NO" (without quotes).
Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where:
- *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*).
If there are multiple answers, print any of them.
Demo Input:
['2 162\n', '4 42\n', '100 40021\n']
Demo Output:
['YES\n5\n2 4 8 81 162 \n', 'NO\n', 'YES\n5\n100 200 2001 4002 40021 \n']
Note:
none | ```python
import copy
ab = input()
ab = list(map(int, ab.split()))
a,b = ab[0],ab[1]
vector = [a]
answer = []
def a_to_b(a,b,vector,answer):
if a>b:
#print('No')
return False
elif a==b:
answer = copy.deepcopy(vector)
print('Yes')
for i in answer:
print(i, end=' ')
#print(answer)
#print('pahuch gye!')
return True
vector.append(2*a)
p1 = a_to_b(2*a,b,vector,answer)
vector.remove(vector[-1])
vector.append(10*a + 1)
p2 = a_to_b(10*a + 1,b,vector,answer)
vector.remove(vector[-1])
return p1 or p2
p = a_to_b(a,b,vector,answer)#==True:
#print()
if p==False:
print('No')
``` | 0 | |
409 | H | A + B Strikes Back | PROGRAMMING | 1,500 | [
"*special",
"brute force",
"constructive algorithms",
"dsu",
"implementation"
] | null | null | A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientists have observed that sometimes this problem is not so easy to get accepted. Want to try? | The input contains two integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=103), separated by a single space. | Output the sum of the given integers. | [
"5 14\n",
"381 492\n"
] | [
"19\n",
"873\n"
] | none | 0 | [
{
"input": "5 14",
"output": "19"
},
{
"input": "381 492",
"output": "873"
},
{
"input": "536 298",
"output": "834"
},
{
"input": "143 522",
"output": "665"
},
{
"input": "433 126",
"output": "559"
},
{
"input": "723 350",
"output": "1073"
},
{
"input": "632 264",
"output": "896"
},
{
"input": "923 488",
"output": "1411"
},
{
"input": "522 92",
"output": "614"
},
{
"input": "953 553",
"output": "1506"
},
{
"input": "242 86",
"output": "328"
},
{
"input": "151 691",
"output": "842"
},
{
"input": "441 915",
"output": "1356"
},
{
"input": "740 519",
"output": "1259"
},
{
"input": "339 433",
"output": "772"
},
{
"input": "629 657",
"output": "1286"
},
{
"input": "538 261",
"output": "799"
},
{
"input": "828 485",
"output": "1313"
},
{
"input": "117 399",
"output": "516"
},
{
"input": "399 280",
"output": "679"
}
] | 1,598,577,578 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 0 | 0 | aa,b=map(int,input().split(" "))
print(sum([aa,b])) | Title: A + B Strikes Back
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientists have observed that sometimes this problem is not so easy to get accepted. Want to try?
Input Specification:
The input contains two integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=103), separated by a single space.
Output Specification:
Output the sum of the given integers.
Demo Input:
['5 14\n', '381 492\n']
Demo Output:
['19\n', '873\n']
Note:
none | ```python
aa,b=map(int,input().split(" "))
print(sum([aa,b]))
``` | 0 | |
405 | A | Gravity Flip | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch! | The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column. | Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch. | [
"4\n3 2 1 2\n",
"3\n2 3 8\n"
] | [
"1 2 2 3 \n",
"2 3 8 \n"
] | The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | 500 | [
{
"input": "4\n3 2 1 2",
"output": "1 2 2 3 "
},
{
"input": "3\n2 3 8",
"output": "2 3 8 "
},
{
"input": "5\n2 1 2 1 2",
"output": "1 1 2 2 2 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n4 3",
"output": "3 4 "
},
{
"input": "6\n100 40 60 20 1 80",
"output": "1 20 40 60 80 100 "
},
{
"input": "10\n10 8 6 7 5 3 4 2 9 1",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "100\n82 51 81 14 37 17 78 92 64 15 8 86 89 8 87 77 66 10 15 12 100 25 92 47 21 78 20 63 13 49 41 36 41 79 16 87 87 69 3 76 80 60 100 49 70 59 72 8 38 71 45 97 71 14 76 54 81 4 59 46 39 29 92 3 49 22 53 99 59 52 74 31 92 43 42 23 44 9 82 47 7 40 12 9 3 55 37 85 46 22 84 52 98 41 21 77 63 17 62 91",
"output": "3 3 3 4 7 8 8 8 9 9 10 12 12 13 14 14 15 15 16 17 17 20 21 21 22 22 23 25 29 31 36 37 37 38 39 40 41 41 41 42 43 44 45 46 46 47 47 49 49 49 51 52 52 53 54 55 59 59 59 60 62 63 63 64 66 69 70 71 71 72 74 76 76 77 77 78 78 79 80 81 81 82 82 84 85 86 87 87 87 89 91 92 92 92 92 97 98 99 100 100 "
},
{
"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": "100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 "
},
{
"input": "10\n1 9 7 6 2 4 7 8 1 3",
"output": "1 1 2 3 4 6 7 7 8 9 "
},
{
"input": "20\n53 32 64 20 41 97 50 20 66 68 22 60 74 61 97 54 80 30 72 59",
"output": "20 20 22 30 32 41 50 53 54 59 60 61 64 66 68 72 74 80 97 97 "
},
{
"input": "30\n7 17 4 18 16 12 14 10 1 13 2 16 13 17 8 16 13 14 9 17 17 5 13 5 1 7 6 20 18 12",
"output": "1 1 2 4 5 5 6 7 7 8 9 10 12 12 13 13 13 13 14 14 16 16 16 17 17 17 17 18 18 20 "
},
{
"input": "40\n22 58 68 58 48 53 52 1 16 78 75 17 63 15 36 32 78 75 49 14 42 46 66 54 49 82 40 43 46 55 12 73 5 45 61 60 1 11 31 84",
"output": "1 1 5 11 12 14 15 16 17 22 31 32 36 40 42 43 45 46 46 48 49 49 52 53 54 55 58 58 60 61 63 66 68 73 75 75 78 78 82 84 "
},
{
"input": "70\n1 3 3 1 3 3 1 1 1 3 3 2 3 3 1 1 1 2 3 1 3 2 3 3 3 2 2 3 1 3 3 2 1 1 2 1 2 1 2 2 1 1 1 3 3 2 3 2 3 2 3 3 2 2 2 3 2 3 3 3 1 1 3 3 1 1 1 1 3 1",
"output": "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 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "90\n17 75 51 30 100 5 50 95 51 73 66 5 7 76 43 49 23 55 3 24 95 79 10 11 44 93 17 99 53 66 82 66 63 76 19 4 51 71 75 43 27 5 24 19 48 7 91 15 55 21 7 6 27 10 2 91 64 58 18 21 16 71 90 88 21 20 6 6 95 85 11 7 40 65 52 49 92 98 46 88 17 48 85 96 77 46 100 34 67 52",
"output": "2 3 4 5 5 5 6 6 6 7 7 7 7 10 10 11 11 15 16 17 17 17 18 19 19 20 21 21 21 23 24 24 27 27 30 34 40 43 43 44 46 46 48 48 49 49 50 51 51 51 52 52 53 55 55 58 63 64 65 66 66 66 67 71 71 73 75 75 76 76 77 79 82 85 85 88 88 90 91 91 92 93 95 95 95 96 98 99 100 100 "
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "100\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100\n2 1 1 1 3 2 3 3 2 3 3 1 3 3 1 3 3 1 1 1 2 3 1 2 3 1 2 3 3 1 3 1 1 2 3 2 3 3 2 3 3 1 2 2 1 2 3 2 3 2 2 1 1 3 1 3 2 1 3 1 3 1 3 1 1 3 3 3 2 3 2 2 2 2 1 3 3 3 1 2 1 2 3 2 1 3 1 3 2 1 3 1 2 1 2 3 1 3 2 3",
"output": "1 1 1 1 1 1 1 1 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 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 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "100\n7 4 5 5 10 10 5 8 5 7 4 5 4 6 8 8 2 6 3 3 10 7 10 8 6 2 7 3 9 7 7 2 4 5 2 4 9 5 10 1 10 5 10 4 1 3 4 2 6 9 9 9 10 6 2 5 6 1 8 10 4 10 3 4 10 5 5 4 10 4 5 3 7 10 2 7 3 6 9 6 1 6 5 5 4 6 6 4 4 1 5 1 6 6 6 8 8 6 2 6",
"output": "1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 10 10 10 "
},
{
"input": "100\n12 10 5 11 13 12 14 13 7 15 15 12 13 19 12 18 14 10 10 3 1 10 16 11 19 8 10 15 5 10 12 16 11 13 11 15 14 12 16 8 11 8 15 2 18 2 14 13 15 20 8 8 4 12 14 7 10 3 9 1 7 19 6 7 2 14 8 20 7 17 18 20 3 18 18 9 6 10 4 1 4 19 9 13 3 3 12 11 11 20 8 2 13 6 7 12 1 4 17 3",
"output": "1 1 1 1 2 2 2 2 3 3 3 3 3 3 4 4 4 4 5 5 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 8 9 9 9 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 14 14 14 14 14 14 15 15 15 15 15 15 16 16 16 17 17 18 18 18 18 18 19 19 19 19 20 20 20 20 "
},
{
"input": "100\n5 13 1 40 30 10 23 32 33 12 6 4 15 29 31 17 23 5 36 31 32 38 24 11 34 39 19 21 6 19 31 35 1 15 6 29 22 15 17 15 1 17 2 34 20 8 27 2 29 26 13 9 22 27 27 3 20 40 4 40 33 29 36 30 35 16 19 28 26 11 36 24 29 5 40 10 38 34 33 23 34 39 31 7 10 31 22 6 36 24 14 31 34 23 2 4 26 16 2 32",
"output": "1 1 1 2 2 2 2 3 4 4 4 5 5 5 6 6 6 6 7 8 9 10 10 10 11 11 12 13 13 14 15 15 15 15 16 16 17 17 17 19 19 19 20 20 21 22 22 22 23 23 23 23 24 24 24 26 26 26 27 27 27 28 29 29 29 29 29 30 30 31 31 31 31 31 31 32 32 32 33 33 33 34 34 34 34 34 35 35 36 36 36 36 38 38 39 39 40 40 40 40 "
},
{
"input": "100\n72 44 34 74 9 60 26 37 55 77 74 69 28 66 54 55 8 36 57 31 31 48 32 66 40 70 77 43 64 28 37 10 21 58 51 32 60 28 51 52 28 35 7 33 1 68 38 70 57 71 8 20 42 57 59 4 58 10 17 47 22 48 16 3 76 67 32 37 64 47 33 41 75 69 2 76 39 9 27 75 20 21 52 25 71 21 11 29 38 10 3 1 45 55 63 36 27 7 59 41",
"output": "1 1 2 3 3 4 7 7 8 8 9 9 10 10 10 11 16 17 20 20 21 21 21 22 25 26 27 27 28 28 28 28 29 31 31 32 32 32 33 33 34 35 36 36 37 37 37 38 38 39 40 41 41 42 43 44 45 47 47 48 48 51 51 52 52 54 55 55 55 57 57 57 58 58 59 59 60 60 63 64 64 66 66 67 68 69 69 70 70 71 71 72 74 74 75 75 76 76 77 77 "
},
{
"input": "100\n75 18 61 10 56 53 42 57 79 80 31 2 50 45 54 99 84 52 71 21 86 3 19 98 14 37 40 62 63 68 5 10 87 8 81 85 52 52 57 94 2 7 56 96 19 76 1 13 81 6 80 47 22 59 99 32 9 5 36 88 98 91 70 70 12 93 12 22 85 1 97 48 94 16 84 84 51 34 62 7 68 51 30 2 37 82 4 7 27 1 80 9 61 16 59 55 12 96 94 82",
"output": "1 1 1 2 2 2 3 4 5 5 6 7 7 7 8 9 9 10 10 12 12 12 13 14 16 16 18 19 19 21 22 22 27 30 31 32 34 36 37 37 40 42 45 47 48 50 51 51 52 52 52 53 54 55 56 56 57 57 59 59 61 61 62 62 63 68 68 70 70 71 75 76 79 80 80 80 81 81 82 82 84 84 84 85 85 86 87 88 91 93 94 94 94 96 96 97 98 98 99 99 "
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "100\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50",
"output": "50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 "
},
{
"input": "49\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97",
"output": "1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 "
},
{
"input": "30\n1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79 82 85 88",
"output": "1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79 82 85 88 "
},
{
"input": "100\n100 51 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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 51 100 "
},
{
"input": "10\n100 90 80 70 60 50 40 30 20 10",
"output": "10 20 30 40 50 60 70 80 90 100 "
},
{
"input": "1\n10",
"output": "10 "
}
] | 1,697,804,990 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 46 | 0 | n = int(input())
a = list(map(int, input().split()))
a = sorted(a)
final = ""
for i in a:
final = final + str(i) + " "
print(final[:-1]) | Title: Gravity Flip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch!
Input Specification:
The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column.
Output Specification:
Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch.
Demo Input:
['4\n3 2 1 2\n', '3\n2 3 8\n']
Demo Output:
['1 2 2 3 \n', '2 3 8 \n']
Note:
The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | ```python
n = int(input())
a = list(map(int, input().split()))
a = sorted(a)
final = ""
for i in a:
final = final + str(i) + " "
print(final[:-1])
``` | 3 | |
962 | A | Equator | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator. | The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day. | Print the index of the day when Polycarp will celebrate the equator. | [
"4\n1 3 2 1\n",
"6\n2 2 2 2 2 2\n"
] | [
"2\n",
"3\n"
] | In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training. | 0 | [
{
"input": "4\n1 3 2 1",
"output": "2"
},
{
"input": "6\n2 2 2 2 2 2",
"output": "3"
},
{
"input": "1\n10000",
"output": "1"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "2\n1 3",
"output": "2"
},
{
"input": "4\n2 1 1 3",
"output": "3"
},
{
"input": "3\n1 1 3",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "2"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "5\n1 2 4 3 5",
"output": "4"
},
{
"input": "5\n2 2 2 4 3",
"output": "4"
},
{
"input": "4\n1 2 3 1",
"output": "3"
},
{
"input": "6\n7 3 10 7 3 11",
"output": "4"
},
{
"input": "2\n3 4",
"output": "2"
},
{
"input": "5\n1 1 1 1 1",
"output": "3"
},
{
"input": "4\n1 3 2 3",
"output": "3"
},
{
"input": "2\n2 3",
"output": "2"
},
{
"input": "3\n32 10 23",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "4"
},
{
"input": "3\n1 2 4",
"output": "3"
},
{
"input": "6\n3 3 3 2 4 4",
"output": "4"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "5"
},
{
"input": "5\n1 3 3 1 1",
"output": "3"
},
{
"input": "4\n1 1 1 2",
"output": "3"
},
{
"input": "4\n1 2 1 3",
"output": "3"
},
{
"input": "3\n2 2 1",
"output": "2"
},
{
"input": "4\n2 3 3 3",
"output": "3"
},
{
"input": "4\n3 2 3 3",
"output": "3"
},
{
"input": "4\n2 1 1 1",
"output": "2"
},
{
"input": "3\n2 1 4",
"output": "3"
},
{
"input": "2\n6 7",
"output": "2"
},
{
"input": "4\n3 3 4 3",
"output": "3"
},
{
"input": "4\n1 1 2 5",
"output": "4"
},
{
"input": "4\n1 8 7 3",
"output": "3"
},
{
"input": "6\n2 2 2 2 2 3",
"output": "4"
},
{
"input": "3\n2 2 5",
"output": "3"
},
{
"input": "4\n1 1 2 1",
"output": "3"
},
{
"input": "5\n1 1 2 2 3",
"output": "4"
},
{
"input": "5\n9 5 3 4 8",
"output": "3"
},
{
"input": "3\n3 3 1",
"output": "2"
},
{
"input": "4\n1 2 2 2",
"output": "3"
},
{
"input": "3\n1 3 5",
"output": "3"
},
{
"input": "4\n1 1 3 6",
"output": "4"
},
{
"input": "6\n1 2 1 1 1 1",
"output": "3"
},
{
"input": "3\n3 1 3",
"output": "2"
},
{
"input": "5\n3 4 5 1 2",
"output": "3"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "5\n3 1 2 5 2",
"output": "4"
},
{
"input": "4\n1 1 1 4",
"output": "4"
},
{
"input": "4\n2 6 1 10",
"output": "4"
},
{
"input": "4\n2 2 3 2",
"output": "3"
},
{
"input": "4\n4 2 2 1",
"output": "2"
},
{
"input": "6\n1 1 1 1 1 4",
"output": "5"
},
{
"input": "3\n3 2 2",
"output": "2"
},
{
"input": "6\n1 3 5 1 7 4",
"output": "5"
},
{
"input": "5\n1 2 4 8 16",
"output": "5"
},
{
"input": "5\n1 2 4 4 4",
"output": "4"
},
{
"input": "6\n4 2 1 2 3 1",
"output": "3"
},
{
"input": "4\n3 2 1 5",
"output": "3"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "3\n2 4 7",
"output": "3"
},
{
"input": "5\n1 1 1 1 3",
"output": "4"
},
{
"input": "3\n3 1 5",
"output": "3"
},
{
"input": "4\n1 2 3 7",
"output": "4"
},
{
"input": "3\n1 4 6",
"output": "3"
},
{
"input": "4\n2 1 2 2",
"output": "3"
},
{
"input": "2\n4 5",
"output": "2"
},
{
"input": "5\n1 2 1 2 1",
"output": "3"
},
{
"input": "3\n2 3 6",
"output": "3"
},
{
"input": "6\n1 1 4 1 1 5",
"output": "4"
},
{
"input": "5\n2 2 2 2 1",
"output": "3"
},
{
"input": "2\n5 6",
"output": "2"
},
{
"input": "4\n2 2 1 4",
"output": "3"
},
{
"input": "5\n2 2 3 4 4",
"output": "4"
},
{
"input": "4\n3 1 1 2",
"output": "2"
},
{
"input": "5\n3 4 1 4 5",
"output": "4"
},
{
"input": "4\n1 3 1 6",
"output": "4"
},
{
"input": "5\n1 1 1 2 2",
"output": "4"
},
{
"input": "4\n1 4 2 4",
"output": "3"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 8",
"output": "9"
},
{
"input": "4\n1 4 5 1",
"output": "3"
},
{
"input": "5\n1 1 1 1 5",
"output": "5"
},
{
"input": "4\n1 3 4 1",
"output": "3"
},
{
"input": "4\n2 2 2 3",
"output": "3"
},
{
"input": "4\n2 3 2 4",
"output": "3"
},
{
"input": "5\n2 2 1 2 2",
"output": "3"
},
{
"input": "3\n4 3 2",
"output": "2"
},
{
"input": "3\n6 5 2",
"output": "2"
},
{
"input": "69\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",
"output": "35"
},
{
"input": "6\n1 1 1 1 1 2",
"output": "4"
},
{
"input": "5\n1 2 5 4 5",
"output": "4"
},
{
"input": "2\n9 10",
"output": "2"
},
{
"input": "3\n1 1 5",
"output": "3"
},
{
"input": "4\n3 4 3 5",
"output": "3"
},
{
"input": "4\n1 4 3 3",
"output": "3"
},
{
"input": "4\n7 1 3 4",
"output": "2"
},
{
"input": "3\n100 100 1",
"output": "2"
},
{
"input": "4\n5 2 2 2",
"output": "2"
}
] | 1,666,161,527 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 108 | 13,209,600 | n=int(input())
x=list(map(int,input().split()))
z=sum(x)//2
y=0
for i in range(n):
y+=x[i]
if y>=z+1 and z%2!=0:
print(i+1)
break
elif y>=z:
print(i+1)
break | Title: Equator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day.
Output Specification:
Print the index of the day when Polycarp will celebrate the equator.
Demo Input:
['4\n1 3 2 1\n', '6\n2 2 2 2 2 2\n']
Demo Output:
['2\n', '3\n']
Note:
In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training. | ```python
n=int(input())
x=list(map(int,input().split()))
z=sum(x)//2
y=0
for i in range(n):
y+=x[i]
if y>=z+1 and z%2!=0:
print(i+1)
break
elif y>=z:
print(i+1)
break
``` | 0 | |
334 | A | Candy Bags | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies. | The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers. | Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits. | [
"2\n"
] | [
"1 4\n2 3\n"
] | The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother. | 500 | [
{
"input": "2",
"output": "1 4\n2 3"
},
{
"input": "4",
"output": "1 16 2 15\n3 14 4 13\n5 12 6 11\n7 10 8 9"
},
{
"input": "6",
"output": "1 36 2 35 3 34\n4 33 5 32 6 31\n7 30 8 29 9 28\n10 27 11 26 12 25\n13 24 14 23 15 22\n16 21 17 20 18 19"
},
{
"input": "8",
"output": "1 64 2 63 3 62 4 61\n5 60 6 59 7 58 8 57\n9 56 10 55 11 54 12 53\n13 52 14 51 15 50 16 49\n17 48 18 47 19 46 20 45\n21 44 22 43 23 42 24 41\n25 40 26 39 27 38 28 37\n29 36 30 35 31 34 32 33"
},
{
"input": "10",
"output": "1 100 2 99 3 98 4 97 5 96\n6 95 7 94 8 93 9 92 10 91\n11 90 12 89 13 88 14 87 15 86\n16 85 17 84 18 83 19 82 20 81\n21 80 22 79 23 78 24 77 25 76\n26 75 27 74 28 73 29 72 30 71\n31 70 32 69 33 68 34 67 35 66\n36 65 37 64 38 63 39 62 40 61\n41 60 42 59 43 58 44 57 45 56\n46 55 47 54 48 53 49 52 50 51"
},
{
"input": "100",
"output": "1 10000 2 9999 3 9998 4 9997 5 9996 6 9995 7 9994 8 9993 9 9992 10 9991 11 9990 12 9989 13 9988 14 9987 15 9986 16 9985 17 9984 18 9983 19 9982 20 9981 21 9980 22 9979 23 9978 24 9977 25 9976 26 9975 27 9974 28 9973 29 9972 30 9971 31 9970 32 9969 33 9968 34 9967 35 9966 36 9965 37 9964 38 9963 39 9962 40 9961 41 9960 42 9959 43 9958 44 9957 45 9956 46 9955 47 9954 48 9953 49 9952 50 9951\n51 9950 52 9949 53 9948 54 9947 55 9946 56 9945 57 9944 58 9943 59 9942 60 9941 61 9940 62 9939 63 9938 64 9937 65 993..."
},
{
"input": "62",
"output": "1 3844 2 3843 3 3842 4 3841 5 3840 6 3839 7 3838 8 3837 9 3836 10 3835 11 3834 12 3833 13 3832 14 3831 15 3830 16 3829 17 3828 18 3827 19 3826 20 3825 21 3824 22 3823 23 3822 24 3821 25 3820 26 3819 27 3818 28 3817 29 3816 30 3815 31 3814\n32 3813 33 3812 34 3811 35 3810 36 3809 37 3808 38 3807 39 3806 40 3805 41 3804 42 3803 43 3802 44 3801 45 3800 46 3799 47 3798 48 3797 49 3796 50 3795 51 3794 52 3793 53 3792 54 3791 55 3790 56 3789 57 3788 58 3787 59 3786 60 3785 61 3784 62 3783\n63 3782 64 3781 65 378..."
},
{
"input": "66",
"output": "1 4356 2 4355 3 4354 4 4353 5 4352 6 4351 7 4350 8 4349 9 4348 10 4347 11 4346 12 4345 13 4344 14 4343 15 4342 16 4341 17 4340 18 4339 19 4338 20 4337 21 4336 22 4335 23 4334 24 4333 25 4332 26 4331 27 4330 28 4329 29 4328 30 4327 31 4326 32 4325 33 4324\n34 4323 35 4322 36 4321 37 4320 38 4319 39 4318 40 4317 41 4316 42 4315 43 4314 44 4313 45 4312 46 4311 47 4310 48 4309 49 4308 50 4307 51 4306 52 4305 53 4304 54 4303 55 4302 56 4301 57 4300 58 4299 59 4298 60 4297 61 4296 62 4295 63 4294 64 4293 65 4292..."
},
{
"input": "18",
"output": "1 324 2 323 3 322 4 321 5 320 6 319 7 318 8 317 9 316\n10 315 11 314 12 313 13 312 14 311 15 310 16 309 17 308 18 307\n19 306 20 305 21 304 22 303 23 302 24 301 25 300 26 299 27 298\n28 297 29 296 30 295 31 294 32 293 33 292 34 291 35 290 36 289\n37 288 38 287 39 286 40 285 41 284 42 283 43 282 44 281 45 280\n46 279 47 278 48 277 49 276 50 275 51 274 52 273 53 272 54 271\n55 270 56 269 57 268 58 267 59 266 60 265 61 264 62 263 63 262\n64 261 65 260 66 259 67 258 68 257 69 256 70 255 71 254 72 253\n73 252 7..."
},
{
"input": "68",
"output": "1 4624 2 4623 3 4622 4 4621 5 4620 6 4619 7 4618 8 4617 9 4616 10 4615 11 4614 12 4613 13 4612 14 4611 15 4610 16 4609 17 4608 18 4607 19 4606 20 4605 21 4604 22 4603 23 4602 24 4601 25 4600 26 4599 27 4598 28 4597 29 4596 30 4595 31 4594 32 4593 33 4592 34 4591\n35 4590 36 4589 37 4588 38 4587 39 4586 40 4585 41 4584 42 4583 43 4582 44 4581 45 4580 46 4579 47 4578 48 4577 49 4576 50 4575 51 4574 52 4573 53 4572 54 4571 55 4570 56 4569 57 4568 58 4567 59 4566 60 4565 61 4564 62 4563 63 4562 64 4561 65 4560..."
},
{
"input": "86",
"output": "1 7396 2 7395 3 7394 4 7393 5 7392 6 7391 7 7390 8 7389 9 7388 10 7387 11 7386 12 7385 13 7384 14 7383 15 7382 16 7381 17 7380 18 7379 19 7378 20 7377 21 7376 22 7375 23 7374 24 7373 25 7372 26 7371 27 7370 28 7369 29 7368 30 7367 31 7366 32 7365 33 7364 34 7363 35 7362 36 7361 37 7360 38 7359 39 7358 40 7357 41 7356 42 7355 43 7354\n44 7353 45 7352 46 7351 47 7350 48 7349 49 7348 50 7347 51 7346 52 7345 53 7344 54 7343 55 7342 56 7341 57 7340 58 7339 59 7338 60 7337 61 7336 62 7335 63 7334 64 7333 65 7332..."
},
{
"input": "96",
"output": "1 9216 2 9215 3 9214 4 9213 5 9212 6 9211 7 9210 8 9209 9 9208 10 9207 11 9206 12 9205 13 9204 14 9203 15 9202 16 9201 17 9200 18 9199 19 9198 20 9197 21 9196 22 9195 23 9194 24 9193 25 9192 26 9191 27 9190 28 9189 29 9188 30 9187 31 9186 32 9185 33 9184 34 9183 35 9182 36 9181 37 9180 38 9179 39 9178 40 9177 41 9176 42 9175 43 9174 44 9173 45 9172 46 9171 47 9170 48 9169\n49 9168 50 9167 51 9166 52 9165 53 9164 54 9163 55 9162 56 9161 57 9160 58 9159 59 9158 60 9157 61 9156 62 9155 63 9154 64 9153 65 9152..."
},
{
"input": "12",
"output": "1 144 2 143 3 142 4 141 5 140 6 139\n7 138 8 137 9 136 10 135 11 134 12 133\n13 132 14 131 15 130 16 129 17 128 18 127\n19 126 20 125 21 124 22 123 23 122 24 121\n25 120 26 119 27 118 28 117 29 116 30 115\n31 114 32 113 33 112 34 111 35 110 36 109\n37 108 38 107 39 106 40 105 41 104 42 103\n43 102 44 101 45 100 46 99 47 98 48 97\n49 96 50 95 51 94 52 93 53 92 54 91\n55 90 56 89 57 88 58 87 59 86 60 85\n61 84 62 83 63 82 64 81 65 80 66 79\n67 78 68 77 69 76 70 75 71 74 72 73"
},
{
"input": "88",
"output": "1 7744 2 7743 3 7742 4 7741 5 7740 6 7739 7 7738 8 7737 9 7736 10 7735 11 7734 12 7733 13 7732 14 7731 15 7730 16 7729 17 7728 18 7727 19 7726 20 7725 21 7724 22 7723 23 7722 24 7721 25 7720 26 7719 27 7718 28 7717 29 7716 30 7715 31 7714 32 7713 33 7712 34 7711 35 7710 36 7709 37 7708 38 7707 39 7706 40 7705 41 7704 42 7703 43 7702 44 7701\n45 7700 46 7699 47 7698 48 7697 49 7696 50 7695 51 7694 52 7693 53 7692 54 7691 55 7690 56 7689 57 7688 58 7687 59 7686 60 7685 61 7684 62 7683 63 7682 64 7681 65 7680..."
},
{
"input": "28",
"output": "1 784 2 783 3 782 4 781 5 780 6 779 7 778 8 777 9 776 10 775 11 774 12 773 13 772 14 771\n15 770 16 769 17 768 18 767 19 766 20 765 21 764 22 763 23 762 24 761 25 760 26 759 27 758 28 757\n29 756 30 755 31 754 32 753 33 752 34 751 35 750 36 749 37 748 38 747 39 746 40 745 41 744 42 743\n43 742 44 741 45 740 46 739 47 738 48 737 49 736 50 735 51 734 52 733 53 732 54 731 55 730 56 729\n57 728 58 727 59 726 60 725 61 724 62 723 63 722 64 721 65 720 66 719 67 718 68 717 69 716 70 715\n71 714 72 713 73 712 74 7..."
},
{
"input": "80",
"output": "1 6400 2 6399 3 6398 4 6397 5 6396 6 6395 7 6394 8 6393 9 6392 10 6391 11 6390 12 6389 13 6388 14 6387 15 6386 16 6385 17 6384 18 6383 19 6382 20 6381 21 6380 22 6379 23 6378 24 6377 25 6376 26 6375 27 6374 28 6373 29 6372 30 6371 31 6370 32 6369 33 6368 34 6367 35 6366 36 6365 37 6364 38 6363 39 6362 40 6361\n41 6360 42 6359 43 6358 44 6357 45 6356 46 6355 47 6354 48 6353 49 6352 50 6351 51 6350 52 6349 53 6348 54 6347 55 6346 56 6345 57 6344 58 6343 59 6342 60 6341 61 6340 62 6339 63 6338 64 6337 65 6336..."
},
{
"input": "48",
"output": "1 2304 2 2303 3 2302 4 2301 5 2300 6 2299 7 2298 8 2297 9 2296 10 2295 11 2294 12 2293 13 2292 14 2291 15 2290 16 2289 17 2288 18 2287 19 2286 20 2285 21 2284 22 2283 23 2282 24 2281\n25 2280 26 2279 27 2278 28 2277 29 2276 30 2275 31 2274 32 2273 33 2272 34 2271 35 2270 36 2269 37 2268 38 2267 39 2266 40 2265 41 2264 42 2263 43 2262 44 2261 45 2260 46 2259 47 2258 48 2257\n49 2256 50 2255 51 2254 52 2253 53 2252 54 2251 55 2250 56 2249 57 2248 58 2247 59 2246 60 2245 61 2244 62 2243 63 2242 64 2241 65 224..."
},
{
"input": "54",
"output": "1 2916 2 2915 3 2914 4 2913 5 2912 6 2911 7 2910 8 2909 9 2908 10 2907 11 2906 12 2905 13 2904 14 2903 15 2902 16 2901 17 2900 18 2899 19 2898 20 2897 21 2896 22 2895 23 2894 24 2893 25 2892 26 2891 27 2890\n28 2889 29 2888 30 2887 31 2886 32 2885 33 2884 34 2883 35 2882 36 2881 37 2880 38 2879 39 2878 40 2877 41 2876 42 2875 43 2874 44 2873 45 2872 46 2871 47 2870 48 2869 49 2868 50 2867 51 2866 52 2865 53 2864 54 2863\n55 2862 56 2861 57 2860 58 2859 59 2858 60 2857 61 2856 62 2855 63 2854 64 2853 65 285..."
},
{
"input": "58",
"output": "1 3364 2 3363 3 3362 4 3361 5 3360 6 3359 7 3358 8 3357 9 3356 10 3355 11 3354 12 3353 13 3352 14 3351 15 3350 16 3349 17 3348 18 3347 19 3346 20 3345 21 3344 22 3343 23 3342 24 3341 25 3340 26 3339 27 3338 28 3337 29 3336\n30 3335 31 3334 32 3333 33 3332 34 3331 35 3330 36 3329 37 3328 38 3327 39 3326 40 3325 41 3324 42 3323 43 3322 44 3321 45 3320 46 3319 47 3318 48 3317 49 3316 50 3315 51 3314 52 3313 53 3312 54 3311 55 3310 56 3309 57 3308 58 3307\n59 3306 60 3305 61 3304 62 3303 63 3302 64 3301 65 330..."
},
{
"input": "64",
"output": "1 4096 2 4095 3 4094 4 4093 5 4092 6 4091 7 4090 8 4089 9 4088 10 4087 11 4086 12 4085 13 4084 14 4083 15 4082 16 4081 17 4080 18 4079 19 4078 20 4077 21 4076 22 4075 23 4074 24 4073 25 4072 26 4071 27 4070 28 4069 29 4068 30 4067 31 4066 32 4065\n33 4064 34 4063 35 4062 36 4061 37 4060 38 4059 39 4058 40 4057 41 4056 42 4055 43 4054 44 4053 45 4052 46 4051 47 4050 48 4049 49 4048 50 4047 51 4046 52 4045 53 4044 54 4043 55 4042 56 4041 57 4040 58 4039 59 4038 60 4037 61 4036 62 4035 63 4034 64 4033\n65 403..."
},
{
"input": "44",
"output": "1 1936 2 1935 3 1934 4 1933 5 1932 6 1931 7 1930 8 1929 9 1928 10 1927 11 1926 12 1925 13 1924 14 1923 15 1922 16 1921 17 1920 18 1919 19 1918 20 1917 21 1916 22 1915\n23 1914 24 1913 25 1912 26 1911 27 1910 28 1909 29 1908 30 1907 31 1906 32 1905 33 1904 34 1903 35 1902 36 1901 37 1900 38 1899 39 1898 40 1897 41 1896 42 1895 43 1894 44 1893\n45 1892 46 1891 47 1890 48 1889 49 1888 50 1887 51 1886 52 1885 53 1884 54 1883 55 1882 56 1881 57 1880 58 1879 59 1878 60 1877 61 1876 62 1875 63 1874 64 1873 65 187..."
}
] | 1,671,009,185 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 92 | 0 | n = int(input())
start = 1
end = n * 2
while(start < end):
print(start,end)
start += 1
end -= 1
| Title: Candy Bags
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies.
Input Specification:
The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers.
Output Specification:
Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits.
Demo Input:
['2\n']
Demo Output:
['1 4\n2 3\n']
Note:
The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother. | ```python
n = int(input())
start = 1
end = n * 2
while(start < end):
print(start,end)
start += 1
end -= 1
``` | 0 | |
680 | B | Bear and Finding Criminals | PROGRAMMING | 1,000 | [
"constructive algorithms",
"implementation"
] | null | null | There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD. | The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city. | Print the number of criminals Limak will catch. | [
"6 3\n1 1 1 0 1 0\n",
"5 2\n0 0 0 1 0\n"
] | [
"3\n",
"1\n"
] | In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is. | 1,000 | [
{
"input": "6 3\n1 1 1 0 1 0",
"output": "3"
},
{
"input": "5 2\n0 0 0 1 0",
"output": "1"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "9 3\n1 1 1 1 1 1 1 1 0",
"output": "8"
},
{
"input": "9 5\n1 0 1 0 1 0 1 0 1",
"output": "5"
},
{
"input": "20 17\n1 1 0 1 1 1 1 0 1 0 1 1 1 0 1 1 0 0 0 0",
"output": "10"
},
{
"input": "100 60\n1 1 1 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 1 0 1 1 0 1 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 0 0 1 0 0 0 1 1 1 0 1 0 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 1 1 1 0 1 0 0",
"output": "27"
},
{
"input": "8 1\n1 0 1 1 0 0 1 0",
"output": "4"
},
{
"input": "11 11\n0 1 0 0 1 1 1 0 0 0 0",
"output": "4"
},
{
"input": "19 10\n0 1 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 0 1",
"output": "4"
},
{
"input": "100 38\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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 38\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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "1"
},
{
"input": "100 38\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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "3"
},
{
"input": "99 38\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "25"
},
{
"input": "99 38\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "24"
},
{
"input": "99 38\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "24"
},
{
"input": "98 70\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 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": "41"
},
{
"input": "99 70\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 1 1 1 1 1 1 1 1 1 1 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": "9"
},
{
"input": "99 60\n0 0 0 1 1 0 1 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1",
"output": "34"
},
{
"input": "98 24\n0 0 0 1 1 0 1 1 0 0 0 0 1 0 0 1 0 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 1 0 1 0 0 1 0 0 0 1 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 0 0 0 1 0 1 1 0 1 0 1 1 1 0 0 1 0 0 0 1 1 0 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1",
"output": "39"
},
{
"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": "100"
},
{
"input": "100 1\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": "2 1\n0 1",
"output": "1"
}
] | 1,689,025,586 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 24 | 61 | 0 | import math
ans = 0
def solve(start, arr):
global ans
start -= 1
if arr[start] == 1: ans+=1
dist = 1
for i in range(dist, len(arr)):
one = start + i
two = start - i
if one < len(arr):
if two >= 0:
if arr[one] == arr[two] == 1:
ans += 2
else:
if arr[one] == 1: ans += 1
else:
if two >= 0:
if arr[two] == 1: ans += 1
f = int(input().split(" ")[1])
a = input().split(" ")
a = [int(x) for x in a]
solve(f, a)
print(ans) | Title: Bear and Finding Criminals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.
Input Specification:
The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city.
Output Specification:
Print the number of criminals Limak will catch.
Demo Input:
['6 3\n1 1 1 0 1 0\n', '5 2\n0 0 0 1 0\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is. | ```python
import math
ans = 0
def solve(start, arr):
global ans
start -= 1
if arr[start] == 1: ans+=1
dist = 1
for i in range(dist, len(arr)):
one = start + i
two = start - i
if one < len(arr):
if two >= 0:
if arr[one] == arr[two] == 1:
ans += 2
else:
if arr[one] == 1: ans += 1
else:
if two >= 0:
if arr[two] == 1: ans += 1
f = int(input().split(" ")[1])
a = input().split(" ")
a = [int(x) for x in a]
solve(f, a)
print(ans)
``` | 3 | |
129 | A | Cookies | PROGRAMMING | 900 | [
"implementation"
] | null | null | Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even? | The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag. | Print in the only line the only number — the sought number of ways. If there are no such ways print 0. | [
"1\n1\n",
"10\n1 2 2 3 4 4 4 2 2 2\n",
"11\n2 2 2 2 2 2 2 2 2 2 99\n"
] | [
"1\n",
"8\n",
"1\n"
] | In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies. | 500 | [
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 2 2 3 4 4 4 2 2 2",
"output": "8"
},
{
"input": "11\n2 2 2 2 2 2 2 2 2 2 99",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "7\n7 7 7 7 7 7 7",
"output": "7"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "4"
},
{
"input": "100\n1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2",
"output": "50"
},
{
"input": "99\n99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99",
"output": "49"
},
{
"input": "82\n43 44 96 33 23 42 33 66 53 87 8 90 43 91 40 88 51 18 48 62 59 10 22 20 54 6 13 63 2 56 31 52 98 42 54 32 26 77 9 24 33 91 16 30 39 34 78 82 73 90 12 15 67 76 30 18 44 86 84 98 65 54 100 79 28 34 40 56 11 43 72 35 86 59 89 40 30 33 7 19 44 15",
"output": "50"
},
{
"input": "17\n50 14 17 77 74 74 38 76 41 27 45 29 66 98 38 73 38",
"output": "7"
},
{
"input": "94\n81 19 90 99 26 11 86 44 78 36 80 59 99 90 78 72 71 20 94 56 42 40 71 84 10 85 10 70 52 27 39 55 90 16 48 25 7 79 99 100 38 10 99 56 3 4 78 9 16 57 14 40 52 54 57 70 30 86 56 84 97 60 59 69 49 66 23 92 90 46 86 73 53 47 1 83 14 20 24 66 13 45 41 14 86 75 55 88 48 95 82 24 47 87",
"output": "39"
},
{
"input": "88\n64 95 12 90 40 65 98 45 52 54 79 7 81 25 98 19 68 82 41 53 35 50 5 22 32 21 8 39 8 6 72 27 81 30 12 79 21 42 60 2 66 87 46 93 62 78 52 71 76 32 78 94 86 85 55 15 34 76 41 20 32 26 94 81 89 45 74 49 11 40 40 39 49 46 80 85 90 23 80 40 86 58 70 26 48 93 23 53",
"output": "37"
},
{
"input": "84\n95 9 43 43 13 84 60 90 1 8 97 99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32",
"output": "51"
},
{
"input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80",
"output": "21"
},
{
"input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81",
"output": "37"
},
{
"input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59",
"output": "28"
},
{
"input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94",
"output": "47"
},
{
"input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28",
"output": "15"
},
{
"input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42",
"output": "11"
},
{
"input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35",
"output": "33"
},
{
"input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22",
"output": "37"
},
{
"input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14",
"output": "15"
},
{
"input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77",
"output": "35"
},
{
"input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75",
"output": "42"
},
{
"input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26",
"output": "23"
},
{
"input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78",
"output": "15"
},
{
"input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78",
"output": "13"
},
{
"input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51",
"output": "17"
},
{
"input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59",
"output": "45"
},
{
"input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43",
"output": "53"
},
{
"input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8",
"output": "40"
},
{
"input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54",
"output": "53"
},
{
"input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77",
"output": "55"
},
{
"input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66",
"output": "52"
},
{
"input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89",
"output": "51"
},
{
"input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2",
"output": "53"
},
{
"input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23",
"output": "48"
},
{
"input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25",
"output": "48"
},
{
"input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71",
"output": "58"
},
{
"input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45",
"output": "53"
},
{
"input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17",
"output": "54"
},
{
"input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52",
"output": "49"
},
{
"input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31",
"output": "44"
},
{
"input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74",
"output": "46"
},
{
"input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75",
"output": "53"
},
{
"input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37",
"output": "43"
},
{
"input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82",
"output": "51"
},
{
"input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5",
"output": "47"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "5\n1 1 3 2 2",
"output": "3"
}
] | 1,551,182,189 | 2,147,483,647 | Python 3 | OK | TESTS | 52 | 218 | 0 | def separate_nums(text):
split_text = text.split(" ")
total_sum = 0
num_array = []
for num in split_text:
num_array.append(int(num))
total_sum += int(num)
return {"total_sum": total_sum, "num_array": num_array,}
def main():
total_of_bags = int(input())
cookies_in_each_bag = input()
num_of_ways = 0
separate_nums_result = separate_nums(cookies_in_each_bag)
for num in separate_nums_result["num_array"]:
if(separate_nums_result["total_sum"] - num)%2 == 0:
num_of_ways+=1
print(num_of_ways)
if __name__ == "__main__":
main() | Title: Cookies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
Output Specification:
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
Demo Input:
['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n']
Demo Output:
['1\n', '8\n', '1\n']
Note:
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies. | ```python
def separate_nums(text):
split_text = text.split(" ")
total_sum = 0
num_array = []
for num in split_text:
num_array.append(int(num))
total_sum += int(num)
return {"total_sum": total_sum, "num_array": num_array,}
def main():
total_of_bags = int(input())
cookies_in_each_bag = input()
num_of_ways = 0
separate_nums_result = separate_nums(cookies_in_each_bag)
for num in separate_nums_result["num_array"]:
if(separate_nums_result["total_sum"] - num)%2 == 0:
num_of_ways+=1
print(num_of_ways)
if __name__ == "__main__":
main()
``` | 3 | |
873 | B | Balanced Substring | PROGRAMMING | 1,500 | [
"dp",
"implementation"
] | null | null | You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of *s*. | The first line contains *n* (1<=≤<=*n*<=≤<=100000) — the number of characters in *s*.
The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*. | If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring. | [
"8\n11010111\n",
"3\n111\n"
] | [
"4\n",
"0\n"
] | In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | 0 | [
{
"input": "8\n11010111",
"output": "4"
},
{
"input": "3\n111",
"output": "0"
},
{
"input": "11\n00001000100",
"output": "2"
},
{
"input": "10\n0100000000",
"output": "2"
},
{
"input": "13\n0001000011010",
"output": "6"
},
{
"input": "14\n00000100101011",
"output": "10"
},
{
"input": "14\n01111101111111",
"output": "2"
},
{
"input": "18\n110010101101111111",
"output": "10"
},
{
"input": "11\n00010000011",
"output": "4"
},
{
"input": "10\n1000010110",
"output": "6"
},
{
"input": "15\n100000100000011",
"output": "4"
},
{
"input": "18\n011010101110111101",
"output": "8"
},
{
"input": "10\n0011011111",
"output": "6"
},
{
"input": "3\n011",
"output": "2"
},
{
"input": "14\n11111111111111",
"output": "0"
},
{
"input": "65\n11001000110001001011011110111100000000101001001010101111000100010",
"output": "48"
},
{
"input": "10\n0011111000",
"output": "10"
},
{
"input": "13\n1110000001110",
"output": "12"
},
{
"input": "3\n110",
"output": "2"
},
{
"input": "4\n1110",
"output": "2"
},
{
"input": "9\n001011001",
"output": "8"
},
{
"input": "2\n10",
"output": "2"
},
{
"input": "2\n01",
"output": "2"
},
{
"input": "12\n110000110000",
"output": "8"
},
{
"input": "3\n001",
"output": "2"
},
{
"input": "14\n11000011000000",
"output": "8"
},
{
"input": "19\n0000011111111100000",
"output": "18"
},
{
"input": "45\n011010001100001111110001011100000001101100111",
"output": "44"
},
{
"input": "18\n000001111111100000",
"output": "16"
},
{
"input": "4\n0101",
"output": "4"
},
{
"input": "12\n000101011001",
"output": "10"
}
] | 1,507,819,195 | 2,095 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 0 | n = int(input())
s = str(input())
count0 = s.count('0')
count1 = s.count('1')
if count0 == 0 or count1 == 0:
print('0')
elif count0 == count1:
print(n)
elif count0>count1:
print(2*count1)
else:
print(2*count0)
| Title: Balanced Substring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of *s*.
Input Specification:
The first line contains *n* (1<=≤<=*n*<=≤<=100000) — the number of characters in *s*.
The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*.
Output Specification:
If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring.
Demo Input:
['8\n11010111\n', '3\n111\n']
Demo Output:
['4\n', '0\n']
Note:
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | ```python
n = int(input())
s = str(input())
count0 = s.count('0')
count1 = s.count('1')
if count0 == 0 or count1 == 0:
print('0')
elif count0 == count1:
print(n)
elif count0>count1:
print(2*count1)
else:
print(2*count0)
``` | 0 | |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,668,753,645 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | result = str(input())
if len(result) <= 10:
print(result)
else:
print(result[0] + str(len(result)-2) + result[-1])
| Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
result = str(input())
if len(result) <= 10:
print(result)
else:
print(result[0] + str(len(result)-2) + result[-1])
``` | 0 |
787 | A | The Monster | PROGRAMMING | 1,200 | [
"brute force",
"math",
"number theory"
] | null | null | A monster is chasing after Rick and Morty on another planet. They're so frightened that sometimes they scream. More accurately, Rick screams at times *b*,<=*b*<=+<=*a*,<=*b*<=+<=2*a*,<=*b*<=+<=3*a*,<=... and Morty screams at times *d*,<=*d*<=+<=*c*,<=*d*<=+<=2*c*,<=*d*<=+<=3*c*,<=....
The Monster will catch them if at any point they scream at the same time, so it wants to know when it will catch them (the first time they scream at the same time) or that they will never scream at the same time. | The first line of input contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100).
The second line contains two integers *c* and *d* (1<=≤<=*c*,<=*d*<=≤<=100). | Print the first time Rick and Morty will scream at the same time, or <=-<=1 if they will never scream at the same time. | [
"20 2\n9 19\n",
"2 1\n16 12\n"
] | [
"82\n",
"-1\n"
] | In the first sample testcase, Rick's 5th scream and Morty's 8th time are at time 82.
In the second sample testcase, all Rick's screams will be at odd times and Morty's will be at even times, so they will never scream at the same time. | 500 | [
{
"input": "20 2\n9 19",
"output": "82"
},
{
"input": "2 1\n16 12",
"output": "-1"
},
{
"input": "39 52\n88 78",
"output": "1222"
},
{
"input": "59 96\n34 48",
"output": "1748"
},
{
"input": "87 37\n91 29",
"output": "211"
},
{
"input": "11 81\n49 7",
"output": "301"
},
{
"input": "39 21\n95 89",
"output": "3414"
},
{
"input": "59 70\n48 54",
"output": "1014"
},
{
"input": "87 22\n98 32",
"output": "718"
},
{
"input": "15 63\n51 13",
"output": "-1"
},
{
"input": "39 7\n97 91",
"output": "1255"
},
{
"input": "18 18\n71 71",
"output": "1278"
},
{
"input": "46 71\n16 49",
"output": "209"
},
{
"input": "70 11\n74 27",
"output": "2321"
},
{
"input": "94 55\n20 96",
"output": "-1"
},
{
"input": "18 4\n77 78",
"output": "1156"
},
{
"input": "46 44\n23 55",
"output": "-1"
},
{
"input": "74 88\n77 37",
"output": "1346"
},
{
"input": "94 37\n34 7",
"output": "789"
},
{
"input": "22 81\n80 88",
"output": "-1"
},
{
"input": "46 30\n34 62",
"output": "674"
},
{
"input": "40 4\n81 40",
"output": "364"
},
{
"input": "69 48\n39 9",
"output": "48"
},
{
"input": "89 93\n84 87",
"output": "5967"
},
{
"input": "17 45\n42 65",
"output": "317"
},
{
"input": "41 85\n95 46",
"output": "331"
},
{
"input": "69 30\n41 16",
"output": "1410"
},
{
"input": "93 74\n99 93",
"output": "-1"
},
{
"input": "17 19\n44 75",
"output": "427"
},
{
"input": "45 63\n98 53",
"output": "3483"
},
{
"input": "69 11\n48 34",
"output": "-1"
},
{
"input": "55 94\n3 96",
"output": "204"
},
{
"input": "100 100\n100 100",
"output": "100"
},
{
"input": "1 1\n1 1",
"output": "1"
},
{
"input": "1 1\n1 100",
"output": "100"
},
{
"input": "1 100\n100 1",
"output": "101"
},
{
"input": "98 1\n99 100",
"output": "9703"
},
{
"input": "98 1\n99 2",
"output": "9605"
},
{
"input": "97 2\n99 100",
"output": "4852"
},
{
"input": "3 3\n3 1",
"output": "-1"
},
{
"input": "3 2\n7 2",
"output": "2"
},
{
"input": "2 3\n2 5",
"output": "5"
},
{
"input": "2 3\n2 3",
"output": "3"
},
{
"input": "100 3\n100 5",
"output": "-1"
},
{
"input": "6 10\n12 14",
"output": "-1"
},
{
"input": "4 2\n4 4",
"output": "-1"
},
{
"input": "2 3\n2 2",
"output": "-1"
},
{
"input": "2 3\n4 99",
"output": "99"
},
{
"input": "1 5\n1 5",
"output": "5"
},
{
"input": "1 100\n3 1",
"output": "100"
},
{
"input": "2 2\n2 1",
"output": "-1"
},
{
"input": "2 10\n6 20",
"output": "20"
},
{
"input": "2 2\n2 10",
"output": "10"
},
{
"input": "3 7\n3 6",
"output": "-1"
},
{
"input": "1 100\n1 100",
"output": "100"
},
{
"input": "7 25\n39 85",
"output": "319"
},
{
"input": "84 82\n38 6",
"output": "82"
},
{
"input": "7 7\n7 14",
"output": "14"
}
] | 1,655,197,403 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 1,000 | 1,331,200 | import sys
input = sys.stdin.readline
a, b = map(int, input().split())
c, d = map(int, input().split())
if a < c:
a, b, c, d = c, d, a, b
x = b - d
if a % c == 0 and x % c != 0:
print(-1)
else:
i = 0
while 1:
if (i*a + x) % c == 0:
print(i*a + x + d)
break
i += 1 | Title: The Monster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A monster is chasing after Rick and Morty on another planet. They're so frightened that sometimes they scream. More accurately, Rick screams at times *b*,<=*b*<=+<=*a*,<=*b*<=+<=2*a*,<=*b*<=+<=3*a*,<=... and Morty screams at times *d*,<=*d*<=+<=*c*,<=*d*<=+<=2*c*,<=*d*<=+<=3*c*,<=....
The Monster will catch them if at any point they scream at the same time, so it wants to know when it will catch them (the first time they scream at the same time) or that they will never scream at the same time.
Input Specification:
The first line of input contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100).
The second line contains two integers *c* and *d* (1<=≤<=*c*,<=*d*<=≤<=100).
Output Specification:
Print the first time Rick and Morty will scream at the same time, or <=-<=1 if they will never scream at the same time.
Demo Input:
['20 2\n9 19\n', '2 1\n16 12\n']
Demo Output:
['82\n', '-1\n']
Note:
In the first sample testcase, Rick's 5th scream and Morty's 8th time are at time 82.
In the second sample testcase, all Rick's screams will be at odd times and Morty's will be at even times, so they will never scream at the same time. | ```python
import sys
input = sys.stdin.readline
a, b = map(int, input().split())
c, d = map(int, input().split())
if a < c:
a, b, c, d = c, d, a, b
x = b - d
if a % c == 0 and x % c != 0:
print(-1)
else:
i = 0
while 1:
if (i*a + x) % c == 0:
print(i*a + x + d)
break
i += 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"
}
] | 1,638,679,360 | 2,147,483,647 | PyPy 3 | OK | TESTS | 91 | 1,029 | 5,734,400 | n,m=map(int,input().split())
p=[0]+list(map(int,input().split()))
for _ in range(m):
l,r,x=map(int,input().split())
c=0
for i in range(l,r+1):
if(p[i]<p[x]):
c+=1
if(l+c==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
n,m=map(int,input().split())
p=[0]+list(map(int,input().split()))
for _ in range(m):
l,r,x=map(int,input().split())
c=0
for i in range(l,r+1):
if(p[i]<p[x]):
c+=1
if(l+c==x):
print('Yes')
else:
print('No')
``` | 3 | |
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,510,314,066 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 124 | 0 | n=int(input())
i=lambda:map(int,input().split())
a=list(i())
even=[]
odd=[]
for j in range(n):
if (a[j]%2==0):
even.append(a[j])
else:
odd.append(a[j])
if(len(even)>len(odd)):
print (a.index(odd[0])+1)
else:
print (a.index(even[0])+1) | Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
n=int(input())
i=lambda:map(int,input().split())
a=list(i())
even=[]
odd=[]
for j in range(n):
if (a[j]%2==0):
even.append(a[j])
else:
odd.append(a[j])
if(len(even)>len(odd)):
print (a.index(odd[0])+1)
else:
print (a.index(even[0])+1)
``` | 3.969 |
940 | B | Our Tanya is Crying Out Loud | PROGRAMMING | 1,400 | [
"dp",
"greedy"
] | null | null | Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109). | Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1. | [
"9\n2\n3\n1\n",
"5\n5\n2\n20\n",
"19\n3\n4\n2\n"
] | [
"6\n",
"8\n",
"12\n"
] | In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total. | 1,250 | [
{
"input": "9\n2\n3\n1",
"output": "6"
},
{
"input": "5\n5\n2\n20",
"output": "8"
},
{
"input": "19\n3\n4\n2",
"output": "12"
},
{
"input": "1845999546\n999435865\n1234234\n2323423",
"output": "1044857680578777"
},
{
"input": "1604353664\n1604353665\n9993432\n1",
"output": "16032999235141416"
},
{
"input": "777888456\n1\n98\n43",
"output": "76233068590"
},
{
"input": "1162261467\n3\n1\n2000000000",
"output": "1162261466"
},
{
"input": "1000000000\n1999999999\n789987\n184569875",
"output": "789986999210013"
},
{
"input": "2000000000\n2\n1\n2000000000",
"output": "1999999999"
},
{
"input": "1999888325\n3\n2\n2000000000",
"output": "3333258884"
},
{
"input": "1897546487\n687\n89798979\n879876541",
"output": "110398404423"
},
{
"input": "20\n1\n20\n1",
"output": "380"
},
{
"input": "16\n5\n17\n3",
"output": "54"
},
{
"input": "19\n19\n19\n1",
"output": "1"
},
{
"input": "18\n2\n3\n16",
"output": "40"
},
{
"input": "1\n11\n8\n9",
"output": "0"
},
{
"input": "9\n10\n1\n20",
"output": "8"
},
{
"input": "19\n10\n19\n2",
"output": "173"
},
{
"input": "16\n9\n14\n2",
"output": "100"
},
{
"input": "15\n2\n5\n2",
"output": "21"
},
{
"input": "14\n7\n13\n1",
"output": "14"
},
{
"input": "43\n3\n45\n3",
"output": "189"
},
{
"input": "99\n1\n98\n1",
"output": "9604"
},
{
"input": "77\n93\n100\n77",
"output": "7600"
},
{
"input": "81\n3\n91\n95",
"output": "380"
},
{
"input": "78\n53\n87\n34",
"output": "2209"
},
{
"input": "80\n3\n15\n1",
"output": "108"
},
{
"input": "97\n24\n4\n24",
"output": "40"
},
{
"input": "100\n100\n1\n100",
"output": "99"
},
{
"input": "87\n4\n17\n7",
"output": "106"
},
{
"input": "65\n2\n3\n6",
"output": "36"
},
{
"input": "1000000\n1435\n3\n999999",
"output": "1005804"
},
{
"input": "783464\n483464\n2\n966928",
"output": "1566926"
},
{
"input": "248035\n11\n3\n20",
"output": "202"
},
{
"input": "524287\n2\n945658\n999756",
"output": "34963354"
},
{
"input": "947352\n78946\n85\n789654",
"output": "790589"
},
{
"input": "1000000\n1\n999899\n60",
"output": "999898000101"
},
{
"input": "753687\n977456\n6547\n456",
"output": "4934382242"
},
{
"input": "1000000\n500000\n1\n999997",
"output": "999998"
},
{
"input": "997458\n843596\n1\n843596",
"output": "997457"
},
{
"input": "821109\n92\n6547\n98787",
"output": "394566"
},
{
"input": "1073741823\n2\n9543\n8923453",
"output": "188412866"
},
{
"input": "1000999777\n1934999345\n2356346\n34534565",
"output": "2358701818178496"
},
{
"input": "2000000000\n1\n2000000000\n98",
"output": "3999999998000000000"
},
{
"input": "1999324353\n978435356\n1\n978435356",
"output": "1020888998"
},
{
"input": "2000000000\n2\n2000000000\n2000000000",
"output": "84000000000"
},
{
"input": "241375690\n17\n2\n1998789654",
"output": "482751378"
},
{
"input": "171507000\n350\n789\n6548687",
"output": "14216965"
},
{
"input": "1100220011\n10001\n2\n1999778654",
"output": "1999998674"
},
{
"input": "1867622656\n43216\n789644\n12315468",
"output": "24630936"
},
{
"input": "1867622656\n43216\n1\n1879865413",
"output": "1867622655"
},
{
"input": "1999999999\n1000000000\n789987\n184569875",
"output": "789987183779888"
},
{
"input": "1987987897\n103546\n7\n98754563",
"output": "98946650"
},
{
"input": "10\n2\n2\n5",
"output": "13"
},
{
"input": "7\n2\n1\n100000",
"output": "6"
},
{
"input": "7\n2\n3\n1",
"output": "8"
},
{
"input": "2000000000\n666666667\n1\n1",
"output": "666666668"
},
{
"input": "1999999997\n666666666\n2\n2",
"output": "1333333334"
}
] | 1,519,570,546 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 9 | 62 | 5,632,000 | RealAnswer = 1e18
def KhelteThako(n, k, A, B, ans):
if n <= 1:
if n == 1:
global RealAnswer
RealAnswer = min(RealAnswer, ans)
return
return
if n%k == 0:
KhelteThako(n/k,k,A,B,ans+B)
else:
if n > k:
xx = n%k
KhelteThako(n-xx,k,A,B,ans+xx*A)
else:
xx = n%k
KhelteThako(n-xx+1,k, A, B, ans+(xx-1)*A)
# KhelteThako(1, k, A, B, ans+(n-1)*A)
#import sys
#sys.stdin = open('in.txt','r')
n = int(input())
k = int(input())
A = int(input())
B = int(input())
ans = (n-1)*A
if k == 1:
print(ans)
else:
KhelteThako(n,k,A,B,0)
print(min(ans,int(RealAnswer))) | Title: Our Tanya is Crying Out Loud
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109).
Output Specification:
Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1.
Demo Input:
['9\n2\n3\n1\n', '5\n5\n2\n20\n', '19\n3\n4\n2\n']
Demo Output:
['6\n', '8\n', '12\n']
Note:
In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total. | ```python
RealAnswer = 1e18
def KhelteThako(n, k, A, B, ans):
if n <= 1:
if n == 1:
global RealAnswer
RealAnswer = min(RealAnswer, ans)
return
return
if n%k == 0:
KhelteThako(n/k,k,A,B,ans+B)
else:
if n > k:
xx = n%k
KhelteThako(n-xx,k,A,B,ans+xx*A)
else:
xx = n%k
KhelteThako(n-xx+1,k, A, B, ans+(xx-1)*A)
# KhelteThako(1, k, A, B, ans+(n-1)*A)
#import sys
#sys.stdin = open('in.txt','r')
n = int(input())
k = int(input())
A = int(input())
B = int(input())
ans = (n-1)*A
if k == 1:
print(ans)
else:
KhelteThako(n,k,A,B,0)
print(min(ans,int(RealAnswer)))
``` | 0 | |
245 | B | Internet Address | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | Vasya is an active Internet user. One day he came across an Internet resource he liked, so he wrote its address in the notebook. We know that the address of the written resource has format:
where:
- <protocol> can equal either "http" (without the quotes) or "ftp" (without the quotes), - <domain> is a non-empty string, consisting of lowercase English letters, - the /<context> part may not be present. If it is present, then <context> is a non-empty string, consisting of lowercase English letters.
If string <context> isn't present in the address, then the additional character "/" isn't written. Thus, the address has either two characters "/" (the ones that go before the domain), or three (an extra one in front of the context).
When the boy came home, he found out that the address he wrote in his notebook had no punctuation marks. Vasya must have been in a lot of hurry and didn't write characters ":", "/", ".".
Help Vasya to restore the possible address of the recorded Internet resource. | The first line contains a non-empty string that Vasya wrote out in his notebook. This line consists of lowercase English letters only.
It is guaranteed that the given string contains at most 50 letters. It is guaranteed that the given string can be obtained from some correct Internet resource address, described above. | Print a single line — the address of the Internet resource that Vasya liked. If there are several addresses that meet the problem limitations, you are allowed to print any of them. | [
"httpsunrux\n",
"ftphttprururu\n"
] | [
"http://sun.ru/x\n",
"ftp://http.ru/ruru\n"
] | In the second sample there are two more possible answers: "ftp://httpruru.ru" and "ftp://httpru.ru/ru". | 0 | [
{
"input": "httpsunrux",
"output": "http://sun.ru/x"
},
{
"input": "ftphttprururu",
"output": "ftp://http.ru/ruru"
},
{
"input": "httpuururrururruruurururrrrrurrurrurruruuruuu",
"output": "http://uu.ru/rrururruruurururrrrrurrurrurruruuruuu"
},
{
"input": "httpabuaruauabbaruru",
"output": "http://abua.ru/auabbaruru"
},
{
"input": "httpuurrruurruuruuruuurrrurururuurruuuuuuruurr",
"output": "http://uurr.ru/urruuruuruuurrrurururuurruuuuuuruurr"
},
{
"input": "httpruhhphhhpuhruruhhpruhhphruhhru",
"output": "http://ruhhphhhpuh.ru/ruhhpruhhphruhhru"
},
{
"input": "httpftprftprutprururftruruftptp",
"output": "http://ftprftp.ru/tprururftruruftptp"
},
{
"input": "httpfttpftpfttftpftpftppfrurururu",
"output": "http://fttpftpfttftpftpftppf.ru/rururu"
},
{
"input": "httpruhttttpruhttprupruhttpruhtturuhttphtruuru",
"output": "http://ruhttttp.ru/httprupruhttpruhtturuhttphtruuru"
},
{
"input": "httpsjkazaaghasjkasjkabruru",
"output": "http://sjkazaaghasjkasjkab.ru/ru"
},
{
"input": "httpftphttptphttphrururuhpftphtpftphtpftphtptpft",
"output": "http://ftphttptphttph.ru/ruruhpftphtpftphtpftphtptpft"
},
{
"input": "httpppppru",
"output": "http://pppp.ru"
},
{
"input": "ftprrurururrurururuurrururruuru",
"output": "ftp://r.ru/rururrurururuurrururruuru"
},
{
"input": "ftpabaruru",
"output": "ftp://aba.ru/ru"
},
{
"input": "ftpruurruurururururuuruuur",
"output": "ftp://ruur.ru/urururururuuruuur"
},
{
"input": "ftphhphruhhpruhhpuhhpuruhhphruhhruhhpuhhru",
"output": "ftp://hhph.ru/hhpruhhpuhhpuruhhphruhhruhhpuhhru"
},
{
"input": "ftparua",
"output": "ftp://a.ru/a"
},
{
"input": "httpzru",
"output": "http://z.ru"
},
{
"input": "httprrur",
"output": "http://r.ru/r"
},
{
"input": "ftprru",
"output": "ftp://r.ru"
}
] | 1,696,436,845 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | def restore_address(s):
# Determine protocol
if s.startswith("ftp"):
protocol = "ftp"
s = s[3:]
else:
protocol = "http"
s = s[4:]
# Determine domain
domain = s[:s.index("ru")]
s = s[s.index("ru")+2:]
# Determine context (if present)
if "/" in s:
context = s[1:]
else:
context = ""
# Reconstruct and print the address
address = f"{protocol}://{domain}.ru/{context}"
return address
# Read the input string
s = input().strip()
# Restore and print the address
print(restore_address(s))
| Title: Internet Address
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya is an active Internet user. One day he came across an Internet resource he liked, so he wrote its address in the notebook. We know that the address of the written resource has format:
where:
- <protocol> can equal either "http" (without the quotes) or "ftp" (without the quotes), - <domain> is a non-empty string, consisting of lowercase English letters, - the /<context> part may not be present. If it is present, then <context> is a non-empty string, consisting of lowercase English letters.
If string <context> isn't present in the address, then the additional character "/" isn't written. Thus, the address has either two characters "/" (the ones that go before the domain), or three (an extra one in front of the context).
When the boy came home, he found out that the address he wrote in his notebook had no punctuation marks. Vasya must have been in a lot of hurry and didn't write characters ":", "/", ".".
Help Vasya to restore the possible address of the recorded Internet resource.
Input Specification:
The first line contains a non-empty string that Vasya wrote out in his notebook. This line consists of lowercase English letters only.
It is guaranteed that the given string contains at most 50 letters. It is guaranteed that the given string can be obtained from some correct Internet resource address, described above.
Output Specification:
Print a single line — the address of the Internet resource that Vasya liked. If there are several addresses that meet the problem limitations, you are allowed to print any of them.
Demo Input:
['httpsunrux\n', 'ftphttprururu\n']
Demo Output:
['http://sun.ru/x\n', 'ftp://http.ru/ruru\n']
Note:
In the second sample there are two more possible answers: "ftp://httpruru.ru" and "ftp://httpru.ru/ru". | ```python
def restore_address(s):
# Determine protocol
if s.startswith("ftp"):
protocol = "ftp"
s = s[3:]
else:
protocol = "http"
s = s[4:]
# Determine domain
domain = s[:s.index("ru")]
s = s[s.index("ru")+2:]
# Determine context (if present)
if "/" in s:
context = s[1:]
else:
context = ""
# Reconstruct and print the address
address = f"{protocol}://{domain}.ru/{context}"
return address
# Read the input string
s = input().strip()
# Restore and print the address
print(restore_address(s))
``` | 0 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,622,183,650 | 2,147,483,647 | PyPy 3 | OK | TESTS | 35 | 216 | 0 | n1, n2 = map(int, input().split())
print(n1*n2//2) | 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
n1, n2 = map(int, input().split())
print(n1*n2//2)
``` | 3.946 |
981 | A | Antipalindrome | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into? | The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only. | If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique. | [
"mew\n",
"wuffuw\n",
"qqqqqqqq\n"
] | [
"3\n",
"5\n",
"0\n"
] | "mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$. | 500 | [
{
"input": "mew",
"output": "3"
},
{
"input": "wuffuw",
"output": "5"
},
{
"input": "qqqqqqqq",
"output": "0"
},
{
"input": "ijvji",
"output": "4"
},
{
"input": "iiiiiii",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvvgiaahqmtjyuchikvvhvrebow",
"output": "49"
},
{
"input": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvahheoqleromusrartldojsjvy",
"output": "50"
},
{
"input": "ijvxljt",
"output": "7"
},
{
"input": "fyhcncnchyf",
"output": "10"
},
{
"input": "ffffffffffff",
"output": "0"
},
{
"input": "fyhcncfsepqj",
"output": "12"
},
{
"input": "ybejrrlbcinttnicblrrjeby",
"output": "23"
},
{
"input": "yyyyyyyyyyyyyyyyyyyyyyyyy",
"output": "0"
},
{
"input": "ybejrrlbcintahovgjddrqatv",
"output": "25"
},
{
"input": "oftmhcmclgyqaojljoaqyglcmchmtfo",
"output": "30"
},
{
"input": "oooooooooooooooooooooooooooooooo",
"output": "0"
},
{
"input": "oftmhcmclgyqaojllbotztajglsmcilv",
"output": "32"
},
{
"input": "gxandbtgpbknxvnkjaajknvxnkbpgtbdnaxg",
"output": "35"
},
{
"input": "gggggggggggggggggggggggggggggggggggg",
"output": "0"
},
{
"input": "gxandbtgpbknxvnkjaygommzqitqzjfalfkk",
"output": "36"
},
{
"input": "fcliblymyqckxvieotjooojtoeivxkcqymylbilcf",
"output": "40"
},
{
"input": "fffffffffffffffffffffffffffffffffffffffffff",
"output": "0"
},
{
"input": "fcliblymyqckxvieotjootiqwtyznhhvuhbaixwqnsy",
"output": "43"
},
{
"input": "rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr",
"output": "0"
},
{
"input": "rajccqwqnqmshmerpvjyfepxwpxyldzpzhctqjnstxyfmlhiy",
"output": "49"
},
{
"input": "a",
"output": "0"
},
{
"input": "abca",
"output": "4"
},
{
"input": "aaaaabaaaaa",
"output": "10"
},
{
"input": "aba",
"output": "2"
},
{
"input": "asaa",
"output": "4"
},
{
"input": "aabaa",
"output": "4"
},
{
"input": "aabbaa",
"output": "5"
},
{
"input": "abcdaaa",
"output": "7"
},
{
"input": "aaholaa",
"output": "7"
},
{
"input": "abcdefghijka",
"output": "12"
},
{
"input": "aaadcba",
"output": "7"
},
{
"input": "aaaabaaaa",
"output": "8"
},
{
"input": "abaa",
"output": "4"
},
{
"input": "abcbaa",
"output": "6"
},
{
"input": "ab",
"output": "2"
},
{
"input": "l",
"output": "0"
},
{
"input": "aaaabcaaaa",
"output": "10"
},
{
"input": "abbaaaaaabba",
"output": "11"
},
{
"input": "abaaa",
"output": "5"
},
{
"input": "baa",
"output": "3"
},
{
"input": "aaaaaaabbba",
"output": "11"
},
{
"input": "ccbcc",
"output": "4"
},
{
"input": "bbbaaab",
"output": "7"
},
{
"input": "abaaaaaaaa",
"output": "10"
},
{
"input": "abaaba",
"output": "5"
},
{
"input": "aabsdfaaaa",
"output": "10"
},
{
"input": "aaaba",
"output": "5"
},
{
"input": "aaabaaa",
"output": "6"
},
{
"input": "baaabbb",
"output": "7"
},
{
"input": "ccbbabbcc",
"output": "8"
},
{
"input": "cabc",
"output": "4"
},
{
"input": "aabcd",
"output": "5"
},
{
"input": "abcdea",
"output": "6"
},
{
"input": "bbabb",
"output": "4"
},
{
"input": "aaaaabababaaaaa",
"output": "14"
},
{
"input": "bbabbb",
"output": "6"
},
{
"input": "aababd",
"output": "6"
},
{
"input": "abaaaa",
"output": "6"
},
{
"input": "aaaaaaaabbba",
"output": "12"
},
{
"input": "aabca",
"output": "5"
},
{
"input": "aaabccbaaa",
"output": "9"
},
{
"input": "aaaaaaaaaaaaaaaaaaaab",
"output": "21"
},
{
"input": "babb",
"output": "4"
},
{
"input": "abcaa",
"output": "5"
},
{
"input": "qwqq",
"output": "4"
},
{
"input": "aaaaaaaaaaabbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaa",
"output": "48"
},
{
"input": "aaab",
"output": "4"
},
{
"input": "aaaaaabaaaaa",
"output": "12"
},
{
"input": "wwuww",
"output": "4"
},
{
"input": "aaaaabcbaaaaa",
"output": "12"
},
{
"input": "aaabbbaaa",
"output": "8"
},
{
"input": "aabcbaa",
"output": "6"
},
{
"input": "abccdefccba",
"output": "11"
},
{
"input": "aabbcbbaa",
"output": "8"
},
{
"input": "aaaabbaaaa",
"output": "9"
},
{
"input": "aabcda",
"output": "6"
},
{
"input": "abbca",
"output": "5"
},
{
"input": "aaaaaabbaaa",
"output": "11"
},
{
"input": "sssssspssssss",
"output": "12"
},
{
"input": "sdnmsdcs",
"output": "8"
},
{
"input": "aaabbbccbbbaaa",
"output": "13"
},
{
"input": "cbdbdc",
"output": "6"
},
{
"input": "abb",
"output": "3"
},
{
"input": "abcdefaaaa",
"output": "10"
},
{
"input": "abbbaaa",
"output": "7"
},
{
"input": "v",
"output": "0"
},
{
"input": "abccbba",
"output": "7"
},
{
"input": "axyza",
"output": "5"
},
{
"input": "abcdefgaaaa",
"output": "11"
},
{
"input": "aaabcdaaa",
"output": "9"
},
{
"input": "aaaacaaaa",
"output": "8"
},
{
"input": "aaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaaaa",
"output": "42"
},
{
"input": "abbbaa",
"output": "6"
},
{
"input": "abcdee",
"output": "6"
},
{
"input": "oom",
"output": "3"
},
{
"input": "aabcaa",
"output": "6"
},
{
"input": "abba",
"output": "3"
},
{
"input": "aaca",
"output": "4"
},
{
"input": "aacbca",
"output": "6"
},
{
"input": "ababa",
"output": "4"
},
{
"input": "abcda",
"output": "5"
},
{
"input": "cccaaccc",
"output": "7"
},
{
"input": "aaabcda",
"output": "7"
},
{
"input": "aa",
"output": "0"
},
{
"input": "aabaaaa",
"output": "7"
},
{
"input": "abbaaaa",
"output": "7"
},
{
"input": "aaabcbaaa",
"output": "8"
},
{
"input": "aabba",
"output": "5"
},
{
"input": "xyxx",
"output": "4"
},
{
"input": "aaaaaaaaaaaabc",
"output": "14"
},
{
"input": "bbaaaabb",
"output": "7"
},
{
"input": "aaabaa",
"output": "6"
},
{
"input": "sssssabsssss",
"output": "12"
},
{
"input": "bbbaaaabbb",
"output": "9"
},
{
"input": "abbbbaaaa",
"output": "9"
},
{
"input": "wwufuww",
"output": "6"
},
{
"input": "oowoo",
"output": "4"
},
{
"input": "cccaccc",
"output": "6"
},
{
"input": "aaa",
"output": "0"
},
{
"input": "bbbcc",
"output": "5"
},
{
"input": "abcdef",
"output": "6"
},
{
"input": "abbba",
"output": "4"
},
{
"input": "aab",
"output": "3"
},
{
"input": "aaba",
"output": "4"
},
{
"input": "azbyaaa",
"output": "7"
},
{
"input": "oooooiooooo",
"output": "10"
},
{
"input": "aabbbbbaaaaaa",
"output": "13"
}
] | 1,611,905,754 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | s=input()
if(len(s)==1):
print(0)
break
elif(len(set(s))==1):
print(0)
break
elif(s==s[::-1]):
print(len(s)-1)
else:
print(len(s))
break | Title: Antipalindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
Input Specification:
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
Output Specification:
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
Demo Input:
['mew\n', 'wuffuw\n', 'qqqqqqqq\n']
Demo Output:
['3\n', '5\n', '0\n']
Note:
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$. | ```python
s=input()
if(len(s)==1):
print(0)
break
elif(len(set(s))==1):
print(0)
break
elif(s==s[::-1]):
print(len(s)-1)
else:
print(len(s))
break
``` | -1 | |
415 | B | Mashmokh and Tokens | PROGRAMMING | 1,500 | [
"binary search",
"greedy",
"implementation",
"math"
] | null | null | Bimokh is Mashmokh's boss. For the following *n* days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives back *w* tokens then he'll get dollars.
Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has *n* numbers *x*1,<=*x*2,<=...,<=*x**n*. Number *x**i* is the number of tokens given to each worker on the *i*-th day. Help him calculate for each of *n* days the number of tokens he can save. | The first line of input contains three space-separated integers *n*,<=*a*,<=*b* (1<=≤<=*n*<=≤<=105; 1<=≤<=*a*,<=*b*<=≤<=109). The second line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109). | Output *n* space-separated integers. The *i*-th of them is the number of tokens Mashmokh can save on the *i*-th day. | [
"5 1 4\n12 6 11 9 1\n",
"3 1 2\n1 2 3\n",
"1 1 1\n1\n"
] | [
"0 2 3 1 1 ",
"1 0 1 ",
"0 "
] | none | 1,000 | [
{
"input": "5 1 4\n12 6 11 9 1",
"output": "0 2 3 1 1 "
},
{
"input": "3 1 2\n1 2 3",
"output": "1 0 1 "
},
{
"input": "1 1 1\n1",
"output": "0 "
},
{
"input": "1 1 1000000000\n1000000000",
"output": "0 "
},
{
"input": "1 1 1000000000\n999999999",
"output": "999999999 "
},
{
"input": "10 1 100000000\n999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999",
"output": "99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 "
}
] | 1,609,833,446 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 77 | 0 | n,a2,b = map(int, input().split())
a = list(map(int, input().split()))
for i in a:
x = i-a2
print(i-((x//b)*b+a2),end=" ")
print() | Title: Mashmokh and Tokens
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bimokh is Mashmokh's boss. For the following *n* days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives back *w* tokens then he'll get dollars.
Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has *n* numbers *x*1,<=*x*2,<=...,<=*x**n*. Number *x**i* is the number of tokens given to each worker on the *i*-th day. Help him calculate for each of *n* days the number of tokens he can save.
Input Specification:
The first line of input contains three space-separated integers *n*,<=*a*,<=*b* (1<=≤<=*n*<=≤<=105; 1<=≤<=*a*,<=*b*<=≤<=109). The second line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109).
Output Specification:
Output *n* space-separated integers. The *i*-th of them is the number of tokens Mashmokh can save on the *i*-th day.
Demo Input:
['5 1 4\n12 6 11 9 1\n', '3 1 2\n1 2 3\n', '1 1 1\n1\n']
Demo Output:
['0 2 3 1 1 ', '1 0 1 ', '0 ']
Note:
none | ```python
n,a2,b = map(int, input().split())
a = list(map(int, input().split()))
for i in a:
x = i-a2
print(i-((x//b)*b+a2),end=" ")
print()
``` | 0 | |
854 | A | Fraction | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"math"
] | null | null | Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem. | In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction. | Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum. | [
"3\n",
"4\n",
"12\n"
] | [
"1 2\n",
"1 3\n",
"5 7\n"
] | none | 500 | [
{
"input": "3",
"output": "1 2"
},
{
"input": "4",
"output": "1 3"
},
{
"input": "12",
"output": "5 7"
},
{
"input": "34",
"output": "15 19"
},
{
"input": "13",
"output": "6 7"
},
{
"input": "11",
"output": "5 6"
},
{
"input": "24",
"output": "11 13"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "10",
"output": "3 7"
},
{
"input": "69",
"output": "34 35"
},
{
"input": "100",
"output": "49 51"
},
{
"input": "57",
"output": "28 29"
},
{
"input": "1000",
"output": "499 501"
},
{
"input": "999",
"output": "499 500"
},
{
"input": "998",
"output": "497 501"
},
{
"input": "997",
"output": "498 499"
},
{
"input": "996",
"output": "497 499"
},
{
"input": "995",
"output": "497 498"
},
{
"input": "994",
"output": "495 499"
},
{
"input": "5",
"output": "2 3"
},
{
"input": "6",
"output": "1 5"
},
{
"input": "8",
"output": "3 5"
},
{
"input": "9",
"output": "4 5"
},
{
"input": "423",
"output": "211 212"
},
{
"input": "876",
"output": "437 439"
},
{
"input": "29",
"output": "14 15"
}
] | 1,540,577,690 | 1,490 | Python 3 | WRONG_ANSWER | TESTS | 3 | 109 | 0 | n=int(input())
l="2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499"
m=[int(x) for x in l.split()]
j=n//2
if(n%2==0):
j=j-1
if(n==3):
print(1,2)
elif(n==4):
print(1,3)
else:
for k in range(len(m)):
if(j>m[k]):
continue
elif(j==m[k]):
print(j,n-j)
break
else:
print(m[k-1],n-m[k-1])
break
| Title: Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
Input Specification:
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
Output Specification:
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
Demo Input:
['3\n', '4\n', '12\n']
Demo Output:
['1 2\n', '1 3\n', '5 7\n']
Note:
none | ```python
n=int(input())
l="2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499"
m=[int(x) for x in l.split()]
j=n//2
if(n%2==0):
j=j-1
if(n==3):
print(1,2)
elif(n==4):
print(1,3)
else:
for k in range(len(m)):
if(j>m[k]):
continue
elif(j==m[k]):
print(j,n-j)
break
else:
print(m[k-1],n-m[k-1])
break
``` | 0 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,649,169,179 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 186 | 2,150,400 | i=int(input('M'))
j=int(input('N'))
a=i*j/2
print('a') | 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
i=int(input('M'))
j=int(input('N'))
a=i*j/2
print('a')
``` | -1 |
386 | A | Second-Price Auction | PROGRAMMING | 800 | [
"implementation"
] | null | null | In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different. | The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder. | The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based. | [
"2\n5 7\n",
"3\n10 2 8\n",
"6\n3 8 2 9 4 14\n"
] | [
"2 5\n",
"1 8\n",
"6 9\n"
] | none | 500 | [
{
"input": "2\n5 7",
"output": "2 5"
},
{
"input": "3\n10 2 8",
"output": "1 8"
},
{
"input": "6\n3 8 2 9 4 14",
"output": "6 9"
},
{
"input": "4\n4707 7586 4221 5842",
"output": "2 5842"
},
{
"input": "5\n3304 4227 4869 6937 6002",
"output": "4 6002"
},
{
"input": "6\n5083 3289 7708 5362 9031 7458",
"output": "5 7708"
},
{
"input": "7\n9038 6222 3392 1706 3778 1807 2657",
"output": "1 6222"
},
{
"input": "8\n7062 2194 4481 3864 7470 1814 8091 733",
"output": "7 7470"
},
{
"input": "9\n2678 5659 9199 2628 7906 7496 4524 2663 3408",
"output": "3 7906"
},
{
"input": "2\n3458 1504",
"output": "1 1504"
},
{
"input": "50\n9237 3904 407 9052 6657 9229 9752 3888 7732 2512 4614 1055 2355 7108 6506 6849 2529 8862 159 8630 7906 7941 960 8470 333 8659 54 9475 3163 5625 6393 6814 2656 3388 169 7918 4881 8468 9983 6281 6340 280 5108 2996 101 7617 3313 8172 326 1991",
"output": "39 9752"
},
{
"input": "100\n2515 3324 7975 6171 4240 1217 4829 5203 8603 6900 3031 4699 4732 6070 4221 3228 6497 7359 9130 4346 4619 1109 3945 5442 3271 16 9711 2045 6410 2301 3406 8125 6003 1892 1260 9661 3940 6692 4708 7027 4930 6925 1979 5361 4263 3144 867 8639 6230 5562 9714 3676 4231 3347 4733 4920 4881 3431 1059 7313 8912 3038 9308 72 9583 7009 3034 7425 2398 6141 3245 2495 2933 6710 8289 9806 1226 8393 7349 6462 1618 9613 3546 6012 2964 9995 1578 210 2123 4874 1252 8625 348 8020 803 7244 9080 5088 706 2602",
"output": "86 9806"
},
{
"input": "2\n2 1",
"output": "1 1"
},
{
"input": "2\n1 2",
"output": "2 1"
},
{
"input": "3\n10 20 30",
"output": "3 20"
},
{
"input": "3\n10 30 20",
"output": "2 20"
},
{
"input": "3\n20 10 30",
"output": "3 20"
},
{
"input": "3\n20 30 10",
"output": "2 20"
},
{
"input": "3\n30 10 20",
"output": "1 20"
},
{
"input": "3\n30 20 10",
"output": "1 20"
},
{
"input": "2\n1 10000",
"output": "2 1"
},
{
"input": "2\n10000 999",
"output": "1 999"
},
{
"input": "3\n3 4 1",
"output": "2 3"
},
{
"input": "6\n1 2 24 6 5 7",
"output": "3 7"
}
] | 1,594,189,679 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 109 | 6,963,200 | n = int(input())
l = list(map(int, input().split(" ")))
print(l.index(max(l))+1, sorted(l)[-2]) | Title: Second-Price Auction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
Output Specification:
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
Demo Input:
['2\n5 7\n', '3\n10 2 8\n', '6\n3 8 2 9 4 14\n']
Demo Output:
['2 5\n', '1 8\n', '6 9\n']
Note:
none | ```python
n = int(input())
l = list(map(int, input().split(" ")))
print(l.index(max(l))+1, sorted(l)[-2])
``` | 3 | |
620 | B | Grandfather Dovlet’s calculator | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Once Max found an electronic calculator from his grandfather Dovlet's chest. He noticed that the numbers were written with seven-segment indicators ([https://en.wikipedia.org/wiki/Seven-segment_display](https://en.wikipedia.org/wiki/Seven-segment_display)).
Max starts to type all the values from *a* to *b*. After typing each number Max resets the calculator. Find the total number of segments printed on the calculator.
For example if *a*<==<=1 and *b*<==<=3 then at first the calculator will print 2 segments, then — 5 segments and at last it will print 5 segments. So the total number of printed segments is 12. | The only line contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*b*<=≤<=106) — the first and the last number typed by Max. | Print the only integer *a* — the total number of printed segments. | [
"1 3\n",
"10 15\n"
] | [
"12\n",
"39\n"
] | none | 0 | [
{
"input": "1 3",
"output": "12"
},
{
"input": "10 15",
"output": "39"
},
{
"input": "1 100",
"output": "928"
},
{
"input": "100 10000",
"output": "188446"
},
{
"input": "213 221442",
"output": "5645356"
},
{
"input": "1 1000000",
"output": "28733372"
},
{
"input": "1000000 1000000",
"output": "38"
},
{
"input": "222145 353252",
"output": "3860750"
},
{
"input": "2 1000000",
"output": "28733370"
},
{
"input": "1 999999",
"output": "28733334"
},
{
"input": "192 200",
"output": "122"
}
] | 1,453,391,635 | 3,235 | PyPy 3 | OK | TESTS | 11 | 155 | 204,800 | def main():
a, b = map(int, input().split())
arr = [6, 2, 5, 5, 4, 5, 6, 3, 7, 6]
ans = 0
for i in range(a, b+1):
x = i
while x != 0:
ans += arr[x%10]
x //= 10
print(ans)
main() | Title: Grandfather Dovlet’s calculator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once Max found an electronic calculator from his grandfather Dovlet's chest. He noticed that the numbers were written with seven-segment indicators ([https://en.wikipedia.org/wiki/Seven-segment_display](https://en.wikipedia.org/wiki/Seven-segment_display)).
Max starts to type all the values from *a* to *b*. After typing each number Max resets the calculator. Find the total number of segments printed on the calculator.
For example if *a*<==<=1 and *b*<==<=3 then at first the calculator will print 2 segments, then — 5 segments and at last it will print 5 segments. So the total number of printed segments is 12.
Input Specification:
The only line contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*b*<=≤<=106) — the first and the last number typed by Max.
Output Specification:
Print the only integer *a* — the total number of printed segments.
Demo Input:
['1 3\n', '10 15\n']
Demo Output:
['12\n', '39\n']
Note:
none | ```python
def main():
a, b = map(int, input().split())
arr = [6, 2, 5, 5, 4, 5, 6, 3, 7, 6]
ans = 0
for i in range(a, b+1):
x = i
while x != 0:
ans += arr[x%10]
x //= 10
print(ans)
main()
``` | 3 | |
381 | A | Sereja and Dima | PROGRAMMING | 800 | [
"greedy",
"implementation",
"two pointers"
] | null | null | Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her. | The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000. | On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game. | [
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] | [
"12 5\n",
"16 12\n"
] | In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | 500 | [
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24",
"output": "644 500"
},
{
"input": "1\n3",
"output": "3 0"
},
{
"input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646",
"output": "6848 6568"
},
{
"input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727",
"output": "9562 9561"
},
{
"input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12",
"output": "315 315"
},
{
"input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304",
"output": "3238 2222"
},
{
"input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325",
"output": "5246 4864"
},
{
"input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71",
"output": "8147 7807"
},
{
"input": "1\n1",
"output": "1 0"
}
] | 1,686,058,082 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 46 | 0 | n = int(input())
a = list(map(int, input().split()))
a.sort()
print(max(sum(a[::2]), sum(a) - sum(a[::2])), min(sum(a[::2]), sum(a) - sum(a[::2]))) | Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | ```python
n = int(input())
a = list(map(int, input().split()))
a.sort()
print(max(sum(a[::2]), sum(a) - sum(a[::2])), min(sum(a[::2]), sum(a) - sum(a[::2])))
``` | 0 | |
769 | D | k-Interesting Pairs Of Integers | PROGRAMMING | 1,700 | [
"*special",
"bitmasks",
"brute force",
"meet-in-the-middle"
] | null | null | Vasya has the sequence consisting of *n* integers. Vasya consider the pair of integers *x* and *y* k-interesting, if their binary representation differs from each other exactly in *k* bits. For example, if *k*<==<=2, the pair of integers *x*<==<=5 and *y*<==<=3 is k-interesting, because their binary representation *x*=101 and *y*=011 differs exactly in two bits.
Vasya wants to know how many pairs of indexes (*i*, *j*) are in his sequence so that *i*<=<<=*j* and the pair of integers *a**i* and *a**j* is k-interesting. Your task is to help Vasya and determine this number. | The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=14) — the number of integers in Vasya's sequence and the number of bits in which integers in k-interesting pair should differ.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=104), which Vasya has. | Print the number of pairs (*i*, *j*) so that *i*<=<<=*j* and the pair of integers *a**i* and *a**j* is k-interesting. | [
"4 1\n0 3 2 1\n",
"6 0\n200 100 100 100 200 200\n"
] | [
"4\n",
"6\n"
] | In the first test there are 4 k-interesting pairs:
- (1, 3), - (1, 4), - (2, 3), - (2, 4).
In the second test *k* = 0. Consequently, integers in any k-interesting pair should be equal to themselves. Thus, for the second test there are 6 k-interesting pairs:
- (1, 5), - (1, 6), - (2, 3), - (2, 4), - (3, 4), - (5, 6). | 2,000 | [
{
"input": "4 1\n0 3 2 1",
"output": "4"
},
{
"input": "6 0\n200 100 100 100 200 200",
"output": "6"
},
{
"input": "2 0\n1 1",
"output": "1"
},
{
"input": "2 0\n0 0",
"output": "1"
},
{
"input": "2 0\n10000 10000",
"output": "1"
},
{
"input": "2 0\n0 10000",
"output": "0"
},
{
"input": "2 1\n0 1",
"output": "1"
},
{
"input": "2 1\n0 2",
"output": "1"
},
{
"input": "3 1\n0 1 2",
"output": "2"
},
{
"input": "3 2\n0 3 3",
"output": "2"
},
{
"input": "3 2\n3 3 3",
"output": "0"
},
{
"input": "10 0\n1 1 1 1 1 1 1 1 1 1",
"output": "45"
},
{
"input": "100 14\n8192 8192 8192 8192 8191 8192 8192 8192 8192 8192 8191 8191 8191 8192 8191 8191 8191 8192 8192 8192 8192 8192 8191 8191 8191 8192 8191 8192 8192 8192 8192 8192 8192 8191 8191 8192 8192 8191 8191 8192 8192 8192 8191 8191 8192 8191 8191 8191 8191 8191 8191 8192 8191 8191 8192 8191 8191 8192 8192 8191 8192 8192 8192 8192 8192 8192 8192 8191 8192 8192 8192 8191 8191 8192 8192 8192 8191 8192 8192 8192 8192 8192 8191 8192 8192 8191 8192 8192 8192 8192 8191 8192 8191 8191 8192 8191 8192 8192 8191 8191",
"output": "2400"
}
] | 1,488,709,179 | 80,379 | Python 3 | TIME_LIMIT_EXCEEDED | PRETESTS | 12 | 2,000 | 9,011,200 | def countBits(n):
count = 0
while (0 != n):
n &= (n - 1)
count += 1
return count
def f(num1, num2):
a = int(num1) ^ int(num2)
return countBits(a)
def main():
count, delta = map(int, input().split())
numbers = [x for x in input().split()]
cache = {}
result = 0
for i in range(count - 1):
exclude = set()
item = numbers[i]
arr = numbers[i + 1:]
for obj in arr:
if obj in exclude:
continue
elif item + '_' + obj in cache:
if cache[item + '_' + obj]:
result += arr.count(obj)
exclude.add(obj)
continue
else:
exclude.add(obj)
continue
if not (delta ^ f(item, obj)):
result += arr.count(obj)
cache[item + '_' + obj] = True
cache[obj + '_' + item] = True
exclude.add(obj)
else:
cache[item + '_' + obj] = False
cache[obj + '_' + item] = False
exclude.add(obj)
print(result)
if __name__ == "__main__":
main() | Title: k-Interesting Pairs Of Integers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has the sequence consisting of *n* integers. Vasya consider the pair of integers *x* and *y* k-interesting, if their binary representation differs from each other exactly in *k* bits. For example, if *k*<==<=2, the pair of integers *x*<==<=5 and *y*<==<=3 is k-interesting, because their binary representation *x*=101 and *y*=011 differs exactly in two bits.
Vasya wants to know how many pairs of indexes (*i*, *j*) are in his sequence so that *i*<=<<=*j* and the pair of integers *a**i* and *a**j* is k-interesting. Your task is to help Vasya and determine this number.
Input Specification:
The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=14) — the number of integers in Vasya's sequence and the number of bits in which integers in k-interesting pair should differ.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=104), which Vasya has.
Output Specification:
Print the number of pairs (*i*, *j*) so that *i*<=<<=*j* and the pair of integers *a**i* and *a**j* is k-interesting.
Demo Input:
['4 1\n0 3 2 1\n', '6 0\n200 100 100 100 200 200\n']
Demo Output:
['4\n', '6\n']
Note:
In the first test there are 4 k-interesting pairs:
- (1, 3), - (1, 4), - (2, 3), - (2, 4).
In the second test *k* = 0. Consequently, integers in any k-interesting pair should be equal to themselves. Thus, for the second test there are 6 k-interesting pairs:
- (1, 5), - (1, 6), - (2, 3), - (2, 4), - (3, 4), - (5, 6). | ```python
def countBits(n):
count = 0
while (0 != n):
n &= (n - 1)
count += 1
return count
def f(num1, num2):
a = int(num1) ^ int(num2)
return countBits(a)
def main():
count, delta = map(int, input().split())
numbers = [x for x in input().split()]
cache = {}
result = 0
for i in range(count - 1):
exclude = set()
item = numbers[i]
arr = numbers[i + 1:]
for obj in arr:
if obj in exclude:
continue
elif item + '_' + obj in cache:
if cache[item + '_' + obj]:
result += arr.count(obj)
exclude.add(obj)
continue
else:
exclude.add(obj)
continue
if not (delta ^ f(item, obj)):
result += arr.count(obj)
cache[item + '_' + obj] = True
cache[obj + '_' + item] = True
exclude.add(obj)
else:
cache[item + '_' + obj] = False
cache[obj + '_' + item] = False
exclude.add(obj)
print(result)
if __name__ == "__main__":
main()
``` | 0 | |
644 | A | Parliament of Berland | PROGRAMMING | 1,000 | [
"*special",
"constructive algorithms"
] | null | null | There are *n* parliamentarians in Berland. They are numbered with integers from 1 to *n*. It happened that all parliamentarians with odd indices are Democrats and all parliamentarians with even indices are Republicans.
New parliament assembly hall is a rectangle consisting of *a*<=×<=*b* chairs — *a* rows of *b* chairs each. Two chairs are considered neighbouring if they share as side. For example, chair number 5 in row number 2 is neighbouring to chairs number 4 and 6 in this row and chairs with number 5 in rows 1 and 3. Thus, chairs have four neighbours in general, except for the chairs on the border of the hall
We know that if two parliamentarians from one political party (that is two Democrats or two Republicans) seat nearby they spent all time discussing internal party issues.
Write the program that given the number of parliamentarians and the sizes of the hall determine if there is a way to find a seat for any parliamentarian, such that no two members of the same party share neighbouring seats. | The first line of the input contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=10<=000, 1<=≤<=*a*,<=*b*<=≤<=100) — the number of parliamentarians, the number of rows in the assembly hall and the number of seats in each row, respectively. | If there is no way to assigns seats to parliamentarians in a proper way print -1.
Otherwise print the solution in *a* lines, each containing *b* integers. The *j*-th integer of the *i*-th line should be equal to the index of parliamentarian occupying this seat, or 0 if this seat should remain empty. If there are multiple possible solution, you may print any of them. | [
"3 2 2\n",
"8 4 3\n",
"10 2 2\n"
] | [
"0 3\n1 2\n",
"7 8 3\n0 1 4\n6 0 5\n0 2 0\n",
"-1\n"
] | In the first sample there are many other possible solutions. For example,
and
The following assignment
is incorrect, because parliamentarians 1 and 3 are both from Democrats party but will occupy neighbouring seats. | 500 | [
{
"input": "3 2 2",
"output": "1 2 \n0 3 "
},
{
"input": "8 4 3",
"output": "1 2 3 \n4 5 6 \n7 8 0 \n0 0 0 "
},
{
"input": "10 2 2",
"output": "-1"
},
{
"input": "1 1 1",
"output": "1 "
},
{
"input": "8 3 3",
"output": "1 2 3 \n4 5 6 \n7 8 0 "
},
{
"input": "1 1 100",
"output": "1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "1 100 1",
"output": "1 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 \n0 "
},
{
"input": "12 4 3",
"output": "1 2 3 \n4 5 6 \n7 8 9 \n10 11 12 "
},
{
"input": "64 8 9",
"output": "1 2 3 4 5 6 7 8 9 \n10 11 12 13 14 15 16 17 18 \n19 20 21 22 23 24 25 26 27 \n28 29 30 31 32 33 34 35 36 \n37 38 39 40 41 42 43 44 45 \n46 47 48 49 50 51 52 53 54 \n55 56 57 58 59 60 61 62 63 \n64 0 0 0 0 0 0 0 0 "
},
{
"input": "13 2 6",
"output": "-1"
},
{
"input": "41 6 7",
"output": "1 2 3 4 5 6 7 \n8 9 10 11 12 13 14 \n15 16 17 18 19 20 21 \n22 23 24 25 26 27 28 \n29 30 31 32 33 34 35 \n36 37 38 39 40 41 0 "
},
{
"input": "9999 100 100",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 \n102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "10000 100 100",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 \n102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "2099 70 30",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 \n32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 \n61 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 \n92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 \n121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 \n152 151 1..."
},
{
"input": "2098 30 70",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 \n72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 \n141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "10000 1 1",
"output": "-1"
},
{
"input": "1583 49 36",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 \n38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 \n73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 \n110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 \n145 146 147 148 149 150 151 152 153..."
},
{
"input": "4825 77 88",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 \n90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "26 1 33",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 0 0 0 0 0 0 0 "
},
{
"input": "274 25 77",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 \n78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 \n..."
},
{
"input": "694 49 22",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 \n45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 \n68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 \n89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 \n112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 \n133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152..."
},
{
"input": "3585 77 62",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 \n64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 \n125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "3 1 6",
"output": "1 2 3 0 0 0 "
},
{
"input": "352 25 59",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 \n60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 \n119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "150 53 3",
"output": "1 2 3 \n4 5 6 \n7 8 9 \n10 11 12 \n13 14 15 \n16 17 18 \n19 20 21 \n22 23 24 \n25 26 27 \n28 29 30 \n31 32 33 \n34 35 36 \n37 38 39 \n40 41 42 \n43 44 45 \n46 47 48 \n49 50 51 \n52 53 54 \n55 56 57 \n58 59 60 \n61 62 63 \n64 65 66 \n67 68 69 \n70 71 72 \n73 74 75 \n76 77 78 \n79 80 81 \n82 83 84 \n85 86 87 \n88 89 90 \n91 92 93 \n94 95 96 \n97 98 99 \n100 101 102 \n103 104 105 \n106 107 108 \n109 110 111 \n112 113 114 \n115 116 117 \n118 119 120 \n121 122 123 \n124 125 126 \n127 128 129 \n130 131 132 \n133..."
},
{
"input": "4227 91 80",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 \n82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "378 19 25",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 \n26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 \n51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 \n76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 \n101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 \n126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 \n151 152..."
},
{
"input": "2357 43 65",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 \n66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 \n131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "232 71 9",
"output": "1 2 3 4 5 6 7 8 9 \n10 11 12 13 14 15 16 17 18 \n19 20 21 22 23 24 25 26 27 \n28 29 30 31 32 33 34 35 36 \n37 38 39 40 41 42 43 44 45 \n46 47 48 49 50 51 52 53 54 \n55 56 57 58 59 60 61 62 63 \n64 65 66 67 68 69 70 71 72 \n73 74 75 76 77 78 79 80 81 \n82 83 84 85 86 87 88 89 90 \n91 92 93 94 95 96 97 98 99 \n100 101 102 103 104 105 106 107 108 \n109 110 111 112 113 114 115 116 117 \n118 119 120 121 122 123 124 125 126 \n127 128 129 130 131 132 133 134 135 \n136 137 138 139 140 141 142 143 144 \n145 146 147..."
},
{
"input": "2362 91 62",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 \n64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 \n125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "4601 59 78",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 \n80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "4439 74 60",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 \n62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 \n121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "3733 89 42",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 \n44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 \n85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 \n128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 1..."
},
{
"input": "335 12 28",
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{
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{
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{
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{
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},
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},
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},
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},
{
"input": "4056 52 78",
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"input": "3822 42 91",
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},
{
"input": "5358 57 94",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 \n96 95 98 97 100 99 102 101 104 103 106 105 108 107 110 109 112 111 114 113 116 115 118 117 120 119 122 121 124 123 126 125 128 127 130 129 132 131 134 133 136 135 138 137 140 139 142 141 144 143 146 145 148 147 150 149 152 151 154 153 1..."
},
{
"input": "2961 47 63",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 \n64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 \n127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
}
] | 1,461,774,360 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 5,222,400 | n, a, b = map(int, input().split())
if n > a * b:
print(-1)
else:
A = [[0 for i in range(b)] for j in range(a)]
s = 1
cht = 0
for i in range(a):
if cht % 2 == 0:
K = [i for i in range(s, s + b)]
else:
K = [i for i in range(s + b - 1, s - 1, -1)]
cht += 1
for j in range(b):
if s <= n:
A[i][j] = K[j]
s += 1
else:
break
for i in range(a):
for j in range(b):
print(A[i][j], end = ' ')
print() | Title: Parliament of Berland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* parliamentarians in Berland. They are numbered with integers from 1 to *n*. It happened that all parliamentarians with odd indices are Democrats and all parliamentarians with even indices are Republicans.
New parliament assembly hall is a rectangle consisting of *a*<=×<=*b* chairs — *a* rows of *b* chairs each. Two chairs are considered neighbouring if they share as side. For example, chair number 5 in row number 2 is neighbouring to chairs number 4 and 6 in this row and chairs with number 5 in rows 1 and 3. Thus, chairs have four neighbours in general, except for the chairs on the border of the hall
We know that if two parliamentarians from one political party (that is two Democrats or two Republicans) seat nearby they spent all time discussing internal party issues.
Write the program that given the number of parliamentarians and the sizes of the hall determine if there is a way to find a seat for any parliamentarian, such that no two members of the same party share neighbouring seats.
Input Specification:
The first line of the input contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=10<=000, 1<=≤<=*a*,<=*b*<=≤<=100) — the number of parliamentarians, the number of rows in the assembly hall and the number of seats in each row, respectively.
Output Specification:
If there is no way to assigns seats to parliamentarians in a proper way print -1.
Otherwise print the solution in *a* lines, each containing *b* integers. The *j*-th integer of the *i*-th line should be equal to the index of parliamentarian occupying this seat, or 0 if this seat should remain empty. If there are multiple possible solution, you may print any of them.
Demo Input:
['3 2 2\n', '8 4 3\n', '10 2 2\n']
Demo Output:
['0 3\n1 2\n', '7 8 3\n0 1 4\n6 0 5\n0 2 0\n', '-1\n']
Note:
In the first sample there are many other possible solutions. For example,
and
The following assignment
is incorrect, because parliamentarians 1 and 3 are both from Democrats party but will occupy neighbouring seats. | ```python
n, a, b = map(int, input().split())
if n > a * b:
print(-1)
else:
A = [[0 for i in range(b)] for j in range(a)]
s = 1
cht = 0
for i in range(a):
if cht % 2 == 0:
K = [i for i in range(s, s + b)]
else:
K = [i for i in range(s + b - 1, s - 1, -1)]
cht += 1
for j in range(b):
if s <= n:
A[i][j] = K[j]
s += 1
else:
break
for i in range(a):
for j in range(b):
print(A[i][j], end = ' ')
print()
``` | 0 | |
92 | A | Chips | PROGRAMMING | 800 | [
"implementation",
"math"
] | A. Chips | 2 | 256 | There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. | Print the number of chips the presenter ended up with. | [
"4 11\n",
"17 107\n",
"3 8\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | 500 | [
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "25 9712",
"output": "11"
},
{
"input": "9 7601",
"output": "5"
},
{
"input": "1 9058",
"output": "0"
},
{
"input": "29 7772",
"output": "26"
},
{
"input": "45 9465",
"output": "14"
},
{
"input": "46 866",
"output": "5"
},
{
"input": "29 1241",
"output": "20"
},
{
"input": "17 4248",
"output": "12"
},
{
"input": "20 8082",
"output": "11"
},
{
"input": "50 9555",
"output": "0"
},
{
"input": "4 7455",
"output": "2"
},
{
"input": "36 880",
"output": "4"
},
{
"input": "24 7440",
"output": "9"
},
{
"input": "44 7888",
"output": "12"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "50 10000",
"output": "40"
},
{
"input": "1 10000",
"output": "0"
},
{
"input": "50 1",
"output": "0"
},
{
"input": "50 50",
"output": "5"
}
] | 1,616,018,700 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 124 | 0 | if __name__ == '__main__':
n, m = map(int, input().split())
m %= n * (n + 1) // 2
for a in range(n):
a += 1
if m < a:
print(m)
break
m -= a
| Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | ```python
if __name__ == '__main__':
n, m = map(int, input().split())
m %= n * (n + 1) // 2
for a in range(n):
a += 1
if m < a:
print(m)
break
m -= a
``` | 3.969 |
988 | A | Diverse Team | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct.
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them. | The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student. | If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them.
Assume that the students are numbered from $1$ to $n$. | [
"5 3\n15 13 15 15 12\n",
"5 4\n15 13 15 15 12\n",
"4 4\n20 10 40 30\n"
] | [
"YES\n1 2 5 \n",
"NO\n",
"YES\n1 2 3 4 \n"
] | All possible answers for the first example:
- {1 2 5} - {2 3 5} - {2 4 5}
Note that the order does not matter. | 0 | [
{
"input": "5 3\n15 13 15 15 12",
"output": "YES\n1 2 5 "
},
{
"input": "5 4\n15 13 15 15 12",
"output": "NO"
},
{
"input": "4 4\n20 10 40 30",
"output": "YES\n1 2 3 4 "
},
{
"input": "1 1\n1",
"output": "YES\n1 "
},
{
"input": "100 53\n16 17 1 2 27 5 9 9 53 24 17 33 35 24 20 48 56 73 12 14 39 55 58 13 59 73 29 26 40 33 22 29 34 22 55 38 63 66 36 13 60 42 10 15 21 9 11 5 23 37 79 47 26 3 79 53 44 8 71 75 42 11 34 39 79 33 10 26 23 23 17 14 54 41 60 31 83 5 45 4 14 35 6 60 28 48 23 18 60 36 21 28 7 34 9 25 52 43 54 19",
"output": "YES\n1 2 3 4 5 6 7 9 10 12 13 15 16 17 18 19 20 21 22 23 24 25 27 28 29 31 33 36 37 38 39 41 42 43 44 45 47 49 50 51 52 54 57 58 59 60 73 74 76 77 79 80 83 "
},
{
"input": "2 2\n100 100",
"output": "NO"
},
{
"input": "2 2\n100 99",
"output": "YES\n1 2 "
},
{
"input": "100 100\n63 100 75 32 53 24 73 98 76 15 70 48 8 81 88 58 95 78 27 92 14 16 72 43 46 39 66 38 64 42 59 9 22 51 4 6 10 94 28 99 68 80 35 50 45 20 47 7 30 26 49 91 77 19 96 57 65 1 11 13 31 12 82 87 93 34 62 3 21 79 56 41 89 18 44 23 74 86 2 33 69 36 61 67 25 83 5 84 90 37 40 29 97 60 52 55 54 71 17 85",
"output": "YES\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "100 41\n54 16 42 3 45 6 9 72 100 13 24 57 35 5 89 13 97 27 43 9 73 89 48 16 48 55 18 15 55 28 30 6 18 41 100 61 9 42 35 54 57 25 73 15 42 54 49 5 72 48 30 55 4 43 94 5 60 92 93 23 89 75 53 92 74 93 89 28 69 6 3 49 15 28 49 57 54 55 30 57 69 18 89 6 25 23 93 74 30 13 87 53 6 42 4 54 60 30 4 35",
"output": "NO"
},
{
"input": "100 2\n70 64 70 32 70 64 32 70 64 32 32 64 70 64 64 32 64 64 64 70 70 64 64 64 64 70 32 64 70 64 32 70 70 70 64 70 64 70 64 32 70 32 70 64 64 64 32 70 64 70 70 32 70 32 32 32 70 32 70 32 64 64 70 32 32 64 70 64 32 32 64 64 32 32 70 70 32 70 32 64 32 70 64 64 32 64 32 64 70 32 70 32 70 64 64 64 70 70 64 70",
"output": "YES\n1 2 "
}
] | 1,623,913,973 | 173 | PyPy 3 | OK | TESTS | 10 | 93 | 0 | from sys import stdin,stdout
from itertools import accumulate
nmbr = lambda: int(input())
lst = lambda: list(map(int, input().split()))
for _ in range(1):#nmbr()):
# n=nmbr()
n,k=lst()
a=lst()
# print(fn(a,w))
s=set()
ans=[]
for i in range(n):
if a[i] in s:continue
else:
ans+=[i+1]
s.add(a[i])
if len(s)>=k:
print('YES')
print(*ans[:k])
else:print('NO') | Title: Diverse Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct.
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.
Input Specification:
The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student.
Output Specification:
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them.
Assume that the students are numbered from $1$ to $n$.
Demo Input:
['5 3\n15 13 15 15 12\n', '5 4\n15 13 15 15 12\n', '4 4\n20 10 40 30\n']
Demo Output:
['YES\n1 2 5 \n', 'NO\n', 'YES\n1 2 3 4 \n']
Note:
All possible answers for the first example:
- {1 2 5} - {2 3 5} - {2 4 5}
Note that the order does not matter. | ```python
from sys import stdin,stdout
from itertools import accumulate
nmbr = lambda: int(input())
lst = lambda: list(map(int, input().split()))
for _ in range(1):#nmbr()):
# n=nmbr()
n,k=lst()
a=lst()
# print(fn(a,w))
s=set()
ans=[]
for i in range(n):
if a[i] in s:continue
else:
ans+=[i+1]
s.add(a[i])
if len(s)>=k:
print('YES')
print(*ans[:k])
else:print('NO')
``` | 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"
}
] | 1,616,757,820 | 2,147,483,647 | PyPy 3 | OK | TESTS | 35 | 218 | 102,400 | inp = input().split()
m = int(inp[0])
n = int(inp[1])
print((n * m - ((n * m) % 2)) // 2)
| 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
inp = input().split()
m = int(inp[0])
n = int(inp[1])
print((n * m - ((n * m) % 2)) // 2)
``` | 3.945309 |
618 | A | Slime Combining | PROGRAMMING | 800 | [
"implementation"
] | null | null | Your friend recently gave you some slimes for your birthday. You have *n* slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other *n*<=-<=1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value *v*, you combine them together to create a slime with value *v*<=+<=1.
You would like to see what the final state of the row is after you've added all *n* slimes. Please print the values of the slimes in the row from left to right. | The first line of the input will contain a single integer, *n* (1<=≤<=*n*<=≤<=100<=000). | Output a single line with *k* integers, where *k* is the number of slimes in the row after you've finished the procedure described in the problem statement. The *i*-th of these numbers should be the value of the *i*-th slime from the left. | [
"1\n",
"2\n",
"3\n",
"8\n"
] | [
"1\n",
"2\n",
"2 1\n",
"4\n"
] | In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1 1. 2 1. 2 1 1. 3 1. 3 1 1. 3 2 1. 3 2 1 1. 4 | 500 | [
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "2 1"
},
{
"input": "8",
"output": "4"
},
{
"input": "100000",
"output": "17 16 11 10 8 6"
},
{
"input": "12345",
"output": "14 13 6 5 4 1"
},
{
"input": "32",
"output": "6"
},
{
"input": "70958",
"output": "17 13 11 9 6 4 3 2"
},
{
"input": "97593",
"output": "17 15 14 13 12 11 9 6 5 4 1"
},
{
"input": "91706",
"output": "17 15 14 11 10 6 5 4 2"
},
{
"input": "85371",
"output": "17 15 12 11 9 7 6 5 4 2 1"
},
{
"input": "97205",
"output": "17 15 14 13 12 10 9 8 6 5 3 1"
},
{
"input": "34768",
"output": "16 11 10 9 8 7 5"
},
{
"input": "12705",
"output": "14 13 9 8 6 1"
},
{
"input": "30151",
"output": "15 14 13 11 9 8 7 3 2 1"
},
{
"input": "4974",
"output": "13 10 9 7 6 4 3 2"
},
{
"input": "32728",
"output": "15 14 13 12 11 10 9 8 7 5 4"
},
{
"input": "8192",
"output": "14"
},
{
"input": "65536",
"output": "17"
},
{
"input": "32",
"output": "6"
},
{
"input": "256",
"output": "9"
},
{
"input": "4096",
"output": "13"
},
{
"input": "33301",
"output": "16 10 5 3 1"
},
{
"input": "16725",
"output": "15 9 7 5 3 1"
},
{
"input": "149",
"output": "8 5 3 1"
},
{
"input": "16277",
"output": "14 13 12 11 10 9 8 5 3 1"
},
{
"input": "99701",
"output": "17 16 11 9 7 6 5 3 1"
}
] | 1,454,087,659 | 259 | Python 3 | COMPILATION_ERROR | PRETESTS | 0 | 0 | 0 | #include <cstdio>
int main()
{
int N;
scanf("%d", &N);
for (int i=20; i>=0; i--) {
if (N>=(1<<i)) {
printf("%d ", i+1);
N -= 1<<i;
}
}
}
| Title: Slime Combining
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend recently gave you some slimes for your birthday. You have *n* slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other *n*<=-<=1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value *v*, you combine them together to create a slime with value *v*<=+<=1.
You would like to see what the final state of the row is after you've added all *n* slimes. Please print the values of the slimes in the row from left to right.
Input Specification:
The first line of the input will contain a single integer, *n* (1<=≤<=*n*<=≤<=100<=000).
Output Specification:
Output a single line with *k* integers, where *k* is the number of slimes in the row after you've finished the procedure described in the problem statement. The *i*-th of these numbers should be the value of the *i*-th slime from the left.
Demo Input:
['1\n', '2\n', '3\n', '8\n']
Demo Output:
['1\n', '2\n', '2 1\n', '4\n']
Note:
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1 1. 2 1. 2 1 1. 3 1. 3 1 1. 3 2 1. 3 2 1 1. 4 | ```python
#include <cstdio>
int main()
{
int N;
scanf("%d", &N);
for (int i=20; i>=0; i--) {
if (N>=(1<<i)) {
printf("%d ", i+1);
N -= 1<<i;
}
}
}
``` | -1 | |
443 | A | Anton and Letters | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set. | The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space. | Print a single number — the number of distinct letters in Anton's set. | [
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] | [
"3\n",
"2\n",
"0\n"
] | none | 500 | [
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,684,541,545 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | def count_distinct_letters(s):
letters = s[1:-1].split(', ')
distinct_letters = set(letters)
print(len(distinct_letters))
| Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none | ```python
def count_distinct_letters(s):
letters = s[1:-1].split(', ')
distinct_letters = set(letters)
print(len(distinct_letters))
``` | 0 | |
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..."
}
] | 1,664,909,629 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 7 | 108 | 1,331,200 | import sys
# sys.stdin = open("input.txt", "r")
n = int(input())
fib = [1, 1]
for i in range(2, n):
fib.append(fib[i - 1] + fib[i - 2])
res = ""
for i in range(1, n+1):
if i in fib:
res += "O"
else:
res += "o"
print(res) | 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
import sys
# sys.stdin = open("input.txt", "r")
n = int(input())
fib = [1, 1]
for i in range(2, n):
fib.append(fib[i - 1] + fib[i - 2])
res = ""
for i in range(1, n+1):
if i in fib:
res += "O"
else:
res += "o"
print(res)
``` | 0 | |
794 | B | Cutting Carrot | PROGRAMMING | 1,200 | [
"geometry",
"math"
] | null | null | Igor the analyst has adopted *n* little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into *n* pieces of equal area.
Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to *h*. Igor wants to make *n*<=-<=1 cuts parallel to the base to cut the carrot into *n* pieces. He wants to make sure that all *n* pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area? | The first and only line of input contains two space-separated integers, *n* and *h* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=105). | The output should contain *n*<=-<=1 real numbers *x*1,<=*x*2,<=...,<=*x**n*<=-<=1. The number *x**i* denotes that the *i*-th cut must be made *x**i* units away from the apex of the carrot. In addition, 0<=<<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=-<=1<=<<=*h* must hold.
Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10<=-<=6.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if . | [
"3 2\n",
"2 100000\n"
] | [
"1.154700538379 1.632993161855\n",
"70710.678118654752\n"
] | Definition of isosceles triangle: [https://en.wikipedia.org/wiki/Isosceles_triangle](https://en.wikipedia.org/wiki/Isosceles_triangle). | 1,000 | [
{
"input": "3 2",
"output": "1.154700538379 1.632993161855"
},
{
"input": "2 100000",
"output": "70710.678118654752"
},
{
"input": "1000 100000",
"output": "3162.277660168379 4472.135954999579 5477.225575051661 6324.555320336759 7071.067811865475 7745.966692414834 8366.600265340755 8944.271909999159 9486.832980505138 10000.000000000000 10488.088481701515 10954.451150103322 11401.754250991380 11832.159566199232 12247.448713915890 12649.110640673517 13038.404810405297 13416.407864998738 13784.048752090222 14142.135623730950 14491.376746189439 14832.396974191326 15165.750888103101 15491.933384829668 15811.388300841897 16124.515496597099 16431.676725154983 16733.2..."
},
{
"input": "2 1",
"output": "0.707106781187"
},
{
"input": "1000 1",
"output": "0.031622776602 0.044721359550 0.054772255751 0.063245553203 0.070710678119 0.077459666924 0.083666002653 0.089442719100 0.094868329805 0.100000000000 0.104880884817 0.109544511501 0.114017542510 0.118321595662 0.122474487139 0.126491106407 0.130384048104 0.134164078650 0.137840487521 0.141421356237 0.144913767462 0.148323969742 0.151657508881 0.154919333848 0.158113883008 0.161245154966 0.164316767252 0.167332005307 0.170293863659 0.173205080757 0.176068168617 0.178885438200 0.181659021246 0.184390889146 0..."
},
{
"input": "20 17",
"output": "3.801315561750 5.375872022286 6.584071688553 7.602631123499 8.500000000000 9.311283477588 10.057335631269 10.751744044572 11.403946685249 12.020815280171 12.607537428063 13.168143377105 13.705838172108 14.223220451079 14.722431864335 15.205262246999 15.673225577398 16.127616066859 16.569550386175"
},
{
"input": "999 1",
"output": "0.031638599858 0.044743737014 0.054799662435 0.063277199717 0.070746059996 0.077498425829 0.083707867056 0.089487474029 0.094915799575 0.100050037531 0.104933364623 0.109599324870 0.114074594073 0.118380800867 0.122535770349 0.126554399434 0.130449289063 0.134231211043 0.137909459498 0.141492119993 0.144986278734 0.148398187395 0.151733394554 0.154996851658 0.158192999292 0.161325838061 0.164398987305 0.167415734111 0.170379074505 0.173291748303 0.176156268782 0.178974948057 0.181749918935 0.184483153795 0..."
},
{
"input": "998 99999",
"output": "3165.413034717700 4476.570044210349 5482.656203071844 6330.826069435401 7078.078722492680 7753.646760213179 8374.895686665300 8953.140088420697 9496.239104153101 10009.914924893578 10498.487342658843 10965.312406143687 11413.059004696742 11843.891063542002 12259.591967329534 12661.652138870802 13051.332290848021 13429.710132631046 13797.715532900862 14156.157444985360 14505.744837393740 14847.103184390411 15180.787616204127 15507.293520426358 15827.065173588502 16140.502832606510 16447.968609215531 16749.7..."
},
{
"input": "574 29184",
"output": "1218.116624752432 1722.677051277028 2109.839883615525 2436.233249504864 2723.791577469041 2983.764177844748 3222.833656968322 3445.354102554056 3654.349874257297 3852.022989934325 4040.035795197963 4219.679767231051 4391.981950040022 4557.775066957079 4717.745401404559 4872.466499009729 5022.423508175150 5168.031153831084 5309.647268742708 5447.583154938083 5582.111638212139 5713.473414041731 5841.882108059006 5967.528355689497 6090.583123762161 6211.200439444432 6329.519650846576 6445.667313936643 6559.75..."
},
{
"input": "2 5713",
"output": "4039.701040918746"
},
{
"input": "937 23565",
"output": "769.834993893392 1088.711089153444 1333.393322867831 1539.669987786784 1721.403377803760 1885.702921177414 2036.791944396843 2177.422178306887 2309.504981680176 2434.432003204934 2553.253825229922 2666.786645735663 2775.679544129132 2880.458791498282 2981.558110676796 3079.339975573568 3174.110994119182 3266.133267460331 3355.632941582547 3442.806755607520 3527.827132142336 3610.846187821139 3691.998931463184 3771.405842354828 3849.174969466960 3925.403656108988 4000.179968603494 4073.583888793686 4145.688..."
},
{
"input": "693 39706",
"output": "1508.306216302128 2133.067107306117 2612.463000007259 3016.612432604256 3372.675230537060 3694.580605808168 3990.603149268227 4266.134214612233 4524.918648906384 4769.683052505315 5002.485788434792 5224.926000014517 5438.275401978402 5643.565095743912 5841.644856719264 6033.224865208513 6218.905845589392 6399.201321918350 6574.554372775177 6745.350461074120 6911.927407376938 7074.583247583148 7233.582498950279 7389.161211616337 7541.531081510641 7690.882829397851 7837.389000021776 7981.206298536455 8122.47..."
},
{
"input": "449 88550",
"output": "4178.932872810542 5909.903544975429 7238.124057127628 8357.865745621084 9344.377977012855 10236.253207728862 11056.417127089408 11819.807089950858 12536.798618431626 13214.946067032045 13859.952363194553 14476.248114255256 15067.356749640443 15636.135052384012 16184.937421313947 16715.731491242168 17230.181636963718 17729.710634926286 18215.546084421264 18688.755954025709 19150.276213793575 19600.932605874766 20041.458005232581 20472.506415457724 20894.664364052710 21308.460264455309 21714.372171382883 221..."
},
{
"input": "642 37394",
"output": "1475.823459881026 2087.129552632132 2556.201215516026 2951.646919762052 3300.041579082908 3615.014427137354 3904.661853880105 4174.259105264265 4427.470379643078 4666.963557534173 4894.752673229489 5112.402431032051 5321.157158133711 5522.025750238117 5715.839682061424 5903.293839524104 6084.976009853978 6261.388657896397 6432.965320127946 6600.083158165816 6763.072717296425 6922.225614943105 7077.800671741869 7230.028854274709 7379.117299405130 7525.252620551370 7668.603646548077 7809.323707760210 7947.55..."
},
{
"input": "961 53535",
"output": "1726.935483870968 2442.255582633666 2991.139999458060 3453.870967741935 3861.545134691976 4230.110754190240 4569.041820575576 4884.511165267332 5180.806451612903 5461.049501197232 5727.597037150849 5982.279998916119 6226.554436514989 6461.600909707837 6688.392369006905 6907.741935483871 7120.337408627144 7326.766747900998 7527.537256208063 7723.090269383951 7913.812575143900 8100.045409746687 8282.091632275692 8460.221508380480 8634.677419354839 8805.677730973862 8973.419998374179 9138.083641151152 9299.83..."
},
{
"input": "4 31901",
"output": "15950.500000000000 22557.413426632053 27627.076406127377"
},
{
"input": "4 23850",
"output": "11925.000000000000 16864.496731299158 20654.705880258862"
},
{
"input": "4 72694",
"output": "36347.000000000000 51402.420351574886 62954.850702705983"
},
{
"input": "4 21538",
"output": "10769.000000000000 15229.665853195861 18652.455146709240"
},
{
"input": "4 70383",
"output": "35191.500000000000 49768.296580252774 60953.465994560145"
},
{
"input": "5 1",
"output": "0.447213595500 0.632455532034 0.774596669241 0.894427191000"
},
{
"input": "5 1",
"output": "0.447213595500 0.632455532034 0.774596669241 0.894427191000"
},
{
"input": "5 1",
"output": "0.447213595500 0.632455532034 0.774596669241 0.894427191000"
},
{
"input": "5 1",
"output": "0.447213595500 0.632455532034 0.774596669241 0.894427191000"
},
{
"input": "5 1",
"output": "0.447213595500 0.632455532034 0.774596669241 0.894427191000"
},
{
"input": "20 1",
"output": "0.223606797750 0.316227766017 0.387298334621 0.447213595500 0.500000000000 0.547722557505 0.591607978310 0.632455532034 0.670820393250 0.707106781187 0.741619848710 0.774596669241 0.806225774830 0.836660026534 0.866025403784 0.894427191000 0.921954445729 0.948683298051 0.974679434481"
},
{
"input": "775 1",
"output": "0.035921060405 0.050800050800 0.062217101684 0.071842120811 0.080321932890 0.087988269013 0.095038192662 0.101600101600 0.107763181216 0.113592366849 0.119136679436 0.124434203368 0.129515225161 0.134404301006 0.139121668728 0.143684241621 0.148106326235 0.152400152400 0.156576272252 0.160643865780 0.164610978351 0.168484707835 0.172271353843 0.175976538026 0.179605302027 0.183162187956 0.186651305051 0.190076385325 0.193440830330 0.196747750735 0.200000000000 0.203200203200 0.206350781829 0.209453975235 0..."
},
{
"input": "531 1",
"output": "0.043396303660 0.061371641193 0.075164602800 0.086792607321 0.097037084957 0.106298800691 0.114815827305 0.122743282386 0.130188910981 0.137231161599 0.143929256529 0.150329205601 0.156467598013 0.162374100149 0.168073161363 0.173585214641 0.178927543753 0.184114923580 0.189160102178 0.194074169913 0.198866846404 0.203546706606 0.208121361089 0.212597601381 0.216981518301 0.221278599182 0.225493808401 0.229631654609 0.233696247231 0.237691344271 0.241620392998 0.245486564773 0.249292785005 0.253041759057 0..."
},
{
"input": "724 1",
"output": "0.037164707312 0.052558833123 0.064371161313 0.074329414625 0.083102811914 0.091034569355 0.098328573097 0.105117666246 0.111494121937 0.117525123681 0.123261389598 0.128742322627 0.133999257852 0.139057601643 0.143938292487 0.148658829249 0.153234013794 0.157676499368 0.161997203441 0.166205623829 0.170310084440 0.174317928887 0.178235674883 0.182069138710 0.185823536562 0.189503567803 0.193113483940 0.196657146194 0.200138073886 0.203559485381 0.206924332929 0.210235332491 0.213494989396 0.216705620524 0..."
},
{
"input": "917 1",
"output": "0.033022909334 0.046701446249 0.057197356781 0.066045818668 0.073841470086 0.080889277691 0.087370405666 0.093402892499 0.099068728003 0.104427608461 0.109524599747 0.114394713561 0.119065792869 0.123560412643 0.127897177895 0.132091637337 0.136156943250 0.140104338748 0.143943524609 0.147682940172 0.151329981692 0.154891174376 0.158372309576 0.161778555382 0.165114546671 0.168384459091 0.171592070342 0.174740811332 0.177833809176 0.180873923568 0.183863777748 0.186805784998 0.189702171441 0.192554995756 0..."
},
{
"input": "458 100",
"output": "4.672693135160 6.608186004551 8.093341918275 9.345386270320 10.448459488214 11.445713905748 12.362783988552 13.216372009102 14.018079405480 14.776353114139 15.497569889795 16.186683836551 16.847634693328 17.483616785299 18.097262694412 18.690772540640 19.266007352363 19.824558013653 20.367797170339 20.896918976429 21.412969991171 21.916873521973 22.409449036367 22.891427811495 23.363465675800 23.826153477212 24.280025754826 24.725567977104 25.163222626003 25.593394344267 26.016454316384 26.432744018204 26...."
}
] | 1,573,475,255 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 109 | 0 | n,h=map(int,input().split())
for i in range(n-1,0,-1):
t=((h*h*(n-i))/n)**0.5
print(t,end=' ') | Title: Cutting Carrot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Igor the analyst has adopted *n* little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into *n* pieces of equal area.
Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to *h*. Igor wants to make *n*<=-<=1 cuts parallel to the base to cut the carrot into *n* pieces. He wants to make sure that all *n* pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?
Input Specification:
The first and only line of input contains two space-separated integers, *n* and *h* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=105).
Output Specification:
The output should contain *n*<=-<=1 real numbers *x*1,<=*x*2,<=...,<=*x**n*<=-<=1. The number *x**i* denotes that the *i*-th cut must be made *x**i* units away from the apex of the carrot. In addition, 0<=<<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=-<=1<=<<=*h* must hold.
Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10<=-<=6.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
Demo Input:
['3 2\n', '2 100000\n']
Demo Output:
['1.154700538379 1.632993161855\n', '70710.678118654752\n']
Note:
Definition of isosceles triangle: [https://en.wikipedia.org/wiki/Isosceles_triangle](https://en.wikipedia.org/wiki/Isosceles_triangle). | ```python
n,h=map(int,input().split())
for i in range(n-1,0,-1):
t=((h*h*(n-i))/n)**0.5
print(t,end=' ')
``` | 3 | |
946 | C | String Transformation | PROGRAMMING | 1,300 | [
"greedy",
"strings"
] | null | null | You are given a string *s* consisting of |*s*| small english letters.
In one move you can replace any character of this string to the next character in alphabetical order (a will be replaced with b, s will be replaced with t, etc.). You cannot replace letter z with any other letter.
Your target is to make some number of moves (not necessary minimal) to get string abcdefghijklmnopqrstuvwxyz (english alphabet) as a subsequence. Subsequence of the string is the string that is obtained by deleting characters at some positions. You need to print the string that will be obtained from the given string and will be contain english alphabet as a subsequence or say that it is impossible. | The only one line of the input consisting of the string *s* consisting of |*s*| (1<=≤<=|*s*|<=≤<=105) small english letters. | If you can get a string that can be obtained from the given string and will contain english alphabet as a subsequence, print it. Otherwise print «-1» (without quotes). | [
"aacceeggiikkmmooqqssuuwwyy\n",
"thereisnoanswer\n"
] | [
"abcdefghijklmnopqrstuvwxyz\n",
"-1\n"
] | none | 0 | [
{
"input": "aacceeggiikkmmooqqssuuwwyy",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "thereisnoanswer",
"output": "-1"
},
{
"input": "jqcfvsaveaixhioaaeephbmsmfcgdyawscpyioybkgxlcrhaxs",
"output": "-1"
},
{
"input": "rtdacjpsjjmjdhcoprjhaenlwuvpfqzurnrswngmpnkdnunaendlpbfuylqgxtndhmhqgbsknsy",
"output": "-1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaa"
},
{
"input": "abcdefghijklmnopqrstuvwxxx",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrstuvwxya",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "cdaaaaaaaaabcdjklmnopqrstuvwxyzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "cdabcdefghijklmnopqrstuvwxyzxyzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"
},
{
"input": "zazaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "zazbcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "abbbefghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaa"
},
{
"input": "abcdefghijklmaopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrstuvwxyx",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaz",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaz"
},
{
"input": "zaaaazaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "zabcdzefghijklmnopqrstuvwxyzaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaa"
},
{
"input": "aaaaaafghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyzz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaz",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaz"
},
{
"input": "abcdefghijklmnopqrstuvwaxy",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaa"
},
{
"input": "abcdefghijklmnapqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrstuvnxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaa"
},
{
"input": "abcdefghijklmnopqrstuvwxyzzzz",
"output": "abcdefghijklmnopqrstuvwxyzzzz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aacceeggiikkmmooqqssuuwwya",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aacdefghijklmnopqrstuvwxyyy",
"output": "abcdefghijklmnopqrstuvwxyzy"
},
{
"input": "abcaefghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "zaaacaaaaaaaaaaaaaaaaaaaayy",
"output": "zabcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdedccdcdccdcdcdcdcdcddccdcdcdc",
"output": "abcdefghijklmnopqrstuvwxyzcdcdcdc"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "abcdecdcdcddcdcdcdcdcdcdcd",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "abaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "a",
"output": "-1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaa"
},
{
"input": "aaadefghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaa"
},
{
"input": "abbbbbbbbbbbbbbbbbbbbbbbbz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aacceeggiikkmmaacceeggiikkmmooaacceeggiikkmmaacceeggiikkmmooqqssuuwwzy",
"output": "abcdefghijklmnopqrstuvwxyzmmooaacceeggiikkmmaacceeggiikkmmooqqssuuwwzy"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "phqghumeaylnlfdxfircvscxggbwkfnqduxwfnfozvsrtkjprepggxrpnrvystmwcysyycqpevikeffmznimkkasvwsrenzkycxf",
"output": "-1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaap",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "zabcdefghijklmnopqrstuvwxyz",
"output": "zabcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyza"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzabcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "rveviaomdienfygifatviahordebxazoxflfgzslhyzowhxbhqzpsgellkoimnwkvhpbijorhpggwfjexivpqbcbmqjyghkbq",
"output": "rveviaomdienfygifbtvichordefxgzoxhlijzslkyzowlxmnqzpsopqrstuvwxyzhpbijorhpggwfjexivpqbcbmqjyghkbq"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "xtlsgypsfadpooefxzbcoejuvpvaboygpoeylfpbnpljvrvipyamyehwqnqrqpmxujjloovaowuxwhmsncbxcoksfzkvatxdknly",
"output": "xtlsgypsfadpooefxzbcoejuvpvdeoygpofylgphnpljvrvipyjmyklwqnqrqpmxunopqrvstwuxwvwxyzbxcoksfzkvatxdknly"
},
{
"input": "jqcfvsaveaixhioaaeephbmsmfcgdyawscpyioybkgxlcrhaxsa",
"output": "jqcfvsavebixhiocdefphgmsmhijkylwsmpynoypqrxstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "wlrbbmqbhcdarzowkkyhiddqscdxrjmowfrxsjybldbefsarcbynecdyggxxpklorellnmpapqfwkhopkmcoqh",
"output": "wlrbbmqbhcdarzowkkyhiddqscdxrjmowfrxsjybldcefsdrefynghiyjkxxplmornopqrstuvwxyzopkmcoqh"
},
{
"input": "abadefghijklmnopqrstuvwxyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "zazsazcbbbbbbbbbbbbbbbbbbbbbbb",
"output": "zazsbzcdefghijklmnopqrstuvwxyz"
},
{
"input": "zazsazcbbbbbbbbbbbbbbbbbbbbbyb",
"output": "zazsbzcdefghijklmnopqrstuvwxyz"
},
{
"input": "bbcdefghijklmnopqrstuvwxyzzz",
"output": "-1"
},
{
"input": "zaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "zabcdefghijklmnopqrstuvwxyz"
},
{
"input": "zzzzzaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "zzzzzabcdefghijklmnopqrstuvwxyza"
},
{
"input": "kkimnfjbbgggicykcciwtoazomcvisigagkjwhyrmojmoebnqoadpmockfjxibdtvrbedrsdoundbcpkfdqdidqdmxdltink",
"output": "kkimnfjbbgggicykcciwtoazomcvisigbgkjwhyrmojmoecnqodepmofkgjxihitvrjklrsmounopqrstuvwxyzdmxdltink"
},
{
"input": "cawgathqceccscakbazmhwbefvygjbcfyihcbgga",
"output": "-1"
},
{
"input": "acrsbyszsbfslzbqzzamcmrypictkcheddehvxdipaxaannjodzyfxgtfnwababzjraapqbqbfzhbiewlzz",
"output": "acrsbyszscfslzdqzzemfmrypigtkhijklmnvxopqrxstuvwxyzyfxgtfnwababzjraapqbqbfzhbiewlzz"
},
{
"input": "ggcebbheeblbioxdvtlrtkxeuilonazpebcbqpzz",
"output": "-1"
},
{
"input": "zzzzabcdefghijklmnopqrstuvwxy",
"output": "-1"
},
{
"input": "zabcdefghijklmnopqrstuvwxy",
"output": "-1"
},
{
"input": "babcdefghijklmnopqrstuvwxyz",
"output": "babcdefghijklmnopqrstuvwxyz"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "-1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "abcdefghijklmnopqrstuvwxyzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
}
] | 1,617,079,415 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cstring>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <algorithm>
using namespace std;
#define LL long long
const int maxn = 100000 + 100;
char str[maxn];
int main() {
while(scanf("%s", str) != EOF) {
char ch = 'a';
for(int i = 0; str[i]; ++i) {
if(str[i] <= ch) {
if(ch <= 'z') {
str[i] = ch;
++ch;
}
}
}
if(ch == 'z' + 1) {
printf("%s\n", str);
} else {
printf("-1\n");
}
}
return 0;
}
| Title: String Transformation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s* consisting of |*s*| small english letters.
In one move you can replace any character of this string to the next character in alphabetical order (a will be replaced with b, s will be replaced with t, etc.). You cannot replace letter z with any other letter.
Your target is to make some number of moves (not necessary minimal) to get string abcdefghijklmnopqrstuvwxyz (english alphabet) as a subsequence. Subsequence of the string is the string that is obtained by deleting characters at some positions. You need to print the string that will be obtained from the given string and will be contain english alphabet as a subsequence or say that it is impossible.
Input Specification:
The only one line of the input consisting of the string *s* consisting of |*s*| (1<=≤<=|*s*|<=≤<=105) small english letters.
Output Specification:
If you can get a string that can be obtained from the given string and will contain english alphabet as a subsequence, print it. Otherwise print «-1» (without quotes).
Demo Input:
['aacceeggiikkmmooqqssuuwwyy\n', 'thereisnoanswer\n']
Demo Output:
['abcdefghijklmnopqrstuvwxyz\n', '-1\n']
Note:
none | ```python
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cstring>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <algorithm>
using namespace std;
#define LL long long
const int maxn = 100000 + 100;
char str[maxn];
int main() {
while(scanf("%s", str) != EOF) {
char ch = 'a';
for(int i = 0; str[i]; ++i) {
if(str[i] <= ch) {
if(ch <= 'z') {
str[i] = ch;
++ch;
}
}
}
if(ch == 'z' + 1) {
printf("%s\n", str);
} else {
printf("-1\n");
}
}
return 0;
}
``` | -1 | |
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"
}
] | 1,617,866,517 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 124 | 0 | user_input = input()
countlower = 0
countupper = 0
for i in user_input:
if i.isupper():
countupper += 1
else:
countlower += 1
if countupper > countlower:
print(user_input.upper())
else:
print(user_input.lower()) | 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
user_input = input()
countlower = 0
countupper = 0
for i in user_input:
if i.isupper():
countupper += 1
else:
countlower += 1
if countupper > countlower:
print(user_input.upper())
else:
print(user_input.lower())
``` | 3.969 |
984 | B | Minesweeper | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | One day Alex decided to remember childhood when computers were not too powerful and lots of people played only default games. Alex enjoyed playing Minesweeper that time. He imagined that he saved world from bombs planted by terrorists, but he rarely won.
Alex has grown up since then, so he easily wins the most difficult levels. This quickly bored him, and he thought: what if the computer gave him invalid fields in the childhood and Alex could not win because of it?
He needs your help to check it.
A Minesweeper field is a rectangle $n \times m$, where each cell is either empty, or contains a digit from $1$ to $8$, or a bomb. The field is valid if for each cell:
- if there is a digit $k$ in the cell, then exactly $k$ neighboring cells have bombs. - if the cell is empty, then all neighboring cells have no bombs.
Two cells are neighbors if they have a common side or a corner (i. e. a cell has at most $8$ neighboring cells). | The first line contains two integers $n$ and $m$ ($1 \le n, m \le 100$) — the sizes of the field.
The next $n$ lines contain the description of the field. Each line contains $m$ characters, each of them is "." (if this cell is empty), "*" (if there is bomb in this cell), or a digit from $1$ to $8$, inclusive. | Print "YES", if the field is valid and "NO" otherwise.
You can choose the case (lower or upper) for each letter arbitrarily. | [
"3 3\n111\n1*1\n111\n",
"2 4\n*.*.\n1211\n"
] | [
"YES",
"NO"
] | In the second example the answer is "NO" because, if the positions of the bombs are preserved, the first line of the field should be *2*1.
You can read more about Minesweeper in [Wikipedia's article](https://en.wikipedia.org/wiki/Minesweeper_(video_game)). | 1,000 | [
{
"input": "3 3\n111\n1*1\n111",
"output": "YES"
},
{
"input": "2 4\n*.*.\n1211",
"output": "NO"
},
{
"input": "1 10\n.....1*1..",
"output": "YES"
},
{
"input": "1 1\n4",
"output": "NO"
},
{
"input": "10 10\n..........\n...111111.\n..13*21*1.\n.12**2111.\n.1*542..11\n.13**1..1*\n..2*31..11\n..111..111\n.......1*1\n.......111",
"output": "YES"
},
{
"input": "10 17\n12*2*22123*31....\n2*333*3*4***3211.\n*22*213**4***3*1.\n11111.12224*6*21.\n221..111.14**4311\n**2233*212****2*1\n*55***4*13*544421\n2***54*322*21**31\n13*4*33*221114*4*\n.1122*22*1...2*31",
"output": "YES"
},
{
"input": "10 10\n**********\n**********\n**********\n**********\n**********\n******3***\n**********\n**********\n**********\n***3.5****",
"output": "NO"
},
{
"input": "21 10\n62637783*1\n23*51**531\n35*7*6.**.\n.*3***581*\n2.32*745**\n83*7*6*6*5\n*74.**6**3\n323*6**7*6\n3454*67.*1\n**63265*6*\n3725*4553*\n24****5**4\n23.34****4\n55257*1*4*\n4*3253*456\n**.3*45488\n*7318**4*5\n234.*4557*\n12..21*.*3\n286.225*4*\n834*11*.3*",
"output": "NO"
},
{
"input": "10 10\n**********\n*********6\n*********5\n**********\n**********\n**********\n**********\n**********\n**********\n**********",
"output": "NO"
},
{
"input": "100 1\n.\n.\n.\n.\n1\n*\n2\n*\n1\n.\n.\n.\n.\n.\n.\n1\n*\n1\n1\n*\n1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n1\n*\n2\n*\n*\n*\n1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n1\n*\n2\n*\n1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.",
"output": "YES"
},
{
"input": "1 100\n*************5****5****************************************************4****************************",
"output": "NO"
},
{
"input": "1 100\n.....1*1........1*1................................1*1...1**11*1.......1*1....1.....1*1.....1*1...1*",
"output": "NO"
},
{
"input": "1 10\n881111882*",
"output": "NO"
},
{
"input": "5 5\n*2221\n24**2\n*3*5*\n3425*\n**12*",
"output": "NO"
},
{
"input": "5 5\n****2\n4***4\n3****\n3*563\n*22**",
"output": "NO"
},
{
"input": "5 5\n***2.\n5**31\n**6**\n***43\n**31*",
"output": "NO"
},
{
"input": "5 5\n*32**\n4*3*4\n**44*\n**45*\n*4***",
"output": "NO"
},
{
"input": "3 3\n***\n*2*\n***",
"output": "NO"
},
{
"input": "1 1\n*",
"output": "YES"
},
{
"input": "1 2\n*1",
"output": "YES"
},
{
"input": "1 2\n*2",
"output": "NO"
},
{
"input": "2 2\n32\n**",
"output": "NO"
},
{
"input": "3 3\n...\n232\n***",
"output": "YES"
},
{
"input": "3 2\n..\n11\n.*",
"output": "NO"
},
{
"input": "2 3\n1*2\n3*2",
"output": "NO"
},
{
"input": "1 3\n.*.",
"output": "NO"
},
{
"input": "3 1\n.\n*\n.",
"output": "NO"
},
{
"input": "3 1\n1\n*\n1",
"output": "YES"
},
{
"input": "3 1\n*\n1\n*",
"output": "NO"
},
{
"input": "1 3\n1**",
"output": "YES"
},
{
"input": "1 1\n8",
"output": "NO"
},
{
"input": "1 1\n.",
"output": "YES"
},
{
"input": "1 2\n2*",
"output": "NO"
},
{
"input": "2 1\n*\n2",
"output": "NO"
},
{
"input": "2 1\n*\n*",
"output": "YES"
},
{
"input": "2 1\n.\n1",
"output": "NO"
},
{
"input": "1 3\n..1",
"output": "NO"
},
{
"input": "3 3\n112\n1*1\n111",
"output": "NO"
},
{
"input": "3 3\n11.\n1*1\n111",
"output": "NO"
},
{
"input": "3 3\n151\n1*1\n111",
"output": "NO"
},
{
"input": "3 3\n1.1\n1*1\n111",
"output": "NO"
},
{
"input": "3 3\n611\n1*1\n111",
"output": "NO"
},
{
"input": "3 3\n.11\n1*1\n111",
"output": "NO"
},
{
"input": "3 3\n111\n2*1\n111",
"output": "NO"
},
{
"input": "3 3\n111\n**1\n111",
"output": "NO"
},
{
"input": "3 3\n111\n5*1\n111",
"output": "NO"
},
{
"input": "3 3\n111\n.*1\n111",
"output": "NO"
},
{
"input": "3 3\n111\n1*1\n411",
"output": "NO"
},
{
"input": "3 3\n111\n1*1\n.11",
"output": "NO"
},
{
"input": "3 3\n111\n1*1\n121",
"output": "NO"
},
{
"input": "3 3\n111\n1*1\n1.1",
"output": "NO"
},
{
"input": "3 3\n111\n1*1\n115",
"output": "NO"
},
{
"input": "3 3\n111\n1*1\n11.",
"output": "NO"
},
{
"input": "3 3\n111\n1*4\n111",
"output": "NO"
},
{
"input": "3 3\n111\n1*.\n111",
"output": "NO"
},
{
"input": "1 4\n*222",
"output": "NO"
},
{
"input": "1 1\n*",
"output": "YES"
},
{
"input": "2 2\n**\n**",
"output": "YES"
},
{
"input": "3 3\n***\n*.*\n***",
"output": "NO"
},
{
"input": "3 3\n***\n***\n***",
"output": "YES"
},
{
"input": "3 3\n***\n*4*\n***",
"output": "NO"
},
{
"input": "2 6\n....1.\n.....*",
"output": "NO"
},
{
"input": "5 5\n*****\n*****\n*****\n*****\n*****",
"output": "YES"
},
{
"input": "4 1\n.\n*\n.\n.",
"output": "NO"
},
{
"input": "2 2\n2*\n11",
"output": "NO"
},
{
"input": "3 3\n888\n888\n888",
"output": "NO"
},
{
"input": "1 4\n*22*",
"output": "NO"
}
] | 1,592,053,133 | 2,147,483,647 | PyPy 3 | OK | TESTS | 84 | 217 | 3,481,600 | import math
from collections import defaultdict
ml=lambda:map(int,input().split())
ll=lambda:list(map(int,input().split()))
ii=lambda:int(input())
ip=lambda:list(input())
"""========main code==============="""
a,b=ml()
lol=[]
for i in range(a):
w=list(input())
lol.append(w)
flag=0
for i in range(a):
for j in range(b):
k=0
if(i-1>=0):
if(lol[i-1][j]=='*'):
k+=1
if(j-1>=0):
if(lol[i-1][j-1]=='*'):
k+=1
if(j+1<b):
if(lol[i-1][j+1]=='*'):
k+=1
if((i+1)<a):
if(lol[i+1][j]=='*'):
k+=1
if(j-1>=0):
if(lol[i+1][j-1]=='*'):
k+=1
if((j+1)<b):
if(lol[i+1][j+1]=='*'):
k+=1
if(j-1>=0):
if(lol[i][j-1]=='*'):
k+=1
if(j+1<b):
if(lol[i][j+1]=='*'):
k+=1
if(lol[i][j] in {'1','2','3','4','5','6','7','8'}):
if(int(lol[i][j])!=k):
flag=1
elif(lol[i][j]=='.' and k!=0):
flag=1
if( flag):
print("NO")
else:
print("YES")
| Title: Minesweeper
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Alex decided to remember childhood when computers were not too powerful and lots of people played only default games. Alex enjoyed playing Minesweeper that time. He imagined that he saved world from bombs planted by terrorists, but he rarely won.
Alex has grown up since then, so he easily wins the most difficult levels. This quickly bored him, and he thought: what if the computer gave him invalid fields in the childhood and Alex could not win because of it?
He needs your help to check it.
A Minesweeper field is a rectangle $n \times m$, where each cell is either empty, or contains a digit from $1$ to $8$, or a bomb. The field is valid if for each cell:
- if there is a digit $k$ in the cell, then exactly $k$ neighboring cells have bombs. - if the cell is empty, then all neighboring cells have no bombs.
Two cells are neighbors if they have a common side or a corner (i. e. a cell has at most $8$ neighboring cells).
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 100$) — the sizes of the field.
The next $n$ lines contain the description of the field. Each line contains $m$ characters, each of them is "." (if this cell is empty), "*" (if there is bomb in this cell), or a digit from $1$ to $8$, inclusive.
Output Specification:
Print "YES", if the field is valid and "NO" otherwise.
You can choose the case (lower or upper) for each letter arbitrarily.
Demo Input:
['3 3\n111\n1*1\n111\n', '2 4\n*.*.\n1211\n']
Demo Output:
['YES', 'NO']
Note:
In the second example the answer is "NO" because, if the positions of the bombs are preserved, the first line of the field should be *2*1.
You can read more about Minesweeper in [Wikipedia's article](https://en.wikipedia.org/wiki/Minesweeper_(video_game)). | ```python
import math
from collections import defaultdict
ml=lambda:map(int,input().split())
ll=lambda:list(map(int,input().split()))
ii=lambda:int(input())
ip=lambda:list(input())
"""========main code==============="""
a,b=ml()
lol=[]
for i in range(a):
w=list(input())
lol.append(w)
flag=0
for i in range(a):
for j in range(b):
k=0
if(i-1>=0):
if(lol[i-1][j]=='*'):
k+=1
if(j-1>=0):
if(lol[i-1][j-1]=='*'):
k+=1
if(j+1<b):
if(lol[i-1][j+1]=='*'):
k+=1
if((i+1)<a):
if(lol[i+1][j]=='*'):
k+=1
if(j-1>=0):
if(lol[i+1][j-1]=='*'):
k+=1
if((j+1)<b):
if(lol[i+1][j+1]=='*'):
k+=1
if(j-1>=0):
if(lol[i][j-1]=='*'):
k+=1
if(j+1<b):
if(lol[i][j+1]=='*'):
k+=1
if(lol[i][j] in {'1','2','3','4','5','6','7','8'}):
if(int(lol[i][j])!=k):
flag=1
elif(lol[i][j]=='.' and k!=0):
flag=1
if( flag):
print("NO")
else:
print("YES")
``` | 3 | |
603 | A | Alternative Thinking | PROGRAMMING | 1,600 | [
"dp",
"greedy",
"math"
] | null | null | Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have. | The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000).
The following line contains a binary string of length *n* representing Kevin's results on the USAICO. | Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring. | [
"8\n10000011\n",
"2\n01\n"
] | [
"5\n",
"2\n"
] | In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score. | 500 | [
{
"input": "8\n10000011",
"output": "5"
},
{
"input": "2\n01",
"output": "2"
},
{
"input": "5\n10101",
"output": "5"
},
{
"input": "75\n010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "75"
},
{
"input": "11\n00000000000",
"output": "3"
},
{
"input": "56\n10101011010101010101010101010101010101011010101010101010",
"output": "56"
},
{
"input": "50\n01011010110101010101010101010101010101010101010100",
"output": "49"
},
{
"input": "7\n0110100",
"output": "7"
},
{
"input": "8\n11011111",
"output": "5"
},
{
"input": "6\n000000",
"output": "3"
},
{
"input": "5\n01000",
"output": "5"
},
{
"input": "59\n10101010101010101010101010101010101010101010101010101010101",
"output": "59"
},
{
"input": "88\n1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "88"
},
{
"input": "93\n010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "93"
},
{
"input": "70\n0101010101010101010101010101010101010101010101010101010101010101010101",
"output": "70"
},
{
"input": "78\n010101010101010101010101010101101010101010101010101010101010101010101010101010",
"output": "78"
},
{
"input": "83\n10101010101010101010101010101010101010101010101010110101010101010101010101010101010",
"output": "83"
},
{
"input": "87\n101010101010101010101010101010101010101010101010101010101010101010101010101010010101010",
"output": "87"
},
{
"input": "65\n01010101101010101010101010101010101010101010101010101010101010101",
"output": "65"
},
{
"input": "69\n010101010101010101101010101010101010101010101010101010101010101010101",
"output": "69"
},
{
"input": "74\n01010101010101010101010101010101010101010101010101010101010101000101010101",
"output": "74"
},
{
"input": "77\n01010101010101001010101010101010100101010101010101010101010101010101010101010",
"output": "77"
},
{
"input": "60\n101010110101010101010101010110101010101010101010101010101010",
"output": "60"
},
{
"input": "89\n01010101010101010101010101010101010101010101010101010101101010101010101010100101010101010",
"output": "89"
},
{
"input": "68\n01010101010101010101010101010101010100101010100101010101010100101010",
"output": "67"
},
{
"input": "73\n0101010101010101010101010101010101010101010111011010101010101010101010101",
"output": "72"
},
{
"input": "55\n1010101010101010010101010101101010101010101010100101010",
"output": "54"
},
{
"input": "85\n1010101010101010101010101010010101010101010101101010101010101010101011010101010101010",
"output": "84"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1111111111",
"output": "3"
},
{
"input": "2\n10",
"output": "2"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "2\n00",
"output": "2"
},
{
"input": "3\n000",
"output": "3"
},
{
"input": "3\n001",
"output": "3"
},
{
"input": "3\n010",
"output": "3"
},
{
"input": "3\n011",
"output": "3"
},
{
"input": "3\n100",
"output": "3"
},
{
"input": "3\n101",
"output": "3"
},
{
"input": "3\n110",
"output": "3"
},
{
"input": "3\n111",
"output": "3"
},
{
"input": "4\n0000",
"output": "3"
},
{
"input": "4\n0001",
"output": "4"
},
{
"input": "4\n0010",
"output": "4"
},
{
"input": "4\n0011",
"output": "4"
},
{
"input": "4\n0100",
"output": "4"
},
{
"input": "4\n0101",
"output": "4"
},
{
"input": "4\n0110",
"output": "4"
},
{
"input": "4\n0111",
"output": "4"
},
{
"input": "4\n1000",
"output": "4"
},
{
"input": "4\n1001",
"output": "4"
},
{
"input": "4\n1010",
"output": "4"
},
{
"input": "4\n1011",
"output": "4"
},
{
"input": "4\n1100",
"output": "4"
},
{
"input": "4\n1101",
"output": "4"
},
{
"input": "4\n1110",
"output": "4"
},
{
"input": "4\n1111",
"output": "3"
},
{
"input": "5\n00000",
"output": "3"
},
{
"input": "5\n00001",
"output": "4"
},
{
"input": "5\n00010",
"output": "5"
},
{
"input": "5\n00011",
"output": "4"
},
{
"input": "5\n00100",
"output": "5"
},
{
"input": "5\n00101",
"output": "5"
},
{
"input": "5\n00110",
"output": "5"
},
{
"input": "5\n00111",
"output": "4"
},
{
"input": "5\n01000",
"output": "5"
},
{
"input": "5\n01001",
"output": "5"
},
{
"input": "5\n01010",
"output": "5"
},
{
"input": "5\n01011",
"output": "5"
},
{
"input": "5\n01100",
"output": "5"
},
{
"input": "5\n01101",
"output": "5"
},
{
"input": "5\n01110",
"output": "5"
},
{
"input": "5\n01111",
"output": "4"
},
{
"input": "5\n10000",
"output": "4"
},
{
"input": "5\n10001",
"output": "5"
},
{
"input": "5\n10010",
"output": "5"
},
{
"input": "5\n10100",
"output": "5"
},
{
"input": "5\n10101",
"output": "5"
},
{
"input": "5\n10110",
"output": "5"
},
{
"input": "5\n10111",
"output": "5"
},
{
"input": "5\n11000",
"output": "4"
},
{
"input": "5\n11001",
"output": "5"
},
{
"input": "5\n11010",
"output": "5"
},
{
"input": "5\n11011",
"output": "5"
},
{
"input": "5\n11100",
"output": "4"
},
{
"input": "5\n11101",
"output": "5"
},
{
"input": "5\n11110",
"output": "4"
},
{
"input": "5\n11111",
"output": "3"
}
] | 1,519,210,582 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 5,632,000 | n = int(input())
a = input()
l = [1]
ans = 0
for i in range(1, n):
l.append((l[i-1] + 1 if a[i-1] == a[i] else 1))
if max(l) >= 3:
ans += 2
s0 = [1 if a[0] == '0' else 0]
s1 = [1 if a[0] == '1' else 1]
for i in range(1, n):
if a[i] == '1':
s1.append(1 + s0[-1])
s0.append(s0[-1])
else:
s0.append(1 + s1[-1])
s1.append(s1[-1])
print(ans + max(s0[-1], s1[-1]))
| Title: Alternative Thinking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input Specification:
The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000).
The following line contains a binary string of length *n* representing Kevin's results on the USAICO.
Output Specification:
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Demo Input:
['8\n10000011\n', '2\n01\n']
Demo Output:
['5\n', '2\n']
Note:
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score. | ```python
n = int(input())
a = input()
l = [1]
ans = 0
for i in range(1, n):
l.append((l[i-1] + 1 if a[i-1] == a[i] else 1))
if max(l) >= 3:
ans += 2
s0 = [1 if a[0] == '0' else 0]
s1 = [1 if a[0] == '1' else 1]
for i in range(1, n):
if a[i] == '1':
s1.append(1 + s0[-1])
s0.append(s0[-1])
else:
s0.append(1 + s1[-1])
s1.append(s1[-1])
print(ans + max(s0[-1], s1[-1]))
``` | 0 | |
858 | A | k-rounding | PROGRAMMING | 1,100 | [
"brute force",
"math",
"number theory"
] | null | null | For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*. | The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8). | Print the *k*-rounding of *n*. | [
"375 4\n",
"10000 1\n",
"38101 0\n",
"123456789 8\n"
] | [
"30000\n",
"10000\n",
"38101\n",
"12345678900000000\n"
] | none | 750 | [
{
"input": "375 4",
"output": "30000"
},
{
"input": "10000 1",
"output": "10000"
},
{
"input": "38101 0",
"output": "38101"
},
{
"input": "123456789 8",
"output": "12345678900000000"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output": "2"
},
{
"input": "100 0",
"output": "100"
},
{
"input": "1000000000 0",
"output": "1000000000"
},
{
"input": "160 2",
"output": "800"
},
{
"input": "3 0",
"output": "3"
},
{
"input": "10 0",
"output": "10"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "3 1",
"output": "30"
},
{
"input": "4 1",
"output": "20"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "6 1",
"output": "30"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "8 1",
"output": "40"
},
{
"input": "9 1",
"output": "90"
},
{
"input": "10 1",
"output": "10"
},
{
"input": "11 1",
"output": "110"
},
{
"input": "12 1",
"output": "60"
},
{
"input": "16 2",
"output": "400"
},
{
"input": "2 2",
"output": "100"
},
{
"input": "1 2",
"output": "100"
},
{
"input": "5 2",
"output": "100"
},
{
"input": "15 2",
"output": "300"
},
{
"input": "36 2",
"output": "900"
},
{
"input": "1 8",
"output": "100000000"
},
{
"input": "8 8",
"output": "100000000"
},
{
"input": "96 8",
"output": "300000000"
},
{
"input": "175 8",
"output": "700000000"
},
{
"input": "9999995 8",
"output": "199999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "12345678 8",
"output": "617283900000000"
},
{
"input": "78125 8",
"output": "100000000"
},
{
"input": "390625 8",
"output": "100000000"
},
{
"input": "1953125 8",
"output": "500000000"
},
{
"input": "9765625 8",
"output": "2500000000"
},
{
"input": "68359375 8",
"output": "17500000000"
},
{
"input": "268435456 8",
"output": "104857600000000"
},
{
"input": "125829120 8",
"output": "9830400000000"
},
{
"input": "128000 8",
"output": "400000000"
},
{
"input": "300000 8",
"output": "300000000"
},
{
"input": "3711871 8",
"output": "371187100000000"
},
{
"input": "55555 8",
"output": "1111100000000"
},
{
"input": "222222222 8",
"output": "11111111100000000"
},
{
"input": "479001600 8",
"output": "7484400000000"
},
{
"input": "655360001 7",
"output": "6553600010000000"
},
{
"input": "655360001 8",
"output": "65536000100000000"
},
{
"input": "1000000000 1",
"output": "1000000000"
},
{
"input": "1000000000 7",
"output": "1000000000"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "10000000 8",
"output": "100000000"
},
{
"input": "1000000 8",
"output": "100000000"
},
{
"input": "10000009 8",
"output": "1000000900000000"
},
{
"input": "10000005 8",
"output": "200000100000000"
},
{
"input": "10000002 8",
"output": "500000100000000"
},
{
"input": "999999997 8",
"output": "99999999700000000"
},
{
"input": "999999997 7",
"output": "9999999970000000"
},
{
"input": "999999995 8",
"output": "19999999900000000"
},
{
"input": "123 8",
"output": "12300000000"
},
{
"input": "24 2",
"output": "600"
},
{
"input": "16 4",
"output": "10000"
},
{
"input": "123456787 8",
"output": "12345678700000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "101 1",
"output": "1010"
},
{
"input": "50 2",
"output": "100"
},
{
"input": "999999818 1",
"output": "4999999090"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "123 1",
"output": "1230"
},
{
"input": "16 1",
"output": "80"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "15304 6",
"output": "1913000000"
},
{
"input": "3 8",
"output": "300000000"
},
{
"input": "4 2",
"output": "100"
},
{
"input": "100000 7",
"output": "10000000"
},
{
"input": "5 8",
"output": "100000000"
},
{
"input": "16724 6",
"output": "4181000000"
},
{
"input": "999999999 1",
"output": "9999999990"
},
{
"input": "999999990 8",
"output": "9999999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "100000 1",
"output": "100000"
},
{
"input": "8 3",
"output": "1000"
},
{
"input": "16768 6",
"output": "262000000"
},
{
"input": "123456789 1",
"output": "1234567890"
},
{
"input": "2 8",
"output": "100000000"
},
{
"input": "999999937 8",
"output": "99999993700000000"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "2000000 7",
"output": "10000000"
},
{
"input": "1999998 2",
"output": "99999900"
},
{
"input": "125 3",
"output": "1000"
}
] | 1,609,610,208 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 8 | 171 | 0 | n,k=input().split()
if k=='0':print(n)
elif n[-1]=='5':
n=str(int(n)*8)
count=0
for x in range(len(n)-1,-1,-1):
if n[x]=='0':count+=1
else:break
if count>=int(k):print(n)
else:print(n+'0'*(int(k)-count))
elif n[-1]=='0':
count=0
for x in range(len(n)-1,-1,-1):
if n[x]=='0':count+=1
else:break
if count>=int(k):print(n)
else:print(n+'0'*(int(k)-count))
else:print(str(n)+'0'*int(k)) | Title: k-rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
Output Specification:
Print the *k*-rounding of *n*.
Demo Input:
['375 4\n', '10000 1\n', '38101 0\n', '123456789 8\n']
Demo Output:
['30000\n', '10000\n', '38101\n', '12345678900000000\n']
Note:
none | ```python
n,k=input().split()
if k=='0':print(n)
elif n[-1]=='5':
n=str(int(n)*8)
count=0
for x in range(len(n)-1,-1,-1):
if n[x]=='0':count+=1
else:break
if count>=int(k):print(n)
else:print(n+'0'*(int(k)-count))
elif n[-1]=='0':
count=0
for x in range(len(n)-1,-1,-1):
if n[x]=='0':count+=1
else:break
if count>=int(k):print(n)
else:print(n+'0'*(int(k)-count))
else:print(str(n)+'0'*int(k))
``` | 0 | |
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"
}
] | 1,645,237,245 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 1,000 | 0 | n = int(input())
suma, counter = 0, 0
pos = 2
for i in range(1,n+1):
for j in range(1,n+1):
suma = i*j+i
if suma == n:
counter += 1
suma = 0
print(counter) | 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
n = int(input())
suma, counter = 0, 0
pos = 2
for i in range(1,n+1):
for j in range(1,n+1):
suma = i*j+i
if suma == n:
counter += 1
suma = 0
print(counter)
``` | 0 | |
527 | C | Glass Carving | PROGRAMMING | 1,500 | [
"binary search",
"data structures",
"implementation"
] | null | null | Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree? | The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts. | After each cut print on a single line the area of the maximum available glass fragment in mm2. | [
"4 3 4\nH 2\nV 2\nV 3\nV 1\n",
"7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n"
] | [
"8\n4\n4\n2\n",
"28\n16\n12\n6\n4\n"
] | Picture for the first sample test: | 1,500 | [
{
"input": "4 3 4\nH 2\nV 2\nV 3\nV 1",
"output": "8\n4\n4\n2"
},
{
"input": "7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1",
"output": "28\n16\n12\n6\n4"
},
{
"input": "2 2 1\nV 1",
"output": "2"
},
{
"input": "2 2 1\nH 1",
"output": "2"
},
{
"input": "2 2 2\nV 1\nH 1",
"output": "2\n1"
},
{
"input": "2 2 2\nH 1\nV 1",
"output": "2\n1"
},
{
"input": "10 10 10\nV 6\nH 8\nV 4\nV 8\nH 2\nH 5\nV 9\nH 7\nH 3\nV 7",
"output": "60\n48\n32\n32\n24\n12\n12\n12\n8\n8"
},
{
"input": "5 15 10\nH 8\nH 9\nV 1\nH 2\nH 6\nH 4\nH 1\nV 2\nH 13\nV 3",
"output": "40\n40\n32\n24\n24\n24\n24\n18\n12\n8"
},
{
"input": "15 5 10\nV 13\nV 10\nV 3\nH 2\nV 9\nV 7\nV 2\nH 1\nV 4\nH 3",
"output": "65\n50\n35\n21\n18\n12\n12\n12\n9\n6"
},
{
"input": "2 3 1\nH 1",
"output": "4"
},
{
"input": "200000 200000 1\nH 1",
"output": "39999800000"
},
{
"input": "2 4 1\nH 2",
"output": "4"
}
] | 1,585,269,550 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 2,000 | 1,843,200 | w, h, n = map(int, input().split())
h_lengths_record = [h]
w_lengths_record = [w]
max_h_idx = 0
max_w_idx = 0
max_h = h
max_w = w
def get_index_split(arr, start, stop, dist):
mid = (start + stop) // 2
if mid == 0:
return 0
if dist > arr[mid - 1] and dist < arr[mid]:
return mid
if dist > arr[mid]:
return get_index_split(arr, mid + 1, stop, dist)
else:
return get_index_split(arr, start, mid, dist)
for i in range(n):
cut, dist = input().split()
dist = int(dist)
if cut == 'H':
cut_idx = get_index_split(h_lengths_record, 0, len(h_lengths_record), dist)
new_h_lengths_record = h_lengths_record[:cut_idx] + [dist] + h_lengths_record[cut_idx:]
h_lengths_record = new_h_lengths_record
if cut_idx == max_h_idx:
max_h = h_lengths_record[0]
max_h_idx = 0
for j in range(1, len(h_lengths_record)):
h_gap = h_lengths_record[j] - h_lengths_record[j - 1]
if h_gap > max_h:
max_h = h_gap
max_h_idx = j
elif cut_idx < max_h_idx:
max_h_idx += 1
if cut == 'V':
cut_idx = get_index_split(w_lengths_record, 0, len(w_lengths_record), dist)
new_w_lengths_record = w_lengths_record[:cut_idx] + [dist] + w_lengths_record[cut_idx:]
w_lengths_record = new_w_lengths_record
if cut_idx == max_w_idx:
max_w = w_lengths_record[0]
max_w_idx = 0
for j in range(1, len(w_lengths_record)):
w_gap = w_lengths_record[j] - w_lengths_record[j - 1]
if w_gap > max_w:
max_w = w_gap
max_w_idx = j
elif cut_idx < max_w_idx:
max_w_idx += 1
# print(cut_idx, max_h_idx, max_w_idx, h_lengths_record, w_lengths_record)
print(max_h * max_w)
| Title: Glass Carving
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree?
Input Specification:
The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.
Output Specification:
After each cut print on a single line the area of the maximum available glass fragment in mm2.
Demo Input:
['4 3 4\nH 2\nV 2\nV 3\nV 1\n', '7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n']
Demo Output:
['8\n4\n4\n2\n', '28\n16\n12\n6\n4\n']
Note:
Picture for the first sample test: | ```python
w, h, n = map(int, input().split())
h_lengths_record = [h]
w_lengths_record = [w]
max_h_idx = 0
max_w_idx = 0
max_h = h
max_w = w
def get_index_split(arr, start, stop, dist):
mid = (start + stop) // 2
if mid == 0:
return 0
if dist > arr[mid - 1] and dist < arr[mid]:
return mid
if dist > arr[mid]:
return get_index_split(arr, mid + 1, stop, dist)
else:
return get_index_split(arr, start, mid, dist)
for i in range(n):
cut, dist = input().split()
dist = int(dist)
if cut == 'H':
cut_idx = get_index_split(h_lengths_record, 0, len(h_lengths_record), dist)
new_h_lengths_record = h_lengths_record[:cut_idx] + [dist] + h_lengths_record[cut_idx:]
h_lengths_record = new_h_lengths_record
if cut_idx == max_h_idx:
max_h = h_lengths_record[0]
max_h_idx = 0
for j in range(1, len(h_lengths_record)):
h_gap = h_lengths_record[j] - h_lengths_record[j - 1]
if h_gap > max_h:
max_h = h_gap
max_h_idx = j
elif cut_idx < max_h_idx:
max_h_idx += 1
if cut == 'V':
cut_idx = get_index_split(w_lengths_record, 0, len(w_lengths_record), dist)
new_w_lengths_record = w_lengths_record[:cut_idx] + [dist] + w_lengths_record[cut_idx:]
w_lengths_record = new_w_lengths_record
if cut_idx == max_w_idx:
max_w = w_lengths_record[0]
max_w_idx = 0
for j in range(1, len(w_lengths_record)):
w_gap = w_lengths_record[j] - w_lengths_record[j - 1]
if w_gap > max_w:
max_w = w_gap
max_w_idx = j
elif cut_idx < max_w_idx:
max_w_idx += 1
# print(cut_idx, max_h_idx, max_w_idx, h_lengths_record, w_lengths_record)
print(max_h * max_w)
``` | 0 | |
456 | A | Laptops | PROGRAMMING | 1,100 | [
"sortings"
] | null | null | One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct. | If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes). | [
"2\n1 2\n2 1\n"
] | [
"Happy Alex\n"
] | none | 500 | [
{
"input": "2\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "2\n1 1\n2 2",
"output": "Poor Alex"
},
{
"input": "3\n2 2\n3 3\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n3 3\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "1\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n2 3\n1 1\n3 2",
"output": "Happy Alex"
},
{
"input": "4\n4 1\n3 2\n2 3\n1 4",
"output": "Happy Alex"
},
{
"input": "2\n2 1\n1 2",
"output": "Happy Alex"
},
{
"input": "3\n3 2\n1 1\n2 3",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 3\n3 1",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 1\n3 3",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 3\n3 1",
"output": "Happy Alex"
}
] | 1,691,066,966 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 62 | 0 | import sys
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = 0
for i in range(len(a)):
for j in range(i, len(a)):
if a[i] > a[j]:
c = a[i]
a[i] = a[j]
a[j] = c
c = b[i]
b[i] = b[j]
b[j] = c
for i in range(len(a)):
for j in range(i, len(a)):
if b[i] > b[j]:
print('Happy Alex')
sys.exit()
print('Poor Alex')
| Title: Laptops
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
Output Specification:
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Demo Input:
['2\n1 2\n2 1\n']
Demo Output:
['Happy Alex\n']
Note:
none | ```python
import sys
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = 0
for i in range(len(a)):
for j in range(i, len(a)):
if a[i] > a[j]:
c = a[i]
a[i] = a[j]
a[j] = c
c = b[i]
b[i] = b[j]
b[j] = c
for i in range(len(a)):
for j in range(i, len(a)):
if b[i] > b[j]:
print('Happy Alex')
sys.exit()
print('Poor Alex')
``` | 0 | |
579 | A | Raising Bacteria | PROGRAMMING | 1,000 | [
"bitmasks"
] | null | null | You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days? | The only line containing one integer *x* (1<=≤<=*x*<=≤<=109). | The only line containing one integer: the answer. | [
"5\n",
"8\n"
] | [
"2\n",
"1\n"
] | For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1. | 250 | [
{
"input": "5",
"output": "2"
},
{
"input": "8",
"output": "1"
},
{
"input": "536870911",
"output": "29"
},
{
"input": "1",
"output": "1"
},
{
"input": "343000816",
"output": "14"
},
{
"input": "559980448",
"output": "12"
},
{
"input": "697681824",
"output": "14"
},
{
"input": "41313494",
"output": "14"
},
{
"input": "673935585",
"output": "13"
},
{
"input": "117422204",
"output": "19"
},
{
"input": "954746654",
"output": "15"
},
{
"input": "536838144",
"output": "14"
},
{
"input": "1000000000",
"output": "13"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "6",
"output": "2"
},
{
"input": "7",
"output": "3"
},
{
"input": "9",
"output": "2"
},
{
"input": "10",
"output": "2"
},
{
"input": "999999999",
"output": "21"
},
{
"input": "999999998",
"output": "20"
},
{
"input": "999999997",
"output": "20"
},
{
"input": "999999996",
"output": "19"
},
{
"input": "999999995",
"output": "20"
},
{
"input": "999999994",
"output": "19"
},
{
"input": "999999993",
"output": "19"
},
{
"input": "999999992",
"output": "18"
},
{
"input": "999999991",
"output": "20"
},
{
"input": "999999990",
"output": "19"
},
{
"input": "536870910",
"output": "28"
},
{
"input": "536870912",
"output": "1"
},
{
"input": "536870913",
"output": "2"
},
{
"input": "536870914",
"output": "2"
},
{
"input": "22",
"output": "3"
},
{
"input": "14",
"output": "3"
}
] | 1,684,594,325 | 2,147,483,647 | PyPy 3 | OK | TESTS | 36 | 93 | 0 | def f(x, k):
if x == 1:
return k + 1
else:
if x % 2 == 0:
return f(x // 2, k)
else:
return f(x - 1, k + 1)
print(f(int(input()),0))
| Title: Raising Bacteria
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input Specification:
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
Output Specification:
The only line containing one integer: the answer.
Demo Input:
['5\n', '8\n']
Demo Output:
['2\n', '1\n']
Note:
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1. | ```python
def f(x, k):
if x == 1:
return k + 1
else:
if x % 2 == 0:
return f(x // 2, k)
else:
return f(x - 1, k + 1)
print(f(int(input()),0))
``` | 3 |
Subsets and Splits
Successful Python Submissions
Retrieves all records from the train dataset where the verdict is 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Retrieves records of users with a rating of 1600 or higher and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a rating above 2000 and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a 'OK' verdict, providing a basic overview of a specific category within the dataset.