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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists know that the Old Berland language had exactly n words. Those words had lengths of l1, l2, ..., ln letters. Every word consisted of two letters, 0 and 1. Ancient Berland people spoke quickly and didn’t make pauses between the words, but at the same time they could always understand each other perfectly. It was possible because no word was a prefix of another one. The prefix of a string is considered to be one of its substrings that starts from the initial symbol.
Help the scientists determine whether all the words of the Old Berland language can be reconstructed and if they can, output the words themselves.
Input
The first line contains one integer N (1 ≤ N ≤ 1000) — the number of words in Old Berland language. The second line contains N space-separated integers — the lengths of these words. All the lengths are natural numbers not exceeding 1000.
Output
If there’s no such set of words, in the single line output NO. Otherwise, in the first line output YES, and in the next N lines output the words themselves in the order their lengths were given in the input file. If the answer is not unique, output any.
Examples
Input
3
1 2 3
Output
YES
0
10
110
Input
3
1 1 1
Output
NO
Submitted Solution:
```
from sys import stdin
n = int(stdin.readline())
a = [int(x) for x in stdin.readline().split()]
a.sort()
high = a[-1]
total = 1<<high
a = [high-x for x in a]
current = 0
words = []
for x in a:
words.append(current>>x)
current += 1<<x
if current > total:
print('NO')
else:
print('YES')
print('\n'.join(["{0:b}".format(x) for x in words]))
``` | instruction | 0 | 23,210 | 6 | 46,420 |
No | output | 1 | 23,210 | 6 | 46,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists know that the Old Berland language had exactly n words. Those words had lengths of l1, l2, ..., ln letters. Every word consisted of two letters, 0 and 1. Ancient Berland people spoke quickly and didn’t make pauses between the words, but at the same time they could always understand each other perfectly. It was possible because no word was a prefix of another one. The prefix of a string is considered to be one of its substrings that starts from the initial symbol.
Help the scientists determine whether all the words of the Old Berland language can be reconstructed and if they can, output the words themselves.
Input
The first line contains one integer N (1 ≤ N ≤ 1000) — the number of words in Old Berland language. The second line contains N space-separated integers — the lengths of these words. All the lengths are natural numbers not exceeding 1000.
Output
If there’s no such set of words, in the single line output NO. Otherwise, in the first line output YES, and in the next N lines output the words themselves in the order their lengths were given in the input file. If the answer is not unique, output any.
Examples
Input
3
1 2 3
Output
YES
0
10
110
Input
3
1 1 1
Output
NO
Submitted Solution:
```
# l1, l2, ln
# 1 2 3
# 0 10 110
# 1 01 001
# 0 1 ?
# N
# 1 2 3
# root
# 0 1
# 0 in children and this node is end of word => 1 ? , 1 is also end of word: yes: return NO
# else: add children in
# 1: add 0
# prefix is when end of word
# 0 1 1 0
# 0 1 1 1
class Node:
def __init__(self):
self.endWord = False
self.children = {}
# add(1) # 0
# add(2) # 10
# add(3) # 110
# add(2) # 00
# add(2) # 01
# add(2) # 0
def addNode(root, num): # num = 1, 2 ,3
s = ''
cur = root
for i in range(num):
c = '0'
if c not in cur.children:
cur.children[c] = Node()
elif cur.children[c].endWord == True:
c = '1'
if c not in cur.children:
cur.children[c] = Node()
elif cur.children[c].endWord == True:
return ''
cur = cur.children[c]
# print(cur)
s += c
# print(s)
cur.endWord = True
return s
def main():
n = int(input())
nums = map(int, input().split())
res = []
root = Node()
# print(res)
for num in nums:
s = addNode(root, num)
# print(s)
if s == '':
return []
res.append(s)
return res
ans = main()
if len(ans) == 0:
print('NO')
else:
print('YES')
print('\n'.join(ans))
``` | instruction | 0 | 23,211 | 6 | 46,422 |
No | output | 1 | 23,211 | 6 | 46,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists know that the Old Berland language had exactly n words. Those words had lengths of l1, l2, ..., ln letters. Every word consisted of two letters, 0 and 1. Ancient Berland people spoke quickly and didn’t make pauses between the words, but at the same time they could always understand each other perfectly. It was possible because no word was a prefix of another one. The prefix of a string is considered to be one of its substrings that starts from the initial symbol.
Help the scientists determine whether all the words of the Old Berland language can be reconstructed and if they can, output the words themselves.
Input
The first line contains one integer N (1 ≤ N ≤ 1000) — the number of words in Old Berland language. The second line contains N space-separated integers — the lengths of these words. All the lengths are natural numbers not exceeding 1000.
Output
If there’s no such set of words, in the single line output NO. Otherwise, in the first line output YES, and in the next N lines output the words themselves in the order their lengths were given in the input file. If the answer is not unique, output any.
Examples
Input
3
1 2 3
Output
YES
0
10
110
Input
3
1 1 1
Output
NO
Submitted Solution:
```
# Problem from Codeforces
# http://codeforces.com/contest/37/problem/C
import sys
class input_tokenizer:
__tokens = None
def has_next(self):
return self.__tokens != [] and self.__tokens != None
def next(self):
token = self.__tokens[-1]
self.__tokens.pop()
return token
def __init__(self):
self.__tokens = sys.stdin.read().split()[::-1]
inp = input_tokenizer()
class Node:
def __init__(self, index, depth):
self.index = index
self.depth = depth
def __lt__(self, other):
return self.depth < other.depth
def dfs(s, depth, n, curr, word, arr):
if curr >= n:
return curr, word
if arr[curr].depth == depth:
word[arr[curr].index] = s
curr += 1
return curr, word
s += '0'
curr, word = dfs(s, depth + 1, n, curr, word, arr)
s = s[:len(s) - 1] + '1'
curr, word = dfs(s, depth + 1, n, curr, word, arr)
return curr, word
def solution():
try:
n = int(inp.next())
arr = []
for i in range(n):
word_len = int(inp.next())
arr.append(Node(i, word_len))
arr.sort()
word = ['' for i in range(n)]
curr = 0
results = dfs('', 0, n, curr, word, arr)
word = results[1]
curr = results[0]
if curr < n:
print('NO')
else:
print('YES')
print(*word, sep='\n')
except Exception:
import traceback
print(traceback.format_exc())
solution()
``` | instruction | 0 | 23,212 | 6 | 46,424 |
No | output | 1 | 23,212 | 6 | 46,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists know that the Old Berland language had exactly n words. Those words had lengths of l1, l2, ..., ln letters. Every word consisted of two letters, 0 and 1. Ancient Berland people spoke quickly and didn’t make pauses between the words, but at the same time they could always understand each other perfectly. It was possible because no word was a prefix of another one. The prefix of a string is considered to be one of its substrings that starts from the initial symbol.
Help the scientists determine whether all the words of the Old Berland language can be reconstructed and if they can, output the words themselves.
Input
The first line contains one integer N (1 ≤ N ≤ 1000) — the number of words in Old Berland language. The second line contains N space-separated integers — the lengths of these words. All the lengths are natural numbers not exceeding 1000.
Output
If there’s no such set of words, in the single line output NO. Otherwise, in the first line output YES, and in the next N lines output the words themselves in the order their lengths were given in the input file. If the answer is not unique, output any.
Examples
Input
3
1 2 3
Output
YES
0
10
110
Input
3
1 1 1
Output
NO
Submitted Solution:
```
import sys
class Node:
def __init__(self, depth, parent):
self.depth = depth
self.parent = parent
self.left = None
self.right = None
self.leaf = False
def __str__(self):
return 'depth = %d, left? %s, right? %s' % (self.depth,
self.left != None, self.right != None)
def traverse(node, symbols):
if node.leaf:
print(''.join(symbols))
return
if node.left != None:
symbols.append('0')
traverse(node.left, symbols)
symbols.pop()
if node.right != None:
symbols.append('1')
traverse(node.right, symbols)
symbols.pop()
n = int(input())
lengths = sorted(list(map(int, input().split())))
root = Node(0, None)
curr = root
for i in range(lengths[0]):
curr.left = Node(curr.depth+1, curr)
curr = curr.left
curr.leaf = True
for length in lengths[1:]:
curr = curr.parent
while curr.right != None:
if curr == root:
print('NO')
sys.exit()
curr = curr.parent
curr.right = Node(curr.depth+1, curr)
curr = curr.right
while curr.depth != length:
curr.left = Node(curr.depth+1, curr)
curr = curr.left
curr.leaf = True
print('YES')
traverse(root, [])
``` | instruction | 0 | 23,213 | 6 | 46,426 |
No | output | 1 | 23,213 | 6 | 46,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t. | instruction | 0 | 23,278 | 6 | 46,556 |
Tags: brute force, constructive algorithms, implementation
Correct Solution:
```
# cook your dish here
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,7)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
def pmat(A):
for ele in A:
print(*ele,end="\n")
def seive():
prime=[1 for i in range(10**6+1)]
prime[0]=0
prime[1]=0
for i in range(10**6+1):
if(prime[i]):
for j in range(2*i,10**6+1,i):
prime[j]=0
return prime
# exit()
# from sys import stdin
# input = stdin.readline
n,m=L()
A=[input() for i in range(n)]
t=["" for i in range(n)]
ans=0
for i in range(m):
f=[t[j]+A[j][i] for j in range(n)]
if f==sorted(f):
t=f
else:
ans+=1
print(ans)
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
``` | output | 1 | 23,278 | 6 | 46,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t. | instruction | 0 | 23,279 | 6 | 46,558 |
Tags: brute force, constructive algorithms, implementation
Correct Solution:
```
# cook your dish here
from sys import stdin,stdout
from collections import Counter
from itertools import permutations
import bisect
import math
I=lambda: map(int,stdin.readline().split())
I1=lambda: stdin.readline()
n,m=I()
if n==1:
print(0)
exit()
s=[I1() for i in range(n)]
if m==1:
if s==sorted(s):
print(0)
else:
print(1)
exit()
c,i=0,1
while True:
if i>len(s[0]): break
for j in range(n-1):
if s[j][:i]>s[j+1][:i]:
c+=1
s=[x[:(i-1)]+x[i:] for x in s]
i-=1
break
i+=1
print(c)
``` | output | 1 | 23,279 | 6 | 46,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t. | instruction | 0 | 23,280 | 6 | 46,560 |
Tags: brute force, constructive algorithms, implementation
Correct Solution:
```
n, m = list(map(int, input().rstrip().split()))
words = []
op = 0
for i in range(n):
words += [input()]
final = [""]*n
for j in range(m):
flag = 1
for i in range(n - 1):
x = final[i] + words[i][j]
y = final[i + 1] + words[i + 1][j]
if x > y:
op += 1
flag = 0
break
if flag == 1:
for i in range(n):
final[i] += words[i][j]
print(op)
``` | output | 1 | 23,280 | 6 | 46,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t. | instruction | 0 | 23,281 | 6 | 46,562 |
Tags: brute force, constructive algorithms, implementation
Correct Solution:
```
from sys import maxsize, stdout, stdin,stderr
mod = int(1e9+7)
import sys
def I(): return int(stdin.readline())
def lint(): return [int(x) for x in stdin.readline().split()]
def S(): return list(map(str,input().strip()))
def grid(r, c): return [lint() for i in range(r)]
from collections import defaultdict, Counter, deque
import math
import heapq
from heapq import heappop , heappush
import bisect
from itertools import groupby
from itertools import permutations as comb
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 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
def primefactors(n):
num=0
while n % 2 == 0:
num+=1
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
num+=1
n = n // i
if n > 2:
num+=1
return num
'''
def iter_ds(src):
store=[src]
while len(store):
tmp=store.pop()
if not vis[tmp]:
vis[tmp]=True
for j in ar[tmp]:
store.append(j)
'''
def ask(a):
print('? {}'.format(a),flush=True)
n=I()
return n
def dfs(i,p):
a,tmp=0,0
for j in d[i]:
if j!=p:
a+=1
tmp+=dfs(j,i)
if a==0:
return 0
return tmp/a + 1
def primeFactors(n):
l=[]
while n % 2 == 0:
l.append(2)
n = n // 2
if n > 2:
l.append(n)
return l
r,c=lint()
s=[]
s.append(['a']*(c+1))
for _ in range(r):
s.append(['a']+S())
cnt=0
k=0
t=[0]*(r+1)
for i in range(1,c+1):
flag=0
for j in range(1,r):
if not t[j]:
if ord(s[j+1][i])<ord(s[j][i]):
cnt+=1
flag=1
break
if not flag:
for j in range(1,r):
if s[j][i]<s[j+1][i]:
t[j]=1
print(cnt)
``` | output | 1 | 23,281 | 6 | 46,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t. | instruction | 0 | 23,282 | 6 | 46,564 |
Tags: brute force, constructive algorithms, implementation
Correct Solution:
```
from operator import itemgetter
n, m = map(int, input().strip().split())
table = []
bad_cols = [False] * m
for _ in range(n):
table.append(input().strip())
for j in range(m):
indices = [idx for idx in range(j + 1) if not bad_cols[idx]]
for i in range(1, n):
if itemgetter(*indices)(table[i]) < itemgetter(*indices)(table[i - 1]):
bad_cols[j] = True
break
print(sum(bad_cols))
``` | output | 1 | 23,282 | 6 | 46,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t. | instruction | 0 | 23,283 | 6 | 46,566 |
Tags: brute force, constructive algorithms, implementation
Correct Solution:
```
n, m = list(map(int, input().split()))
ls = [input() for _ in range(n)]
sorted_inds = [0] * n
ans = 0
for i in range(m):
new_sorted_inds = set()
for j in range(1, n):
if sorted_inds[j] == 0:
if ls[j][i] < ls[j - 1][i]:
ans += 1
break
else:
for j in range(1, n):
if ls[j][i] > ls[j - 1][i]:
sorted_inds[j] = 1
print(ans)
``` | output | 1 | 23,283 | 6 | 46,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t. | instruction | 0 | 23,284 | 6 | 46,568 |
Tags: brute force, constructive algorithms, implementation
Correct Solution:
```
import sys
import math
MAXNUM = math.inf
MINNUM = -1 * math.inf
ASCIILOWER = 97
ASCIIUPPER = 65
def getInt():
return int(sys.stdin.readline().rstrip())
def getInts():
return map(int, sys.stdin.readline().rstrip().split(" "))
def getString():
return sys.stdin.readline().rstrip()
def printOutput(ans):
sys.stdout.write()
pass
def orderedVal(grid, curValues, column):
if column == 0:
return 0
for i in range(len(grid)):
if grid[i][curValues] > grid[i][column]:
return math.inf
return 0
def isLex(grid, column):
for i in range(1, len(grid)):
if grid[i][column] < grid[i - 1][column]:
return False
return True
def getLexPairs(grid, column):
lPairs = set()
for i in range(1, len(grid)):
if grid[i][column] == grid[i - 1][column]:
lPairs.add((i - 1, i))
return lPairs
def compare(grid, column, lPairs):
nextlPairs = set()
"""either returns remaining lPairs, True, or False"""
for a, b in lPairs:
# print("hi", a, b)
# print(grid[a][column], grid[b][column])
if grid[a][column] > grid[b][column]:
return False
elif grid[a][column] == grid[b][column]:
nextlPairs.add((a, b))
return nextlPairs
def solve(grid):
ans = len(grid[0])
for column in range(len(grid[0])):
if ans <= column:
break
if isLex(grid, column):
removeCol = 0
lPairs = getLexPairs(grid, column)
# print(column, lPairs)
for nextColumn in range(column + 1, len(grid[0])):
k = compare(grid, nextColumn, lPairs)
# print('comp',k)
if k == False:
removeCol += 1
elif len(k) == 0:
break
else:
lPairs = k
ans = min(ans, column + removeCol)
return ans
def readinput():
n, m = getInts()
grid = []
for _ in range(n):
grid.append(getString())
print(solve(grid))
readinput()
``` | output | 1 | 23,284 | 6 | 46,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t. | instruction | 0 | 23,285 | 6 | 46,570 |
Tags: brute force, constructive algorithms, implementation
Correct Solution:
```
line = input().split()
n = int(line[0])
m = int(line[1])
data = []
for i in range(n):
data.append(input())
x = [1 for i in range(n)]
delete = 0
for i in range(m):
dum = [1 for j in range(n)]
status = 1
for j in range(1,n):
if data[j][i] > data[j-1][i]:
dum[j] = 0
elif data[j][i] == data[j-1][i]:
dum[j] = 1
else:
if x[j] == 1:
delete += 1
status = 0
break
if status == 1:
for j in range(len(dum)):
if dum[j] == 0:
x[j] = 0
print(delete)
``` | output | 1 | 23,285 | 6 | 46,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t.
Submitted Solution:
```
n, m = [int(x) for x in input().split()]
rs = [input() for _ in range(n)]
mask = [1] * m
def get_masked(x):
return ''.join(y for y, m in zip(x, mask) if m)
res = 0
for i in range(m):
l = [get_masked(x)[:i - res + 1] for x in rs]
if not all(l[i] <= l[i+1] for i in range(len(l)-1)):
mask[i] = 0
res += 1
print(res)
``` | instruction | 0 | 23,286 | 6 | 46,572 |
Yes | output | 1 | 23,286 | 6 | 46,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t.
Submitted Solution:
```
n, m = map(int, input().split())
grid = []
for _ in range(n):
grid.append(list(input()))
deleted = 0
last = []
for j in range(m):
for i in range(1,n):
if grid[i][j] < grid[i-1][j] and len(last) == 0:
deleted += 1
break
elif len(last) and grid[i][j] < grid[i-1][j]:
if ''.join(grid[i-1][x] for x in last+[j]) > ''.join(grid[i][x] for x in last+[j]):
deleted += 1
break
else:
last.append(j)
print(deleted)
``` | instruction | 0 | 23,287 | 6 | 46,574 |
Yes | output | 1 | 23,287 | 6 | 46,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t.
Submitted Solution:
```
import sys
from collections import deque
import math
input_ = lambda: sys.stdin.readline().strip("\r\n")
ii = lambda : int(input_())
il = lambda : list(map(int, input_().split()))
ilf = lambda : list(map(float, input_().split()))
ip = lambda : input_()
fi = lambda : float(input_())
li = lambda : list(input_())
pr = lambda x : print(x)
f = lambda : sys.stdout.flush()
n,m = il()
ans = 0
a = []
for _ in range(n) :
a.append(ip())
b = ['' for _ in range (n)]
for i in range (m) :
c = []
for j in range(n) :
c.append(b[j]+a[j][i])
if (c == sorted(c)) :
if len(c) == len(set(c)) :
break
else :
b = c
else :
ans += 1
print(ans)
``` | instruction | 0 | 23,288 | 6 | 46,576 |
Yes | output | 1 | 23,288 | 6 | 46,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t.
Submitted Solution:
```
def remove(p, a):
for i in range(len(a)):
a[i] = a[i][: p] + a[i][p + 1: ]
n, m = [int(x) for x in input().split()]
a = []
for i in range(n):
a.append(input())
removed = 0
for i in range(m):
i -= removed
for j in range(n - 1):
if a[j][: i + 1] > a[j + 1][: i + 1]:
remove(i, a)
removed += 1
break
print(removed)
# Made By Mostafa_Khaled
``` | instruction | 0 | 23,289 | 6 | 46,578 |
Yes | output | 1 | 23,289 | 6 | 46,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t.
Submitted Solution:
```
#!/usr/bin/env python3
n, m = map(int, input().split())
table = [[c for c in input()] for i in range(n)]
def is_good_column(c, adj):
flag = False
for i in range(n - 1):
if (not adj or (adj and table[i][adj] == table[i+1][adj])) and table[i][c] > table[i+1][c]:
return False, False
elif table[i][c] == table[i+1][c]:
flag = True
return True, flag
i = 0
need = m
count = 0
adj = None
while i < need:
flag, dup = is_good_column(i, adj)
if flag:
adj = i
if not dup:
break
else:
count += 1
need = min(i + 2, m)
i += 1
print(count)
``` | instruction | 0 | 23,290 | 6 | 46,580 |
No | output | 1 | 23,290 | 6 | 46,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t.
Submitted Solution:
```
# from itertools import combinations
# from bisect import bisect_left
# from functools import *
# from collections import Counter
def is_lexigraphic(s):
for i in range(len(s) - 1):
if s[i] > s[i + 1]:
return False
return True
I = lambda: list(map(int, input().split()))
n, m = I()
a = [input() for i in range(n)]
mn = m
columns = list(zip(*a))
for i in range(m):
curCol = columns[i]
if not is_lexigraphic(curCol):
continue
l = []
for j in range(n - 1):
if curCol[j] == curCol[j + 1]:
l.append((j, j + 1))
x = 0
for k in range(i + 1, m):
if any([columns[k][a] > columns[k][b] for a, b in l]):
x += 1
else:
break
mn = min(x + i, mn)
print(mn)
``` | instruction | 0 | 23,291 | 6 | 46,582 |
No | output | 1 | 23,291 | 6 | 46,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t.
Submitted Solution:
```
from itertools import combinations, permutations
def is_good(s,n,m):
for i in range(n-1):
if s[i] > s[i+1]: return False
return True
def main2(n,m,a):
if n <= 0 or m<=0 or is_good(a,n,m):
print(0)
return
for i in range(1,m):
p = combinations(range(m),i)
for j in p:
s = ["".join([k for b,k in enumerate(x) if b not in j]) for x in a]
if is_good(s,n,m-i):
print(i)
return
print(m)
def main3(n,m,a):
if n <= 1 or m<=0 or is_good(a,n,m):
print(0)
return
bad = []
for i in range(0,m):
if not is_good(["".join([k for b,k in enumerate(x) if b!=i])
for x in a],n,m):
bad.append(i)
for i in range(1,len(bad)+1):
p = combinations(bad,i)
for j in p:
s = ["".join([k for b,k in enumerate(x) if b not in j]) for x in a]
if is_good(s,n,m-i):
print(i)
return
print(m)
def main4(n,m,a):
if n <= 1 or m<=0 or is_good(a,n,m):
print(0)
return
bad = []
for i in range(0,m):
if not is_good(["".join([k for b,k in enumerate(x) if b!=i])
for x in a],n,m):
bad.append(i)
p = []
for i in bad:
p.append(i)
s = ["".join([k for b,k in enumerate(x) if b not in p]) for x in a]
if is_good(s,n,m-i):
print(i)
return
print(m)
def line_is_good(a,j):
for i in range(len(a)-1):
if a[i][j] > a[i+1][j]: return False
return True
def main(n,m,a):
if n <= 1 or m<=0 or is_good(a,n,m):
print(0)
return
bad = []
for i in range(0,m):
if line_is_good(a,i):
continue
bad.append(i)
print(i)
s = ["".join([k for b,k in enumerate(x) if b not in bad]) for x in a]
if is_good(s,n,m-len(bad)):
print(len(bad))
return
print(m)
def main_input():
n,m = [int(i) for i in input().split()]
a = [input() for s in range(n)]
main(n,m,a)
if __name__ == "__main__":
main_input()
``` | instruction | 0 | 23,292 | 6 | 46,584 |
No | output | 1 | 23,292 | 6 | 46,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd
edfg
hijk
we obtain the table:
acd
efg
hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Output
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Examples
Input
1 10
codeforces
Output
0
Input
4 4
case
care
test
code
Output
2
Input
5 4
code
forc
esco
defo
rces
Output
4
Note
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t.
Submitted Solution:
```
from itertools import combinations, permutations
def is_good(s,n,m):
for i in range(n-1):
if s[i] > s[i+1]: return False
return True
def main2(n,m,a):
if n <= 0 or m<=0 or is_good(a,n,m):
print(0)
return
for i in range(1,m):
p = combinations(range(m),i)
for j in p:
s = ["".join([k for b,k in enumerate(x) if b not in j]) for x in a]
if is_good(s,n,m-i):
print(i)
return
print(m)
def main(n,m,a):
if n <= 0 or m<=0 or is_good(a,n,m):
print(0)
return
bad = []
for i in range(0,m):
if not is_good(["".join([k for b,k in enumerate(x) if b!=i])
for x in a],n,m):
bad.append(i)
for i in range(1,m):
p = combinations(bad,i)
for j in p:
s = ["".join([k for b,k in enumerate(x) if b not in j]) for x in a]
if is_good(s,n,m-i):
print(i)
return
print(m)
def main_input():
n,m = [int(i) for i in input().split()]
a = [input() for s in range(n)]
main(n,m,a)
if __name__ == "__main__":
main_input()
``` | instruction | 0 | 23,293 | 6 | 46,586 |
No | output | 1 | 23,293 | 6 | 46,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is working on a new operating system called BerOS. He asks you to help with implementation of a file suggestion feature.
There are n files on hard drive and their names are f_1, f_2, ..., f_n. Any file name contains between 1 and 8 characters, inclusive. All file names are unique.
The file suggestion feature handles queries, each represented by a string s. For each query s it should count number of files containing s as a substring (i.e. some continuous segment of characters in a file name equals s) and suggest any such file name.
For example, if file names are "read.me", "hosts", "ops", and "beros.18", and the query is "os", the number of matched files is 2 (two file names contain "os" as a substring) and suggested file name can be either "hosts" or "beros.18".
Input
The first line of the input contains integer n (1 ≤ n ≤ 10000) — the total number of files.
The following n lines contain file names, one per line. The i-th line contains f_i — the name of the i-th file. Each file name contains between 1 and 8 characters, inclusive. File names contain only lowercase Latin letters, digits and dot characters ('.'). Any sequence of valid characters can be a file name (for example, in BerOS ".", ".." and "..." are valid file names). All file names are unique.
The following line contains integer q (1 ≤ q ≤ 50000) — the total number of queries.
The following q lines contain queries s_1, s_2, ..., s_q, one per line. Each s_j has length between 1 and 8 characters, inclusive. It contains only lowercase Latin letters, digits and dot characters ('.').
Output
Print q lines, one per query. The j-th line should contain the response on the j-th query — two values c_j and t_j, where
* c_j is the number of matched files for the j-th query,
* t_j is the name of any file matched by the j-th query. If there is no such file, print a single character '-' instead. If there are multiple matched files, print any.
Example
Input
4
test
contests
test.
.test
6
ts
.
st.
.test
contes.
st
Output
1 contests
2 .test
1 test.
1 .test
0 -
4 test. | instruction | 0 | 23,865 | 6 | 47,730 |
Tags: brute force, implementation
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()
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)]
########################################################################
n=ri()
d=ddict(lambda:[0,''])
for _ in range(n):
s=rl()
k=len(s)
S=set()
for i in range(k):
for j in range(i,k):
S.add(s[i:j+1])
for t in S:
d[t][0]+=1
if d[t][0]==1:
d[t][1]=s
q=ri()
for _ in range(q):
s=rl()
if s not in d:
print(0,'-')
continue
print(*d[s])
``` | output | 1 | 23,865 | 6 | 47,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is working on a new operating system called BerOS. He asks you to help with implementation of a file suggestion feature.
There are n files on hard drive and their names are f_1, f_2, ..., f_n. Any file name contains between 1 and 8 characters, inclusive. All file names are unique.
The file suggestion feature handles queries, each represented by a string s. For each query s it should count number of files containing s as a substring (i.e. some continuous segment of characters in a file name equals s) and suggest any such file name.
For example, if file names are "read.me", "hosts", "ops", and "beros.18", and the query is "os", the number of matched files is 2 (two file names contain "os" as a substring) and suggested file name can be either "hosts" or "beros.18".
Input
The first line of the input contains integer n (1 ≤ n ≤ 10000) — the total number of files.
The following n lines contain file names, one per line. The i-th line contains f_i — the name of the i-th file. Each file name contains between 1 and 8 characters, inclusive. File names contain only lowercase Latin letters, digits and dot characters ('.'). Any sequence of valid characters can be a file name (for example, in BerOS ".", ".." and "..." are valid file names). All file names are unique.
The following line contains integer q (1 ≤ q ≤ 50000) — the total number of queries.
The following q lines contain queries s_1, s_2, ..., s_q, one per line. Each s_j has length between 1 and 8 characters, inclusive. It contains only lowercase Latin letters, digits and dot characters ('.').
Output
Print q lines, one per query. The j-th line should contain the response on the j-th query — two values c_j and t_j, where
* c_j is the number of matched files for the j-th query,
* t_j is the name of any file matched by the j-th query. If there is no such file, print a single character '-' instead. If there are multiple matched files, print any.
Example
Input
4
test
contests
test.
.test
6
ts
.
st.
.test
contes.
st
Output
1 contests
2 .test
1 test.
1 .test
0 -
4 test. | instruction | 0 | 23,866 | 6 | 47,732 |
Tags: brute force, implementation
Correct Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 11/3/18
"""
import collections
N = int(input())
subs = collections.defaultdict(int)
subi = collections.defaultdict(int)
files = []
for fi in range(N):
f = input()
files.append(f)
ss = set()
for i in range(len(f)):
for j in range(i+1, len(f)+1):
s = f[i: j]
ss.add(s)
subi[s] = fi
for s in ss:
subs[s] += 1
Q = int(input())
qs = [input() for _ in range(Q)]
ans = []
for s in qs:
count = subs[s]
af = files[subi[s]] if count > 0 else '-'
ans.append('{} {}'.format(count, af))
print('\n'.join(ans))
``` | output | 1 | 23,866 | 6 | 47,733 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is working on a new operating system called BerOS. He asks you to help with implementation of a file suggestion feature.
There are n files on hard drive and their names are f_1, f_2, ..., f_n. Any file name contains between 1 and 8 characters, inclusive. All file names are unique.
The file suggestion feature handles queries, each represented by a string s. For each query s it should count number of files containing s as a substring (i.e. some continuous segment of characters in a file name equals s) and suggest any such file name.
For example, if file names are "read.me", "hosts", "ops", and "beros.18", and the query is "os", the number of matched files is 2 (two file names contain "os" as a substring) and suggested file name can be either "hosts" or "beros.18".
Input
The first line of the input contains integer n (1 ≤ n ≤ 10000) — the total number of files.
The following n lines contain file names, one per line. The i-th line contains f_i — the name of the i-th file. Each file name contains between 1 and 8 characters, inclusive. File names contain only lowercase Latin letters, digits and dot characters ('.'). Any sequence of valid characters can be a file name (for example, in BerOS ".", ".." and "..." are valid file names). All file names are unique.
The following line contains integer q (1 ≤ q ≤ 50000) — the total number of queries.
The following q lines contain queries s_1, s_2, ..., s_q, one per line. Each s_j has length between 1 and 8 characters, inclusive. It contains only lowercase Latin letters, digits and dot characters ('.').
Output
Print q lines, one per query. The j-th line should contain the response on the j-th query — two values c_j and t_j, where
* c_j is the number of matched files for the j-th query,
* t_j is the name of any file matched by the j-th query. If there is no such file, print a single character '-' instead. If there are multiple matched files, print any.
Example
Input
4
test
contests
test.
.test
6
ts
.
st.
.test
contes.
st
Output
1 contests
2 .test
1 test.
1 .test
0 -
4 test. | instruction | 0 | 23,867 | 6 | 47,734 |
Tags: brute force, implementation
Correct Solution:
```
def func1(s,i,index,dict1):
if(i==len(s)):
return
sx=''
for j in range(i,len(s)):
sx+=s[j]
try:
if(dict1[sx][1]!=index):
dict1[sx][0]+=1
dict1[sx][1]=index
except:
KeyError
dict1[sx]=[1,index]
func1(s,i+1,index,dict1)
return
n=int(input())
arr=[]
dict1={}
for i in range(n):
s=str(input())
arr.append(s)
func1(s,0,i,dict1)
q=int(input())
for i in range(q):
s=str(input())
#print(dict1['ts'])
try:
val1=dict1[s][0]
val2=arr[dict1[s][1]]
print(val1,val2)
except:
KeyError
print(0,'-')
``` | output | 1 | 23,867 | 6 | 47,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is working on a new operating system called BerOS. He asks you to help with implementation of a file suggestion feature.
There are n files on hard drive and their names are f_1, f_2, ..., f_n. Any file name contains between 1 and 8 characters, inclusive. All file names are unique.
The file suggestion feature handles queries, each represented by a string s. For each query s it should count number of files containing s as a substring (i.e. some continuous segment of characters in a file name equals s) and suggest any such file name.
For example, if file names are "read.me", "hosts", "ops", and "beros.18", and the query is "os", the number of matched files is 2 (two file names contain "os" as a substring) and suggested file name can be either "hosts" or "beros.18".
Input
The first line of the input contains integer n (1 ≤ n ≤ 10000) — the total number of files.
The following n lines contain file names, one per line. The i-th line contains f_i — the name of the i-th file. Each file name contains between 1 and 8 characters, inclusive. File names contain only lowercase Latin letters, digits and dot characters ('.'). Any sequence of valid characters can be a file name (for example, in BerOS ".", ".." and "..." are valid file names). All file names are unique.
The following line contains integer q (1 ≤ q ≤ 50000) — the total number of queries.
The following q lines contain queries s_1, s_2, ..., s_q, one per line. Each s_j has length between 1 and 8 characters, inclusive. It contains only lowercase Latin letters, digits and dot characters ('.').
Output
Print q lines, one per query. The j-th line should contain the response on the j-th query — two values c_j and t_j, where
* c_j is the number of matched files for the j-th query,
* t_j is the name of any file matched by the j-th query. If there is no such file, print a single character '-' instead. If there are multiple matched files, print any.
Example
Input
4
test
contests
test.
.test
6
ts
.
st.
.test
contes.
st
Output
1 contests
2 .test
1 test.
1 .test
0 -
4 test. | instruction | 0 | 23,868 | 6 | 47,736 |
Tags: brute force, implementation
Correct Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
from math import ceil, floor, factorial;
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp
## gcd function
def gcd(a,b):
if a == 0:
return b
return gcd(b%a, a)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if a[mid] < x:
lo = mid+1
else:
hi = mid
return lo
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()))
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
count = {}; dick = {};
n = int(input());
for __ in range(n):
s = input().strip();
this = set();
for i in range(len(s)):
x = '';
for j in range(i, len(s)):
x += s[j];
if x in this:
continue;
this.add(x);
dick[x] = s;
count[x] = count.get(x, 0) + 1;
q = int(input());
for __ in range(q):
s = input().strip();
if s in dick:
x = count[s];
y = dick[s];
else:
x = 0;
y = '-';
print(*(x, y));
``` | output | 1 | 23,868 | 6 | 47,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is working on a new operating system called BerOS. He asks you to help with implementation of a file suggestion feature.
There are n files on hard drive and their names are f_1, f_2, ..., f_n. Any file name contains between 1 and 8 characters, inclusive. All file names are unique.
The file suggestion feature handles queries, each represented by a string s. For each query s it should count number of files containing s as a substring (i.e. some continuous segment of characters in a file name equals s) and suggest any such file name.
For example, if file names are "read.me", "hosts", "ops", and "beros.18", and the query is "os", the number of matched files is 2 (two file names contain "os" as a substring) and suggested file name can be either "hosts" or "beros.18".
Input
The first line of the input contains integer n (1 ≤ n ≤ 10000) — the total number of files.
The following n lines contain file names, one per line. The i-th line contains f_i — the name of the i-th file. Each file name contains between 1 and 8 characters, inclusive. File names contain only lowercase Latin letters, digits and dot characters ('.'). Any sequence of valid characters can be a file name (for example, in BerOS ".", ".." and "..." are valid file names). All file names are unique.
The following line contains integer q (1 ≤ q ≤ 50000) — the total number of queries.
The following q lines contain queries s_1, s_2, ..., s_q, one per line. Each s_j has length between 1 and 8 characters, inclusive. It contains only lowercase Latin letters, digits and dot characters ('.').
Output
Print q lines, one per query. The j-th line should contain the response on the j-th query — two values c_j and t_j, where
* c_j is the number of matched files for the j-th query,
* t_j is the name of any file matched by the j-th query. If there is no such file, print a single character '-' instead. If there are multiple matched files, print any.
Example
Input
4
test
contests
test.
.test
6
ts
.
st.
.test
contes.
st
Output
1 contests
2 .test
1 test.
1 .test
0 -
4 test. | instruction | 0 | 23,869 | 6 | 47,738 |
Tags: brute force, implementation
Correct Solution:
```
if __name__ == "__main__":
Map = {}
siz = {}
n = int(input())
for i in range(n):
str = input()
MMap = {}
for j in range(0,len(str)):
t = ""
for k in range(j,len(str)):
t += str[k]
MMap[t] = 1
for x in MMap:
if not x in Map:
Map[x] = str
siz[x] = 0
siz[x] +=1
#for x in Map:
#print(x,Map[x])
q = int(input())
for i in range(q):
str = input()
if str in Map:
print(siz[str], Map[str])
else:
print("0 -")
``` | output | 1 | 23,869 | 6 | 47,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is working on a new operating system called BerOS. He asks you to help with implementation of a file suggestion feature.
There are n files on hard drive and their names are f_1, f_2, ..., f_n. Any file name contains between 1 and 8 characters, inclusive. All file names are unique.
The file suggestion feature handles queries, each represented by a string s. For each query s it should count number of files containing s as a substring (i.e. some continuous segment of characters in a file name equals s) and suggest any such file name.
For example, if file names are "read.me", "hosts", "ops", and "beros.18", and the query is "os", the number of matched files is 2 (two file names contain "os" as a substring) and suggested file name can be either "hosts" or "beros.18".
Input
The first line of the input contains integer n (1 ≤ n ≤ 10000) — the total number of files.
The following n lines contain file names, one per line. The i-th line contains f_i — the name of the i-th file. Each file name contains between 1 and 8 characters, inclusive. File names contain only lowercase Latin letters, digits and dot characters ('.'). Any sequence of valid characters can be a file name (for example, in BerOS ".", ".." and "..." are valid file names). All file names are unique.
The following line contains integer q (1 ≤ q ≤ 50000) — the total number of queries.
The following q lines contain queries s_1, s_2, ..., s_q, one per line. Each s_j has length between 1 and 8 characters, inclusive. It contains only lowercase Latin letters, digits and dot characters ('.').
Output
Print q lines, one per query. The j-th line should contain the response on the j-th query — two values c_j and t_j, where
* c_j is the number of matched files for the j-th query,
* t_j is the name of any file matched by the j-th query. If there is no such file, print a single character '-' instead. If there are multiple matched files, print any.
Example
Input
4
test
contests
test.
.test
6
ts
.
st.
.test
contes.
st
Output
1 contests
2 .test
1 test.
1 .test
0 -
4 test. | instruction | 0 | 23,870 | 6 | 47,740 |
Tags: brute force, implementation
Correct Solution:
```
###################
# 18th August 2019.
###################
##########################################################################
# Getting fileCount from Codeforces.
fileCount = int(input())
# Method to generate all substrings of given string.
from itertools import combinations as cmb
def generateSubstringsFor(string):
length = len(string) + 1
return [string[x:y] for x, y in cmb(range(length), r=2)]
# HashMap mapping substrings to set of containing strings.
stringMap = {}
for i in range(fileCount):
newFile = input()
# Generating entry for new file.
for string in generateSubstringsFor(newFile):
# Case Set has been created.
if not string in stringMap:
stringMap[string] = set([newFile])
# Case First Entry.
else: stringMap[string].add(newFile)
# Method to entertain a single query.
def outputForQuery(query):
if query in stringMap:
return str(len(stringMap[query]))+" "+next(iter(stringMap[query]))
else: return "0 - "
# We now entertain all requests from Codeforces.
queryCount = int(input())
for i in range(queryCount): print(outputForQuery(input()))
##########################################################################
########################################
# Programming-Credits atifcppprogrammer.
########################################
``` | output | 1 | 23,870 | 6 | 47,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is working on a new operating system called BerOS. He asks you to help with implementation of a file suggestion feature.
There are n files on hard drive and their names are f_1, f_2, ..., f_n. Any file name contains between 1 and 8 characters, inclusive. All file names are unique.
The file suggestion feature handles queries, each represented by a string s. For each query s it should count number of files containing s as a substring (i.e. some continuous segment of characters in a file name equals s) and suggest any such file name.
For example, if file names are "read.me", "hosts", "ops", and "beros.18", and the query is "os", the number of matched files is 2 (two file names contain "os" as a substring) and suggested file name can be either "hosts" or "beros.18".
Input
The first line of the input contains integer n (1 ≤ n ≤ 10000) — the total number of files.
The following n lines contain file names, one per line. The i-th line contains f_i — the name of the i-th file. Each file name contains between 1 and 8 characters, inclusive. File names contain only lowercase Latin letters, digits and dot characters ('.'). Any sequence of valid characters can be a file name (for example, in BerOS ".", ".." and "..." are valid file names). All file names are unique.
The following line contains integer q (1 ≤ q ≤ 50000) — the total number of queries.
The following q lines contain queries s_1, s_2, ..., s_q, one per line. Each s_j has length between 1 and 8 characters, inclusive. It contains only lowercase Latin letters, digits and dot characters ('.').
Output
Print q lines, one per query. The j-th line should contain the response on the j-th query — two values c_j and t_j, where
* c_j is the number of matched files for the j-th query,
* t_j is the name of any file matched by the j-th query. If there is no such file, print a single character '-' instead. If there are multiple matched files, print any.
Example
Input
4
test
contests
test.
.test
6
ts
.
st.
.test
contes.
st
Output
1 contests
2 .test
1 test.
1 .test
0 -
4 test. | instruction | 0 | 23,871 | 6 | 47,742 |
Tags: brute force, implementation
Correct Solution:
```
from collections import defaultdict
N = int(input())
F = []
for i in range(N):
F.append(input())
dic = defaultdict(set)
for i in range(N):
t = F[i]
M = len(t)
for j in range(M):
for k in range(j + 1, M + 1):
dic[t[j: k]].add(F[i])
Q = int(input())
for i in range(Q):
s = input()
if s in dic:
print(len(dic[s]), list(dic[s])[0])
else:
print("0 -")
``` | output | 1 | 23,871 | 6 | 47,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is working on a new operating system called BerOS. He asks you to help with implementation of a file suggestion feature.
There are n files on hard drive and their names are f_1, f_2, ..., f_n. Any file name contains between 1 and 8 characters, inclusive. All file names are unique.
The file suggestion feature handles queries, each represented by a string s. For each query s it should count number of files containing s as a substring (i.e. some continuous segment of characters in a file name equals s) and suggest any such file name.
For example, if file names are "read.me", "hosts", "ops", and "beros.18", and the query is "os", the number of matched files is 2 (two file names contain "os" as a substring) and suggested file name can be either "hosts" or "beros.18".
Input
The first line of the input contains integer n (1 ≤ n ≤ 10000) — the total number of files.
The following n lines contain file names, one per line. The i-th line contains f_i — the name of the i-th file. Each file name contains between 1 and 8 characters, inclusive. File names contain only lowercase Latin letters, digits and dot characters ('.'). Any sequence of valid characters can be a file name (for example, in BerOS ".", ".." and "..." are valid file names). All file names are unique.
The following line contains integer q (1 ≤ q ≤ 50000) — the total number of queries.
The following q lines contain queries s_1, s_2, ..., s_q, one per line. Each s_j has length between 1 and 8 characters, inclusive. It contains only lowercase Latin letters, digits and dot characters ('.').
Output
Print q lines, one per query. The j-th line should contain the response on the j-th query — two values c_j and t_j, where
* c_j is the number of matched files for the j-th query,
* t_j is the name of any file matched by the j-th query. If there is no such file, print a single character '-' instead. If there are multiple matched files, print any.
Example
Input
4
test
contests
test.
.test
6
ts
.
st.
.test
contes.
st
Output
1 contests
2 .test
1 test.
1 .test
0 -
4 test. | instruction | 0 | 23,872 | 6 | 47,744 |
Tags: brute force, implementation
Correct Solution:
```
# https://codeforces.com/problemset/problem/1070/H
from sys import stdin, exit
from typing import List, Tuple
sub_strings = dict()
files_num = int(stdin.readline().rstrip())
for _id in range(files_num):
f_name = stdin.readline().rstrip()
for i in range(len(f_name)):
for j in range(i + 1, len(f_name) + 1):
key = f_name[i:j]
if key not in sub_strings:
sub_strings[key] = [1, f_name, _id]
else:
rez = sub_strings[key]
if rez[2] != _id:
rez[2] = _id
sub_strings[key][0] += 1
queries = int(stdin.readline().rstrip())
for _ in range(queries):
query = stdin.readline().rstrip()
if query in sub_strings:
rez = sub_strings[query]
print(rez[0], ' ', rez[1])
else:
print(0, ' ', '-')
``` | output | 1 | 23,872 | 6 | 47,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc.
The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23.
Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example.
Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system.
Input
The first line of the input contains integer number n (1 ≤ n ≤ 105), the number of coordinates in the test. Then there follow n lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 106 .
Output
Write n lines, each line should contain a cell coordinates in the other numeration system.
Examples
Input
2
R23C55
BC23
Output
BC23
R23C55 | instruction | 0 | 24,081 | 6 | 48,162 |
Tags: implementation, math
Correct Solution:
```
import string
import re
alphabet=string.ascii_uppercase
n=input()
n=int(n)
def numer1(inp):
inp=inp.replace("R","")
inp=inp.replace("C"," ")
row,column=inp.split(" ",1)
column=int(column)
columnalpha=[]
over=False
while not over:
a=(column-1)//26
b=column-a*26
if b==0:
b=26
column=a
columnalpha.insert(0,alphabet[b-1])
if column==0:
over=True
print("".join(columnalpha)+row)
def numer2(inp):
row="".join(list(filter(str.isdigit,inp)))
columnalpha=list(filter(str.isalpha,inp))
column=0
for i in range(0,len(columnalpha)):
column+=(alphabet.index(columnalpha[i])+1)*26**(len(columnalpha)-1-i)
column=str(column)
print("R"+row+"C"+column)
for i in range(0,n):
inp=input()
numertypeR=len(re.findall(r"R"+"\d",inp))
numertypeC=len(re.findall(r"C"+"\d",inp))
if numertypeR==1 and numertypeC==1:
numer1(inp)
else:
numer2(inp)
``` | output | 1 | 24,081 | 6 | 48,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution | instruction | 0 | 24,136 | 6 | 48,272 |
Tags: greedy, implementation, strings
Correct Solution:
```
s=input()
au=[pos for pos,char in enumerate(s) if char=='@']
dis=[au[i+1]-au[i] for i in range(len(au)-1)]
dis=[1 for i in dis if i<=2]
if len(au)==0 or au[0]==0 or au[-1]==len(s)-1 or sum(dis)>0:
print("No solution")
else:
if len(au)==1:
print(s)
else:
res=[]
i=0
j=0
m=au[j]+2
while j<len(au):
res.append(s[i:m])
i=m
if i>len(s):
break
j+=1
if j==len(au)-1:
m=len(s)+1
else:
m=au[j]+2
print(*res,sep=",")
``` | output | 1 | 24,136 | 6 | 48,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution | instruction | 0 | 24,137 | 6 | 48,274 |
Tags: greedy, implementation, strings
Correct Solution:
```
#amros
s=input()
t=[]
k=0
y=1
while 1:
a=s.find('@',k+1)
if a<0:break
r=s[k:a+2]
if r.count('@')==0 or r[0]=='@'or r[-1]=='@':y=0
t+=[r];k=a+2
if t and s[k:].count('@')==0:t[-1]+=s[k:]
p=''
for i in t:p+=i
print(['No solution',','.join(t)][p==s and y])
``` | output | 1 | 24,137 | 6 | 48,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution | instruction | 0 | 24,138 | 6 | 48,276 |
Tags: greedy, implementation, strings
Correct Solution:
```
from sys import stdout
a=input()
c=list(a.split('@'))
while '' in c:
c.remove('')
if a.count('@')==0:
exit(print('No solution'))
if a.count('@')!=len(c)-1:
exit(print('No solution'))
if any(len(i)==1 for i in c[1:len(c)-1]):
exit(print('No solution'))
stdout.write(c[0]+'@')
for i in range(1,len(c)-1):
stdout.write(c[i][:len(c[i])//2]+','+c[i][len(c[i])//2:]+'@')
stdout.write(c[-1])
``` | output | 1 | 24,138 | 6 | 48,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution | instruction | 0 | 24,139 | 6 | 48,278 |
Tags: greedy, implementation, strings
Correct Solution:
```
a=input().split('@')
n=len(a)
b=[]
if n>=2:
for i in range(n):
if len(a[i])>int(i!=0 and i!=n-1):
if i!=0 and i!=n-1:
b.append(a[i-1][min(1,i-1):]+'@'+a[i][0])
elif i==n-1:
b.append(a[i-1][min(1,i-1):]+'@'+a[i])
else:
print('No solution')
break
else:
c=b[0]
for i in range(1,len(b)):
c+=','+b[i]
print(c)
else:
print('No solution')
``` | output | 1 | 24,139 | 6 | 48,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution | instruction | 0 | 24,140 | 6 | 48,280 |
Tags: greedy, implementation, strings
Correct Solution:
```
a = input()
b = a.split("@")
ans = []
flag = True
for i in b[1:-1]:
if len(i) < 2:
flag = False
break
if len(b) < 2:
flag = False
if b[0] and b[-1] and flag:
ind = 0
index_list = []
str_len = len(a)
temp = ""
j = 0
ans = []
while j != str_len:
temp += a[j]
if a[j] == "@":
temp += a[j + 1]
ans.append(temp)
temp = ""
j += 1
j += 1
ans[-1] = ans[-1] + temp
print(",".join(ans))
else:
print("No solution")
``` | output | 1 | 24,140 | 6 | 48,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution | instruction | 0 | 24,141 | 6 | 48,282 |
Tags: greedy, implementation, strings
Correct Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
s = input().rstrip()
ans = []
while s and s[0] != '@':
for i in range(len(s) - 1):
if s[i] == '@':
ans.append(s[:i + 2])
s = s[i + 2:]
break
else:
break
if s and '@' not in s and ans:
ans[-1] += s
s = ''
if not s and all(addr.count('@') == 1 for addr in ans):
print(','.join(ans))
else:
print('No solution')
``` | output | 1 | 24,141 | 6 | 48,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution | instruction | 0 | 24,142 | 6 | 48,284 |
Tags: greedy, implementation, strings
Correct Solution:
```
def cf31B():
#print("Hello World")
from sys import stdin,stdout
inp = list(stdin.readline().strip().split('@'))
flag = True
if len(inp)>=2:
if len(inp[0])==0 or len(inp[-1])==0:
flag = False
if flag:
for i in range(1,len(inp)-1):
if len(inp[i]) < 2:
flag = False
break
else:
flag = False
answer = ""
if flag:
for i , j in enumerate(inp):
if i==0:
answer+=j
elif i==len(inp)-1:
answer+='@'+j
else:
answer+='@'+j[:-1]+','+j[-1]
else:
answer = "No solution"
stdout.write(answer+"\n")
if __name__=='__main__':
cf31B()
``` | output | 1 | 24,142 | 6 | 48,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution | instruction | 0 | 24,143 | 6 | 48,286 |
Tags: greedy, implementation, strings
Correct Solution:
```
s=input()
n=len(s)
ans=[]
x=''
c=0
for i in range(n):
if i!=n-1:
if s[i+1]!='@':
x=x+s[i]
else:
if c==0:
x=x+s[i]
c=1
else:
ans.append(x)
x=''
x=x+s[i]
c=1
else:
x=x+s[i]
ans.append(x)
f=0
for i in range(len(ans)):
if ans[i][0]=='@' or ans[i][-1]=='@' or '@' not in ans[i]:
f=1
break
if f:
print('No solution')
else:
for i in range(len(ans)-1):
print(ans[i]+',',end='')
print(ans[-1])
``` | output | 1 | 24,143 | 6 | 48,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution
Submitted Solution:
```
import sys
def input(): return sys.stdin.readline().strip()
def iinput(): return int(input())
def rinput(): return map(int, sys.stdin.readline().strip().split())
def get_list(): return list(map(int, sys.stdin.readline().strip().split()))
mod = int(1e9)+7
s = input()
l = len(s)
ans = []
prev = -1
flag = True
tot = s.count("@")
cnt = 0
if (s[0]=="@" or s[l-1]=="@" or tot==0):
print("No solution")
sys.exit(0)
for i in range(l):
if s[i]=="@":
bef = s[prev+1:i]
aft = s[i+1:]
if "@" in aft:
aft = aft[:aft.index("@")]
if (len(ans)>0 and len(bef)<1) or len(aft)==0:
flag = False
break
if cnt!=tot-1:
ans.append(bef+"@"+aft[0])
else:
ans.append(bef+"@"+aft)
prev = i+1
cnt+=1
# print(bef, aft)
if flag == False:
print("No solution")
else:
print(*ans, sep=",")
``` | instruction | 0 | 24,144 | 6 | 48,288 |
Yes | output | 1 | 24,144 | 6 | 48,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution
Submitted Solution:
```
def solution():
s = input()
if '@' not in s:
return "No solution"
mas = s.split('@')
if len(mas[0]) < 1 or len(mas[-1]) < 1:
return "No solution"
front = mas[0]
back = mas[-1]
mas = mas[1:len(mas) - 1]
for i in range(len(mas)):
if len(mas[i]) < 2:
return "No solution"
else:
mas[i] = mas[i][:len(mas[i]) // 2] + ',' + mas[i][len(mas[i]) // 2:]
return '@'.join([front] + mas + [back])
print(solution())
``` | instruction | 0 | 24,145 | 6 | 48,290 |
Yes | output | 1 | 24,145 | 6 | 48,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution
Submitted Solution:
```
s = input().split('@')
if len(s) > 1 and s[0] and s[-1] and min(map(len, s[1:-1] + ['__'])) > 1:
for i in range(1, len(s) - 1):
s[i] = s[i][0] + ',' + s[i][1:]
print('@'.join(s))
else:
print('No solution')
``` | instruction | 0 | 24,146 | 6 | 48,292 |
Yes | output | 1 | 24,146 | 6 | 48,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution
Submitted Solution:
```
a=input().split("@")
ans=a[0]
if len(a[0])==0 or len(a)<=1 or len(a[-1])==0:
print("No solution")
exit()
for u in range(1,len(a)-1):
if len(a[u])>1:
ans+="@"+a[u][0]+","+a[u][1:]
else:
print("No solution")
exit()
print(ans+"@"+a[-1])
``` | instruction | 0 | 24,147 | 6 | 48,294 |
Yes | output | 1 | 24,147 | 6 | 48,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution
Submitted Solution:
```
s = input()
prev = 0
lis = []
start = 0
flag = 0
for i in range(len(s)):
if s[i] == '@' and prev == 0:
if i-prev<=0:
flag = 1
break
prev= i
elif s[i] == '@':
lis.append(s[start:prev+2])
start = prev+2
prev = i
if flag:
print("No solution")
else:
if '@' in s[start:] and s[len(s)-1]!='@':
lis.append(s[start:])
for i in range(len(lis)-1):
print(lis[i], end = ",")
print(lis[len(lis)-1])
else:
print("No solution")
``` | instruction | 0 | 24,148 | 6 | 48,296 |
No | output | 1 | 24,148 | 6 | 48,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution
Submitted Solution:
```
string=input()
if len(string)==1:
print('No solution')
exit()
lst=string.split('@')
middle=lst[1:-1]
if not( lst[0].isalpha() and lst[-1].isalpha()):
print('No solution')
exit()
if len(lst)==2:
print(string)
exit()
if not all(list(map(lambda x:len(x)>1 and x.isalpha(),middle))):
print('No solution')
exit()
first=lst[0]
emails=[]
for i in middle:
last=i[0]
emails.append(first+'@'+last)
first=i[1:]
emails.append(first+'@'+lst[-1])
print(','.join(emails))
``` | instruction | 0 | 24,149 | 6 | 48,298 |
No | output | 1 | 24,149 | 6 | 48,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution
Submitted Solution:
```
def main():
string = input()
comps = string.split('@')
ans = []
for i in range(len(comps)):
if i == 0 or i == len(comps) - 1:
if len(comps[i]) < 1:
print("No solution")
return
else:
if len(comps[i]) < 2:
print("No solution")
return
for i in range(len(comps) - 1):
if i == 0:
ans.append(comps[i] + '@' + comps[i + 1][:1])
elif i == len(comps) - 2:
ans.append(comps[i][1:] + '@' + comps[i + 1])
else:
ans.append(comps[i][1:] + '@' + comps[i + 1][:1])
print(', '.join(ans))
main()
``` | instruction | 0 | 24,150 | 6 | 48,300 |
No | output | 1 | 24,150 | 6 | 48,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Email address in Berland is a string of the form A@B, where A and B are arbitrary strings consisting of small Latin letters.
Bob is a system administrator in «Bersoft» company. He keeps a list of email addresses of the company's staff. This list is as a large string, where all addresses are written in arbitrary order, separated by commas. The same address can be written more than once.
Suddenly, because of unknown reasons, all commas in Bob's list disappeared. Now Bob has a string, where all addresses are written one after another without any separators, and there is impossible to determine, where the boundaries between addresses are. Unfortunately, on the same day his chief asked him to bring the initial list of addresses. Now Bob wants to disjoin addresses in some valid way. Help him to do that.
Input
The first line contains the list of addresses without separators. The length of this string is between 1 and 200, inclusive. The string consists only from small Latin letters and characters «@».
Output
If there is no list of the valid (according to the Berland rules) email addresses such that after removing all commas it coincides with the given string, output No solution. In the other case, output the list. The same address can be written in this list more than once. If there are several solutions, output any of them.
Examples
Input
a@aa@a
Output
a@a,a@a
Input
a@a@a
Output
No solution
Input
@aa@a
Output
No solution
Submitted Solution:
```
s=input()
if s[0]!="@" and s[len(s)-1]!="@":
exit=s[::-1]
exit=len(exit)-exit.index("@")-1
mas=[]
res=""
t=True
for i in range(len(s)):
res=res+s[i]
if (s[i]=="@"):
if i!=exit:
t=False
if (s[i+1]=="@"):
break
continue
else:
t=True
if t==False:
t=True
mas.append(res)
res=""
if i+1==len(s):
mas.append(res)
if t==True:
print(",".join(mas))
else:
print("No solution")
else:
print("No solution")
``` | instruction | 0 | 24,151 | 6 | 48,302 |
No | output | 1 | 24,151 | 6 | 48,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
Input
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The i-th line contains the i-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
Output
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
Examples
Input
on codeforces
beta round is running
a rustling of keys
Output
YES
Input
how many gallons
of edo s rain did you drink
cuckoo
Output
NO | instruction | 0 | 24,380 | 6 | 48,760 |
Tags: implementation, strings
Correct Solution:
```
print('YES' if [sum(s.count(x) for x in "aeiou") for s in [input() for _ in range(3)]]==[5,7,5] else 'NO')
``` | output | 1 | 24,380 | 6 | 48,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
Input
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The i-th line contains the i-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
Output
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
Examples
Input
on codeforces
beta round is running
a rustling of keys
Output
YES
Input
how many gallons
of edo s rain did you drink
cuckoo
Output
NO | instruction | 0 | 24,381 | 6 | 48,762 |
Tags: implementation, strings
Correct Solution:
```
import sys
import os
from collections import Counter
def changeStdioToFile():
path = os.path.dirname(os.path.abspath(__file__))
sys.stdin = open(f'{path}/input.txt', 'r')
sys.stdout = open(f'{path}/output.txt', 'w')
#changeStdioToFile()
t = 1
# t = int(input())
for _ in range(t):
flag = False
for i in (5, 7, 5):
c = Counter(input())
if sum(c.get(w, 0) for w in 'aeiou') != i:
flag = True
print("YES" if not flag else "NO")
``` | output | 1 | 24,381 | 6 | 48,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
Input
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The i-th line contains the i-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
Output
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
Examples
Input
on codeforces
beta round is running
a rustling of keys
Output
YES
Input
how many gallons
of edo s rain did you drink
cuckoo
Output
NO | instruction | 0 | 24,382 | 6 | 48,764 |
Tags: implementation, strings
Correct Solution:
```
s1=input()
s2=input()
s3=input()
c1=s1.count('a')+s1.count('e')+s1.count('i')+s1.count('o')+s1.count('u')
c2=s2.count('a')+s2.count('e')+s2.count('i')+s2.count('o')+s2.count('u')
c3=s3.count('a')+s3.count('e')+s3.count('i')+s3.count('o')+s3.count('u')
if(c1==5 and c2==7 and c3==5):
print("YES")
else:
print("NO")
``` | output | 1 | 24,382 | 6 | 48,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
Input
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The i-th line contains the i-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
Output
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
Examples
Input
on codeforces
beta round is running
a rustling of keys
Output
YES
Input
how many gallons
of edo s rain did you drink
cuckoo
Output
NO | instruction | 0 | 24,383 | 6 | 48,766 |
Tags: implementation, strings
Correct Solution:
```
a=input().replace(' ','')
b=input().replace(' ','')
c=input().replace(' ','')
l=list('aeiou')
f,d,e=0,0,0
for i in a:
if i in l:
f+=1
for i in b:
if i in l:
d+=1
for i in c:
if i in l:
e+=1
if(f==5 and d==7 and e==5):
print("YES")
else:
print("NO")
``` | output | 1 | 24,383 | 6 | 48,767 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
Input
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The i-th line contains the i-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
Output
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
Examples
Input
on codeforces
beta round is running
a rustling of keys
Output
YES
Input
how many gallons
of edo s rain did you drink
cuckoo
Output
NO | instruction | 0 | 24,384 | 6 | 48,768 |
Tags: implementation, strings
Correct Solution:
```
# import sys
# sys.stdin=open('input.in','r')
# sys.stdout=open('output.out','w')
l=['a','e','i','o','u']
t1=t2=t3=0
s1=input().strip()
s2=input().strip()
s3=input().strip()
for x in s1:
if x in l:
t1+=1
for x in s2:
if x in l:
t2+=1
for x in s3:
if x in l:
t3+=1
if t1==5 and t2==7 and t3==5:
print('YES')
else:
print('NO')
``` | output | 1 | 24,384 | 6 | 48,769 |
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