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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...
Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.
A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next 2T lines contain descriptions of test cases β two lines per test case. The first line contains two integers n and x (1 β€ n β€ 10^5, -10^9 β€ x β€ 10^9) β the length of string s and the desired balance, respectively.
The second line contains the binary string s (|s| = n, s_i β \{0, 1\}).
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
Print T integers β one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.
Example
Input
4
6 10
010010
5 3
10101
1 0
0
2 0
01
Output
3
0
1
-1
Note
In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32. | instruction | 0 | 73,449 | 0 | 146,898 |
Tags: math, strings
Correct Solution:
```
t = int(input())
for _ in range(t):
n, x = list(map(int, input().split()))
s = list(map(int, input()))
bal = [0] * (n + 1)
cnt = 1 if x == 0 else 0
for i in range(n):
if s[i] == 0:
bal[i+1] = bal[i] + 1
else:
bal[i+1] = bal[i] -1
if bal[i+1] == x:
cnt += 1
#print(bal)
b = bal[-1]
if b == 0 and cnt > 0:
print(-1)
elif b == 0 and cnt == 0:
print(0)
elif b != 0:
for i in range(n):
rest = x - bal[i+1]
k = rest // b
#print(x, bal[i+1], rest, rest % b, k, bal[i+1] + k * b)
if rest % b == 0 and k > 0:
cnt += 1
print(cnt)
else:
print(cnt)
``` | output | 1 | 73,449 | 0 | 146,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...
Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.
A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next 2T lines contain descriptions of test cases β two lines per test case. The first line contains two integers n and x (1 β€ n β€ 10^5, -10^9 β€ x β€ 10^9) β the length of string s and the desired balance, respectively.
The second line contains the binary string s (|s| = n, s_i β \{0, 1\}).
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
Print T integers β one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.
Example
Input
4
6 10
010010
5 3
10101
1 0
0
2 0
01
Output
3
0
1
-1
Note
In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.
Submitted Solution:
```
def solve():
n, x = map(int,input().split())
s = input()
if s.count('0') == s.count('1'):
cnt = 0
if x==0:
print(-1)
return
for i in range(n):
if s[i]=='0':
cnt+=1
else:
cnt-=1
if cnt==x:
print('-1')
return
print(0)
return
else:
d = s.count('0') - s.count('1')
cnt = 0
ans = 0
for i in range(n):
if s[i] == '0':
cnt += 1
else:
cnt -= 1
if (x-cnt)*d>0:
if abs(cnt-x)%d==0:
ans+=1
if (x-cnt)==0:
ans+=1
if x==0:
ans+=1
print(ans)
return
test = int(input())
for _ in range(test):
solve()
``` | instruction | 0 | 73,450 | 0 | 146,900 |
Yes | output | 1 | 73,450 | 0 | 146,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...
Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.
A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next 2T lines contain descriptions of test cases β two lines per test case. The first line contains two integers n and x (1 β€ n β€ 10^5, -10^9 β€ x β€ 10^9) β the length of string s and the desired balance, respectively.
The second line contains the binary string s (|s| = n, s_i β \{0, 1\}).
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
Print T integers β one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.
Example
Input
4
6 10
010010
5 3
10101
1 0
0
2 0
01
Output
3
0
1
-1
Note
In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.
Submitted Solution:
```
# Contest: Educational Codeforces Round 81 (Rated for Div. 2) (https://codeforces.com/contest/1295)
# Problem: B: Infinite Prefixes (https://codeforces.com/contest/1295/problem/B)
def rint():
return int(input())
def rints():
return list(map(int, input().split()))
t = rint()
for _ in range(t):
n, x = rints()
s = input()
bal, min_bal, max_bal = 0, 0, 0
from collections import defaultdict
counts = defaultdict(int)
for c in s:
bal += 1 if c == '0' else -1
min_bal = min(min_bal, bal)
max_bal = max(max_bal, bal)
counts[bal] += 1
bal_range = max_bal - min_bal
if bal == 0:
if min_bal <= x <= max_bal:
print(-1)
else:
print(0)
else:
b = 0
d = 1 if x == 0 else 0
for c in s:
b += 1 if c == '0' else -1
d += (x - b) % bal == 0 and (x - b) // bal >= 0
print(d)
``` | instruction | 0 | 73,451 | 0 | 146,902 |
Yes | output | 1 | 73,451 | 0 | 146,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...
Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.
A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next 2T lines contain descriptions of test cases β two lines per test case. The first line contains two integers n and x (1 β€ n β€ 10^5, -10^9 β€ x β€ 10^9) β the length of string s and the desired balance, respectively.
The second line contains the binary string s (|s| = n, s_i β \{0, 1\}).
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
Print T integers β one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.
Example
Input
4
6 10
010010
5 3
10101
1 0
0
2 0
01
Output
3
0
1
-1
Note
In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.
Submitted Solution:
```
T = int(input())
for i in range(T):
n, x = map(int, input().split())
s = str(input())
l = [0]*n
l[0] = 1 - 2*int(s[0])
for j in range(1, n):
if s[j] == "0":
l[j] = l[j-1] + 1
else:
l[j] = l[j-1] - 1
tmp = l[-1]
ans = 0
if x == 0:
ans = 1
if tmp != 0:
for j in range(n):
if (x - l[j]) % tmp == 0 and (x - l[j]) * tmp >= 0:
ans += 1
print(ans)
else:
flag = True
for j in range(n):
if x - l[j] == 0:
flag = False
print(-1)
break
if flag:
print(ans)
``` | instruction | 0 | 73,452 | 0 | 146,904 |
Yes | output | 1 | 73,452 | 0 | 146,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...
Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.
A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next 2T lines contain descriptions of test cases β two lines per test case. The first line contains two integers n and x (1 β€ n β€ 10^5, -10^9 β€ x β€ 10^9) β the length of string s and the desired balance, respectively.
The second line contains the binary string s (|s| = n, s_i β \{0, 1\}).
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
Print T integers β one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.
Example
Input
4
6 10
010010
5 3
10101
1 0
0
2 0
01
Output
3
0
1
-1
Note
In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.
Submitted Solution:
```
T = int(input())
for _ in range(T):
n, x = map(int, input().split())
s = input()
b = [1] if s[0]=='0' else [-1]
for i in range(1, n):
b.append(b[-1]+(1 if s[i]=='0' else -1))
zero = s.count('0')
one = n-zero
if x==0:
ans = 1
else:
ans = 0
if zero==one:
if x in b:
print(-1)
else:
print(0)
elif zero>one:
for bi in b:
if x>=bi and (x-bi)%(zero-one)==0:
ans += 1
print(ans)
else:
for bi in b:
if x<=bi and (bi-x)%(one-zero)==0:
ans += 1
print(ans)
``` | instruction | 0 | 73,453 | 0 | 146,906 |
Yes | output | 1 | 73,453 | 0 | 146,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...
Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.
A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next 2T lines contain descriptions of test cases β two lines per test case. The first line contains two integers n and x (1 β€ n β€ 10^5, -10^9 β€ x β€ 10^9) β the length of string s and the desired balance, respectively.
The second line contains the binary string s (|s| = n, s_i β \{0, 1\}).
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
Print T integers β one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.
Example
Input
4
6 10
010010
5 3
10101
1 0
0
2 0
01
Output
3
0
1
-1
Note
In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.
Submitted Solution:
```
for _ in range(int(input())):
n,x=map(int,input().split())
a=input()
l=[];c=0
for i in a:
if(i=='0'):
c+=1
if(i=='1'):
c-=1
l.append(c)
z=l.count(0)
#print(z,l)
l=[];c=0
for i in range(n-1,-1,-1):
if(a[i]=='0'):
c+=1
if(a[i]=='1'):
c-=1
l.append(c)
z+=l.count(0)
#print(z,l)
diff=max(l)
#print(l,diff)
ans=1+z if(diff!=0 and x%diff==0) else 0
if(x in l):
ans=-1
print(ans)
``` | instruction | 0 | 73,454 | 0 | 146,908 |
No | output | 1 | 73,454 | 0 | 146,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...
Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.
A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next 2T lines contain descriptions of test cases β two lines per test case. The first line contains two integers n and x (1 β€ n β€ 10^5, -10^9 β€ x β€ 10^9) β the length of string s and the desired balance, respectively.
The second line contains the binary string s (|s| = n, s_i β \{0, 1\}).
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
Print T integers β one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.
Example
Input
4
6 10
010010
5 3
10101
1 0
0
2 0
01
Output
3
0
1
-1
Note
In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.
Submitted Solution:
```
from sys import stdin
def input():
return stdin.readline()[:-1]
def intput():
return int(input())
def sinput():
return input().split()
def intsput():
return map(int, sinput())
debugging = False
def dprint(*args):
if debugging:
print(*args)
else:
pass
shift = 100001
# Code
t = intput()
for _ in range(t):
n, x = intsput()
s = input()
balances = [0] * 200002
balances[0] += 1
b = 0
for i in range(n):
if s[i] == '1':
b -= 1
else:
b += 1
balances[b + shift] += 1
finalb = b
if finalb == 0:
print(-1)
continue
k = x
ans = 0
if len(balances) > k + shift >= 0:
while len(balances) > k + shift >= 0:
ans += balances[k + shift]
k -= finalb
print(ans)
else:
if k + shift >= len(balances) and finalb > 0:
k -= ((k + shift) - len(balances)) // finalb * finalb
while k >= len(balances):
k -= finalb
while len(balances) > k + shift >= 0:
ans += balances[k + shift]
k -= finalb
print(ans)
elif k + shift < 0 and finalb < 0:
k += (0 - (k + shift)) // finalb * finalb
while k < 0:
k += - finalb
while len(balances) > k + shift >= 0:
ans += balances[k + shift]
k -= finalb
print(ans)
else:
print(0)
``` | instruction | 0 | 73,455 | 0 | 146,910 |
No | output | 1 | 73,455 | 0 | 146,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...
Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.
A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next 2T lines contain descriptions of test cases β two lines per test case. The first line contains two integers n and x (1 β€ n β€ 10^5, -10^9 β€ x β€ 10^9) β the length of string s and the desired balance, respectively.
The second line contains the binary string s (|s| = n, s_i β \{0, 1\}).
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
Print T integers β one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.
Example
Input
4
6 10
010010
5 3
10101
1 0
0
2 0
01
Output
3
0
1
-1
Note
In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.
Submitted Solution:
```
nt=int(input())
for qz in range(nt):
m,n=map(int,input().split())
s=list(input())
dif=s.count('0')-s.count('1')
cnt=0
d=0
if(n==dif):
print(-1)
elif(dif<0 and n>0):
print(0)
elif(n==0):
cnt+=1
for i in s:
i=int(i)
if(i==0):
d+=1
else:
d-=1
if(d<0):
cnt+=1
print(cnt)
elif(n<0 and dif<0):
dif=abs(dif)
for i in range(len(s)):
if(s[i]=='0'):
s[i]=='1'
else:
s[i]=='0'
for i in s:
i=int(i)
if(i==0):
d+=1
else:
d-=1
try:
if((n+d)%dif==0 and d>=0):
cnt+=1
# print(i,n+d,cnt)
except:pass
print(cnt)
else:
for i in s:
i=int(i)
if(i==0):
d+=1
else:
d-=1
try:
if((n+d)%dif==0 and d>=0):
cnt+=1
# print(i,n+d,cnt)
except:pass
print(cnt)
``` | instruction | 0 | 73,456 | 0 | 146,912 |
No | output | 1 | 73,456 | 0 | 146,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...
Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.
A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next 2T lines contain descriptions of test cases β two lines per test case. The first line contains two integers n and x (1 β€ n β€ 10^5, -10^9 β€ x β€ 10^9) β the length of string s and the desired balance, respectively.
The second line contains the binary string s (|s| = n, s_i β \{0, 1\}).
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
Print T integers β one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.
Example
Input
4
6 10
010010
5 3
10101
1 0
0
2 0
01
Output
3
0
1
-1
Note
In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.
Submitted Solution:
```
from math import *
zzz = int(input())
for zz in range(zzz):
n, x = map(int, input().split())
s = [int(i) for i in input()]
a = []
bal = 0
for i in range(n):
if s[i] > 0:
bal -= 1
else:
bal += 1
a.append(bal)
if x > 0 and a[-1] < 0 and max(a) < x:
print(0)
elif x < 0 and a[-1] > 0 and min(a) > x:
print(0)
elif a[-1] != 0:
b = []
for i in range(n):
if (x - a[i]) % a[-1] == 0:
b.append(i)
print(len(b))
else:
if x in a:
print(-1)
else:
print(0)
``` | instruction | 0 | 73,457 | 0 | 146,914 |
No | output | 1 | 73,457 | 0 | 146,915 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1. | instruction | 0 | 73,458 | 0 | 146,916 |
Tags: brute force, constructive algorithms, implementation, sortings, strings
Correct Solution:
```
t=eval(input())
for i in range(t):
n=eval(input())
st=input()
min='z'
for i in range(n):
if(st[i]<min):
min=st[i]
li=[]
for i in range(n):
if(st[i]==min):
li=li+[(i+1)]
ans=st
j=1
for k in li:
temp=st[k-1:]
#print(len(temp))
if((len(temp))%2==0):
temp=temp+st[0:k-1]
else:
stm=st[::-1]
temp=temp+stm[-k+1:]
if(temp<ans):
ans=temp
j=k
print(ans)
print(j)
``` | output | 1 | 73,458 | 0 | 146,917 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1. | instruction | 0 | 73,459 | 0 | 146,918 |
Tags: brute force, constructive algorithms, implementation, sortings, strings
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
s=input()
s=' '+s
L=list(s)
m=min(L[1:])
p=(L[1:].count(m))
if p==n:
print(s[1:])
print('1')
else:
k=0
f=''
k=L[k+1:].index(m)
k+=1
te=L[1:k]
if (n-k)%2!=0:
Li=L[k:]+te
else:
te.reverse()
Li=L[k:]+te
f=''.join(Li)
ki=k
p-=1
while p!=0:
k+=L[k+1:].index(m)
k+=1
te=L[1:k]
if (n-k)%2!=0:
Li=L[k:]+te
else:
te.reverse()
Li=L[k:]+te
x=''.join(Li)
if f>x:
f=x
ki=k
p-=1
print(f)
print(ki)
``` | output | 1 | 73,459 | 0 | 146,919 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1. | instruction | 0 | 73,460 | 0 | 146,920 |
Tags: brute force, constructive algorithms, implementation, sortings, strings
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
from collections import Counter
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
# sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7
for _ in range(INT()):
N = INT()
S = input()
ans = [(S, 1)]
for k in range(2, N+1):
s, t = S[:k-1], S[k-1:]
if (k % 2) ^ (N % 2):
s = t + s
else:
s = t + s[::-1]
ans.append((s, k))
ans.sort()
print(ans[0][0])
print(ans[0][1])
``` | output | 1 | 73,460 | 0 | 146,921 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1. | instruction | 0 | 73,461 | 0 | 146,922 |
Tags: brute force, constructive algorithms, implementation, sortings, strings
Correct Solution:
```
t = int(input())
while (t):
n = int(input())
s = input()
m = s
mk = 1
for k in range(2,n+1):
tmp = s[k-n-1:] + (s[k-2::-1] if ((n-k+1) % 2 == 1) else s[:k-1])
if (tmp < m):
m = tmp
mk = k
print(m)
print(mk)
t -= 1
``` | output | 1 | 73,461 | 0 | 146,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1. | instruction | 0 | 73,462 | 0 | 146,924 |
Tags: brute force, constructive algorithms, implementation, sortings, strings
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
s = input()
ok = min(s)
hehe = ""
dapan, vitri = s, 1
for i in range(1, n+1):
if s[i-1] == ok:
if (n-i+1) % 2 == 0:
hehe = s[i-1:] + s[:i-1]
else:
hehe = s[i-1:] + s[:i-1][::-1]
if hehe < dapan:
dapan = hehe
vitri = i
print(dapan)
print(vitri)
``` | output | 1 | 73,462 | 0 | 146,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1. | instruction | 0 | 73,463 | 0 | 146,926 |
Tags: brute force, constructive algorithms, implementation, sortings, strings
Correct Solution:
```
for _ in range(int(input())):
k = int(input())
word = input()
l = []
for x in range(k+1):
if (k-x) % 2 == 0:
l.append([word[x:] + word[:x], x+1])
else:
l.append([word[x:] + word[:x][::-1], x+1])
l.sort()
print(l[0][0])
print(l[0][1])
``` | output | 1 | 73,463 | 0 | 146,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1. | instruction | 0 | 73,464 | 0 | 146,928 |
Tags: brute force, constructive algorithms, implementation, sortings, strings
Correct Solution:
```
import math
from math import gcd,floor,sqrt,log
def iin(): return int(input())
def sin(): return input().strip()
def listin(): return list(map(int,input().strip().split()))
def liststr(): return list(map(str,input().strip().split()))
def ceill(x): return int(x) if(x==int(x)) else int(x)+1
def ceilldiv(x,d): x//d if(x%d==0) else x//d+1
def LCM(a,b): return (a*b)//gcd(a,b)
def solve():
n = iin()
a = sin()
z = []
for k in range(n-1):
if n%2==k%2:
z.append([a[k:] + a[:k],k+1])
continue
z.append([a[k:] + a[:k][::-1],k+1])
z.append([a[::-1],n])
# print(z)
for i in min(z):
print(i)
t = 1
t = int(input())
for hula in range(t):
solve()
``` | output | 1 | 73,464 | 0 | 146,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1. | instruction | 0 | 73,465 | 0 | 146,930 |
Tags: brute force, constructive algorithms, implementation, sortings, strings
Correct Solution:
```
for u in range(int(input())):
n=int(input())
s=input()
r=''
x=s
c=1
for i in range(n):
r=r+s[i:n]
h=s[0:i]
if((n-i)%2==1):
h=h[::-1]
r=r+h
if(r<x):
x=r
c=1+i
r=''
print(x)
print(c)
``` | output | 1 | 73,465 | 0 | 146,931 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1.
Submitted Solution:
```
'''input
6
4
abab
10
qwertyuiop
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
'''
from collections import defaultdict
import math
from functools import reduce
for _ in range(int(input())):
# n, m = map(int, input().split())
n = int(input())
string = input()
ans = [[string, 1]]
ans = [[string, 1]]
for k in range(2, n + 1):
new = string
new = new[k - 1:] + new[:k - 1][::-
1] if (n - k + 1) % 2 else new[k - 1:] + new[:k - 1]
ans.append([new, k])
# print(ans)
print(*min(ans), sep="\n")
``` | instruction | 0 | 73,466 | 0 | 146,932 |
Yes | output | 1 | 73,466 | 0 | 146,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1.
Submitted Solution:
```
def f():
n = int(input())
s = input()
rk = 0
rs = s
for k in range(1,n):
if (n-k)%2:
cur = s[k:]+s[:k][::-1]
else:
cur = s[k:]+s[:k]
if cur < rs:
rs = cur
rk = k
print(rs)
print(rk+1)
t = int(input())
for i in range(t):
f()
``` | instruction | 0 | 73,467 | 0 | 146,934 |
Yes | output | 1 | 73,467 | 0 | 146,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1.
Submitted Solution:
```
t = int(input())
def replace_k(s, k):
coef = len(s) - k + 1
if k == len(s):
return s[::-1]
return s[k - 1:] + ( s[:k - 1] if coef % 2 == 0 else s[:k-1][::-1])
def solve(s):
min_s, res = s, 1
for k in range(1, len(s) + 1):
replaced = replace_k(s, k)
if min_s > replaced:
min_s = replaced
res = k
return res, min_s
for i in range(t):
_ = input()
s = input()
res, min_s = solve(s)
print(min_s)
print(res)
``` | instruction | 0 | 73,468 | 0 | 146,936 |
Yes | output | 1 | 73,468 | 0 | 146,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
s = input()
x = min(s)
minm = "zzzzzzzzzzzzzzzzzzzzzzzzzzz"
pos = 0
res=''
for i in range(n):
if s[i]==x:
temp=s[i:n]
if len(temp)%2==0:
res=temp+s[0:i]
else:
res=temp+s[0:i][::-1]
#print(res)
if res!='' and res < minm:
minm=res
pos=i
#print(minm)
# print(pos)
print(minm)
print(pos+1)
``` | instruction | 0 | 73,469 | 0 | 146,938 |
Yes | output | 1 | 73,469 | 0 | 146,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1.
Submitted Solution:
```
# import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
t = 1
t = int(input())
while t:
t -= 1
n = int(input())
# a = list(map(int, input().split()))
s = input()
minStr, minIdx = s[:], 0
for i in range(n-1):
if s[i:] + s[:i] < minStr:
minStr = s[i:]+s[:i]
minIdx = i
if s[::-1] < minStr:
minStr = s[::-1]
minIdx = n-1
print(minStr)
print(minIdx+1)
# erqw
``` | instruction | 0 | 73,470 | 0 | 146,940 |
No | output | 1 | 73,470 | 0 | 146,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1.
Submitted Solution:
```
from sys import stdin
t=int(stdin.readline())
for i in range(t):
n=int(stdin.readline())
string=stdin.readline()[:-1]
k=1
small_string=string
for j in range(2,n+1):
cur_string=string[j-1:]+string[:j-1]
if small_string>cur_string:
small_string=cur_string
k=j
print(small_string)
print(k)
``` | instruction | 0 | 73,471 | 0 | 146,942 |
No | output | 1 | 73,471 | 0 | 146,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1.
Submitted Solution:
```
import math
from math import gcd,floor,sqrt,log
def iin(): return int(input())
def sin(): return input().strip()
def listin(): return list(map(int,input().strip().split()))
def liststr(): return list(map(str,input().strip().split()))
def ceill(x): return int(x) if(x==int(x)) else int(x)+1
def ceilldiv(x,d): x//d if(x%d==0) else x//d+1
def LCM(a,b): return (a*b)//gcd(a,b)
def solve():
n = iin()
a = sin()
z = []
for k in range(n-1):
if n%2==1:
z.append([a[k:] + a[:k][::-1],k+1])
continue
z.append([a[k:] + a[:k],k+1])
z.append([a[::-1],n])
# print(z)
for i in min(z):
print(i)
t = 1
t = int(input())
for hula in range(t):
solve()
``` | instruction | 0 | 73,472 | 0 | 146,944 |
No | output | 1 | 73,472 | 0 | 146,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has a string s of length n. He decides to make the following modification to the string:
1. Pick an integer k, (1 β€ k β€ n).
2. For i from 1 to n-k+1, reverse the substring s[i:i+k-1] of s. For example, if string s is qwer and k = 2, below is the series of transformations the string goes through:
* qwer (original string)
* wqer (after reversing the first substring of length 2)
* weqr (after reversing the second substring of length 2)
* werq (after reversing the last substring of length 2)
Hence, the resulting string after modifying s with k = 2 is werq.
Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a β b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 5000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 5000) β the length of the string s.
The second line of each test case contains the string s of n lowercase latin letters.
It is guaranteed that the sum of n over all test cases does not exceed 5000.
Output
For each testcase output two lines:
In the first line output the lexicographically smallest string s' achievable after the above-mentioned modification.
In the second line output the appropriate value of k (1 β€ k β€ n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.
Example
Input
6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
Output
abab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
Note
In the first testcase of the first sample, the string modification results for the sample abab are as follows :
* for k = 1 : abab
* for k = 2 : baba
* for k = 3 : abab
* for k = 4 : baba
The lexicographically smallest string achievable through modification is abab for k = 1 and 3. Smallest value of k needed to achieve is hence 1.
Submitted Solution:
```
# -*- coding: utf-8 -*-
t = int(input())
def mk(k_list, k):
k_list0 = list(k_list)
for i in range(len(k_list) - k + 1):
a0 = k_list0[i:i+k]
a0.reverse()
for j in range(k):
k_list0[i+j] = a0[j]
return k_list0
for t0 in range(t):
n0 = int(input())
s = input()
S = [ord(g) for g in s]
smin = min(S)
ind = []
counter = 0
for h in S:
if h == smin:
ind.append(counter)
counter += 1
if len(ind) > 1:
sx0 = S[ind[0]:ind[1]]
rx0 = ind[0]+1
for v in range(len(ind) - 1):
sx = S[ind[v]:ind[v+1]]
if sx < sx0:
sx0 = list(sx)
rx0 = ind[v]+1
if ind[-1] == n0-1:
d = list(S)
d.reverse()
if d < sx0:
sx0 = list(d)
rx0 = n0
else:
d0 = S[ind[-1]:]
d1 = S[:ind[0]]
if d0+d1 < sx0:
sx0 = list(d0+d1)
rx0 = ind[-1]
else:
rx0 = S.index(smin) + 1
res0 = mk(S, rx0)
res = [chr(x) for x in res0]
print(''.join(res))
print(rx0)
``` | instruction | 0 | 73,473 | 0 | 146,946 |
No | output | 1 | 73,473 | 0 | 146,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A non-empty string s is called binary, if it consists only of characters "0" and "1". Let's number the characters of binary string s from 1 to the string's length and let's denote the i-th character in string s as si.
Binary string s with length n is periodical, if there is an integer 1 β€ k < n such that:
* k is a divisor of number n
* for all 1 β€ i β€ n - k, the following condition fulfills: si = si + k
For example, binary strings "101010" and "11" are periodical and "10" and "10010" are not.
A positive integer x is periodical, if its binary representation (without leading zeroes) is a periodic string.
Your task is to calculate, how many periodic numbers are in the interval from l to r (both ends are included).
Input
The single input line contains two integers l and r (1 β€ l β€ r β€ 1018). The numbers are separated by a space.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print a single integer, showing how many periodic numbers are in the interval from l to r (both ends are included).
Examples
Input
1 10
Output
3
Input
25 38
Output
2
Note
In the first sample periodic numbers are 3, 7 and 10.
In the second sample periodic numbers are 31 and 36. | instruction | 0 | 73,608 | 0 | 147,216 |
Tags: combinatorics, dp, number theory
Correct Solution:
```
from itertools import combinations
from fractions import gcd
primos = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61]
def ToBinary(x):
if x == 0:
return "0"
if x == 1:
return "1"
if x % 2 == 0:
return ToBinary(x>>1)+ "0"
else:
return ToBinary(x>>1)+ "1"
def CountPeriodic(length):
p = [x for x in primos if length % x == 0]
ans = 0
for x in p:
d = length/x
ans += 2**(d-1)
for x,y in combinations(p,2):
d = gcd(length/x, length/y)
ans -= 2**(d-1)
for x,y,z in combinations(p,3):
d = gcd(gcd(length/x, length/y), length/z)
ans += 2**(d-1)
return ans
def L(l,x,d):
ans = 0
d = int(d)
x = int(x)
p = 1
while True:
q = l[p:d].find('0')
if q == -1:
break
ans += 2**(d-(p+q)-1)
p += q+1
for k in range(1,x):
if l[0:d] > l[d*k:d*(k+1)]:
return ans+1
elif l[0:d] == l[d*k:d*(k+1)]:
continue
else:
return ans
return ans+1
def CountPeriodicL(l):
length = len(l)
p = [x for x in primos if length % x == 0]
ans = 0
for x in p:
d = length/x
ans += L(l,x,d)
for x,y in combinations(p,2):
d = gcd(length/x, length/y)
ans -= L(l,length/d,d)
for x,y,z in combinations(p,3):
d = gcd(gcd(length/x, length/y), length/z)
ans += L(l,length/d,d)
return ans
def R(r,x,d,eq=True):
ans = 0
d = int(d)
x = int(x)
p = 1
while True:
q = r[p:d].find('1')
if q == -1:
break
ans += 2**(d-(p+q)-1)
p += q+1
for k in range(1,x):
if r[0:d] < r[d*k:d*(k+1)]:
return ans+1
elif r[0:d] == r[d*k:d*(k+1)]:
continue
else:
return ans
return ans+(1 if eq else 0)
def CountPeriodicR(r,eq=True):
length = len(r)
p = [x for x in primos if length % x == 0]
ans = 0
for x in p:
d = length/x
ans += R(r,x,d,eq)
for x,y in combinations(p,2):
d = gcd(length/x, length/y)
ans -= R(r,length/d,d,eq)
for x,y,z in combinations(p,3):
d = gcd(gcd(length/x, length/y), length/z)
ans += R(r,length/d,d,eq)
return ans
l,r = map(int,input().split())
l_binary = ToBinary(l)
r_binary = ToBinary(r)
ans = 0
for i in range(len(l_binary)+1,len(r_binary)):
ans += CountPeriodic(i)
if len(l_binary)==len(r_binary):
ans += CountPeriodicR(r_binary)
ans -= CountPeriodicR(l_binary,False)
else:
ans += CountPeriodicL(l_binary)
ans += CountPeriodicR(r_binary)
print(int(ans))
``` | output | 1 | 73,608 | 0 | 147,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A non-empty string s is called binary, if it consists only of characters "0" and "1". Let's number the characters of binary string s from 1 to the string's length and let's denote the i-th character in string s as si.
Binary string s with length n is periodical, if there is an integer 1 β€ k < n such that:
* k is a divisor of number n
* for all 1 β€ i β€ n - k, the following condition fulfills: si = si + k
For example, binary strings "101010" and "11" are periodical and "10" and "10010" are not.
A positive integer x is periodical, if its binary representation (without leading zeroes) is a periodic string.
Your task is to calculate, how many periodic numbers are in the interval from l to r (both ends are included).
Input
The single input line contains two integers l and r (1 β€ l β€ r β€ 1018). The numbers are separated by a space.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print a single integer, showing how many periodic numbers are in the interval from l to r (both ends are included).
Examples
Input
1 10
Output
3
Input
25 38
Output
2
Note
In the first sample periodic numbers are 3, 7 and 10.
In the second sample periodic numbers are 31 and 36. | instruction | 0 | 73,609 | 0 | 147,218 |
Tags: combinatorics, dp, number theory
Correct Solution:
```
from itertools import combinations
from fractions import gcd
#Numeros primos hasta el 59
primos = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59]
#Para obtener la cadena binaria de x.
def ToBinary(x):
if x == 0:
return "0"
if x == 1:
return "1"
if x % 2 == 0:
return ToBinary(x>>1)+ "0"
else:
return ToBinary(x>>1)+ "1"
#Obtiene la cantidad de numeros periodicos con longitud de cadena binaria
#igual a "length".
def CountPeriodic(length):
#Contiene los numeros primos que dividen a "length".
p = [x for x in primos if length % x == 0]
ans = 0
for x in p:
#"d" cantidad de repeticiones en las variaciones de "0" y "1", y "x" cantidad de subcadenas.
d = length/x
#Se agrega a la respuesta la cantidad de variaciones de "0" y "1" con "d-1" repeticiones
#porque el primer elemento siempre es "1".
ans += 2**(d-1)
for x,y in combinations(p,2):
#Se calcula el mcd de todos los primos 2 a 2.
d = gcd(length/x, length/y)
#Se elimina de la respuesta el total de variaciones con cantidad de repeticiones igual al
#"mcd -1", mcd de dos primos diferentes dichas variaciones se contaron doble, es decir,
#una vez por cada uno de los primos.
ans -= 2**(d-1)
for x,y,z in combinations(p,3):
d = gcd(gcd(length/x, length/y), length/z)
#Aplicando el principio de inclusion-exclusion, si en el paso anterior se sustrajo
#el total de variaciones de "0" y "1" con cantidad de repeticiones asociadas al mismo mcd de tres primos distintos, significa que se quitaron
#dos veces, luego se debe agregar a la respuesta la cardinalidad del conjunto de esas
#variaciones (solo una vez).
ans += 2**(d-1)
return ans
#Chequea las variaciones para las cadenas binarias cuyos numeros son mayores o iguales
#que el de la cadena "l".
def L(l,x,d):
ans = 0
d = int(d)
x = int(x)
p = 1
while True:
#"q" es la posicion de "0" en el substring "l[p:d]".
q = l[p:d].find('0')
if q == -1:
break
#Se aΓ±ade a la respuesta el total de variacines de "0" y "1" con "d" repeticiones, pero con todas las
#posiciones fijadas desde la primera hasta la del "p"-esimo "0" encontrado en el substring.
ans += 2**(d-(p+q)-1)
p += q+1
for k in range(1,x):
#Si todos los digitos en el substring "l[0:d]" son mayores que los del siguente k substring, se debe
#aumentar la respuesta en 1 porque es posible crear una cadena binaria de periodo "x" con "l[0:d]".
if l[0:d] > l[d*k:d*(k+1)]:
return ans+1
elif l[0:d] == l[d*k:d*(k+1)]:
continue
else:
#No es posible obtener una cadena binaria de periodo "x" con "l[0:d]".
return ans
#Si el programa llega hasta aqui implica que todos los substring son iguales
#por tanto la cadena binaria pertenece a un numero periodico entonces hay que
#aumentar la respuesta en 1.
return ans+1
def CountPeriodicL(l):
length = len(l)
p = [x for x in primos if length % x == 0]
ans = 0
for x in p:
d = length/x
ans += L(l,x,d)
for x,y in combinations(p,2):
d = gcd(length/x, length/y)
ans -= L(l,length/d,d)
for x,y,z in combinations(p,3):
d = gcd(gcd(length/x, length/y), length/z)
ans += L(l,length/d,d)
return ans
#Chequea las variaciones para las cadenas binarias cuyos numeros son menores o iguales
#que el de la cadena "r".
def R(r,x,d,eq=False):
ans = 0
d = int(d)
x = int(x)
p = 1
while True:
q = r[p:d].find('1')
if q == -1:
break
ans += 2**(d-(p+q)-1)
p += q+1
for k in range(1,x):
if r[0:d] < r[d*k:d*(k+1)]:
return ans+1
elif r[0:d] == r[d*k:d*(k+1)]:
continue
else:
return ans
return ans+(1 if not eq else 0)
def CountPeriodicR(r,eq=False):
length = len(r)
p = [x for x in primos if length % x == 0]
ans = 0
for x in p:
d = length/x
ans += R(r,x,d,eq)
for x,y in combinations(p,2):
d = gcd(length/x, length/y)
ans -= R(r,length/d,d,eq)
for x,y,z in combinations(p,3):
d = gcd(gcd(length/x, length/y), length/z)
ans += R(r,length/d,d,eq)
return ans
l,r = map(int,input().split())
l_binary = ToBinary(l)
r_binary = ToBinary(r)
ans = 0
#Chequea la cantidad de numeros periodicos cuya longitud de cadena binaria se encuentra en el
#interior del intervalo conformado por las longitudes de las cadenas binaras de los numeros del intervalo del
#ejercicio.
for i in range(len(l_binary)+1,len(r_binary)):
ans += CountPeriodic(i)
#Chequea la cantidad de numeros periodicos con longitud de cadenas binarias igual a la de los extremos
#del intervalo del ejercicio, en el caso del extremo izquierdo solo se calculan aquellos numeros mayores o
#iguales a "l", y en el caso del derecho los menores o iguales a "r".
if len(l_binary)==len(r_binary):
ans += CountPeriodicR(r_binary)
ans -= CountPeriodicR(l_binary,True)
else:
ans += CountPeriodicL(l_binary)
ans += CountPeriodicR(r_binary)
print(int(ans))
``` | output | 1 | 73,609 | 0 | 147,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A non-empty string s is called binary, if it consists only of characters "0" and "1". Let's number the characters of binary string s from 1 to the string's length and let's denote the i-th character in string s as si.
Binary string s with length n is periodical, if there is an integer 1 β€ k < n such that:
* k is a divisor of number n
* for all 1 β€ i β€ n - k, the following condition fulfills: si = si + k
For example, binary strings "101010" and "11" are periodical and "10" and "10010" are not.
A positive integer x is periodical, if its binary representation (without leading zeroes) is a periodic string.
Your task is to calculate, how many periodic numbers are in the interval from l to r (both ends are included).
Input
The single input line contains two integers l and r (1 β€ l β€ r β€ 1018). The numbers are separated by a space.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print a single integer, showing how many periodic numbers are in the interval from l to r (both ends are included).
Examples
Input
1 10
Output
3
Input
25 38
Output
2
Note
In the first sample periodic numbers are 3, 7 and 10.
In the second sample periodic numbers are 31 and 36.
Submitted Solution:
```
from itertools import combinations
from fractions import gcd
primos = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61]
def ToBinary(x):
if x == 0:
return "0"
if x == 1:
return "1"
if x % 2 == 0:
return ToBinary(x>>1)+ "0"
else:
return ToBinary(x>>1)+ "1"
def CountPeriodic(length):
p = [x for x in primos if length % x == 0]
ans = 0
for x in p:
d = length/x
ans += 2**(d-1)
for x,y in combinations(p,2):
d = gcd(length/x, length/y)
ans -= 2**(d-1)
for x,y,z in combinations(p,3):
d = gcd(gcd(length/x, length/y), length/z)
ans += 2**(d-1)
return ans
def L(l,x,d):
ans = 0
d = int(d)
x = int(x)
p = 1
while True:
q = l[p:d].find('0')
if q == -1:
break
ans += 2**(d-(p+q)-1)
p += q+1
for k in range(1,x):
if l[0:d] > l[d*k:d*(k+1)]:
return ans+1
elif l[0:d] == l[d*k:d*(k+1)]:
continue
else:
return ans
return ans+1
def CountPeriodicL(l):
length = len(l)
p = [x for x in primos if length % x == 0]
ans = 0
for x in p:
d = length/x
ans += L(l,x,d)
for x,y in combinations(p,2):
d = gcd(length/x, length/y)
ans -= L(l,length/d,d)
for x,y,z in combinations(p,3):
d = gcd(gcd(length/x, length/y), length/z)
ans += L(l,length/d,d)
return ans
def R(r,x,d,eq=True):
ans = 0
d = int(d)
x = int(x)
p = 1
while True:
q = r[p:d].find('1')
if q == -1:
break
ans += 2**(d-(p+q)-1)
p += q+1
for k in range(1,x):
if r[0:d] < r[d*k:d*(k+1)]:
return ans+1
elif r[0:d] == r[d*k:d*(k+1)]:
continue
else:
return ans
return ans+(1 if eq else 0)
def CountPeriodicR(r,eq=True):
length = len(r)
p = [x for x in primos if length % x == 0]
ans = 0
for x in p:
d = length/x
ans += R(r,x,d,eq)
for x,y in combinations(p,2):
d = gcd(length/x, length/y)
ans -= R(r,length/d,d,eq)
for x,y,z in combinations(p,3):
d = gcd(gcd(length/x, length/y), length/z)
ans += R(r,length/d,d,eq)
return ans
l,r = map(int,input().split())
l_binary = ToBinary(l)
r_binary = ToBinary(r)
ans = 0
for i in range(len(l_binary)+1,len(r_binary)):
ans += CountPeriodic(i)
if len(l_binary)==len(r_binary):
ans += CountPeriodicR(r_binary)
ans -= CountPeriodicR(r_binary,False)
else:
ans += CountPeriodicL(l_binary)
ans += CountPeriodicR(r_binary)
print(int(ans))
``` | instruction | 0 | 73,610 | 0 | 147,220 |
No | output | 1 | 73,610 | 0 | 147,221 |
Provide a correct Python 3 solution for this coding contest problem.
G: Palindromic Subsequences
problem
Given a string S consisting only of lowercase letters, find out how many subsequences of this string S are not necessarily continuous and are palindromes.
Here, a subsequence that is not necessarily continuous with S is an arbitrary selection of one or more characters | S | characters or less from the original character string S (the position of each character to be selected may be discontinuous), and they are used. Refers to a character string created by concatenating the characters in the original order. Note that an empty string is not allowed as a subsequence in this issue.
Also, if the string X is a palindrome, it means that the original string X and the inverted X string X'are equal.
Also note that even if the same palindrome is generated as a result of different subsequence arrangements, it will not be counted twice. For example, if S = `acpc`, both the subsequence consisting of only the second character and the subsequence consisting of only the fourth character are palindromes` c`, but this cannot be repeated multiple times, only once in total. I will count it.
The answer can be very large, so print the remainder divided by 1,000,000,007.
Input format
S
Constraint
* 1 \ leq | S | \ leq 2,000
* S contains only lowercase letters
Output format
* Output the remainder of the answer divided by 1,000,000,007 on one line.
Input example 1
acpc
Output example 1
Five
There are five types of subsequences of the string `acpc` that are not necessarily continuous and are palindromes:` a`, `c`,` cc`, `cpc`, and` p`. Note that we count the number of substring types.
Input example 2
z
Output example 2
1
The only subsequence that meets the condition is `z`. Note that an empty string is not allowed as a subsequence.
Input example 3
madokamagica
Output example 3
28
Example
Input
acpc
Output
5 | instruction | 0 | 74,146 | 0 | 148,292 |
"Correct Solution:
```
def main():
from bisect import bisect_left as bl
from bisect import bisect_right as br
s=input()
alpha="abcdefghijklmnopqrstuvwxyz"
l=len(s)
alpha2={j:i for i,j in enumerate(alpha)}
memo=[[] for _ in [0]*26]
mod=10**9+7
for i in range(l):
memo[alpha2[s[i]]].append(i)
dp=[[0]*(l+1) for _ in range(l+1)]
dp[0][l]=1
ans=0
for i in range(l):
for j in range(i+1,l+1):
p=dp[i][j]
if p==0:
continue
for k in range(26):
x=bl(memo[k],i)
y=br(memo[k],j-1)-1
if x>y:
continue
ans+=p
if x<y:
mx,my=memo[k][x],memo[k][y]
dp[mx+1][my]=(dp[mx+1][my]+p)%mod
ans=ans%mod
ans=(ans+sum([sum([i for i in j])%mod for j in dp])+mod-1)%mod
print(ans)
if __name__=='__main__':
main()
``` | output | 1 | 74,146 | 0 | 148,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters. | instruction | 0 | 74,292 | 0 | 148,584 |
Tags: constructive algorithms
Correct Solution:
```
l="abcdefghijklmnopqrstuvwxyz"
T=int(input())
for t in range(T):
n,a,b=[int(x) for x in input().split()]
unit=l[:b]
rem=n%b
req_str=unit*(n//b)
req_str+=unit[:rem]
print(req_str)
``` | output | 1 | 74,292 | 0 | 148,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters. | instruction | 0 | 74,293 | 0 | 148,586 |
Tags: constructive algorithms
Correct Solution:
```
t = int(input())
alpha = 'abcdefghijklmnopqrstuvwxyz'
for _ in range(t):
n, a, b = map(int, input().split())
ans = []
for i in range(n):
ans.append(alpha[i%b])
print(''.join(ans))
``` | output | 1 | 74,293 | 0 | 148,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters. | instruction | 0 | 74,294 | 0 | 148,588 |
Tags: constructive algorithms
Correct Solution:
```
t = int(input())
for i in range(t):
n,a,b = list(map(int,input().split()))
rem = a//b
str = ""
i = 97
for j in range(b):
str1 = ""
for k in range(rem):
str1 += chr(i)
i += 1
str += str1
res = ""
while(len(res)<=n):
res += str[0:a:1]
res = res[0:n:1]
print(res)
``` | output | 1 | 74,294 | 0 | 148,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters. | instruction | 0 | 74,295 | 0 | 148,590 |
Tags: constructive algorithms
Correct Solution:
```
t = int(input())
for _ in range(t):
n, a, b = [int(x) for x in input().split()]
for i in range(n):
print(chr(97 + i % b), end='')
print()
``` | output | 1 | 74,295 | 0 | 148,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters. | instruction | 0 | 74,296 | 0 | 148,592 |
Tags: constructive algorithms
Correct Solution:
```
t = int(input())
letters = ['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']
# main loop
for i in range(t):
n,a,b = input().split()
n = int(n)
a = int(a)
b = int(b)
x = 0
word = ''
for j in range(n):
if x == b:
x = 0
word= word+letters[x]
x = x+1
print(word)
``` | output | 1 | 74,296 | 0 | 148,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters. | instruction | 0 | 74,297 | 0 | 148,594 |
Tags: constructive algorithms
Correct Solution:
```
t = int(input())
s = 'abcdefghijklmnopqrstuvwxyz'
for i in range(t):
n, a, b = map(int, input().split())
k = s[:b] * 2000
print(k[:n])
``` | output | 1 | 74,297 | 0 | 148,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters. | instruction | 0 | 74,298 | 0 | 148,596 |
Tags: constructive algorithms
Correct Solution:
```
number_test = int(input())
for k in range(number_test):
nab = input().split(" ")
n,a,b= int(nab[0]), int(nab[1]), int(nab[2])
tab = [(k+97) for k in range(b)]
string = "".join(map(chr, tab))
mod = n//b
r = n%b
sol = ""
for i in range(mod):
sol +=string
sol += string[:r]
print(sol)
``` | output | 1 | 74,298 | 0 | 148,597 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters. | instruction | 0 | 74,299 | 0 | 148,598 |
Tags: constructive algorithms
Correct Solution:
```
#div2 c yesterday's problem find max of all reguired to make changes compared to previous
import string
for t in range(int(input())):
x=list(string.ascii_lowercase)
n,a,b=map(int,input().strip().split())
ans=[]
j=0
while(b!=0):
ans.append(x[j])
a-=1
b-=1
j+=1
n-=1
k=0
while(n!=0):
ans.append(ans[k])
k+=1
n-=1
print("".join(ans))
``` | output | 1 | 74,299 | 0 | 148,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters.
Submitted Solution:
```
from math import ceil, sqrt, floor, sqrt
# import sys
# sys.setrecursionlimit(50000)
import collections
from typing import List
from fractions import gcd
from collections import deque
# 1.5m recursive solution
# 7m submit WA
# 4m approach
# 16m submit
# 36 submit, WA
# 38 AC, faster than 19%
from collections import defaultdict
t = int(input())
for _ in range(t):
n, a, b = list(map(int, input().split()))
alphabets = ''.join([chr(ord('a') + i) for i in range(26)])
astr = ''.join([alphabets[i % b] for i in range(a)])
res = astr * int(ceil(n / a))
res = res[:n]
print(res)
``` | instruction | 0 | 74,300 | 0 | 148,600 |
Yes | output | 1 | 74,300 | 0 | 148,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters.
Submitted Solution:
```
for test in range(0,int(input())):
n,a,b=[int(x) for x in input().split()]
for i in range(0,n):
print(chr(97+i%b),end="")
print()
``` | instruction | 0 | 74,301 | 0 | 148,602 |
Yes | output | 1 | 74,301 | 0 | 148,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters.
Submitted Solution:
```
alpha = "abcdefghijklmnopqrstuvwxyz"
T = int(input())
for t in range(T):
N,A,B = [int(x) for x in input().split()]
our = alpha[:B]
thestr = ''
while len(thestr) < N:
thestr += our
print(thestr[:N])
``` | instruction | 0 | 74,302 | 0 | 148,604 |
Yes | output | 1 | 74,302 | 0 | 148,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters.
Submitted Solution:
```
def find_string(n,a,b):
lista=list("abcdefghijklmnopqrstuvwxyz")
string=''
substring=''
for i in range(b):
substring+=lista[i]
substring+=(a-b)*lista[i]
string=(n//a)*substring
for i in range(n%a):
string+=substring[i]
return string
t=int(input())
for i in range(t):
n,a,b=input().split()
n=int(n)
a=int(a)
b=int(b)
print(find_string(n,a,b))
``` | instruction | 0 | 74,303 | 0 | 148,606 |
Yes | output | 1 | 74,303 | 0 | 148,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters.
Submitted Solution:
```
from random import shuffle
def create(n,b,j="",s=""):
for i in range (b):
s+=chr(97+i)
for i in range (0,n,b):
a=list(s)
shuffle(a)
m=''
for k in range (len(a)):
m+=a[k]
j+=m
j+=s[:n-len(j)]
return j
t=int(input(""))
for i in range (t):
n,a,b=input("").split()
print(create(int(n),int(b)))
``` | instruction | 0 | 74,304 | 0 | 148,608 |
No | output | 1 | 74,304 | 0 | 148,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters.
Submitted Solution:
```
t=int(input())
for i in range (t):
n,a,b=map(int,input().split())
z=["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"]
y=z[:b]
y=y*n
y=y[:n]
for j in range(n):
print(y[j],end="")
``` | instruction | 0 | 74,305 | 0 | 148,610 |
No | output | 1 | 74,305 | 0 | 148,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters.
Submitted Solution:
```
letters = ['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']
t = int(input())
for i in range(t):
n,a,b = map(int,input().split())
ans = []
while len(ans) < n:
for j in letters:
for k in range(round(a/b)):
ans.append(j)
print(''.join(ans[:n]))
``` | instruction | 0 | 74,306 | 0 | 148,612 |
No | output | 1 | 74,306 | 0 | 148,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three positive integers n, a and b. You have to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. It is guaranteed that the answer exists.
You have to answer t independent test cases.
Recall that the substring s[l ... r] is the string s_l, s_{l+1}, ..., s_{r} and its length is r - l + 1. In this problem you are only interested in substrings of length a.
Input
The first line of the input contains one integer t (1 β€ t β€ 2000) β the number of test cases. Then t test cases follow.
The only line of a test case contains three space-separated integers n, a and b (1 β€ a β€ n β€ 2000, 1 β€ b β€ min(26, a)), where n is the length of the required string, a is the length of a substring and b is the required number of distinct letters in each substring of length a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (β n β€ 2000).
Output
For each test case, print the answer β such a string s of length n consisting of lowercase Latin letters that each substring of length a has exactly b distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.
Example
Input
4
7 5 3
6 1 1
6 6 1
5 2 2
Output
tleelte
qwerty
vvvvvv
abcde
Note
In the first test case of the example, consider all the substrings of length 5:
* "tleel": it contains 3 distinct (unique) letters,
* "leelt": it contains 3 distinct (unique) letters,
* "eelte": it contains 3 distinct (unique) letters.
Submitted Solution:
```
import sys
def read_line():
return sys.stdin.readline()
def solve(n, a, b):
start, end = 97, 97 + b
lst = [chr(i) for i in range(start, end)]
for i in range(a - len(lst)):
lst.append(chr(end))
l = 0
i = len(lst) - 1
while i <= n:
if i - l > a:
lst.append(lst[l])
l += 1
i += 1
print(''.join(lst))
n = int(read_line())
for i in range(n):
line = read_line().split(' ')
solve(int(line[0]), int(line[1]), int(line[2]))
``` | instruction | 0 | 74,307 | 0 | 148,614 |
No | output | 1 | 74,307 | 0 | 148,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's assume that p and q are strings of positive length, called the container and the key correspondingly, string q only consists of characters 0 and 1. Let's take a look at a simple algorithm that extracts message s from the given container p:
i = 0;
j = 0;
s = <>;
while i is less than the length of the string p
{
if q[j] == 1, then add to the right of string s character p[i];
increase variables i, j by one;
if the value of the variable j equals the length of the string q, then j = 0;
}
In the given pseudocode i, j are integer variables, s is a string, '=' is an assignment operator, '==' is a comparison operation, '[]' is the operation of obtaining the string character with the preset index, '<>' is an empty string. We suppose that in all strings the characters are numbered starting from zero.
We understand that implementing such algorithm is quite easy, so your task is going to be slightly different. You need to construct the lexicographically minimum key of length k, such that when it is used, the algorithm given above extracts message s from container p (otherwise find out that such key doesn't exist).
Input
The first two lines of the input are non-empty strings p and s (1 β€ |p| β€ 106, 1 β€ |s| β€ 200), describing the container and the message, correspondingly. The strings can contain any characters with the ASCII codes from 32 to 126, inclusive.
The third line contains a single integer k (1 β€ k β€ 2000) β the key's length.
Output
Print the required key (string of length k, consisting only of characters 0 and 1). If the key doesn't exist, print the single character 0.
Examples
Input
abacaba
aba
6
Output
100001
Input
abacaba
aba
3
Output
0
Note
String x = x1x2... xp is lexicographically smaller than string y = y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or there exists such integer r (0 β€ r < min(p, q)) that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1. Symbols are compared according to their ASCII codes.
Submitted Solution:
```
#! /usr/bin/env python3
# -*- coding: utf-8 -*-
from sys import stdin
from itertools import *
from operator import eq
from collections import defaultdict
def main():
p = stdin.readline()[:-1]
s = stdin.readline()[:-1]
k = int(stdin.readline())
N = len(s)
for comb in (combinations_with_replacement([0, 1], k)):
for key_ in permutations(comb):
key, d = tee(iter(key_))
key_str = "".join(map(str, d))
#print(key_str)
word, d = tee(compress(p, cycle(key)))
good_word = "".join(d)
#print(good_word)
if good_word == s:
print(key_str)
return 0
#pairs, d = tee(zip(word, s))
#correct, d = tee(map(lambda x: x[0] == x[1], pairs))
#print("".join(map(repr,d)))
#bad = dropwhile(lambda x: x[0] == x[1], pairs)
#try:
#next(bad)
#continue
#except StopIteration:
#print("FOUND", key_str)
#return
#except:
#pass
if __name__ == '__main__':
main()
``` | instruction | 0 | 74,505 | 0 | 149,010 |
No | output | 1 | 74,505 | 0 | 149,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's assume that p and q are strings of positive length, called the container and the key correspondingly, string q only consists of characters 0 and 1. Let's take a look at a simple algorithm that extracts message s from the given container p:
i = 0;
j = 0;
s = <>;
while i is less than the length of the string p
{
if q[j] == 1, then add to the right of string s character p[i];
increase variables i, j by one;
if the value of the variable j equals the length of the string q, then j = 0;
}
In the given pseudocode i, j are integer variables, s is a string, '=' is an assignment operator, '==' is a comparison operation, '[]' is the operation of obtaining the string character with the preset index, '<>' is an empty string. We suppose that in all strings the characters are numbered starting from zero.
We understand that implementing such algorithm is quite easy, so your task is going to be slightly different. You need to construct the lexicographically minimum key of length k, such that when it is used, the algorithm given above extracts message s from container p (otherwise find out that such key doesn't exist).
Input
The first two lines of the input are non-empty strings p and s (1 β€ |p| β€ 106, 1 β€ |s| β€ 200), describing the container and the message, correspondingly. The strings can contain any characters with the ASCII codes from 32 to 126, inclusive.
The third line contains a single integer k (1 β€ k β€ 2000) β the key's length.
Output
Print the required key (string of length k, consisting only of characters 0 and 1). If the key doesn't exist, print the single character 0.
Examples
Input
abacaba
aba
6
Output
100001
Input
abacaba
aba
3
Output
0
Note
String x = x1x2... xp is lexicographically smaller than string y = y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or there exists such integer r (0 β€ r < min(p, q)) that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1. Symbols are compared according to their ASCII codes.
Submitted Solution:
```
#! /usr/bin/env python3
# -*- coding: utf-8 -*-
from sys import stdin
from itertools import *
from operator import eq
from collections import defaultdict
def main():
p = stdin.readline()[:-1]
s = stdin.readline()[:-1]
k = int(stdin.readline())
N = len(s)
if not any(starmap(eq, zip(p, s))):
return 0
for comb in (combinations_with_replacement([0, 1], k)):
for key_ in permutations(comb):
key, d = tee(iter(key_))
key_str = "".join(map(str, d))
#print(key_str)
word, d = tee(compress(p, cycle(key)))
good_word = "".join(d)
#print(good_word)
if good_word == s:
print(key_str)
return 0
print(0)
return 0
if __name__ == '__main__':
main()
``` | instruction | 0 | 74,506 | 0 | 149,012 |
No | output | 1 | 74,506 | 0 | 149,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's assume that p and q are strings of positive length, called the container and the key correspondingly, string q only consists of characters 0 and 1. Let's take a look at a simple algorithm that extracts message s from the given container p:
i = 0;
j = 0;
s = <>;
while i is less than the length of the string p
{
if q[j] == 1, then add to the right of string s character p[i];
increase variables i, j by one;
if the value of the variable j equals the length of the string q, then j = 0;
}
In the given pseudocode i, j are integer variables, s is a string, '=' is an assignment operator, '==' is a comparison operation, '[]' is the operation of obtaining the string character with the preset index, '<>' is an empty string. We suppose that in all strings the characters are numbered starting from zero.
We understand that implementing such algorithm is quite easy, so your task is going to be slightly different. You need to construct the lexicographically minimum key of length k, such that when it is used, the algorithm given above extracts message s from container p (otherwise find out that such key doesn't exist).
Input
The first two lines of the input are non-empty strings p and s (1 β€ |p| β€ 106, 1 β€ |s| β€ 200), describing the container and the message, correspondingly. The strings can contain any characters with the ASCII codes from 32 to 126, inclusive.
The third line contains a single integer k (1 β€ k β€ 2000) β the key's length.
Output
Print the required key (string of length k, consisting only of characters 0 and 1). If the key doesn't exist, print the single character 0.
Examples
Input
abacaba
aba
6
Output
100001
Input
abacaba
aba
3
Output
0
Note
String x = x1x2... xp is lexicographically smaller than string y = y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or there exists such integer r (0 β€ r < min(p, q)) that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1. Symbols are compared according to their ASCII codes.
Submitted Solution:
```
#! /usr/bin/env python3
# -*- coding: utf-8 -*-
from sys import stdin
from itertools import *
from operator import eq
from collections import defaultdict
def main():
p = stdin.readline()[:-1]
s = stdin.readline()[:-1]
k = int(stdin.readline())
N = len(s)
for comb in (combinations_with_replacement([0, 1], k)):
for key_ in permutations(comb):
key, d = tee(iter(key_))
key_str = "".join(map(str, d))
#print(key_str)
word, d = tee(compress(p, cycle(key)))
good_word = "".join(d)
#print(good_word)
if good_word == s:
print(key_str)
return
``` | instruction | 0 | 74,507 | 0 | 149,014 |
No | output | 1 | 74,507 | 0 | 149,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's assume that p and q are strings of positive length, called the container and the key correspondingly, string q only consists of characters 0 and 1. Let's take a look at a simple algorithm that extracts message s from the given container p:
i = 0;
j = 0;
s = <>;
while i is less than the length of the string p
{
if q[j] == 1, then add to the right of string s character p[i];
increase variables i, j by one;
if the value of the variable j equals the length of the string q, then j = 0;
}
In the given pseudocode i, j are integer variables, s is a string, '=' is an assignment operator, '==' is a comparison operation, '[]' is the operation of obtaining the string character with the preset index, '<>' is an empty string. We suppose that in all strings the characters are numbered starting from zero.
We understand that implementing such algorithm is quite easy, so your task is going to be slightly different. You need to construct the lexicographically minimum key of length k, such that when it is used, the algorithm given above extracts message s from container p (otherwise find out that such key doesn't exist).
Input
The first two lines of the input are non-empty strings p and s (1 β€ |p| β€ 106, 1 β€ |s| β€ 200), describing the container and the message, correspondingly. The strings can contain any characters with the ASCII codes from 32 to 126, inclusive.
The third line contains a single integer k (1 β€ k β€ 2000) β the key's length.
Output
Print the required key (string of length k, consisting only of characters 0 and 1). If the key doesn't exist, print the single character 0.
Examples
Input
abacaba
aba
6
Output
100001
Input
abacaba
aba
3
Output
0
Note
String x = x1x2... xp is lexicographically smaller than string y = y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or there exists such integer r (0 β€ r < min(p, q)) that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1. Symbols are compared according to their ASCII codes.
Submitted Solution:
```
#! /usr/bin/env python3
# -*- coding: utf-8 -*-
from sys import stdin
from itertools import *
from operator import eq
from collections import defaultdict
def main():
p = stdin.readline()[:-1]
s = stdin.readline()[:-1]
k = int(stdin.readline())
N = len(s)
if not set(p[:N]).intersection(s):
return 0
for comb in (combinations_with_replacement([0, 1], k)):
for key_ in permutations(comb):
key, d = tee(iter(key_))
key_str = "".join(map(str, d))
#print(key_str)
word, d = tee(compress(p, cycle(key)))
good_word = "".join(d)
#print(good_word)
if good_word == s:
print(key_str)
return 0
print(0)
return 0
if __name__ == '__main__':
main()
``` | instruction | 0 | 74,508 | 0 | 149,016 |
No | output | 1 | 74,508 | 0 | 149,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hamed has recently found a string t and suddenly became quite fond of it. He spent several days trying to find all occurrences of t in other strings he had. Finally he became tired and started thinking about the following problem. Given a string s how many ways are there to extract k β₯ 1 non-overlapping substrings from it such that each of them contains string t as a substring? More formally, you need to calculate the number of ways to choose two sequences a1, a2, ..., ak and b1, b2, ..., bk satisfying the following requirements:
* k β₯ 1
* <image>
* <image>
* <image>
* <image> t is a substring of string saisai + 1... sbi (string s is considered as 1-indexed).
As the number of ways can be rather large print it modulo 109 + 7.
Input
Input consists of two lines containing strings s and t (1 β€ |s|, |t| β€ 105). Each string consists of lowercase Latin letters.
Output
Print the answer in a single line.
Examples
Input
ababa
aba
Output
5
Input
welcometoroundtwohundredandeightytwo
d
Output
274201
Input
ddd
d
Output
12 | instruction | 0 | 74,541 | 0 | 149,082 |
Tags: dp, strings
Correct Solution:
```
#!/usr/bin/env python
# coding=utf-8
s, t = input(), input()
n, m = len(s), len(t)
t = '$'.join((t, s))
p = [0]
k = 0
for i in range(1, n + m + 1):
while k and t[k] != t[i]:
k = p[k - 1]
if t[k] == t[i]:
k += 1
p.append(k)
ans = [0] * n
sums = [0] * (n + 1)
curs = 0
was = False
j = 0
mod = 1000000007
for i in range(n):
if p[i + m + 1] == m:
if not was:
was = True
curs = 1
while j <= i - m:
curs = (curs + sums[j] + 1) % mod
j += 1
ans[i]= curs;
sums[i] = (sums[i - 1] + ans[i]) % mod;
print(sum(ans) % mod)
``` | output | 1 | 74,541 | 0 | 149,083 |
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