message stringlengths 2 23.4k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 129 108k | cluster float64 6 6 | __index_level_0__ int64 258 216k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
A widely known among some people Belarusian sport programmer Lesha decided to make some money to buy a one square meter larger flat. To do this, he wants to make and carry out a Super Rated Match (SRM) on the site Torcoder.com. But there's a problem — a severe torcoder coordinator Ivan does not accept any Lesha's problem, calling each of them an offensive word "duped" (that is, duplicated). And one day they nearely quarrelled over yet another problem Ivan wouldn't accept.
You are invited to act as a fair judge and determine whether the problem is indeed brand new, or Ivan is right and the problem bears some resemblance to those used in the previous SRMs.
You are given the descriptions of Lesha's problem and each of Torcoder.com archive problems. The description of each problem is a sequence of words. Besides, it is guaranteed that Lesha's problem has no repeated words, while the description of an archive problem may contain any number of repeated words.
The "similarity" between Lesha's problem and some archive problem can be found as follows. Among all permutations of words in Lesha's problem we choose the one that occurs in the archive problem as a subsequence. If there are multiple such permutations, we choose the one with the smallest number of inversions. Then the "similarity" of a problem can be written as <image>, where n is the number of words in Lesha's problem and x is the number of inversions in the chosen permutation. Note that the "similarity" p is always a positive integer.
The problem is called brand new if there is not a single problem in Ivan's archive which contains a permutation of words from Lesha's problem as a subsequence.
Help the boys and determine whether the proposed problem is new, or specify the problem from the archive which resembles Lesha's problem the most, otherwise.
Input
The first line contains a single integer n (1 ≤ n ≤ 4) — the number of words in Lesha's problem. The second line contains n space-separated words — the short description of the problem.
The third line contains a single integer m (1 ≤ m ≤ 10) — the number of problems in the Torcoder.com archive. Next m lines contain the descriptions of the problems as "k s1 s2 ... sk", where k (1 ≤ k ≤ 20) is the number of words in the problem and si is a word of the problem description.
All words from all problem descriptions contain no more than 10 lowercase English letters.
Output
If Lesha's problem is brand new, print string "Brand new problem!" (without quotes).
Otherwise, on the first line print the index of the archive problem which resembles Lesha's problem most. If there are multiple such problems, print the one with the smallest index. On the second line print a string consisting of characters [:, character | repeated p times, and characters :], where p is the "similarity" between this problem and Lesha's one. The archive problems are numbered starting from one in the order in which they are given in the input.
Examples
Input
4
find the next palindrome
1
10 find the previous palindrome or print better luck next time
Output
1
[:||||||:]
Input
3
add two numbers
3
1 add
2 two two
3 numbers numbers numbers
Output
Brand new problem!
Input
4
these papers are formulas
3
6 what are these formulas and papers
5 papers are driving me crazy
4 crazy into the night
Output
1
[:||||:]
Input
3
add two decimals
5
4 please two decimals add
5 decimals want to be added
4 two add decimals add
4 add one two three
7 one plus two plus three equals six
Output
3
[:|||:]
Note
Let us remind you that the number of inversions is the number of pairs of words that follow in the permutation not in their original order. Thus, for example, if the original problem is "add two numbers", then permutation "numbers add two" contains two inversions — pairs of words "numbers" and "add", "numbers" and "two".
Sequence b1, b2, ..., bk is a subsequence of sequence a1, a2, ..., an if there exists such a set of indices 1 ≤ i1 < i2 < ... < ik ≤ n that aij = bj (in other words, if sequence b can be obtained from a by deleting some of its elements).
In the first test case the first problem contains the "find the palindrome next" permutation as a subsequence, in which the number of inversions equals 1 (words "palindrome" and "next").
In the second test case there is no problem that contains a permutation of words from Lesha's problem as a subsequence. | instruction | 0 | 52,491 | 6 | 104,982 |
Tags: brute force
Correct Solution:
```
import itertools
def count_inversions(enumerate_seq):
tmp = list(enumerate_seq[:])
result = 0
for i in range(len(tmp)):
for j in range(len(tmp) - 1):
if tmp[j][0] > tmp[j + 1][0]:
result += 1
tmp[j], tmp[j + 1] = tmp[j + 1], tmp[j]
return result
def sub_seq(a, b):
i, j = 0, 0
while i < len(a) and j < len(b):
if a[i] == b[j]:
i += 1
j += 1
return i == len(a)
n = int(input())
A = input().split()
B = []
m = int(input())
for _ in range(m):
B += [input().split()[1:]]
brand_new = True
best_perms = 0
best_seq = []
best_ind = -1
for i, b in enumerate(B):
for tmp in itertools.permutations(enumerate(A)):
if sub_seq([x for y, x in tmp], b):
brand_new = False
inversions = count_inversions(tmp)
similarity = n * (n - 1) // 2 - inversions + 1
if best_perms < similarity:
best_perms = similarity
best_seq = [x for y, x in tmp]
best_ind = i
if not brand_new:
print(best_ind + 1)
print('[:' + '|' * best_perms + ':]')
else:
print("Brand new problem!")
``` | output | 1 | 52,491 | 6 | 104,983 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A widely known among some people Belarusian sport programmer Lesha decided to make some money to buy a one square meter larger flat. To do this, he wants to make and carry out a Super Rated Match (SRM) on the site Torcoder.com. But there's a problem — a severe torcoder coordinator Ivan does not accept any Lesha's problem, calling each of them an offensive word "duped" (that is, duplicated). And one day they nearely quarrelled over yet another problem Ivan wouldn't accept.
You are invited to act as a fair judge and determine whether the problem is indeed brand new, or Ivan is right and the problem bears some resemblance to those used in the previous SRMs.
You are given the descriptions of Lesha's problem and each of Torcoder.com archive problems. The description of each problem is a sequence of words. Besides, it is guaranteed that Lesha's problem has no repeated words, while the description of an archive problem may contain any number of repeated words.
The "similarity" between Lesha's problem and some archive problem can be found as follows. Among all permutations of words in Lesha's problem we choose the one that occurs in the archive problem as a subsequence. If there are multiple such permutations, we choose the one with the smallest number of inversions. Then the "similarity" of a problem can be written as <image>, where n is the number of words in Lesha's problem and x is the number of inversions in the chosen permutation. Note that the "similarity" p is always a positive integer.
The problem is called brand new if there is not a single problem in Ivan's archive which contains a permutation of words from Lesha's problem as a subsequence.
Help the boys and determine whether the proposed problem is new, or specify the problem from the archive which resembles Lesha's problem the most, otherwise.
Input
The first line contains a single integer n (1 ≤ n ≤ 4) — the number of words in Lesha's problem. The second line contains n space-separated words — the short description of the problem.
The third line contains a single integer m (1 ≤ m ≤ 10) — the number of problems in the Torcoder.com archive. Next m lines contain the descriptions of the problems as "k s1 s2 ... sk", where k (1 ≤ k ≤ 20) is the number of words in the problem and si is a word of the problem description.
All words from all problem descriptions contain no more than 10 lowercase English letters.
Output
If Lesha's problem is brand new, print string "Brand new problem!" (without quotes).
Otherwise, on the first line print the index of the archive problem which resembles Lesha's problem most. If there are multiple such problems, print the one with the smallest index. On the second line print a string consisting of characters [:, character | repeated p times, and characters :], where p is the "similarity" between this problem and Lesha's one. The archive problems are numbered starting from one in the order in which they are given in the input.
Examples
Input
4
find the next palindrome
1
10 find the previous palindrome or print better luck next time
Output
1
[:||||||:]
Input
3
add two numbers
3
1 add
2 two two
3 numbers numbers numbers
Output
Brand new problem!
Input
4
these papers are formulas
3
6 what are these formulas and papers
5 papers are driving me crazy
4 crazy into the night
Output
1
[:||||:]
Input
3
add two decimals
5
4 please two decimals add
5 decimals want to be added
4 two add decimals add
4 add one two three
7 one plus two plus three equals six
Output
3
[:|||:]
Note
Let us remind you that the number of inversions is the number of pairs of words that follow in the permutation not in their original order. Thus, for example, if the original problem is "add two numbers", then permutation "numbers add two" contains two inversions — pairs of words "numbers" and "add", "numbers" and "two".
Sequence b1, b2, ..., bk is a subsequence of sequence a1, a2, ..., an if there exists such a set of indices 1 ≤ i1 < i2 < ... < ik ≤ n that aij = bj (in other words, if sequence b can be obtained from a by deleting some of its elements).
In the first test case the first problem contains the "find the palindrome next" permutation as a subsequence, in which the number of inversions equals 1 (words "palindrome" and "next").
In the second test case there is no problem that contains a permutation of words from Lesha's problem as a subsequence. | instruction | 0 | 52,492 | 6 | 104,984 |
Tags: brute force
Correct Solution:
```
import itertools
def check(curr_words, line):
if curr_words == []:
return True
for i in range(len(line)):
if line[i] == curr_words[0]:
return check(curr_words[1:], line[i+1:])
return False
n = int(input())
words = input().split()
m = int(input())
res, idx = 0, 0
for i in range(m):
line = input().split()[1:]
for p in itertools.permutations(range(n)):
curr_words = [words[j] for j in p]
cnt = 0
for j in range(n):
cnt += len([k for k in range(j+1, n) if p[k] < p[j]])
v = n * (n-1) // 2 - cnt + 1
if check(curr_words, line[:]) and v > res:
res, idx = v, i+1
if res > 0:
print(idx)
print('[:'+str('|'*res)+':]')
else:
print('Brand new problem!')
``` | output | 1 | 52,492 | 6 | 104,985 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A one-dimensional Japanese crossword can be represented as a binary string of length x. An encoding of this crossword is an array a of size n, where n is the number of segments formed completely of 1's, and ai is the length of i-th segment. No two segments touch or intersect.
For example:
* If x = 6 and the crossword is 111011, then its encoding is an array {3, 2};
* If x = 8 and the crossword is 01101010, then its encoding is an array {2, 1, 1};
* If x = 5 and the crossword is 11111, then its encoding is an array {5};
* If x = 5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input
The first line contains two integer numbers n and x (1 ≤ n ≤ 100000, 1 ≤ x ≤ 109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 10000) — the encoding.
Output
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Examples
Input
2 4
1 3
Output
NO
Input
3 10
3 3 2
Output
YES
Input
2 10
1 3
Output
NO | instruction | 0 | 52,773 | 6 | 105,546 |
Tags: implementation
Correct Solution:
```
n, x = list(map(int, input().split()))
a = list(map(int, input().split()))
t = sum(a)
if t >= x and n != 1:
print('NO')
else:
if x - n < t and x - t == n - 1:
print('YES')
else:
print('NO')
``` | output | 1 | 52,773 | 6 | 105,547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A one-dimensional Japanese crossword can be represented as a binary string of length x. An encoding of this crossword is an array a of size n, where n is the number of segments formed completely of 1's, and ai is the length of i-th segment. No two segments touch or intersect.
For example:
* If x = 6 and the crossword is 111011, then its encoding is an array {3, 2};
* If x = 8 and the crossword is 01101010, then its encoding is an array {2, 1, 1};
* If x = 5 and the crossword is 11111, then its encoding is an array {5};
* If x = 5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input
The first line contains two integer numbers n and x (1 ≤ n ≤ 100000, 1 ≤ x ≤ 109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 10000) — the encoding.
Output
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Examples
Input
2 4
1 3
Output
NO
Input
3 10
3 3 2
Output
YES
Input
2 10
1 3
Output
NO | instruction | 0 | 52,774 | 6 | 105,548 |
Tags: implementation
Correct Solution:
```
import sys
n, m = map(int, sys.stdin.readline().split())
arr = list(map(int, sys.stdin.readline().split()))
val = sum(arr)
temp = m - val
if temp == n - 1:
print('YES')
else:
print('NO')
``` | output | 1 | 52,774 | 6 | 105,549 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A one-dimensional Japanese crossword can be represented as a binary string of length x. An encoding of this crossword is an array a of size n, where n is the number of segments formed completely of 1's, and ai is the length of i-th segment. No two segments touch or intersect.
For example:
* If x = 6 and the crossword is 111011, then its encoding is an array {3, 2};
* If x = 8 and the crossword is 01101010, then its encoding is an array {2, 1, 1};
* If x = 5 and the crossword is 11111, then its encoding is an array {5};
* If x = 5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input
The first line contains two integer numbers n and x (1 ≤ n ≤ 100000, 1 ≤ x ≤ 109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 10000) — the encoding.
Output
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Examples
Input
2 4
1 3
Output
NO
Input
3 10
3 3 2
Output
YES
Input
2 10
1 3
Output
NO | instruction | 0 | 52,775 | 6 | 105,550 |
Tags: implementation
Correct Solution:
```
import math
for m in range(1,2 ):
#n = int(input())
n, x= list(map(int, input().split()))
a = list(map(int, input().split()))
sum1=sum(a)
if (sum1==x and a[0]==x):
print("YES")
elif sum1==x:
print("NO")
else:
zero = len(a)-1
if sum1+zero ==x:
print("YES")
else:
print("NO")
``` | output | 1 | 52,775 | 6 | 105,551 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A one-dimensional Japanese crossword can be represented as a binary string of length x. An encoding of this crossword is an array a of size n, where n is the number of segments formed completely of 1's, and ai is the length of i-th segment. No two segments touch or intersect.
For example:
* If x = 6 and the crossword is 111011, then its encoding is an array {3, 2};
* If x = 8 and the crossword is 01101010, then its encoding is an array {2, 1, 1};
* If x = 5 and the crossword is 11111, then its encoding is an array {5};
* If x = 5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input
The first line contains two integer numbers n and x (1 ≤ n ≤ 100000, 1 ≤ x ≤ 109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 10000) — the encoding.
Output
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Examples
Input
2 4
1 3
Output
NO
Input
3 10
3 3 2
Output
YES
Input
2 10
1 3
Output
NO | instruction | 0 | 52,776 | 6 | 105,552 |
Tags: implementation
Correct Solution:
```
n=list(map(int,input().split()))
a=list(map(int,input().split()))
x=n[1]
if x==sum(a)+(len(a)-1):
print('YES')
else:
print('NO')
``` | output | 1 | 52,776 | 6 | 105,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A one-dimensional Japanese crossword can be represented as a binary string of length x. An encoding of this crossword is an array a of size n, where n is the number of segments formed completely of 1's, and ai is the length of i-th segment. No two segments touch or intersect.
For example:
* If x = 6 and the crossword is 111011, then its encoding is an array {3, 2};
* If x = 8 and the crossword is 01101010, then its encoding is an array {2, 1, 1};
* If x = 5 and the crossword is 11111, then its encoding is an array {5};
* If x = 5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input
The first line contains two integer numbers n and x (1 ≤ n ≤ 100000, 1 ≤ x ≤ 109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 10000) — the encoding.
Output
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Examples
Input
2 4
1 3
Output
NO
Input
3 10
3 3 2
Output
YES
Input
2 10
1 3
Output
NO | instruction | 0 | 52,777 | 6 | 105,554 |
Tags: implementation
Correct Solution:
```
"""
Author : co_devil Chirag Garg
Institute : JIIT
"""
from __future__ import division, print_function
import itertools, os, sys, threading
from collections import deque, Counter, OrderedDict, defaultdict
import heapq
from math import ceil,floor,log,sqrt,factorial,pow,pi
# from bisect import bisect_left,bisect_right
# from decimal import *,threading
"""from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
else:
from builtins import str as __str__
str = lambda x=b'': x if type(x) is bytes else __str__(x).encode()
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._buffer = BytesIO()
self._fd = file.fileno()
self._writable = 'x' in file.mode or 'r' not in file.mode
self.write = self._buffer.write if self._writable else None
def read(self):
return self._buffer.read() if self._buffer.tell() else os.read(self._fd, os.fstat(self._fd).st_size)
def readline(self):
while self.newlines == 0:
b, ptr = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)), self._buffer.tell()
self._buffer.seek(0, 2), self._buffer.write(b), self._buffer.seek(ptr)
self.newlines += b.count(b'\n') + (not b)
self.newlines -= 1
return self._buffer.readline()
def flush(self):
if self._writable:
os.write(self._fd, self._buffer.getvalue())
self._buffer.truncate(0), self._buffer.seek(0)
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip(b'\r\n')
def print(*args, **kwargs):
sep, file = kwargs.pop('sep', b' '), kwargs.pop('file', sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop('end', b'\n'))
if kwargs.pop('flush', False):
file.flush()
"""
def ii(): return int(input())
def si(): return str(input())
def mi(): return map(int,input().split())
def li(): return list(mi())
abc = 'abcdefghijklmnopqrstuvwxyz'
abd = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12,
'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24,
'z': 25}
mod = 1000000007
dx, dy = [-1, 1, 0, 0], [0, 0, 1, -1]
def getKey(item): return item[0]
def sort2(l): return sorted(l, key=getKey)
def d2(n, m, num): return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo(x): return (x and (not (x & (x - 1))))
def decimalToBinary(n): return bin(n).replace("0b", "")
def ntl(n): return [int(i) for i in str(n)]
def powerMod(x, y, p):
res = 1
x %= p
while y > 0:
if y & 1:
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
def gcd(x, y):
while y:
x, y = y, x % y
return x
# For getting input from input.txt file
# sys.stdin = open('input.txt', 'r')
# Printing the Output to output.txt file
# sys.stdout = open('output.txt', 'w')
graph = defaultdict(list)
visited = [0] * 1000000
col = [-1] * 1000000
def dfs(v, c):
if visited[v]:
if col[v] != c:
print('-1')
exit()
return
col[v] = c
visited[v] = 1
for i in graph[v]:
dfs(i, c ^ 1)
def bfs(d,v):
q=[]
q.append(v)
visited[v]=1
while len(q)!=0:
x=q[0]
q.pop(0)
for i in d[x]:
if visited[i]!=1:
visited[i]=1
q.append(i)
print(x)
print(l)
def make_graph(e):
d={}
for i in range(e):
x,y=mi()
if x not in d.keys():
d[x]=[y]
else:
d[x].append(y)
if y not in d.keys():
d[y] = [x]
else:
d[y].append(x)
return d
def gr2(n):
d={}
for i in range(n):
x,y=mi()
if x not in d.keys():
d[x]=[y]
else:
d[x].append(y)
return d
def connected_components(graph):
seen = set()
def dfs(v):
vs = set([v])
component=[]
while vs:
v = vs.pop()
seen.add(v)
vs |= set(graph[v]) - seen
component.append(v)
return component
ans=[]
for v in graph:
if v not in seen:
d=dfs(v)
ans.append(d)
return ans
n,x=mi()
l=li()
count=0
for i in range(n-1):
count+=l[i]+1
count+=l[-1]
if count==x:
print("YES")
else:
print("NO")
``` | output | 1 | 52,777 | 6 | 105,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A one-dimensional Japanese crossword can be represented as a binary string of length x. An encoding of this crossword is an array a of size n, where n is the number of segments formed completely of 1's, and ai is the length of i-th segment. No two segments touch or intersect.
For example:
* If x = 6 and the crossword is 111011, then its encoding is an array {3, 2};
* If x = 8 and the crossword is 01101010, then its encoding is an array {2, 1, 1};
* If x = 5 and the crossword is 11111, then its encoding is an array {5};
* If x = 5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input
The first line contains two integer numbers n and x (1 ≤ n ≤ 100000, 1 ≤ x ≤ 109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 10000) — the encoding.
Output
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Examples
Input
2 4
1 3
Output
NO
Input
3 10
3 3 2
Output
YES
Input
2 10
1 3
Output
NO | instruction | 0 | 52,778 | 6 | 105,556 |
Tags: implementation
Correct Solution:
```
n, x = map(int, input().split())
a = list(map(int, input().split()))
s = 0
for i in range(n):
s = s + a[i] + 1
if s-1 == x:
print('YES')
else:
print('NO')
``` | output | 1 | 52,778 | 6 | 105,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A one-dimensional Japanese crossword can be represented as a binary string of length x. An encoding of this crossword is an array a of size n, where n is the number of segments formed completely of 1's, and ai is the length of i-th segment. No two segments touch or intersect.
For example:
* If x = 6 and the crossword is 111011, then its encoding is an array {3, 2};
* If x = 8 and the crossword is 01101010, then its encoding is an array {2, 1, 1};
* If x = 5 and the crossword is 11111, then its encoding is an array {5};
* If x = 5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input
The first line contains two integer numbers n and x (1 ≤ n ≤ 100000, 1 ≤ x ≤ 109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 10000) — the encoding.
Output
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Examples
Input
2 4
1 3
Output
NO
Input
3 10
3 3 2
Output
YES
Input
2 10
1 3
Output
NO | instruction | 0 | 52,779 | 6 | 105,558 |
Tags: implementation
Correct Solution:
```
n , t = map(int, input().split())
a = list(map(int, input().split()))
ln = n-1
for i in a:
ln += i
if ln == t:
print("YES")
else :
print("NO")
``` | output | 1 | 52,779 | 6 | 105,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A one-dimensional Japanese crossword can be represented as a binary string of length x. An encoding of this crossword is an array a of size n, where n is the number of segments formed completely of 1's, and ai is the length of i-th segment. No two segments touch or intersect.
For example:
* If x = 6 and the crossword is 111011, then its encoding is an array {3, 2};
* If x = 8 and the crossword is 01101010, then its encoding is an array {2, 1, 1};
* If x = 5 and the crossword is 11111, then its encoding is an array {5};
* If x = 5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input
The first line contains two integer numbers n and x (1 ≤ n ≤ 100000, 1 ≤ x ≤ 109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 10000) — the encoding.
Output
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Examples
Input
2 4
1 3
Output
NO
Input
3 10
3 3 2
Output
YES
Input
2 10
1 3
Output
NO | instruction | 0 | 52,780 | 6 | 105,560 |
Tags: implementation
Correct Solution:
```
n,x=map(int,input().split())
a=[int(i) for i in input().split()]
print(['NO','YES'][sum(a)+n-1==x])
``` | output | 1 | 52,780 | 6 | 105,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A newspaper is published in Walrusland. Its heading is s1, it consists of lowercase Latin letters. Fangy the little walrus wants to buy several such newspapers, cut out their headings, glue them one to another in order to get one big string. After that walrus erase several letters from this string in order to get a new word s2. It is considered that when Fangy erases some letter, there's no whitespace formed instead of the letter. That is, the string remains unbroken and it still only consists of lowercase Latin letters.
For example, the heading is "abc". If we take two such headings and glue them one to the other one, we get "abcabc". If we erase the letters on positions 1 and 5, we get a word "bcac".
Which least number of newspaper headings s1 will Fangy need to glue them, erase several letters and get word s2?
Input
The input data contain two lines. The first line contain the heading s1, the second line contains the word s2. The lines only consist of lowercase Latin letters (1 ≤ |s1| ≤ 104, 1 ≤ |s2| ≤ 106).
Output
If it is impossible to get the word s2 in the above-described manner, print "-1" (without the quotes). Otherwise, print the least number of newspaper headings s1, which Fangy will need to receive the word s2.
Examples
Input
abc
xyz
Output
-1
Input
abcd
dabc
Output
2 | instruction | 0 | 52,799 | 6 | 105,598 |
Tags: greedy, strings
Correct Solution:
```
import bisect
s1,s2,d=input(),input(),{}
for c in set(s1):
d[c]=[]
for i,x in enumerate(s1):
if x==c: d[c].append(i)
try:
ind,ans=-1,1
for c in s2:
ind = bisect.bisect_left(d[c], ind)
if ind >= len(d[c]):
ind, ans = 0, ans + 1
ind = d[c][ind] + 1
print (ans)
except:
print (-1)
'''
from bisect import bisect_left
s1=input()
s1=input()
d={}
for c in set(s1):
d[c]=[]
for i,x in enumerate(s1):
if x==c: d[c].append(i)
ans,ind=1,-1
try:
for c in s2:
ind=bisect.bisect_left(d[c],ind)
if ind>=len(d[c]):
p=0
ans+=1
ind=d[c][ind]+1
print(ans)
except:
print(-1)
'''
``` | output | 1 | 52,799 | 6 | 105,599 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A newspaper is published in Walrusland. Its heading is s1, it consists of lowercase Latin letters. Fangy the little walrus wants to buy several such newspapers, cut out their headings, glue them one to another in order to get one big string. After that walrus erase several letters from this string in order to get a new word s2. It is considered that when Fangy erases some letter, there's no whitespace formed instead of the letter. That is, the string remains unbroken and it still only consists of lowercase Latin letters.
For example, the heading is "abc". If we take two such headings and glue them one to the other one, we get "abcabc". If we erase the letters on positions 1 and 5, we get a word "bcac".
Which least number of newspaper headings s1 will Fangy need to glue them, erase several letters and get word s2?
Input
The input data contain two lines. The first line contain the heading s1, the second line contains the word s2. The lines only consist of lowercase Latin letters (1 ≤ |s1| ≤ 104, 1 ≤ |s2| ≤ 106).
Output
If it is impossible to get the word s2 in the above-described manner, print "-1" (without the quotes). Otherwise, print the least number of newspaper headings s1, which Fangy will need to receive the word s2.
Examples
Input
abc
xyz
Output
-1
Input
abcd
dabc
Output
2 | instruction | 0 | 52,800 | 6 | 105,600 |
Tags: greedy, strings
Correct Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
s = [ord(c) - 97 for c in input().rstrip()]
t = [ord(c) - 97 for c in input().rstrip()]
n, m = len(s), len(t)
next_c = [[-1] * 26 for _ in range(n)]
for _ in range(2):
for i in range(n - 1, -1, -1):
for cc in range(26):
next_c[i][cc] = next_c[(i + 1) % n][cc]
next_c[i][s[(i + 1) % n]] = (i + 1) % n
j = n - 1
ans = 0
for i, cc in enumerate(t):
k = next_c[j][cc]
if k == -1:
print(-1)
exit()
if j >= k:
ans += 1
j = k
print(ans)
``` | output | 1 | 52,800 | 6 | 105,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b | instruction | 0 | 53,319 | 6 | 106,638 |
Tags: brute force, constructive algorithms, strings
Correct Solution:
```
from string import ascii_lowercase as AL
f = list(AL)
for i in range(26):
for j in range(26): f.append(AL[i]+AL[j])
for p in range(26):
for q in range(26):
for r in range(26): f.append(AL[p]+AL[q]+AL[r])
for _ in range(int(input())):
N = int(input())
inpstr = input().strip()
for item in f:
for j in range(len(inpstr)-len(item)+1):
if inpstr[j:j+len(item)] == item:break
else:print(item);break
# for _ in range(int(input())):
# N = int(input())
# a, flag = list(map(int,input().split())), False
# res = [i for i in range(101)]
# for i in a:
# if i<0: flag = True; break
# if flag:print('NO')
# else:print('YES');print(101);print(*res)
# def primeNumbersTill(n):
# res=[0 for i in range(n+1)]
# prime=set([])
# for i in range(2,n+1):
# if not res[i]:
# prime.add(i)
# for j in range(1,n//i+1):
# res[i*j]=1
# return prime
# print('Prime numbers till 10**7 are: ',primeNumbersTill(10**7))
# def primeFactorsOf(n):
# res=[]
# for p in primeNumbersTill(n): #UNCOMMENT ABV FN. IF INVOKING THIS FN.
# if n%p==0:
# while n%p==0:n//=p
# res.append(p)
# if n!=1:res.append(n)
# return res
# print('Prime factors are: ',primeFactorsOf(10**7))
``` | output | 1 | 53,319 | 6 | 106,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b | instruction | 0 | 53,320 | 6 | 106,640 |
Tags: brute force, constructive algorithms, strings
Correct Solution:
```
import string
from itertools import count, product
def words(chars=string.ascii_letters + string.digits):
for n in count(1):
yield from map(''.join, product(chars, repeat=n))
alpha = [chr(i) for i in range(97, 123)]
n = int(input())
arr = []
arr1 = []
for i in range(n):
s = int(input())
arr.append(input())
for i in range(n):
arr1.append(sorted(set(arr[i])))
for i in range(n):
if len(arr1[i]) != len(alpha):
for j in range(len(alpha)):
if alpha[j] not in arr1[i]:
print(alpha[j])
break
else:
for gg in words(''.join(alpha)):
if gg not in arr[i]:
print(gg)
break
``` | output | 1 | 53,320 | 6 | 106,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b | instruction | 0 | 53,321 | 6 | 106,642 |
Tags: brute force, constructive algorithms, strings
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
a=[chr(i) for i in range(97,123)]
for i in range(97,123):
for j in range(97,123):
a.append(chr(i)+chr(j))
for i in range(97, 123):
for j in range(97, 123):
for k in range(97,123):
a.append(chr(i)+chr(j)+chr(k))
for _ in range(int(input())):
n = int(input())
s = input().rstrip()
b = set()
for i in range(n):
for j in range(i + 1, min(i + 6, n + 1)):
b.add(s[i:j])
for i in a:
if i not in b:
print(i)
break
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 53,321 | 6 | 106,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b | instruction | 0 | 53,322 | 6 | 106,644 |
Tags: brute force, constructive algorithms, strings
Correct Solution:
```
import re
import sys
exit=sys.exit
from bisect import bisect_left as bsl,bisect_right as bsr
from collections import Counter,defaultdict as ddict,deque
from functools import lru_cache
cache=lru_cache(None)
from heapq import *
from itertools import *
from math import inf
from pprint import pprint as pp
enum=enumerate
ri=lambda:int(rln())
ris=lambda:list(map(int,rfs()))
rln=sys.stdin.readline
rl=lambda:rln().rstrip('\n')
rfs=lambda:rln().split()
cat=''.join
catn='\n'.join
mod=1000000007
d4=[(0,-1),(1,0),(0,1),(-1,0)]
d8=[(-1,-1),(0,-1),(1,-1),(-1,0),(1,0),(-1,1),(0,1),(1,1)]
########################################################################
from string import ascii_lowercase
def solve():
n=ri()
s=rl()
que=deque([''])
while que:
for _ in range(len(que)):
t=que.popleft()
for c in ascii_lowercase:
if t+c not in s:
return t+c
que.append(t+c)
t=ri()
for _ in range(t):
print(solve())
``` | output | 1 | 53,322 | 6 | 106,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b | instruction | 0 | 53,323 | 6 | 106,646 |
Tags: brute force, constructive algorithms, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
def solve():
n = int(input())
a = input().strip()
for i in range(26):
f = a.find(chr(ord('a')+i))
if f < 0:
print(chr(ord('a')+i))
return
for i in range(26):
for j in range(26):
v = chr(ord('a')+i)+chr(ord('a')+j)
f = a.find(v)
if f < 0:
print(v)
return
for i in range(26):
for j in range(26):
for k in range(26):
v = chr(ord('a')+i)+chr(ord('a')+j)+chr(ord('a')+k)
f = a.find(v)
if f < 0:
print(v)
return
for i in range(int(input())):
solve()
``` | output | 1 | 53,323 | 6 | 106,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b | instruction | 0 | 53,324 | 6 | 106,648 |
Tags: brute force, constructive algorithms, strings
Correct Solution:
```
import string
alphabets = string.ascii_lowercase
def solve():
n = int(input())
s = input()
l = []
for i in alphabets:
l.append(i)
for i in alphabets:
for j in alphabets:
l.append(i + j)
for i in alphabets:
for j in alphabets:
for k in alphabets:
l.append(i+j+k)
for i in l:
if i not in s:return i
for testis in range(int(input())):
print(solve())
``` | output | 1 | 53,324 | 6 | 106,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b | instruction | 0 | 53,325 | 6 | 106,650 |
Tags: brute force, constructive algorithms, strings
Correct Solution:
```
def find_pref(s):
alph = ' abcdefghijklmnopqrstuvwxyz'
for l1 in alph:
for l2 in alph:
for l3 in alph:
pref = ''
if l1 != ' ':
pref = l1
if l2 != ' ':
pref += l2
if l3 != ' ':
pref += l3
if pref not in s and pref != '':
return pref
def main():
for _ in range(int(input())):
n = int(input())
s = input()
print(find_pref(s))
main()
``` | output | 1 | 53,325 | 6 | 106,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b | instruction | 0 | 53,326 | 6 | 106,652 |
Tags: brute force, constructive algorithms, strings
Correct Solution:
```
from itertools import *
from string import ascii_lowercase
for _ in range(int(input())):
input()
S = input()
N = len(S)
se = set()
for i in range(N):
cur = ''
for j in range(i, N):
cur += S[j]
se.add(cur)
for i in range(1, 1000):
for n in product(*([ascii_lowercase] * i)):
if "".join(n) not in se:
print("".join(n))
break
else:
continue
break
``` | output | 1 | 53,326 | 6 | 106,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b
Submitted Solution:
```
import string
alphabets = string.ascii_lowercase
l = []
for i in alphabets:
l.append(i)
for j in alphabets:
l.append(i + j)
for k in alphabets:
l.append(i+j+k)
l.sort(key = len)
for testis in range(int(input())):
n = int(input())
s = input()
for i in l:
if i not in s:
print(i)
break
``` | instruction | 0 | 53,327 | 6 | 106,654 |
Yes | output | 1 | 53,327 | 6 | 106,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b
Submitted Solution:
```
t=int(input())
def MEX(n,s):
a=ord('a')
z=ord('z')
for i in range(a,z+1):
if chr(i) not in s:
return chr(i)
for i in range(a,z+1):
for j in range(a,z+1):
if chr(i)+chr(j) not in s:
return chr(i)+chr(j)
for i in range(a,z+1):
for j in range(a,z+1):
for k in range(a,z+1):
if chr(i)+chr(j)+chr(k) not in s:
return chr(i)+chr(j)+chr(k)
for _ in range(t):
n=int(input())
s=input()
print(MEX(n,s))
``` | instruction | 0 | 53,328 | 6 | 106,656 |
Yes | output | 1 | 53,328 | 6 | 106,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b
Submitted Solution:
```
from itertools import product
from sys import stdin
def getInt():
return int(stdin.readline().strip())
def getInts():
return tuple(int(z) for z in stdin.readline().split())
def getWord():
return stdin.readline().strip()
t = getInt()
for i in range(t):
n = getInt()
k = getWord()
x = {}
for i in range(n):
for j in range(i + 1, n + 1):
x[k[i:j]] = True
ans = None
for l in range(1,5):
for z in product("abcdefghijklmnopqrstuvwxyz",repeat=l):
w = "".join(z)
if w not in x:
ans = w
break
if ans is not None:
break
print(ans)
``` | instruction | 0 | 53,329 | 6 | 106,658 |
Yes | output | 1 | 53,329 | 6 | 106,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b
Submitted Solution:
```
import sys
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(get_ints())
def input(): return sys.stdin.readline().strip()
arr = [chr(i) for i in range(97, 123)]
def find(string):
for i in arr:
if i not in string:
return i
for i in arr:
for j in arr:
if (i+j) not in string:
return (i+j)
for i in arr:
for j in arr:
for k in arr:
if(i + j + k) not in string:
return (i+j+k)
for i in arr:
for j in arr:
for k in arr:
for m in arr:
if (i+j+k+m) not in string:
return (i+j+k+m)
T = int(input())
while T > 0:
n = int(input())
string = input()
ans = find(string)
print(ans)
T -= 1
``` | instruction | 0 | 53,330 | 6 | 106,660 |
Yes | output | 1 | 53,330 | 6 | 106,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b
Submitted Solution:
```
for a0 in range(int(input())):
n=int(input())
# n,k=[int(x) for x in input().split()]
# a=[int(x) for x in input().split()]
s=input()
a=[i for i in 'abcdefghijklmnopqrstuvwxyz'.strip()]
ans=''
f=False
while not f:
for i in a:
tmp=ans+i
if tmp not in s:
ans=ans+i
f=True
break
else:
if ans=='':
ans='a'
else:
if ans[-1]!='z':
ans=ans[:-1]+str(ord(ans[-1])+1)
else:
n=len(ans)
for i in range(n,-1,-1):
t1=ans[:-i]
t2=ans[-i:]
t3='z'*len(t2)
# print(t1,t2,sep=" ")
if t2==t3:
if len(t1)>0:
ans=t1[:-1]+str(ord(t1[-1])+1)+('a'*len(t2))
else:
ans=t1+('a'*len(t2+1))
print(ans)
``` | instruction | 0 | 53,331 | 6 | 106,662 |
No | output | 1 | 53,331 | 6 | 106,663 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
s = input()
st = set()
str = "abc"
for Len in range(1,n + 1):
if(Len==3):
break
for i in range(n - Len + 1):
j = i + Len - 1
st.add(s[i:j+1])
all = []
for i in range(97, 123):
all.append(chr(i))
for i in range(97, 123):
s1 = chr(i)
for j in range(97, 123):
s2 = s1 + chr(j)
all.append(s2)
for i in all:
if(i not in st):
print(i)
break
``` | instruction | 0 | 53,332 | 6 | 106,664 |
No | output | 1 | 53,332 | 6 | 106,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b
Submitted Solution:
```
t = int(input())
while(t > 0):
t-=1
n = int(input())
s = input()
ans = s
lst = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
boole = False
for i in lst:
if i not in s:
ans = i
break
if(ans == s):
for i in lst:
for j in lst:
substr = i + j
if substr not in s:
ans = substr
boole = True
break
if(boole):
break
if(ans == s):
for i in lst:
for j in lst:
for k in lst:
substr = i + j
if substr not in s:
ans = substr
boole = True
break
if(boole):
break
if(boole):
break
print(ans)
``` | instruction | 0 | 53,333 | 6 | 106,666 |
No | output | 1 | 53,333 | 6 | 106,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.
It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?
You are given a string of n lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find out what the MEX of the string is!
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 1000) — the length of the word. The second line for each test case contains a single string of n lowercase Latin letters.
The sum of n over all test cases will not exceed 1000.
Output
For each test case, output the MEX of the string on a new line.
Example
Input
3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn
Output
ac
f
b
Submitted Solution:
```
from sys import maxsize, stdout, stdin,stderr
mod = int(1e9 + 7)
def I(): return int(stdin.readline())
def lint(): return [int(x) for x in stdin.readline().split()]
def S(): return input().strip()
def grid(r, c): return [lint() for i in range(r)]
from collections import defaultdict, Counter
import math
import heapq
from heapq import heappop , heappush
import bisect
from itertools import groupby
def gcd(a,b):
while b:
a %= b
tmp = a
a = b
b = tmp
return a
def lcm(a,b):
return a / gcd(a, b) * b
def check_prime(n):
for i in range(2, int(n ** (1 / 2)) + 1):
if not n % i:
return False
return True
def Bs(a, x):
i=0
j=0
left = 0
right = len(a)
flag=False
while left<right:
mi = (left+right)//2
#print(smi,a[mi],x)
if a[mi]<=x:
left = mi+1
i+=1
else:
right = mi
j+=1
#print(left,right,"----")
#print(i-1,j)
if left>0 and a[left-1]==x:
return i-1, j
else:
return -1, -1
def nCr(n, r):
return (fact(n) / (fact(r)
* fact(n - r)))
# Returns factorial of n
def fact(n):
res = 1
for i in range(2, n+1):
res = res * i
return res
t = I()
for _ in range(t):
n = I()
s = input()
d= Counter(s)
ans=None
for i in range(26):
if d[chr(97+i)]==0:
ans=chr(97+i)
break
if ans!=None:
print(ans)
else:
l = [chr(97+i) for i in range(26)]
#print(l)
for i in l:
for j in l:
tmp= i+j
a=0
for k in range(n-1):
if s[k:k+2]==tmp:
a+=1
if a==0:
ans=tmp
break
if ans!=None:
break
print(ans)
``` | instruction | 0 | 53,334 | 6 | 106,668 |
No | output | 1 | 53,334 | 6 | 106,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.
<image>
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.
Input
The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).
The second line contains integer k (1 ≤ k ≤ 1000).
Output
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Examples
Input
saba
2
Output
NO
Input
saddastavvat
2
Output
YES
Note
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat". | instruction | 0 | 55,174 | 6 | 110,348 |
Tags: brute force, implementation, strings
Correct Solution:
```
from sys import*
def pol(s):
n = len(s)
for i in range(n // 2 + 1):
if s[i] != s[-1 - i]:
return False
return True
s = input()
k = int(input())
if len(s) % k == 0:
n = len(s) // k
for i in range(k):
#print(s[0 : n])
if not pol(s[0 : n]):
print("NO")
exit()
s = s[n:]
print("YES")
else:
print("NO")
``` | output | 1 | 55,174 | 6 | 110,349 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.
<image>
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.
Input
The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).
The second line contains integer k (1 ≤ k ≤ 1000).
Output
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Examples
Input
saba
2
Output
NO
Input
saddastavvat
2
Output
YES
Note
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat". | instruction | 0 | 55,175 | 6 | 110,350 |
Tags: brute force, implementation, strings
Correct Solution:
```
s = input()
k = int(input())
n = len(s)
if (n % k):
print("NO")
else:
le = n // k
ok = 1
for i in range(0, n, le):
for j in range(le):
if(s[i + j] != s[i + le - 1 - j]):
ok = 0
print("YES" if ok else "NO")
``` | output | 1 | 55,175 | 6 | 110,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.
<image>
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.
Input
The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).
The second line contains integer k (1 ≤ k ≤ 1000).
Output
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Examples
Input
saba
2
Output
NO
Input
saddastavvat
2
Output
YES
Note
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat". | instruction | 0 | 55,176 | 6 | 110,352 |
Tags: brute force, implementation, strings
Correct Solution:
```
s = input()
n = int(input())
cnt = 0
if len(s)%n!=0 or len(s)<n:
print("NO")
else:
for i in range(0,len(s),len(s)//n):
if s[i:i+len(s)//n] == s[i:i+len(s)//n][::-1]:
cnt+=1
if cnt==n:
print("YES")
else:
print("NO")
``` | output | 1 | 55,176 | 6 | 110,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.
<image>
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.
Input
The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).
The second line contains integer k (1 ≤ k ≤ 1000).
Output
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Examples
Input
saba
2
Output
NO
Input
saddastavvat
2
Output
YES
Note
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat". | instruction | 0 | 55,177 | 6 | 110,354 |
Tags: brute force, implementation, strings
Correct Solution:
```
s = input()
n = int(input())
msg=len(s)//n
if(len(s)//n == len(s)/n):
l1 = []
for i in range(0, len(s), msg):
l1.append(s[i:i + msg])
count = 0
#print(l1)
for i in l1:
if (i == i[::-1]):
#print(i[::-1])
count += 1
if (count == n):
print("YES")
else:
print("NO")
else:
print("NO")
``` | output | 1 | 55,177 | 6 | 110,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.
<image>
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.
Input
The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).
The second line contains integer k (1 ≤ k ≤ 1000).
Output
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Examples
Input
saba
2
Output
NO
Input
saddastavvat
2
Output
YES
Note
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat". | instruction | 0 | 55,178 | 6 | 110,356 |
Tags: brute force, implementation, strings
Correct Solution:
```
'''input
saba
2
'''
s, k = input(), int(input())
if len(s) % k != 0:
print("NO")
else:
if all(i == i[::-1] for i in [s[x:x+len(s)//k] for x in range(0,len(s),len(s)//k)]):
print("YES")
else:
print("NO")
``` | output | 1 | 55,178 | 6 | 110,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.
<image>
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.
Input
The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).
The second line contains integer k (1 ≤ k ≤ 1000).
Output
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Examples
Input
saba
2
Output
NO
Input
saddastavvat
2
Output
YES
Note
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat". | instruction | 0 | 55,179 | 6 | 110,358 |
Tags: brute force, implementation, strings
Correct Solution:
```
s = input()
n = int(input())
#a = [int(c) for c in input().split()]
wordlength = len(s)//n
if wordlength*n != len(s):
print("NO")
else:
current = wordlength
word = 0
yes = True
for i in range(n):
if not yes:
break
for j in range(wordlength):
if s[i*wordlength + j] != s[(i+1)*wordlength - j - 1]:
yes = False
break
if yes:
print("YES")
else:
print("NO")
``` | output | 1 | 55,179 | 6 | 110,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.
<image>
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.
Input
The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).
The second line contains integer k (1 ≤ k ≤ 1000).
Output
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Examples
Input
saba
2
Output
NO
Input
saddastavvat
2
Output
YES
Note
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat". | instruction | 0 | 55,180 | 6 | 110,360 |
Tags: brute force, implementation, strings
Correct Solution:
```
s = input()
k = int(input())
ans = 'YES'
if len(s)%k ==0:
r = len(s)//k
for i in range(0, len(s), r):
p = s[i:i+r]
if p != p[::-1]:
ans='NO'
else:
ans = 'NO'
print(ans)
``` | output | 1 | 55,180 | 6 | 110,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.
<image>
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.
Input
The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).
The second line contains integer k (1 ≤ k ≤ 1000).
Output
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Examples
Input
saba
2
Output
NO
Input
saddastavvat
2
Output
YES
Note
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat". | instruction | 0 | 55,181 | 6 | 110,362 |
Tags: brute force, implementation, strings
Correct Solution:
```
s = str(input())
n = int(input())
l = []
if(len(s) % n != 0):
print("NO")
exit()
for i in range(0,len(s),(len(s) // n)):
t = s[i:i+(len(s) // n):]
for j in range(len(t)):
if t[::] != t[::-1]:
print("NO")
exit()
print("YES")
``` | output | 1 | 55,181 | 6 | 110,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What are you doing at the end of the world? Are you busy? Will you save us?
<image>
Nephren is playing a game with little leprechauns.
She gives them an infinite array of strings, f0... ∞.
f0 is "What are you doing at the end of the world? Are you busy? Will you save us?".
She wants to let more people know about it, so she defines fi = "What are you doing while sending "fi - 1"? Are you busy? Will you send "fi - 1"?" for all i ≥ 1.
For example, f1 is
"What are you doing while sending "What are you doing at the end of the world? Are you busy? Will you save us?"? Are you busy? Will you send "What are you doing at the end of the world? Are you busy? Will you save us?"?". Note that the quotes in the very beginning and in the very end are for clarity and are not a part of f1.
It can be seen that the characters in fi are letters, question marks, (possibly) quotation marks and spaces.
Nephren will ask the little leprechauns q times. Each time she will let them find the k-th character of fn. The characters are indexed starting from 1. If fn consists of less than k characters, output '.' (without quotes).
Can you answer her queries?
Input
The first line contains one integer q (1 ≤ q ≤ 10) — the number of Nephren's questions.
Each of the next q lines describes Nephren's question and contains two integers n and k (0 ≤ n ≤ 105, 1 ≤ k ≤ 1018).
Output
One line containing q characters. The i-th character in it should be the answer for the i-th query.
Examples
Input
3
1 1
1 2
1 111111111111
Output
Wh.
Input
5
0 69
1 194
1 139
0 47
1 66
Output
abdef
Input
10
4 1825
3 75
3 530
4 1829
4 1651
3 187
4 584
4 255
4 774
2 474
Output
Areyoubusy
Note
For the first two examples, refer to f0 and f1 given in the legend. | instruction | 0 | 55,340 | 6 | 110,680 |
Tags: binary search, dfs and similar
Correct Solution:
```
s = "What are you doing at the end of the world? Are you busy? Will you save us?"
s1 = 'What are you doing while sending "'
s2 = '"? Are you busy? Will you send "'
s3 = '"?'
l1,l2,l3=len(s1),len(s2),len(s3)
def count(n):
if n>=60:return 10**20
return (1<<n)*75+((1<<n)-1)*68
def find(n,k):
if count(n)<k:return '.'
if n==0:return s[k-1]
if k<=l1:return s1[k-1]
c=count(n-1)
k-=l1
if k<=c:
return find(n-1,k)
k-=c
if k<=l2:return s2[k-1]
k-=l2
if k<=c:
return find(n-1,k)
k-=c
if k<=l3:return s3[k-1]
q=int(input())
ans=''
while q:
n,k=map(int,input().split())
while n > 70 and k > 34:
k -= 34
n -= 1
if n > 0 and k <= 34: ans+=s1[k - 1]
else :ans+=find(n,k)
q-=1
print(ans)
``` | output | 1 | 55,340 | 6 | 110,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What are you doing at the end of the world? Are you busy? Will you save us?
<image>
Nephren is playing a game with little leprechauns.
She gives them an infinite array of strings, f0... ∞.
f0 is "What are you doing at the end of the world? Are you busy? Will you save us?".
She wants to let more people know about it, so she defines fi = "What are you doing while sending "fi - 1"? Are you busy? Will you send "fi - 1"?" for all i ≥ 1.
For example, f1 is
"What are you doing while sending "What are you doing at the end of the world? Are you busy? Will you save us?"? Are you busy? Will you send "What are you doing at the end of the world? Are you busy? Will you save us?"?". Note that the quotes in the very beginning and in the very end are for clarity and are not a part of f1.
It can be seen that the characters in fi are letters, question marks, (possibly) quotation marks and spaces.
Nephren will ask the little leprechauns q times. Each time she will let them find the k-th character of fn. The characters are indexed starting from 1. If fn consists of less than k characters, output '.' (without quotes).
Can you answer her queries?
Input
The first line contains one integer q (1 ≤ q ≤ 10) — the number of Nephren's questions.
Each of the next q lines describes Nephren's question and contains two integers n and k (0 ≤ n ≤ 105, 1 ≤ k ≤ 1018).
Output
One line containing q characters. The i-th character in it should be the answer for the i-th query.
Examples
Input
3
1 1
1 2
1 111111111111
Output
Wh.
Input
5
0 69
1 194
1 139
0 47
1 66
Output
abdef
Input
10
4 1825
3 75
3 530
4 1829
4 1651
3 187
4 584
4 255
4 774
2 474
Output
Areyoubusy
Note
For the first two examples, refer to f0 and f1 given in the legend. | instruction | 0 | 55,341 | 6 | 110,682 |
Tags: binary search, dfs and similar
Correct Solution:
```
f0 = "What are you doing at the end of the world? Are you busy? Will you save us?"
f1 = "What are you doing while sending \"{0}\"? Are you busy? Will you send \"{0}\"?"
a = list(f1.split("{0}"))
b = list(map(len, a))
q = int(input())
f = lambda n: 143 * 2**min(n, 54) - 68
for _ in range(q):
n, k = map(int, input().split())
ans = ""
while n > 0 and b[0] < k < b[0] + f(n-1):
k -= b[0]
n -= 1
while not ans:
w = f(n-1)
if k > f(n):
ans = "."
elif n == 0:
ans = f0[k-1]
elif k <= b[0]:
ans = a[0][k-1]
elif k <= b[0] + w:
k -= b[0]
n -= 1
elif k <= b[0] + w + b[1]:
k -= b[0] + w
ans = a[1][k-1]
elif k <= b[0] + w + b[1] + w:
k -= b[0] + w + b[1]
n -= 1
else:
k -= b[0] + w + b[1] + w
ans = a[2][k-1]
print(ans, end="")
``` | output | 1 | 55,341 | 6 | 110,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What are you doing at the end of the world? Are you busy? Will you save us?
<image>
Nephren is playing a game with little leprechauns.
She gives them an infinite array of strings, f0... ∞.
f0 is "What are you doing at the end of the world? Are you busy? Will you save us?".
She wants to let more people know about it, so she defines fi = "What are you doing while sending "fi - 1"? Are you busy? Will you send "fi - 1"?" for all i ≥ 1.
For example, f1 is
"What are you doing while sending "What are you doing at the end of the world? Are you busy? Will you save us?"? Are you busy? Will you send "What are you doing at the end of the world? Are you busy? Will you save us?"?". Note that the quotes in the very beginning and in the very end are for clarity and are not a part of f1.
It can be seen that the characters in fi are letters, question marks, (possibly) quotation marks and spaces.
Nephren will ask the little leprechauns q times. Each time she will let them find the k-th character of fn. The characters are indexed starting from 1. If fn consists of less than k characters, output '.' (without quotes).
Can you answer her queries?
Input
The first line contains one integer q (1 ≤ q ≤ 10) — the number of Nephren's questions.
Each of the next q lines describes Nephren's question and contains two integers n and k (0 ≤ n ≤ 105, 1 ≤ k ≤ 1018).
Output
One line containing q characters. The i-th character in it should be the answer for the i-th query.
Examples
Input
3
1 1
1 2
1 111111111111
Output
Wh.
Input
5
0 69
1 194
1 139
0 47
1 66
Output
abdef
Input
10
4 1825
3 75
3 530
4 1829
4 1651
3 187
4 584
4 255
4 774
2 474
Output
Areyoubusy
Note
For the first two examples, refer to f0 and f1 given in the legend. | instruction | 0 | 55,342 | 6 | 110,684 |
Tags: binary search, dfs and similar
Correct Solution:
```
s0 = 'What are you doing at the end of the world? Are you busy? Will you save us?'
s1 = 'What are you doing while sending "'
s2 = '"? Are you busy? Will you send "'
l0 = len(s0)
l1 = len(s1)
l2 = len(s2)
def get(h):
if h > 55:
return int(1e20)
return (l0 + l1 + l2 + 2 << h) - l1 - l2 - 2
def solve(n, k):
if get(n) <= k:
return '.'
while True:
if n == 0:
return s0[k]
if k < l1:
return s1[k]
k -= l1
if k < get(n-1):
n -= 1
continue
k -= get(n-1)
if k < l2:
return s2[k]
k -= l2
if k < get(n-1):
n -=1
else:
return '"?'[k - get(n - 1)]
q = int(input())
for i in range(q):
n, k = list(map(int, input().split()))
print(solve(n, k-1), end='')
``` | output | 1 | 55,342 | 6 | 110,685 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What are you doing at the end of the world? Are you busy? Will you save us?
<image>
Nephren is playing a game with little leprechauns.
She gives them an infinite array of strings, f0... ∞.
f0 is "What are you doing at the end of the world? Are you busy? Will you save us?".
She wants to let more people know about it, so she defines fi = "What are you doing while sending "fi - 1"? Are you busy? Will you send "fi - 1"?" for all i ≥ 1.
For example, f1 is
"What are you doing while sending "What are you doing at the end of the world? Are you busy? Will you save us?"? Are you busy? Will you send "What are you doing at the end of the world? Are you busy? Will you save us?"?". Note that the quotes in the very beginning and in the very end are for clarity and are not a part of f1.
It can be seen that the characters in fi are letters, question marks, (possibly) quotation marks and spaces.
Nephren will ask the little leprechauns q times. Each time she will let them find the k-th character of fn. The characters are indexed starting from 1. If fn consists of less than k characters, output '.' (without quotes).
Can you answer her queries?
Input
The first line contains one integer q (1 ≤ q ≤ 10) — the number of Nephren's questions.
Each of the next q lines describes Nephren's question and contains two integers n and k (0 ≤ n ≤ 105, 1 ≤ k ≤ 1018).
Output
One line containing q characters. The i-th character in it should be the answer for the i-th query.
Examples
Input
3
1 1
1 2
1 111111111111
Output
Wh.
Input
5
0 69
1 194
1 139
0 47
1 66
Output
abdef
Input
10
4 1825
3 75
3 530
4 1829
4 1651
3 187
4 584
4 255
4 774
2 474
Output
Areyoubusy
Note
For the first two examples, refer to f0 and f1 given in the legend. | instruction | 0 | 55,343 | 6 | 110,686 |
Tags: binary search, dfs and similar
Correct Solution:
```
f0 = 'What are you doing at the end of the world? Are you busy? Will you save us?'
ft1, ft2, ft3 = 'What are you doing while sending "', '"? Are you busy? Will you send "', '"?'
flen = [2 * 10 ** 18] * (10 ** 5 + 1)
flen[0] = len(f0)
for i in range(1, 56):
flen[i] = len(ft1) + len(ft2) + len(ft3) + 2 * flen[i-1]
def ans(n, k):
while True:
if n == 0:
return f0[k]
if k < len(ft1):
return ft1[k]
k -= len(ft1)
if k < flen[n-1]:
n -= 1
continue
k -= flen[n-1]
if k < len(ft2):
return ft2[k]
k -= len(ft2)
if k < flen[n-1]:
n -= 1
continue
k -= flen[n-1]
return ft3[k]
q = int(input())
a = ''
for _ in range(q):
n, k = map(int, input().split())
k -= 1
if k >= flen[n]:
a += '.'
continue
a += ans(n, k)
print(a)
``` | output | 1 | 55,343 | 6 | 110,687 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What are you doing at the end of the world? Are you busy? Will you save us?
<image>
Nephren is playing a game with little leprechauns.
She gives them an infinite array of strings, f0... ∞.
f0 is "What are you doing at the end of the world? Are you busy? Will you save us?".
She wants to let more people know about it, so she defines fi = "What are you doing while sending "fi - 1"? Are you busy? Will you send "fi - 1"?" for all i ≥ 1.
For example, f1 is
"What are you doing while sending "What are you doing at the end of the world? Are you busy? Will you save us?"? Are you busy? Will you send "What are you doing at the end of the world? Are you busy? Will you save us?"?". Note that the quotes in the very beginning and in the very end are for clarity and are not a part of f1.
It can be seen that the characters in fi are letters, question marks, (possibly) quotation marks and spaces.
Nephren will ask the little leprechauns q times. Each time she will let them find the k-th character of fn. The characters are indexed starting from 1. If fn consists of less than k characters, output '.' (without quotes).
Can you answer her queries?
Input
The first line contains one integer q (1 ≤ q ≤ 10) — the number of Nephren's questions.
Each of the next q lines describes Nephren's question and contains two integers n and k (0 ≤ n ≤ 105, 1 ≤ k ≤ 1018).
Output
One line containing q characters. The i-th character in it should be the answer for the i-th query.
Examples
Input
3
1 1
1 2
1 111111111111
Output
Wh.
Input
5
0 69
1 194
1 139
0 47
1 66
Output
abdef
Input
10
4 1825
3 75
3 530
4 1829
4 1651
3 187
4 584
4 255
4 774
2 474
Output
Areyoubusy
Note
For the first two examples, refer to f0 and f1 given in the legend. | instruction | 0 | 55,344 | 6 | 110,688 |
Tags: binary search, dfs and similar
Correct Solution:
```
f0 = "What are you doing at the end of the world? Are you busy? Will you save us?"
first_part = 'What are you doing while sending "'
between1 = '"? Are you busy? Will you send "'
f_lengths = [0 for i in range(64)]
f_lengths[0] = len(f0)
for i in range(1, 64):
f_lengths[i] = 2*f_lengths[i-1] + 68
# print(f_lengths)
def determine_n(n, k):
while n > 64 and k >= 34:
k -= 34
n -= 1
return n
# k is 0 based here
def index_find(n, k) -> str:
# print("index_find(n, k): ", n , k)
if n == 0:
if k >= len(f0):
return "."
return f0[k]
if k < 34:
return first_part[k]
first_end = 34 + f_lengths[n-1]
if 34 <= k < first_end:
return index_find(n-1, k-34)
if first_end <= k < first_end + 32:
return between1[k-first_end]
second_end = f_lengths[n-1] + first_end + 32
if first_end + 32 <= k < second_end:
return index_find(n-1, k-first_end-32)
else:
if k - second_end > 1:
return '.'
return '"?'[k - second_end]
n = int(input())
queries = [map(int, input().split()) for i in range(n)]
r = []
for n, k in queries:
# print(k, n)
if n > 64:
new_n = determine_n(n, k-1)
prefix = (n - new_n) * 34
r.append(index_find(new_n, k-1-prefix))
else:
r.append(index_find(n, k-1))
print("".join(r))
``` | output | 1 | 55,344 | 6 | 110,689 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What are you doing at the end of the world? Are you busy? Will you save us?
<image>
Nephren is playing a game with little leprechauns.
She gives them an infinite array of strings, f0... ∞.
f0 is "What are you doing at the end of the world? Are you busy? Will you save us?".
She wants to let more people know about it, so she defines fi = "What are you doing while sending "fi - 1"? Are you busy? Will you send "fi - 1"?" for all i ≥ 1.
For example, f1 is
"What are you doing while sending "What are you doing at the end of the world? Are you busy? Will you save us?"? Are you busy? Will you send "What are you doing at the end of the world? Are you busy? Will you save us?"?". Note that the quotes in the very beginning and in the very end are for clarity and are not a part of f1.
It can be seen that the characters in fi are letters, question marks, (possibly) quotation marks and spaces.
Nephren will ask the little leprechauns q times. Each time she will let them find the k-th character of fn. The characters are indexed starting from 1. If fn consists of less than k characters, output '.' (without quotes).
Can you answer her queries?
Input
The first line contains one integer q (1 ≤ q ≤ 10) — the number of Nephren's questions.
Each of the next q lines describes Nephren's question and contains two integers n and k (0 ≤ n ≤ 105, 1 ≤ k ≤ 1018).
Output
One line containing q characters. The i-th character in it should be the answer for the i-th query.
Examples
Input
3
1 1
1 2
1 111111111111
Output
Wh.
Input
5
0 69
1 194
1 139
0 47
1 66
Output
abdef
Input
10
4 1825
3 75
3 530
4 1829
4 1651
3 187
4 584
4 255
4 774
2 474
Output
Areyoubusy
Note
For the first two examples, refer to f0 and f1 given in the legend. | instruction | 0 | 55,345 | 6 | 110,690 |
Tags: binary search, dfs and similar
Correct Solution:
```
f0 = "What are you doing at the end of the world? Are you busy? Will you save us?"
f1 = "What are you doing while sending \"{0}\"? Are you busy? Will you send \"{0}\"?"
a = list(f1.split("{0}"))
b = list(map(len, a))
q = int(input())
f = lambda n: 143 * 2**min(n, 54) - 68
for _ in range(q):
n, k = map(int, input().split())
ans = ""
while n > 0 and b[0] < k < b[0] + f(n-1):
k -= b[0]
n -= 1
while not ans:
w = f(n-1)
if k > f(n):
ans = "."
elif n == 0:
ans = f0[k-1]
elif k <= b[0]:
ans = a[0][k-1]
elif k <= b[0] + w:
k -= b[0]
n -= 1
elif k <= b[0] + w + b[1]:
k -= b[0] + w
ans = a[1][k-1]
elif k <= b[0] + w + b[1] + w:
k -= b[0] + w + b[1]
n -= 1
else:
k -= b[0] + w + b[1] + w
ans = a[2][k-1]
print(ans, end="")
# Made By Mostafa_Khaled
``` | output | 1 | 55,345 | 6 | 110,691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What are you doing at the end of the world? Are you busy? Will you save us?
<image>
Nephren is playing a game with little leprechauns.
She gives them an infinite array of strings, f0... ∞.
f0 is "What are you doing at the end of the world? Are you busy? Will you save us?".
She wants to let more people know about it, so she defines fi = "What are you doing while sending "fi - 1"? Are you busy? Will you send "fi - 1"?" for all i ≥ 1.
For example, f1 is
"What are you doing while sending "What are you doing at the end of the world? Are you busy? Will you save us?"? Are you busy? Will you send "What are you doing at the end of the world? Are you busy? Will you save us?"?". Note that the quotes in the very beginning and in the very end are for clarity and are not a part of f1.
It can be seen that the characters in fi are letters, question marks, (possibly) quotation marks and spaces.
Nephren will ask the little leprechauns q times. Each time she will let them find the k-th character of fn. The characters are indexed starting from 1. If fn consists of less than k characters, output '.' (without quotes).
Can you answer her queries?
Input
The first line contains one integer q (1 ≤ q ≤ 10) — the number of Nephren's questions.
Each of the next q lines describes Nephren's question and contains two integers n and k (0 ≤ n ≤ 105, 1 ≤ k ≤ 1018).
Output
One line containing q characters. The i-th character in it should be the answer for the i-th query.
Examples
Input
3
1 1
1 2
1 111111111111
Output
Wh.
Input
5
0 69
1 194
1 139
0 47
1 66
Output
abdef
Input
10
4 1825
3 75
3 530
4 1829
4 1651
3 187
4 584
4 255
4 774
2 474
Output
Areyoubusy
Note
For the first two examples, refer to f0 and f1 given in the legend. | instruction | 0 | 55,346 | 6 | 110,692 |
Tags: binary search, dfs and similar
Correct Solution:
```
from sys import setrecursionlimit
setrecursionlimit(100000000)
def get_int(string, n):
i = j = k = 0
for s in string:
k += 1
for s in string:
if i == n - 1:
break
if s == ' ':
i += 1
j += 1
i = 0
while j < k:
if string[j] == ' ':
break
i = 10 * i + int(string[j])
j += 1
return i
d = 'What are you doing at the end of the world? Are you busy? Will you save us?'
a = 'What are you doing while sending "'
b = '"? Are you busy? Will you send "'
c = '"?'
len_a = len(a)
len_b = len(b)
len_c = len(c)
len_d = len(d)
ln = len_a + len_b + len_c
q = int(input())
ls = [len_d]
ans = ''
for i in range(1, 60):
ls += [ls[i - 1] *2 + ln]
def get_ans(n, k):
if n == 0:
if k > len_d:
return '.'
return d[k - 1]
elif n > 60:
if k > len_a:
return get_ans(n - 1, k - len_a)
else:
return a[k - 1]
elif k > ln + 2 * ls[n - 1]:
return '.'
elif k > len_a + len_b + 2 * ls[n - 1]:
return c[k - (len_a + len_b + 2 * ls[n - 1]) - 1]
elif k > len_a + len_b + ls[n - 1]:
return get_ans(n-1, k - len_a - len_b - ls[n - 1])
elif k > len_a + ls[n-1]:
return b[k - len_a - ls[n-1] - 1]
elif k > len_a:
return get_ans(n - 1, k - len_a)
else:
return a[k - 1]
for i in range(0, q):
x = input()
n = get_int(x, 1)
k = get_int(x, 2)
if n > 8698 and k > 295726:
k -= (n - 1000) * len_a
ans += get_ans(1000, k)
else:
ans += get_ans(n, k)
print(ans)
``` | output | 1 | 55,346 | 6 | 110,693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
What are you doing at the end of the world? Are you busy? Will you save us?
<image>
Nephren is playing a game with little leprechauns.
She gives them an infinite array of strings, f0... ∞.
f0 is "What are you doing at the end of the world? Are you busy? Will you save us?".
She wants to let more people know about it, so she defines fi = "What are you doing while sending "fi - 1"? Are you busy? Will you send "fi - 1"?" for all i ≥ 1.
For example, f1 is
"What are you doing while sending "What are you doing at the end of the world? Are you busy? Will you save us?"? Are you busy? Will you send "What are you doing at the end of the world? Are you busy? Will you save us?"?". Note that the quotes in the very beginning and in the very end are for clarity and are not a part of f1.
It can be seen that the characters in fi are letters, question marks, (possibly) quotation marks and spaces.
Nephren will ask the little leprechauns q times. Each time she will let them find the k-th character of fn. The characters are indexed starting from 1. If fn consists of less than k characters, output '.' (without quotes).
Can you answer her queries?
Input
The first line contains one integer q (1 ≤ q ≤ 10) — the number of Nephren's questions.
Each of the next q lines describes Nephren's question and contains two integers n and k (0 ≤ n ≤ 105, 1 ≤ k ≤ 1018).
Output
One line containing q characters. The i-th character in it should be the answer for the i-th query.
Examples
Input
3
1 1
1 2
1 111111111111
Output
Wh.
Input
5
0 69
1 194
1 139
0 47
1 66
Output
abdef
Input
10
4 1825
3 75
3 530
4 1829
4 1651
3 187
4 584
4 255
4 774
2 474
Output
Areyoubusy
Note
For the first two examples, refer to f0 and f1 given in the legend. | instruction | 0 | 55,347 | 6 | 110,694 |
Tags: binary search, dfs and similar
Correct Solution:
```
from functools import lru_cache
import sys
f0 = "What are you doing at the end of the world? Are you busy? Will you save us?"
fnext = ['What are you doing while sending "', '"? Are you busy? Will you send "', '"?']
BIG = 10**20
@lru_cache(maxsize=None)
def flength(n):
if n == 0:
return len(f0)
else:
sublen = flength(n-1)
return min(BIG, sum([len(part) for part in fnext]) + 2 * sublen)
def get_char(n, k):
while True:
if k >= flength(n):
return '.'
if n == 0:
return f0[k]
for part in fnext:
if k < len(part):
return part[k]
else:
k -= len(part)
if k < flength(n-1):
n -= 1
break
else:
k -= flength(n-1)
q = int(input())
for i in range(2 * 10**5):
flength(i)
for _ in range(q):
n, k = map(int, input().split())
print(get_char(n, k-1), end="")
print("")
``` | output | 1 | 55,347 | 6 | 110,695 |
Provide a correct Python 3 solution for this coding contest problem.
Fukushima Prefecture is also famous for producing fruits, and among them, peaches and apples boast one of the highest production volumes in Japan. By the way, when I made a print manuscript of an English pamphlet for sale, I mistakenly wrote the description about apples and the description about peaches in reverse.
You've been tasked with fixing apple and peach, but it's kind of annoying. Enter a single line of English text and create a program that outputs the English text with all the character strings apple in it replaced with peach and all the character strings peach replaced with apple.
Input
English text (including half-width alphanumeric characters, spaces, and symbols) is given on one line. The length of the string entered is 1000 or less.
Output
Outputs English sentences with the character strings apple and peach exchanged on one line.
Example
Input
the cost of one peach is higher than that of one apple.
Output
the cost of one apple is higher than that of one peach. | instruction | 0 | 55,520 | 6 | 111,040 |
"Correct Solution:
```
print(input().replace("peach","asdf:*O`P*P+").replace("apple","peach").replace("asdf:*O`P*P+","apple"))
``` | output | 1 | 55,520 | 6 | 111,041 |
Provide a correct Python 3 solution for this coding contest problem.
Fukushima Prefecture is also famous for producing fruits, and among them, peaches and apples boast one of the highest production volumes in Japan. By the way, when I made a print manuscript of an English pamphlet for sale, I mistakenly wrote the description about apples and the description about peaches in reverse.
You've been tasked with fixing apple and peach, but it's kind of annoying. Enter a single line of English text and create a program that outputs the English text with all the character strings apple in it replaced with peach and all the character strings peach replaced with apple.
Input
English text (including half-width alphanumeric characters, spaces, and symbols) is given on one line. The length of the string entered is 1000 or less.
Output
Outputs English sentences with the character strings apple and peach exchanged on one line.
Example
Input
the cost of one peach is higher than that of one apple.
Output
the cost of one apple is higher than that of one peach. | instruction | 0 | 55,521 | 6 | 111,042 |
"Correct Solution:
```
print(input().replace('apple', '???').replace('peach', 'apple').replace('???', 'peach'))
``` | output | 1 | 55,521 | 6 | 111,043 |
Provide a correct Python 3 solution for this coding contest problem.
Fukushima Prefecture is also famous for producing fruits, and among them, peaches and apples boast one of the highest production volumes in Japan. By the way, when I made a print manuscript of an English pamphlet for sale, I mistakenly wrote the description about apples and the description about peaches in reverse.
You've been tasked with fixing apple and peach, but it's kind of annoying. Enter a single line of English text and create a program that outputs the English text with all the character strings apple in it replaced with peach and all the character strings peach replaced with apple.
Input
English text (including half-width alphanumeric characters, spaces, and symbols) is given on one line. The length of the string entered is 1000 or less.
Output
Outputs English sentences with the character strings apple and peach exchanged on one line.
Example
Input
the cost of one peach is higher than that of one apple.
Output
the cost of one apple is higher than that of one peach. | instruction | 0 | 55,523 | 6 | 111,046 |
"Correct Solution:
```
def change_apple_peach(sentence):
sentence = sentence.replace('apple','aaa')
sentence = sentence.replace('peach','apple')
sentence = sentence.replace('aaa','peach')
print(sentence)
def main():
str = input()
change_apple_peach(str)
if __name__ == '__main__':
main()
``` | output | 1 | 55,523 | 6 | 111,047 |
Provide a correct Python 3 solution for this coding contest problem.
Fukushima Prefecture is also famous for producing fruits, and among them, peaches and apples boast one of the highest production volumes in Japan. By the way, when I made a print manuscript of an English pamphlet for sale, I mistakenly wrote the description about apples and the description about peaches in reverse.
You've been tasked with fixing apple and peach, but it's kind of annoying. Enter a single line of English text and create a program that outputs the English text with all the character strings apple in it replaced with peach and all the character strings peach replaced with apple.
Input
English text (including half-width alphanumeric characters, spaces, and symbols) is given on one line. The length of the string entered is 1000 or less.
Output
Outputs English sentences with the character strings apple and peach exchanged on one line.
Example
Input
the cost of one peach is higher than that of one apple.
Output
the cost of one apple is higher than that of one peach. | instruction | 0 | 55,525 | 6 | 111,050 |
"Correct Solution:
```
print(input().replace("apple","_").replace("peach","apple").replace("_","peach"))
``` | output | 1 | 55,525 | 6 | 111,051 |
Provide a correct Python 3 solution for this coding contest problem.
We, the researchers who discovered and investigated the ancient nation Iwashiro, finally discovered the temple in the center of Iwashiro. A lithograph dedicated to the god of Iwashiro was stored in the temple. On the lithograph, two strings were written, one for each sentence and one for the spell.
In Iwashiro, how many times a spell appears in a sentence has an important meaning. However, it is considered that all the characters contained in the spell appear in order, and some of them appear in the sentence in a discrete manner once. For example, if the sentence is "abab" and the spell is "ab", then "ab" appears three times in "abab", including non-continuous ones (three ways: abab, abab, and abab).
Create a program that prints how many times a spell appears in a sentence when it is given a sentence and a spell.
Input
The input is given in the following format.
t
b b
The first line is given the string t that represents the text written on the lithograph. The second line is given the string b that represents the spell written on the lithograph. Both strings are composed of only lowercase letters and have a length of 1 or more and 1000 or less.
Output
Prints how many times a spell appears in a sentence on one line. However, the value to be output can be very large, so instead output the remainder divided by 1,000,000,007.
Examples
Input
abab
ab
Output
3
Input
aaaabaaaabaaaabaaaab
aaaaa
Output
4368
Input
data
structure
Output
0 | instruction | 0 | 55,544 | 6 | 111,088 |
"Correct Solution:
```
MOD = 1000000007
t = input()
b = input()
lent = len(t)
lenb = len(b)
dp = [0] * (lenb + 1)
dp[0] = 1
for x in range(1, lent + 1):
ct = t[x - 1]
for y in range(lenb, 0, -1):
if ct == b[y - 1]:
dp[y] += dp[y - 1]
dp[y] %= MOD
print(dp[lenb])
``` | output | 1 | 55,544 | 6 | 111,089 |
Provide a correct Python 3 solution for this coding contest problem.
We, the researchers who discovered and investigated the ancient nation Iwashiro, finally discovered the temple in the center of Iwashiro. A lithograph dedicated to the god of Iwashiro was stored in the temple. On the lithograph, two strings were written, one for each sentence and one for the spell.
In Iwashiro, how many times a spell appears in a sentence has an important meaning. However, it is considered that all the characters contained in the spell appear in order, and some of them appear in the sentence in a discrete manner once. For example, if the sentence is "abab" and the spell is "ab", then "ab" appears three times in "abab", including non-continuous ones (three ways: abab, abab, and abab).
Create a program that prints how many times a spell appears in a sentence when it is given a sentence and a spell.
Input
The input is given in the following format.
t
b b
The first line is given the string t that represents the text written on the lithograph. The second line is given the string b that represents the spell written on the lithograph. Both strings are composed of only lowercase letters and have a length of 1 or more and 1000 or less.
Output
Prints how many times a spell appears in a sentence on one line. However, the value to be output can be very large, so instead output the remainder divided by 1,000,000,007.
Examples
Input
abab
ab
Output
3
Input
aaaabaaaabaaaabaaaab
aaaaa
Output
4368
Input
data
structure
Output
0 | instruction | 0 | 55,545 | 6 | 111,090 |
"Correct Solution:
```
"""
dp[x][y] ... tのx文字目までで, y文字一致した場合の数
dp[x][y] = dp[x - 1][y] + dp[x - 1][y - 1] * (t[x] == b[y])
"""
MOD = 1000000007
t = input()
b = input()
lent = len(t)
lenb = len(b)
dp = [[0] * (lenb + 1) for _ in range(lent + 1)]
for i in range(lent + 1):
dp[i][0] = 1
for x in range(1, lent + 1):
for y in range(1, lenb + 1):
dp[x][y] = (dp[x - 1][y] + dp[x - 1][y - 1] * (t[x - 1] == b[y - 1])) % MOD
print(dp[lent][lenb])
``` | output | 1 | 55,545 | 6 | 111,091 |
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