message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kaavi, the mysterious fortune teller, deeply believes that one's fate is inevitable and unavoidable. Of course, she makes her living by predicting others' future. While doing divination, Kaavi believes that magic spells can provide great power for her to see the future.
<image>
Kaavi has a string T of length m and all the strings with the prefix T are magic spells. Kaavi also has a string S of length n and an empty string A.
During the divination, Kaavi needs to perform a sequence of operations. There are two different operations:
* Delete the first character of S and add it at the front of A.
* Delete the first character of S and add it at the back of A.
Kaavi can perform no more than n operations. To finish the divination, she wants to know the number of different operation sequences to make A a magic spell (i.e. with the prefix T). As her assistant, can you help her? The answer might be huge, so Kaavi only needs to know the answer modulo 998 244 353.
Two operation sequences are considered different if they are different in length or there exists an i that their i-th operation is different.
A substring is a contiguous sequence of characters within a string. A prefix of a string S is a substring of S that occurs at the beginning of S.
Input
The first line contains a string S of length n (1 ≤ n ≤ 3000).
The second line contains a string T of length m (1 ≤ m ≤ n).
Both strings contain only lowercase Latin letters.
Output
The output contains only one integer — the answer modulo 998 244 353.
Examples
Input
abab
ba
Output
12
Input
defineintlonglong
signedmain
Output
0
Input
rotator
rotator
Output
4
Input
cacdcdbbbb
bdcaccdbbb
Output
24
Note
The first test:
<image>
The red ones are the magic spells. In the first operation, Kaavi can either add the first character "a" at the front or the back of A, although the results are the same, they are considered as different operations. So the answer is 6×2=12.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 5)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def main():
memo={}
def dfs(t):
si=len(t)-1
if si==0:return 2
if t in memo:return memo[t]
res=0
if t[0]==s[si]:res+=dfs(t[1:])
if t[-1]==s[si]:res+=dfs(t[:-1])
memo[t]=res
return res
s=SI()
t=SI()
sn=len(s)
tn=len(t)
cnts=[0]*26
cntt=[0]*26
for c in s[:tn]:cnts[ord(c)-97]+=1
for c in t:cntt[ord(c)-97]+=1
ans=0
if cnts==cntt:
ans+=dfs(t)*(sn-tn+1)
rt=t[::-1]
cs=[0]*tn
for si,c in enumerate(s):
for ti in range(tn-1,-1,-1):
if c==rt[ti]:
if ti>0:cs[ti]+=cs[ti-1]
else:cs[ti]+=1
if s[:tn]==rt:cs[-1]-=1
ans+=cs[-1]*(sn-tn+1)
print(ans)
main()
``` | instruction | 0 | 55,763 | 0 | 111,526 |
No | output | 1 | 55,763 | 0 | 111,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kaavi, the mysterious fortune teller, deeply believes that one's fate is inevitable and unavoidable. Of course, she makes her living by predicting others' future. While doing divination, Kaavi believes that magic spells can provide great power for her to see the future.
<image>
Kaavi has a string T of length m and all the strings with the prefix T are magic spells. Kaavi also has a string S of length n and an empty string A.
During the divination, Kaavi needs to perform a sequence of operations. There are two different operations:
* Delete the first character of S and add it at the front of A.
* Delete the first character of S and add it at the back of A.
Kaavi can perform no more than n operations. To finish the divination, she wants to know the number of different operation sequences to make A a magic spell (i.e. with the prefix T). As her assistant, can you help her? The answer might be huge, so Kaavi only needs to know the answer modulo 998 244 353.
Two operation sequences are considered different if they are different in length or there exists an i that their i-th operation is different.
A substring is a contiguous sequence of characters within a string. A prefix of a string S is a substring of S that occurs at the beginning of S.
Input
The first line contains a string S of length n (1 ≤ n ≤ 3000).
The second line contains a string T of length m (1 ≤ m ≤ n).
Both strings contain only lowercase Latin letters.
Output
The output contains only one integer — the answer modulo 998 244 353.
Examples
Input
abab
ba
Output
12
Input
defineintlonglong
signedmain
Output
0
Input
rotator
rotator
Output
4
Input
cacdcdbbbb
bdcaccdbbb
Output
24
Note
The first test:
<image>
The red ones are the magic spells. In the first operation, Kaavi can either add the first character "a" at the front or the back of A, although the results are the same, they are considered as different operations. So the answer is 6×2=12.
Submitted Solution:
```
import sys
input = sys.stdin.readline
s = input()[:-1]
t = input()[:-1]
sn = len(s)
tn = len(t)
MOD = 998244353
# ng pushback
dp = [[0] * (tn + 1) for i in range(sn + 1)]
dp[0][0] = 1
t = t[::-1]
for i in range(sn):
for j in range(tn + 1):
dp[i + 1][j] += dp[i][j]
for j in range(tn)[::-1]:
if s[i] == t[j]:
dp[i + 1][j + 1] += dp[i][j]
dp[i + 1][j + 1] %= MOD
else:
dp[i + 1][0] += dp[i][j]
dp[i + 1][0] %= MOD
ans = 0
for i in range(sn + 1):
ans += 2 * dp[i][-1]
ans %= MOD
# ok pushback
dp = [[0] * (tn + 1) for i in range(tn + 1)]
t = t[::-1]
for si in range(tn):
for ti in range(tn):
if s[si] == t[ti]:
if si == 0:
dp[si + 1][ti] = 1
else:
dp[si + 1][ti] += dp[si][ti + 1]
dp[si + 1][ti] %= MOD
if ti - si < 0:
continue
dp[si + 1][ti - si] += dp[si][ti - si]
dp[si + 1][ti - si] %= MOD
if t[::-1] == s[0:tn]:
ans -= 2
print((ans + 2 * dp[-1][0]) % MOD)
``` | instruction | 0 | 55,764 | 0 | 111,528 |
No | output | 1 | 55,764 | 0 | 111,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2 | instruction | 0 | 55,897 | 0 | 111,794 |
Tags: brute force, greedy
Correct Solution:
```
s = input()
quit = False
n = len(s)-1;
while((not quit) and n):
lis = []
for i in range(0, len(s)-n+1):
if s[i:i+n] in lis:
print(n)
quit = True
break
lis.append(s[i:i+n])
n-=1
if(not quit):
print(0)
``` | output | 1 | 55,897 | 0 | 111,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2 | instruction | 0 | 55,898 | 0 | 111,796 |
Tags: brute force, greedy
Correct Solution:
```
t = input()
n = len(t)
maxi = 0
for i in range(n):
s = t[i]
if t.count(s) > 1:
maxi = max(maxi, 1)
nr = 1
for j in range(i + 1, n):
s += t[j]
nr += 1
g = 0
for h in range(n - nr + 1):
if s == t[h:h + nr]:
g += 1
if g > 1:
maxi = max(nr,maxi)
print(maxi)
``` | output | 1 | 55,898 | 0 | 111,797 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2 | instruction | 0 | 55,899 | 0 | 111,798 |
Tags: brute force, greedy
Correct Solution:
```
str=input()
m=0
n=len(str)
for i in range(n):
for j in range(i,n+1) :
if str[i:j] in str[i+1:n] and len(str[i:j])>m:
m=len(str[i:j])
print(m)
``` | output | 1 | 55,899 | 0 | 111,799 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2 | instruction | 0 | 55,900 | 0 | 111,800 |
Tags: brute force, greedy
Correct Solution:
```
import sys
import logging
logging.root.setLevel(level=logging.DEBUG)
import re
s = sys.stdin.readline().strip()
from collections import defaultdict
substr = defaultdict(int)
for left in range(len(s)):
for right in range(left+1,len(s)+1):
substr[s[left:right]] += 1
max_len = 0
for segment,times in substr.items():
if times >= 2:
max_len = max(max_len,len(segment))
print(max_len)
``` | output | 1 | 55,900 | 0 | 111,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2 | instruction | 0 | 55,901 | 0 | 111,802 |
Tags: brute force, greedy
Correct Solution:
```
L=input()
m=0
n=len(L)
for i in range(n-1):
for j in range(i,n+1) :
if L[i:j] in L[i+1:n] and len(L[i:j])>m:
m=len(L[i:j])
print(m)
``` | output | 1 | 55,901 | 0 | 111,803 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2 | instruction | 0 | 55,902 | 0 | 111,804 |
Tags: brute force, greedy
Correct Solution:
```
s = input()
leng = 0
for i in range(len(s)):
for j in range(i + 1, len(s) + 1):
sub = s[i:j]
if s.count(sub) >= 2 and len(sub) > leng:
leng = len(sub)
elif s.count(sub) == 1:
for k in range(1, len(sub)):
if s[i - k:j - k] == sub and len(sub) > leng:
leng = len(sub)
print(leng)
``` | output | 1 | 55,902 | 0 | 111,805 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2 | instruction | 0 | 55,903 | 0 | 111,806 |
Tags: brute force, greedy
Correct Solution:
```
s=input()
n=len(s)
done=False
for length in range(n-1,0,-1):
if(done):
break
for start in range(n):
if(start+length-1>=n):
break
x=s.count(s[start:start+length])
if(x>1):
print(length)
done=True
break
if(x==1):
i=s.index(s[start:start+length])
h=s[i+1:].count(s[start:start+length])
if(h>0):
print(length)
done=True
break
if(not done):
print(0)
``` | output | 1 | 55,903 | 0 | 111,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2 | instruction | 0 | 55,904 | 0 | 111,808 |
Tags: brute force, greedy
Correct Solution:
```
inputS=input()
ans=0
for i in range (0,len(inputS)-1):
for count in range(1,len(inputS)):
for j in range(i+1, len(inputS)-count+1):
A=inputS[i: i+count]
B=inputS[j: j+count]
if A==B:
ans=count if count>ans else ans
print(ans)
``` | output | 1 | 55,904 | 0 | 111,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2
Submitted Solution:
```
s=input()
length=len(s)
answer=[ ]
for i in range (0,length):
for j in range(i+1,length+1):
k=s[i:j]
co=0
for u in range (0,length):
if(s[u:].startswith(k)):
co+=1
if(co>=2):
#answer=max(answer,len(k))
answer.append(len(k))
if(len(set(s))==length):
print('0')
else:
print(max(answer))
``` | instruction | 0 | 55,905 | 0 | 111,810 |
Yes | output | 1 | 55,905 | 0 | 111,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2
Submitted Solution:
```
S = input()
best = 0
for i in range(len(S)):
for j in range(i+1, len(S)+1):
s = S[i:j]
c = 0
for k in range(len(S)):
if S[k:].startswith(s): c += 1
if c >= 2:
best = max(best, len(s))
print(best)
``` | instruction | 0 | 55,906 | 0 | 111,812 |
Yes | output | 1 | 55,906 | 0 | 111,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2
Submitted Solution:
```
#!/usr/bin/python3
# kkikkkkiii
s = input()
maxx = 0
a = [0] * len(s)
for j in range(len(s)):
k = j
i = j
while i < len(s) - 1:
i += 1
if s[i] == s[k]:
k += 1
a[i] = k
maxx = max(maxx, a[i] - j)
continue
i -= k - j
k = j
a = [0] * len(s)
print(maxx)
``` | instruction | 0 | 55,907 | 0 | 111,814 |
Yes | output | 1 | 55,907 | 0 | 111,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2
Submitted Solution:
```
st=input()
m=0
n=len(st)
for i in range(n):
for j in range(i,n+1) :
if st[i:j] in st[i+1:n] and len(st[i:j])>m:
m=len(st[i:j])
print(m)
``` | instruction | 0 | 55,908 | 0 | 111,816 |
Yes | output | 1 | 55,908 | 0 | 111,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2
Submitted Solution:
```
'''
s = input()
for i in range(len(s)):
for k in range(i + 1, len(s)+1, 1):
print(s[i:k])
'''
'''
s = input()
sLen = len(s)
for startInd in range(sLen):
for endInd in range(startInd + 1, sLen + 1):
print(s[startInd, endInd])
'''
s = input()
slen = len(s)
ans = 0
for st1 in range(slen - 1):
for end1 in range(st1 + 1, slen):
end2 = end1 + 1
for st2 in range(st1 + 1, slen):
if end2 > slen:
if end2 > slen:
break
sub1 = s[st1:end1]
sub2 = s[st2:end2]
subLen = len(sub1)
if sub1 == sub2 and ans < subLen:
ans = subLen
print(ans)
``` | instruction | 0 | 55,909 | 0 | 111,818 |
No | output | 1 | 55,909 | 0 | 111,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2
Submitted Solution:
```
string=input()
n=len(string)
check=True
for sub_len in range(n-1,1,-1):
for starting_index in range(n-sub_len+1):
if string[starting_index:starting_index+sub_len] in string[starting_index+1:]:
print(sub_len)
check=False
break
if check==False:
break
if check:
print(0)
``` | instruction | 0 | 55,910 | 0 | 111,820 |
No | output | 1 | 55,910 | 0 | 111,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2
Submitted Solution:
```
import sys
def countall(string, substring):
total = 0
for i in range(len(string)-len(substring)+1):
if string[i:i+len(substring)] == substring:
total += 1
return total
n = input()
for i in range(len(n)):
for j in range(len(n)-1, i-1, -1):
if countall(n, n[i:j+1]) > 1:
print(j-i+1)
sys.exit()
print(0)
``` | instruction | 0 | 55,911 | 0 | 111,822 |
No | output | 1 | 55,911 | 0 | 111,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output
Output one number — length of the longest substring that can be met in the string at least twice.
Examples
Input
abcd
Output
0
Input
ababa
Output
3
Input
zzz
Output
2
Submitted Solution:
```
#length of the longest substring at least twice
import re
import sys
ex = sys.stdin.readline()
arr = []
for y in range(len(ex)):
for x in range(y,len(ex)):
g = ex[:len(ex)-x-1]
n = len(re.findall('(?='+g+')',ex))
if (n >= 2):
arr.append(len(g))
else:
arr.append(0)
ex = ex[1:]
print(max(arr))
``` | instruction | 0 | 55,912 | 0 | 111,824 |
No | output | 1 | 55,912 | 0 | 111,825 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement. | instruction | 0 | 56,037 | 0 | 112,074 |
Tags: implementation, strings
Correct Solution:
```
import sys
s = list(sys.stdin.readline())
m = int(sys.stdin.readline())
for _ in range(m):
l, r, k = map(int, sys.stdin.readline().split())
l -= 1
k = k % (r - l)
if k:
part = s[l:r]
s[l:r] = part[-k:] + part[:-k]
print("".join(s))
``` | output | 1 | 56,037 | 0 | 112,075 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement. | instruction | 0 | 56,038 | 0 | 112,076 |
Tags: implementation, strings
Correct Solution:
```
s = input()
for i in range( int( input() ) ):
l, r, k = map(int, input().split() )
l -= 1
r -= 1
k %= r - l + 1
s = s[:l] + s[r - k + 1:r + 1] + s[l:r - k + 1] + s[r + 1:]
print(s)
``` | output | 1 | 56,038 | 0 | 112,077 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement. | instruction | 0 | 56,039 | 0 | 112,078 |
Tags: implementation, strings
Correct Solution:
```
s = input()
m = int(input())
for i in range(m):
l,r,k = map(int,input().split())
k %= (r - l + 1)
t = s[l-1:r]
s = s[:l-1] + t[len(t)-k:] + t[:len(t)-k] + s[r:]
print(s)
``` | output | 1 | 56,039 | 0 | 112,079 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement. | instruction | 0 | 56,040 | 0 | 112,080 |
Tags: implementation, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
s = list(input().strip())
m = int(input())
for _ in range(m):
l, r, k = map(int, input().split())
l -= 1
r -= 1
k %= r - l + 1
# s[l : r + 1]
s = s[:l] + s[r + 1 - k: r + 1] + s[l:r+1-k] + s[r+1:]
print(''.join(s))
``` | output | 1 | 56,040 | 0 | 112,081 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement. | instruction | 0 | 56,041 | 0 | 112,082 |
Tags: implementation, strings
Correct Solution:
```
s = list(input())
m = int(input())
for i in range(m):
l, r, k = map(int, input().split())
l += -1
r += -1
ln = (r - l + 1)
k %= ln
tmp = []
for i in s: tmp.append(i)
for j in range(l, r + 1):
# print(j, k, r)
if (j + k) <= r: tmp[j + k] = s[j]
else: tmp[l - 1 + j + k - r] = s[j]
# print(tmp)
for i in range(len(s)): s[i] = tmp[i]
# print(tmp)
print(*s, sep = '')
``` | output | 1 | 56,041 | 0 | 112,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement. | instruction | 0 | 56,042 | 0 | 112,084 |
Tags: implementation, strings
Correct Solution:
```
s = input()
for n in range(int(input())):
l, r, k = map(int, input().split())
l -= 1
r -= 1
if l != r:
k = k % (r - l + 1)
b = s[l:r + 1]
b = b[-k:] + b[:-k]
s = s[:l] + b + s[r + 1:]
print(s)
``` | output | 1 | 56,042 | 0 | 112,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement. | instruction | 0 | 56,043 | 0 | 112,086 |
Tags: implementation, strings
Correct Solution:
```
s=list(input())
d=len(s)
m=int(input())
for i in range(m):
l,r,k=map(int,input().split())
x=k%(r-l+1)
if x:
pre=s[l-1:r-x]
post=s[r-x:r]
s[l-1:r]=post+pre
print(*s,sep="")
``` | output | 1 | 56,043 | 0 | 112,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement. | instruction | 0 | 56,044 | 0 | 112,088 |
Tags: implementation, strings
Correct Solution:
```
s=list(input())
t=int(input());
while(t):
t=t-1;
l,r,k=map(int,input().split(" "));
a=s[0:l-1];
b=s[l-1:r];
c=s[r:];
x=k%len(b);
b=b[len(b)-x:]+b[:len(b)-x]
s=a+b+c
print("".join(s))
``` | output | 1 | 56,044 | 0 | 112,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement.
Submitted Solution:
```
s = input().strip()
n = int(input())
for i in range(n):
f, t, k = map(int, input().split())
l = t - f + 1
f -= 1
k %= l
if k != 0:
#print(s[:f])
#print(s[t - k:t])
#print(s[f:f + l - k])
#print(s[t:])
s = s[:f] + s[t - k:t] + s[f:f + l - k] + s[t:]
#print(s)
print(s)
``` | instruction | 0 | 56,045 | 0 | 112,090 |
Yes | output | 1 | 56,045 | 0 | 112,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement.
Submitted Solution:
```
from collections import deque
s = input()
string = []
for i in range(len(s)):
string.append(s[i])
m = int(input())
for i in range(m):
l, r, k = map(int, input().split())
l -= 1
r -= 1
k %= (r - l + 1)
d = deque()
for j in range(l, r + 1):
d.append(string[j])
d.rotate(k)
for j in range(l, r + 1):
string[j] = d[j - l]
print(''.join(string))
``` | instruction | 0 | 56,046 | 0 | 112,092 |
Yes | output | 1 | 56,046 | 0 | 112,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement.
Submitted Solution:
```
if __name__=='__main__':
string = list(input())
tests = int(input())
for i in range(tests):
l,r,times = map(int,input().split())
l-=1
temp = string[l:r]
#check = [i for i in range(len(temp))]
#print("temp is ",temp)
k = (times-1)%(r-l)
for i in range(len(temp)):
k +=1
k = k%(r-l)
# print("mark-> ",l+k)
string[l+k]=temp[i]
print(''.join(string))
``` | instruction | 0 | 56,047 | 0 | 112,094 |
Yes | output | 1 | 56,047 | 0 | 112,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement.
Submitted Solution:
```
s = input()
m = int(input())
for i in range(m):
l,r,k = [int(c) for c in input().split()]
k = k%(r-l+1)
s = s[:l-1] + s[r-k:r] + s[l-1:r-k] + s[r:]
print(s)
``` | instruction | 0 | 56,048 | 0 | 112,096 |
Yes | output | 1 | 56,048 | 0 | 112,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement.
Submitted Solution:
```
s=list(input())
d=len(s)
m=int(input())
for i in range(m):
l,r,k=map(int,input().split())
x=k%d
if x:
pre=s[l-1:r-x]
post=s[r-x:r]
s[l-1:r]=post+pre
print(*s,sep="")
``` | instruction | 0 | 56,049 | 0 | 112,098 |
No | output | 1 | 56,049 | 0 | 112,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement.
Submitted Solution:
```
import sys
#read = lambda: sys.stdin.buffer.read().rstrip()
input = lambda: sys.stdin.readline().rstrip()
#readlines = lambda: sys.stdin.buffer.readlines().rstrip()
sys.setrecursionlimit(10**5)
from collections import deque
def rolling(s, n):
t = deque(s)
t.rotate(n)
return t
def solve():
S = list(input())
N = int(input())
print("".join(S))
for _ in range(N):
l,r,k = map(int, input().split())
t = rolling(S[l-1:r],k)
S[l-1:r] = t
print("".join(S))
solve()
``` | instruction | 0 | 56,050 | 0 | 112,100 |
No | output | 1 | 56,050 | 0 | 112,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement.
Submitted Solution:
```
s=list('q'+input())
m=int(input())
for i in range(m):
l,r,k=map(int,input().split())
s[l:r+1]=s[r-k+1:r+1]+s[l:r-k+1]
print(''.join(s[1:]))
``` | instruction | 0 | 56,051 | 0 | 112,102 |
No | output | 1 | 56,051 | 0 | 112,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s and should process m queries. Each query is described by two 1-based indices li, ri and integer ki. It means that you should cyclically shift the substring s[li... ri] ki times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string s is abacaba and the query is l1 = 3, r1 = 6, k1 = 1 then the answer is abbacaa. If after that we would process the query l2 = 1, r2 = 4, k2 = 2 then we would get the string baabcaa.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where |s| stands for the length of s. It contains only lowercase English letters.
Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.
The i-th of the next m lines contains three integers li, ri and ki (1 ≤ li ≤ ri ≤ |s|, 1 ≤ ki ≤ 1 000 000) — the description of the i-th query.
Output
Print the resulting string s after processing all m queries.
Examples
Input
abacaba
2
3 6 1
1 4 2
Output
baabcaa
Note
The sample is described in problem statement.
Submitted Solution:
```
#5.5
def ans(s,l,r,k):
if(r==len(s)):
subR=""
# print("bbbbbbb")
else:
subR=s[r:len(s)]
# print("ccccccccc")
sub=s[l-1:r]
subL=s[0:l-1]
L=len(sub)
sub=sub+sub
sub.replace(" ","")
reminder=k%L
# print("reminder",reminder)
# print("subL",subL)
# print("sub",sub)
# print("subR",subR)
if(reminder!=0):
sub=sub[L-reminder:L-reminder+L]
s=(subL+sub+subR).replace(" ","")
# print(s)
return s
else:
# print(s)
return s
s=str(input(""))
n=int(input(""))
for i in range(n):
l,r,k=map(int,input().split(" "))
# l=int(input(""))
# r=int(input(""))
# k=int(input(""))
s=ans(s,l,r,k)
print(s)
``` | instruction | 0 | 56,052 | 0 | 112,104 |
No | output | 1 | 56,052 | 0 | 112,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 56,109 | 0 | 112,218 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
def locos(n):
cos = 0
i = 1
while cos + i <= n:
cos += i
i += 1
return i , cos
rtn = ''
for i in "abcdefghijklmnopqrstuvwxyz":
#print(locos(n))
num , cos = locos(n)
#print(num)
rtn += i * num
n -= cos
if(n == 0): break
print(rtn)
``` | output | 1 | 56,109 | 0 | 112,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 56,110 | 0 | 112,220 |
Tags: constructive algorithms
Correct Solution:
```
from math import sqrt
def readln(): return map(int, input().rstrip().split())
def possible(n): return n * (n + 1) // 2
charset = 'abcdefghijklmnopqrstuvwxyz'
ci = 0
n = int(input())
rs = []
while n > 0:
for i in range(int(sqrt(n) + 1), 0, -1):
if n >= possible(i):
n -= possible(i)
rs.extend([charset[ci]] * (i + 1))
ci += 1
else:
if not rs:
rs.append('a')
print(''.join(rs))
``` | output | 1 | 56,110 | 0 | 112,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 56,111 | 0 | 112,222 |
Tags: constructive algorithms
Correct Solution:
```
# http://codeforces.com/contest/849/problem/C
import math
import string
alpha = string.ascii_lowercase
k = int(input())
def Sn(n):
return (n-1)*(n) // 2
letters = []
if k == 0:
letters = [1]
else:
while k > 0:
# find n st n letters have max cost < k
det = 8*k + 1
n = int((math.sqrt(det) + 1) // 2)
letters.append(n)
k -= Sn(n)
# print(letters)
# print([Sn(n) for n in letters])
res = []
for ll, n in enumerate(letters):
ss = "".join([alpha[ll] for nn in range(n)])
res.append(ss)
print("".join(res))
``` | output | 1 | 56,111 | 0 | 112,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 56,112 | 0 | 112,224 |
Tags: constructive algorithms
Correct Solution:
```
k= int(input())
c='a'
while ( k != 0):
j = 1
while ((j+1)*(j+2) < 2*k):
j = j+1
for i in range (0,j+1):
print(c,end="")
c=chr(ord(c)+1)
k=k-((j+1)*(j))/2
print("z")
# Made By Mostafa_Khaled
``` | output | 1 | 56,112 | 0 | 112,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 56,113 | 0 | 112,226 |
Tags: constructive algorithms
Correct Solution:
```
k=int(input())
alphabet='abcdefghijklmnopqrstuvwxyz'
ans=[]
pointer=0
if k==0:
ans.append('a')
while k>0:
curr=2
while curr*(curr+1)<=2*k:
curr+=1
for j in range(curr):
ans.append(alphabet[pointer])
k-=(curr*(curr-1))//2
pointer+=1
print(''.join(ans))
``` | output | 1 | 56,113 | 0 | 112,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 56,114 | 0 | 112,228 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
lis=[0]*(n+10)
for i in range(1,n+5):
lis[i]=i+lis[i-1]
i=n+1
#print(lis)
if n==0:
print('a')
exit()
cc='abcdefghijklmnopqrstuvwxyz';c=0;ans=''
while n>0:
while n<lis[i]:
i-=1
z=min(lis[i],n)
n-=z
ans+=(cc[c]*(i+1))
c+=1
print(ans)
``` | output | 1 | 56,114 | 0 | 112,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 56,115 | 0 | 112,230 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
a = 'abcdefghijklmnopqrstuvwxyz'
s = ''
i = 0
k = 0
while n:
if n - k >= 0:
s += a[i]
n -= k
k += 1
else:
i += 1
k = 0
if not s:
s = 'a'
print(s)
``` | output | 1 | 56,115 | 0 | 112,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 56,116 | 0 | 112,232 |
Tags: constructive algorithms
Correct Solution:
```
def f(n):
return n * (n + 1) // 2
def rev_f(x):
l = 0
r = 450
y = 0
while l + 1 < r:
m = (l + r) // 2
y = f(m)
if y > x:
r = m
else:
l = m
return l
n = int(input())
for i in range(0, 26):
c = chr(ord('a') + i)
q = rev_f(n)
n -= f(q)
print(c * (q + 1), end = '')
if n == 0:
break
``` | output | 1 | 56,116 | 0 | 112,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
n = int(input())
cost = [None] * 100001
cost[0] = 0
# def get_cost(k):
# global cost
# if cost[k] is None:
# cost[k] = compute_cost(k)
# return cost[k]
# def compute_cost(k):
# global cost
# less = k // 2
# more = k - less
# return get_cost(less) + get_cost(more) + less * more
# for i in range(50000, len(cost)):
# cost[i] = compute_cost(i)
for i in range(1, len(cost)):
cost[i] = cost[i - 1] + i - 1
def compute_ans(k):
global cost
i = 0
while cost[i] < k:
i += 1
if cost[i] == k:
return [i]
return [i - 1] + compute_ans(k - cost[i - 1])
s = compute_ans(n)
res = ''
for i, cnt in zip('abcdefghijklmnopqrstuvwxyz', s):
res += i * cnt
if not res:
res = 'a'
print(res)
``` | instruction | 0 | 56,117 | 0 | 112,234 |
Yes | output | 1 | 56,117 | 0 | 112,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
n=int(input())
a=[]
b='z'
c="abcdefghijklmnopqrstuvwxyz"
l=0
for i in range(500):
a.append(i*(i+1)/2)
for i in range(499,0,-1):
while(n>=a[i]):
b+=c[l]*(i+1)
l+=1
n-=a[i]
print(b)
``` | instruction | 0 | 56,118 | 0 | 112,236 |
Yes | output | 1 | 56,118 | 0 | 112,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
k= int(input())
c='a'
while ( k != 0):
j = 1
while ((j+1)*(j+2) < 2*k):
j = j+1
for i in range (0,j+1):
print(c,end="")
c=chr(ord(c)+1)
k=k-((j+1)*(j))/2
print("z")
``` | instruction | 0 | 56,119 | 0 | 112,238 |
Yes | output | 1 | 56,119 | 0 | 112,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heapify, heappop, heappush
import math
from copy import deepcopy
from itertools import combinations, permutations, product
from bisect import bisect_left, bisect_right
import sys
def input():
return sys.stdin.readline().rstrip()
def getN():
return int(input())
def getNM():
return map(int, input().split())
def getList():
return list(map(int, input().split()))
def getArray(intn):
return [int(input()) for i in range(intn)]
mod = 10 ** 9 + 7
MOD = 998244353
INF = float('inf')
eps = 10 ** (-10)
dx = [1, 0, -1, 0]
dy = [0, 1, 0, -1]
#############
# Main Code #
#############
"""
convineするときの最小値
前からやれば... 使える文字は26個
sにあるaの数 * tにあるaの数 を足し合わせる
1+2+3...の組み合わせで表現したいが
"""
K = getN()
if K == 0:
print('a')
exit()
ans = ''
str = 0
while K > 0:
now = 1
acc = 0
while acc + now <= K:
acc += now
now += 1
K -= acc
ans += chr(str + ord('a')) * now
str += 1
print(ans)
``` | instruction | 0 | 56,120 | 0 | 112,240 |
Yes | output | 1 | 56,120 | 0 | 112,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
k=int(input())
alphabet='abcdefghijklmnopqrstuvwxyz'
ans=[]
pointer=0
while k>0:
curr=2
while curr*(curr+1)<=2*k:
curr+=1
for j in range(curr):
ans.append(alphabet[pointer])
k-=(curr*(curr-1))//2
pointer+=1
print(''.join(ans))
``` | instruction | 0 | 56,121 | 0 | 112,242 |
No | output | 1 | 56,121 | 0 | 112,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
k = int(input())
if k == 1:
print('fall')
exit(0)
if k == 3:
print('september')
exit(0)
# от порядка объединения строк ничего не зависит
# для каждой буквы, итоговая стоимость будет n * (n - 1) / 2
# жадно наберём ответ
op_count = [0, 0]
for i in range(500):
op_count.append(op_count[-1] + i)
if op_count[-1] > k:
break
ans = []
letter = 'a'
for i in range(len(op_count) - 1, 1, -1):
if op_count[i] <= k:
k -= op_count[i]
ans.append(letter * i)
letter = chr(ord(letter) + 1)
print(''.join(ans))
``` | instruction | 0 | 56,122 | 0 | 112,244 |
No | output | 1 | 56,122 | 0 | 112,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
from sys import stdin, stdout
n = int(stdin.readline())
current = 0
ans = ''
while n:
cnt = 0
while cnt * (cnt + 1) // 2 <= n:
cnt += 1
for i in range(cnt):
ans += chr(ord('a') + current)
current += 1
cnt -= 1
n -= cnt * (cnt + 1) // 2
``` | instruction | 0 | 56,123 | 0 | 112,246 |
No | output | 1 | 56,123 | 0 | 112,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
char=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
n=int(input())
s='a'
c=1
k=0
i=0
while k!=n:
s+=char[i]
k+=c
c+=1
if k>n:
k-=c
i+=1
c=0
#c+=1
print(s)
``` | instruction | 0 | 56,124 | 0 | 112,248 |
No | output | 1 | 56,124 | 0 | 112,249 |
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