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Provide a correct Python 3 solution for this coding contest problem.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
|
instruction
| 0
| 98,501
| 0
| 197,002
|
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
W = read().rstrip()
N = len(W)
def Z_algorithm(S):
# 共通接頭辞の長さを返す
N=len(S)
arr = [0]*N
arr[0] = N
i,j = 1,0
while i<N:
while i+j<N and S[j]==S[i+j]:
j += 1
arr[i]=j
if not j:
i += 1
continue
k = 1
while i+k<N and k+arr[k]<j:
arr[i+k] = arr[k]
k += 1
i += k; j -= k
return arr
def is_periodic_left(W):
Z = Z_algorithm(W)
is_periodic = [False] * N
for p in range(1,N//2 + 1):
if is_periodic[p-1]:
continue
for i in range(p,N,p):
if Z[i] >= p:
is_periodic[p + i - 1] = True
else:
break
return is_periodic
L = is_periodic_left(W)
R = is_periodic_left(W[::-1])[::-1]
if not L[-1]:
answer = (1,1)
elif len(set(W)) == 1:
answer = (N,1)
else:
x = sum(not(x or y) for x,y in zip(L,R[1:]))
answer = (2,x)
print('\n'.join(map(str,answer)))
```
|
output
| 1
| 98,501
| 0
| 197,003
|
Provide a correct Python 3 solution for this coding contest problem.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
|
instruction
| 0
| 98,502
| 0
| 197,004
|
"Correct Solution:
```
w=list(input());n=len(w);t=-1
def Z(s):
m=len(s);z=[0]*m;c=0;f=[1]*m;
for i in range(1,m):
if i+z[i-c]<c+z[c]:z[i]=z[i-c]
else:
j=max(0,c+z[c]-i)
while i+j<n and s[j]==s[i+j]:j=j+1
z[i]=j;c=i
for p in range(1,m):
for k in range(2,z[p]//p+2):f[k*p-1]=0
return f
for j in range(1,n//2+1):
if n%j==0 and w[:n-j]==w[j:]:t=j;break;
if t==-1:print ('1\n1')
elif t==1:print (n);print (1)
else:
zl=Z(w)
w.reverse()
zr=Z(w)
cnt=0
for i in range(0,n-1):
if zl[i] and zr[n-2-i]:cnt=cnt+1
print(2);print(cnt);
```
|
output
| 1
| 98,502
| 0
| 197,005
|
Provide a correct Python 3 solution for this coding contest problem.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
|
instruction
| 0
| 98,503
| 0
| 197,006
|
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
W = list(map(int,read().rstrip()))
N = len(W)
def Z_Algorithm(S):
# 共通接頭辞の長さを返す
N=len(S)
arr = [0]*N
arr[0] = N
i,j = 1,0
while i<N:
while i+j<N and S[j]==S[i+j]:
j += 1
arr[i]=j
if not j:
i += 1
continue
k = 1
while i+k<N and k+arr[k]<j:
arr[i+k] = arr[k]
k += 1
i += k; j -= k
return arr
def find_good_substr(W):
Z = Z_Algorithm(W)
is_good = [True]*(N+1) # 長さをインデックス
for n in range(1,N//2+1):
# n文字での周期をもってくる
if not is_good[n]:
continue
for i in range(n+n,Z[n]+n+1,n):
is_good[i] = False
return is_good
def solve():
G1 = find_good_substr(W)
G2 = find_good_substr(W[::-1])[::-1]
if G1[-1]:
# 自分自身がoptimal
return 1,1
if len(set(W)) == 1:
return N,1
# 2個にできるはず
x = sum(x and y for x,y in zip(G1[1:-1],G2[1:-1]))
return 2,x
x,y = solve()
print(x)
print(y)
```
|
output
| 1
| 98,503
| 0
| 197,007
|
Provide a correct Python 3 solution for this coding contest problem.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
|
instruction
| 0
| 98,504
| 0
| 197,008
|
"Correct Solution:
```
import sys
readline = sys.stdin.readline
class Rollinhash:
def __init__(self, S):
N = len(S)
self.mod = 10**9+9
self.base = 2009
self.has = [0]*(N+1)
self.power = [1]*(N+1)
for i in range(N):
s = S[i]
self.has[i+1] = (self.has[i]*self.base + s)%self.mod
self.power[i+1] = self.power[i]*self.base%self.mod
def rh(self, i, j):
return (self.has[j] - self.has[i]*self.power[j-i])%self.mod
MOD = 10**9+7
S = list(map(ord, readline().strip()))
N = len(S)
if len(set(S)) == 1:
print(N)
print(1)
else:
Rs = Rollinhash(S)
tabler = [True]*(N+1)
for d in range(1, 1+N//2):
r = Rs.rh(0, d)
for i in range(1, N//d):
if r != Rs.rh(i*d, (i+1)*d):
break
tabler[(i+1)*d] = False
tablel = [True]*(N+1)
for d in range(1, 1+N//2):
r = Rs.rh(N-d, N)
for i in range(1, N//d):
if r != Rs.rh(N-(i+1)*d, N-i*d):
break
tablel[N-(i+1)*d] = False
if tabler[N]:
print(1)
print(1)
else:
print(2)
ans = 0
for i in range(N+1):
if tabler[i] and tablel[i]:
ans += 1
assert ans > 0, ''
print(ans)
```
|
output
| 1
| 98,504
| 0
| 197,009
|
Provide a correct Python 3 solution for this coding contest problem.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
|
instruction
| 0
| 98,505
| 0
| 197,010
|
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
W = read().rstrip()
N = len(W)
def Z_algorithm(S):
# 共通接頭辞の長さを返す
N=len(S)
arr = [0]*N
arr[0] = N
i,j = 1,0
while i<N:
while i+j<N and S[j]==S[i+j]:
j += 1
arr[i]=j
if not j:
i += 1
continue
k = 1
while i+k<N and k+arr[k]<j:
arr[i+k] = arr[k]
k += 1
i += k; j -= k
return arr
def is_periodic_left(W):
Z = Z_algorithm(W)
is_periodic = [False] * N
for p in range(1,N//2 + 10):
for i in range(p,N,p):
if Z[i] >= p:
is_periodic[p + i - 1] = True
else:
break
return is_periodic
L = is_periodic_left(W)
R = is_periodic_left(W[::-1])[::-1]
if not L[-1]:
answer = (1,1)
elif len(set(W)) == 1:
answer = (N,1)
else:
x = sum(not(x or y) for x,y in zip(L,R[1:]))
answer = (2,x)
print('\n'.join(map(str,answer)))
```
|
output
| 1
| 98,505
| 0
| 197,011
|
Provide a correct Python 3 solution for this coding contest problem.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
|
instruction
| 0
| 98,506
| 0
| 197,012
|
"Correct Solution:
```
from collections import Counter
*W, = map(ord, input())
N = len(W)
C = Counter(W)
if len(C) == 1:
print(N)
print(1)
exit(0)
def z_algo(S):
A = [0]*N
i = 1; j = 0
A[0] = l = len(S)
while i < l:
while i+j < l and S[j] == S[i+j]:
j += 1
A[i] = j
if not j:
i += 1
continue
k = 1
while l-i > k < j - A[k]:
A[i+k] = A[k]
k += 1
i += k; j -= k
return A
def calc(W):
Z = z_algo(W)
G = [0]*N
for i in range(N):
G[i] = 1
for p in range(1, N):
if not G[p-1]:
continue
for k in range(2, Z[p]//p+2):
G[k*p-1] = 0
return G
G0 = calc(W)
W.reverse()
G1 = calc(W)
if G0[N-1]:
print(1)
print(1)
exit(0)
print(2)
print(sum(p and q for p, q in zip(G0[:-1], reversed(G1[:-1]))))
```
|
output
| 1
| 98,506
| 0
| 197,013
|
Provide a correct Python 3 solution for this coding contest problem.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
|
instruction
| 0
| 98,507
| 0
| 197,014
|
"Correct Solution:
```
w=list(input())
n=len(w)
t=-1
def Z(s):
m=len(s);z=[0]*m;c=0;f=[1]*m;
for i in range(1,m):
if i+z[i-c]<c+z[c]:z[i]=z[i-c]
else:
j=max(0,c+z[c]-i)
while i+j<n and s[j]==s[i+j]:j=j+1
z[i]=j;c=i
for p in range(1,m):
for k in range(2,z[p]//p+2):f[k*p-1]=0
return f
for j in range(1,n//2+1):
if n%j==0 and w[:n-j]==w[j:]:t=j;break;
if t==-1:print ('1\n1')
elif t==1:print (n);print (1)
else:
zl=Z(w)
w.reverse()
zr=Z(w)
cnt=0
for i in range(0,n-1):
if zl[i] and zr[n-2-i]:cnt=cnt+1
print(2);print(cnt);
```
|
output
| 1
| 98,507
| 0
| 197,015
|
Provide a correct Python 3 solution for this coding contest problem.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
|
instruction
| 0
| 98,508
| 0
| 197,016
|
"Correct Solution:
```
def Z_algorithm(S):
l=len(S)
A=[0]*l
A[0]=l
i=1; j=0
while i<l:
while i+j<l and S[j]==S[i+j]:
j+=1
if not j:
i+=1
continue
A[i]=j
k=1
while l-i>k<j-A[k]:
A[i+k]=A[k]
k+=1
i+=k; j-=k
return A
def jugde(W):
Z=Z_algorithm(W)
l=len(W)
B=[True]*l
for p in range(1,l):
if not B[p-1]:
continue
k=2
while (k-1)*p<=Z[p]:
B[k*p-1]=False
k+=1
return B
def solve(W):
n=len(W)
if len(set(W))==1:
print(n)
print(1)
return
G=jugde(W)
W.reverse()
G_rev=jugde(W)
if G[-1]:
print(1)
print(1)
return
print(2)
cnt=0
for i in range(n-1):
cnt+=G[i] and G_rev[-(i+2)]
print(cnt)
return
solve(list(input()))
```
|
output
| 1
| 98,508
| 0
| 197,017
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
Submitted Solution:
```
import sys
readline = sys.stdin.readline
class Rollinhash:
def __init__(self, S):
N = len(S)
self.mod = 10**9+9
self.base = 2009
self.has = [0]*(N+1)
self.power = [1]*(N+1)
for i in range(N):
s = S[i]
self.has[i+1] = (self.has[i]*self.base + s)%self.mod
self.power[i+1] = self.power[i]*self.base%self.mod
def rh(self, i, j):
return (self.has[j] - self.has[i]*self.power[j-i])%self.mod
MOD = 10**9+7
S = list(map(ord, readline().strip()))
N = len(S)
if len(set(S)) == 1:
print(N)
print(1)
else:
Rs = Rollinhash(S)
tabler = [True]*(N+1)
for d in range(2, 1+N//2):
r = Rs.rh(0, d)
for i in range(1, N//d):
if r != Rs.rh(i*d, (i+1)*d):
break
tabler[(i+1)*d] = False
tablel = [True]*(N+1)
for d in range(2, 1+N//2):
r = Rs.rh(N-d, N)
for i in range(1, N//d):
if r != Rs.rh(N-(i+1)*d, N-i*d):
break
tablel[N-(i+1)*d] = False
if tabler[N]:
print(1)
print(1)
else:
print(2)
ans = 0
for i in range(N+1):
if tabler[i] and tablel[i]:
ans += 1
assert ans > 0, ''
print(ans)
```
|
instruction
| 0
| 98,509
| 0
| 197,018
|
No
|
output
| 1
| 98,509
| 0
| 197,019
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
W = read().rstrip()
N = len(W)
def Z_algorithm(S):
# 共通接頭辞の長さを返す
N=len(S)
arr = [0]*N
arr[0] = N
i,j = 1,0
while i<N:
while i+j<N and S[j]==S[i+j]:
j += 1
arr[i]=j
if not j:
i += 1
continue
k = 1
while i+k<N and k+arr[k]<j:
arr[i+k] = arr[k]
k += 1
i += k; j -= k
return arr
def is_periodic_left(W):
Z = Z_algorithm(W)
is_periodic = [False] * N
for p in range(1,N//2 + 10):
for i in range(p,N,p):
if Z[i] >= p:
is_periodic[p + i - 1] = True
else:
break
return is_periodic
L = is_periodic_left(W)
R = is_periodic_left(W[::-1])[::-1]
if not L[-1]:
answer = (1,1)
elif len(set(W)) == 1:
answer = (N,1)
else:
x = sum(not(x or y) for x,y in zip(L,R[1:]))
answer = (2,x)
print('\n'.join(map(str,answer)))
```
|
instruction
| 0
| 98,510
| 0
| 197,020
|
No
|
output
| 1
| 98,510
| 0
| 197,021
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
W = list(map(int,read().rstrip()))
N = len(W)
def Z_Algorithm(S):
# 共通接頭辞の長さを返す
N=len(S)
arr = [0]*N
arr[0] = N
i,j = 1,0
while i<N:
while i+j<N and S[j]==S[i+j]:
j += 1
arr[i]=j
if not j:
i += 1
continue
k = 1
while i+k<N and k+arr[k]<j:
arr[i+k] = arr[k]
k += 1
i += k; j -= k
return arr
def find_good_substr(W):
Z = Z_Algorithm(W)
is_good = [True]*N
for n in range(1,N//2+1):
# n文字での周期をもってくる
if not is_good[n]:
continue
for i in range(n+n-1,Z[n]+n,n):
is_good[i] = False
return is_good
def solve():
G1 = find_good_substr(W)
G2 = find_good_substr(W[::-1])
if G1[-1]:
# 自分自身がoptimal
return 1,1
x,y = solve()
print(x)
print(y)
```
|
instruction
| 0
| 98,511
| 0
| 197,022
|
No
|
output
| 1
| 98,511
| 0
| 197,023
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let x be a string of length at least 1. We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x. For example, `a`, `bbc` and `cdcdc` are good strings, while `aa`, `bbbb` and `cdcdcd` are not.
Let w be a string of length at least 1. For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements, we will call F a good representation of w, if the following conditions are both satisfied:
* For any i \, (1 \leq i \leq m), f_i is a good string.
* The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.
For example, when w=`aabb`, there are five good representations of w:
* (`aabb`)
* (`a`,`abb`)
* (`aab`,`b`)
* (`a`,`ab`,`b`)
* (`a`,`a`,`b`,`b`)
Among the good representations of w, the ones with the smallest number of elements are called the best representations of w. For example, there are only one best representation of w=`aabb`: (`aabb`).
You are given a string w. Find the following:
* the number of elements of a best representation of w
* the number of the best representations of w, modulo 1000000007 \, (=10^9+7)
(It is guaranteed that a good representation of w always exists.)
Constraints
* 1 \leq |w| \leq 500000 \, (=5 \times 10^5)
* w consists of lowercase letters (`a`-`z`).
Input
The input is given from Standard Input in the following format:
w
Output
Print 2 lines.
* In the first line, print the number of elements of a best representation of w.
* In the second line, print the number of the best representations of w, modulo 1000000007.
Examples
Input
aab
Output
1
1
Input
bcbc
Output
2
3
Input
ddd
Output
3
1
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
W = list(map(int,read().rstrip()))
N = len(W)
def Z_Algorithm(S):
# 共通接頭辞の長さを返す
N=len(S)
arr = [0]*N
arr[0] = N
i,j = 1,0
while i<N:
while i+j<N and S[j]==S[i+j]:
j += 1
arr[i]=j
if not j:
i += 1
continue
k = 1
while i+k<N and k+arr[k]<j:
arr[i+k] = arr[k]
k += 1
i += k; j -= k
return arr
Z_Algorithm(W)
```
|
instruction
| 0
| 98,512
| 0
| 197,024
|
No
|
output
| 1
| 98,512
| 0
| 197,025
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
|
instruction
| 0
| 98,604
| 0
| 197,208
|
Tags: implementation, strings
Correct Solution:
```
from sys import stdin
trim = lambda s: s[:-1] if s[-1] == "\n" else s
T = int(stdin.readline())
for i in range(0, T):
n = int(stdin.readline())
s = trim(stdin.readline())
possible = True
for j in range(0, n//2 + 1):
if abs(ord(s[j])-ord(s[-(j+1)])) not in [0, 2]:
possible = False
if possible:
print("YES")
else:
print("NO")
```
|
output
| 1
| 98,604
| 0
| 197,209
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
|
instruction
| 0
| 98,605
| 0
| 197,210
|
Tags: implementation, strings
Correct Solution:
```
import sys
import os
def solve(s):
n = len(s)
for i in range(n // 2):
a = ord(s[i])
b = ord(s[n - 1 - i])
diff = abs(a - b)
if diff != 0 and abs(diff) != 2:
return 'NO'
return 'YES'
def main():
t = int(input())
for i in range(t):
_ = int(input())
s = input()
print(solve(s))
if __name__ == '__main__':
main()
```
|
output
| 1
| 98,605
| 0
| 197,211
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
|
instruction
| 0
| 98,606
| 0
| 197,212
|
Tags: implementation, strings
Correct Solution:
```
def is_pal(s):
for i in range(len(s)//2):
dif = abs(ord(s[i]) - ord(s[-i-1]))
if dif != 0 and dif != 2 :
return False
return True
t = int(input())
for i in range(t):
n = int(input())
s = input()
buf = []
print("YES" if is_pal(s) else "NO")
```
|
output
| 1
| 98,606
| 0
| 197,213
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
|
instruction
| 0
| 98,607
| 0
| 197,214
|
Tags: implementation, strings
Correct Solution:
```
t = int(input())
while (t):
n = int(input())
s = input()
ok = 0
for i in range (n):
v = abs(ord(s[i]) - ord(s[n-i-1]))
if (v != 0 and v != 2):
ok = 1
if (ok):print("NO")
else: print("YES")
t -= 1
```
|
output
| 1
| 98,607
| 0
| 197,215
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
|
instruction
| 0
| 98,608
| 0
| 197,216
|
Tags: implementation, strings
Correct Solution:
```
t = int(input())
alphabet = "abcdefghijklmnopqrstuvwxyz"
for i in range(t):
n = int(input())
string = input()
palindrome = True
i = 0
j = len(string) - 1
while(i <= j):
if(string[i] != string[j]):
esq = alphabet.index(string[i])
dir = alphabet.index(string[j])
if(esq + 1 != dir -1 and esq + 1 != dir + 1 and esq - 1 != dir - 1 and esq -1 != dir + 1):
palindrome = False
i+= 1
j-= 1
if(palindrome):
print("YES")
else:
print("NO")
```
|
output
| 1
| 98,608
| 0
| 197,217
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
|
instruction
| 0
| 98,609
| 0
| 197,218
|
Tags: implementation, strings
Correct Solution:
```
import sys
import math
from collections import deque
def scan():
return list(map(int, sys.stdin.readline().strip().split()))
def print_primes_till_n(n):
i, j, flag = 0, 0, 0
a = []
c = 0
for i in range(1, n + 1, 1):
if i == 1 or i == 0:
continue
flag = 1
for j in range(2, ((i // 2) + 1), 1):
if i % j == 0:
flag = 0
break
if flag == 1:
a.append(i)
c += 1
return a, c
def is_square(n):
b = math.sqrt(n)
if n == int(b * b):
return True
return False
def solution():
for _ in range(int(input())):
n = int(input())
s = input()
s = [i for i in s]
c = 0
for i in range(n // 2):
if abs(ord(s[n-i-1])-ord(s[i])) == 2 or abs(ord(s[n-i-1])-ord(s[i])) == 0:
c += 1
else:
c = -1
break
if c == n//2:
print('YES')
else:
print('NO')
if __name__ == '__main__':
solution()
```
|
output
| 1
| 98,609
| 0
| 197,219
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
|
instruction
| 0
| 98,610
| 0
| 197,220
|
Tags: implementation, strings
Correct Solution:
```
def can_be_transformed(word):
for i in range(len(word)//2):
if abs(ord(word[i])-ord(word[-i-1]))>2 or abs(ord(word[i])-ord(word[-i-1])) == 1:
return False
return True
def main():
answers = []
n=int(input())
for i in range(0,n):
m=int(input())
s=input()
if can_be_transformed(s):
answers.append('YES')
else:
answers.append('NO')
for i in range(0,n):
print(answers[i])
if __name__ == '__main__':
main()
```
|
output
| 1
| 98,610
| 0
| 197,221
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
|
instruction
| 0
| 98,611
| 0
| 197,222
|
Tags: implementation, strings
Correct Solution:
```
def search(s,n):
for i in range(n//2):
x=abs(ord(s[i])-ord(s[n-i-1]))
if x!=0 and x!=2:
return(False)
return(True)
t=int(input())
for _ in range(t):
n=int(input())
s=input()
if search(s,n):
print("YES")
else:
print("NO")
```
|
output
| 1
| 98,611
| 0
| 197,223
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
t = int(input())
def get_str(s):
next = ''
prev = ''
if s == 'a':
next = chr(ord(s) + 1)
elif s == 'z':
prev = chr(ord(s) - 1)
else:
next = chr(ord(s) + 1)
prev = chr(ord(s) - 1)
return next + prev
def canPalindrome(n, s):
options = []
for i in range(n // 2):
a = get_str(s[i])
b = get_str(s[(-1 * i) -1])
l = len(options) // 2
options.insert(l,a)
options.insert(l + 1 ,b)
for i in range(n // 2):
if not len(''.join(set(options[i]).intersection(set(options[(-1 * i) - 1])))) > 0:
return False
return True
res = []
for i in range(t):
n = int(input())
s = input()
res.append(canPalindrome(n, s))
for i in res:
if i == True:
print('YES')
else:
print('NO')
```
|
instruction
| 0
| 98,612
| 0
| 197,224
|
Yes
|
output
| 1
| 98,612
| 0
| 197,225
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
import math
from sys import stdin
n_round = int(stdin.readline())
for _ in range(n_round):
n_letter = int(stdin.readline())
string = list(stdin.readline().strip())
numbers = [ord(x) for x in string]
#a-97 z-122
result = True
for _ in range(math.floor(n_letter/2)):
if not (numbers[_] == numbers[-(_+1)] or abs(numbers[_]-numbers[-(_+1)])==2):
result = False
if result == True:
print('YES')
else:
print('NO')
```
|
instruction
| 0
| 98,613
| 0
| 197,226
|
Yes
|
output
| 1
| 98,613
| 0
| 197,227
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
def canbe(c, s):
for i in range(int(c/2), c):
if (abs(ord(s[i]) - ord(s[c-i-1]))>2) or (abs(ord(s[i]) - ord(s[c-i-1]))==1):
return False
return True
t = int(input())
buf = []
for i in range(t):
c = int(input())
s = input()
if canbe(c, s):
buf.append('YES')
else:
buf.append('NO')
for el in buf:
print(el)
```
|
instruction
| 0
| 98,614
| 0
| 197,228
|
Yes
|
output
| 1
| 98,614
| 0
| 197,229
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
n = int(input())
tt = [-2, 0, 2]
for jj in range(n):
i, inp = input(), input()
need = 1
for j in range(len(inp)):
if (ord(inp[j]) - ord(inp[len(inp) - j - 1])) not in tt:
need = 0
if need:
print("YES")
else:
print("NO")
```
|
instruction
| 0
| 98,615
| 0
| 197,230
|
Yes
|
output
| 1
| 98,615
| 0
| 197,231
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
from __future__ import division, print_function
from collections import *
from math import *
from itertools import *
from time import time
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
'''
Notes:
n = number of balls
k = attract power
ni = ball's coords
'''
def main():
n = int(input())
s = input()
if s == s[::-1]:
print('YES')
return
opposite = sorted(s, reverse = True)
for i in range(n // 2):
first = ord(s[i])
second = ord(opposite[i])
if first - 1 == second or first == second or first + 1 == second:
continue
else:
print('NO')
return
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
t = int(input())
while(t):
main()
t -= 1
```
|
instruction
| 0
| 98,616
| 0
| 197,232
|
No
|
output
| 1
| 98,616
| 0
| 197,233
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
T=int(input())
for i in range(T):
n=int(input())
s=input()
l=len(s)-1
for i in range(len(s)):
if s[i]==s[l]:
l=l-1
if i==(len(s)-1):
print ("YES")
break
continue
if abs((ord(s[i])-ord(s[l])))!=2 and (s[i]!='a' or s[i]!='z') and (s[l]!='a' or s[l]!='z'):
print ("NO")
break
if (s[i]=='a' and (s[l] in ['c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'])):
print ("NO")
break
if (s[l]=='a' and (s[i] in ['c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'])):
print ("NO")
break
if (s[i]=='z' and (s[l] in ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x'])):
print ("NO")
break
if (s[l]=='z' and (s[i] in ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x'])):
print ("N0")
break
else:
if i==(len(s)-1):
print ("YES")
break
l=l-1
```
|
instruction
| 0
| 98,617
| 0
| 197,234
|
No
|
output
| 1
| 98,617
| 0
| 197,235
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
i=input
t=int(i())
while t:
t-=1
i();a=[*map(ord,i())]
print("YNEOS"[1-all((x-y)%26in{0,2,24}for x,y in zip(a,a[::-1]))::2])
```
|
instruction
| 0
| 98,618
| 0
| 197,236
|
No
|
output
| 1
| 98,618
| 0
| 197,237
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
def check(n, s):
for i in range(n//2):
c1, c2 = ord(s[i]), ord(s[n-i-1])
if abs(c1 - c2) > 2:
return False
return True
T = int(input())
for _ in range(T):
n = int(input())
s = input()
if check(n, s):
print("YES")
else:
print("NO")
```
|
instruction
| 0
| 98,619
| 0
| 197,238
|
No
|
output
| 1
| 98,619
| 0
| 197,239
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
|
instruction
| 0
| 98,661
| 0
| 197,322
|
Tags: constructive algorithms, math, strings
Correct Solution:
```
import sys
from math import *
from fractions import gcd
readints=lambda:map(int, input().strip('\n').split())
def hasUnique(s,n,k):
freq = {}
for i in range(n-k+1):
x = s[i:i+k]
if x not in freq: freq[x]=0
freq[x]+=1
for k in freq:
if freq[k] == 1:
return True
return False
def good(s,n,k):
for i in range(1,k):
if hasUnique(s,n,i):
return False
freq={}
for i in range(n-k+1):
x = s[i:i+k]
if x not in freq: freq[x]=0
freq[x]+=1
st = set()
for x in freq:
if freq[x]==1:
if len(st)>0: return False
st.add(x)
return len(st) == 1
def gen(i,s,n,k):
if i==n:
if good(s,n,k):
print(s)
else:
gen(i+1,s+'0',n,k)
gen(i+1,s+'1',n,k)
#gen(0,'',5,3)
# for n in range(3,9):
# for k in range(1,n+1):
# if (n%2 != k%2):
# print(n,k,'skip')
# print()
# continue
# print(n,k)
# gen(0,'',n,k)
# print()
n,k = readints()
if n == k:
print('0' * n)
sys.exit(0)
s = ''
while len(s) < n:
r = int((n-k)//2)
s = s + (('0'*r) + '1')
s = s[:n]
print(s)
```
|
output
| 1
| 98,661
| 0
| 197,323
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
|
instruction
| 0
| 98,662
| 0
| 197,324
|
Tags: constructive algorithms, math, strings
Correct Solution:
```
n, k = map(int, input().split())
if k == 1:
print("1", end = "")
for i in range(1, n):
print("0", end = "")
print()
exit()
a = (n + k - 2) // 2
ans = ""
if (a - k + 2) % 2:
for i in range (a - k + 2):
ans += chr(i % 2 + 48)
else:
ans += chr(48)
for i in range(1, a - k + 2):
ans += chr(49)
for i in range(a - k + 2, n):
ans += ans[i - a + k - 2]
print(ans)
```
|
output
| 1
| 98,662
| 0
| 197,325
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
|
instruction
| 0
| 98,663
| 0
| 197,326
|
Tags: constructive algorithms, math, strings
Correct Solution:
```
N, K = map(int, input().split())
if K == 0 or N == K:
print("1" * N)
elif K == 1:
print("0" + "1" * (N - 1))
elif K <= 2//N:
if K & 1:
print("10" * (K//2+1) + "1" * (N-2-K//2*2))
else:
print("0" + "10" * (K//2) + "1" * (N-1-K//2*2))
else:
print((("0"+"1"*((N-K)//2))*(N//(((N-K)//2))+1))[:N])
```
|
output
| 1
| 98,663
| 0
| 197,327
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
|
instruction
| 0
| 98,664
| 0
| 197,328
|
Tags: constructive algorithms, math, strings
Correct Solution:
```
n, k = list(map(int,input().split()))
chuj_twojej_starej = (n - k) // 2 + 1
i = 1
while True:
if i % chuj_twojej_starej == 0:
print(0, end = "")
else:
print(1, end = "")
if i == n:
break
i += 1
```
|
output
| 1
| 98,664
| 0
| 197,329
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
|
instruction
| 0
| 98,665
| 0
| 197,330
|
Tags: constructive algorithms, math, strings
Correct Solution:
```
n,k=map(int,input().split())
d=(n-k)//2
s=0
while s!=n:
if (s+1)%(d+1)==0:
print("1",end="")
else :
print("0",end="")
s+=1
```
|
output
| 1
| 98,665
| 0
| 197,331
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
|
instruction
| 0
| 98,666
| 0
| 197,332
|
Tags: constructive algorithms, math, strings
Correct Solution:
```
N, K = map(int, input().split())
if N == K:
print("0"*N)
elif K == 1:
print("0"*(N-1) + "1")
elif K == 3:
print("1" + "0"*(N-4) + "101")
else:
res = ["0"]*N
for i in range(0, N, N//2-K//2+1):
res[i] = "1"
print(''.join(res))
```
|
output
| 1
| 98,666
| 0
| 197,333
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
|
instruction
| 0
| 98,667
| 0
| 197,334
|
Tags: constructive algorithms, math, strings
Correct Solution:
```
n,k=map(int,input().split())
x=(n-(k-1)+1)//2
STR="0"*(x-1)+"1"
ANS=STR*(n//x+1)
print(ANS[:n])
```
|
output
| 1
| 98,667
| 0
| 197,335
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
|
instruction
| 0
| 98,668
| 0
| 197,336
|
Tags: constructive algorithms, math, strings
Correct Solution:
```
#------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
import math
n,k=map(int,input().split())
a=(n-k)//2
s=""
while(len(s)<n):
s=s+"0"*(a)+"1"
print(s[:n])
```
|
output
| 1
| 98,668
| 0
| 197,337
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
n,k=[int(x) for x in input().split()]
a=(n-k)//2
tot=''
for i in range(n):
if (i+1)%(a+1)==0:
tot+='1'
else:
tot+='0'
print(tot)
```
|
instruction
| 0
| 98,669
| 0
| 197,338
|
Yes
|
output
| 1
| 98,669
| 0
| 197,339
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
from sys import stdin
n,m=map(int,stdin.readline().strip().split())
x=n-m
x=x//2
s=""
y=0
if m==1:
print("0"+(n-1)*"1")
exit(0)
while len(s)<n:
if y==1:
s+="1"*x
else:
s+="0"
y+=1
y=y%2
print(s[0:n])
```
|
instruction
| 0
| 98,670
| 0
| 197,340
|
Yes
|
output
| 1
| 98,670
| 0
| 197,341
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
n, k = map(int, input().split())
# while making the string, k = (n - 2 * length + 2)
length = (n - k + 2) // 2 # length of a cycle
string = "0" * (length - 1) + "1" # make the cycle
answer = string * (n // length + 1) # make the string with length >= n
print(answer[ : n])
```
|
instruction
| 0
| 98,671
| 0
| 197,342
|
Yes
|
output
| 1
| 98,671
| 0
| 197,343
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
n, k = map(int, input().split())
s, v = (n - k) // 2 * '0' + '1', ''
while len(v) < n:
v += s
print(v[:n])
```
|
instruction
| 0
| 98,672
| 0
| 197,344
|
Yes
|
output
| 1
| 98,672
| 0
| 197,345
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
n,k=map(int,input().strip().split())
d=n-k//2+1
x=['1' if (i+1)%d==0 else '0' for i in range(n)]
print(''.join(x))
```
|
instruction
| 0
| 98,673
| 0
| 197,346
|
No
|
output
| 1
| 98,673
| 0
| 197,347
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
def solve(n, k):
if n // 3 >= k:
string1 = ["0"] + ["1"] * (k-2)
string2 = ["1"] * (n + 2 - 2*k)
answer = (string2 + string1 + string1)
return ("".join(answer))
else:
key = (n - k) // 2
a, c = key + 1, key - 1
b = n // a
string1 = ["1"] * (a-1) + ["0"]
string2 = ["1"] * (n - b*a)
answer = (string1 * b + string2)
return ("".join(answer))
return
n, k = map(int, input().split())
print (solve(n, k))
```
|
instruction
| 0
| 98,674
| 0
| 197,348
|
No
|
output
| 1
| 98,674
| 0
| 197,349
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
def solve(n, k):
if n // 2 >= k:
string1 = ["0"] + ["1"] * (k-2)
string2 = ["1"] * (n + 2 - 2*k)
answer = (string2 + string1 + string1)
return ("".join(answer))
else:
key = (n - k) // 2
a, c = key + 1, key - 1
b = n // a
string1 = ["1"] * (a-1) + ["0"]
string2 = ["1"] * (n - b*a)
answer = (string1 * b + string2)
return ("".join(answer))
return
n, k = map(int, input().split())
print (solve(n, k))
```
|
instruction
| 0
| 98,675
| 0
| 197,350
|
No
|
output
| 1
| 98,675
| 0
| 197,351
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
if __name__ == "__main__":
N, K = [int(x) for x in input().split()]
if K * 2 <= N:
print('0' * K + '1' * (N - K))
elif N == K:
print('0' * N)
elif N % 2 == 0 and K == N // 2 + 2:
print('01' * (N // 2))
elif N % 2 == 1 and K == N // 2 + 1:
print('01' * (N // 2) + '0')
else:
print('orzJumpmelonAKCTS2019')
```
|
instruction
| 0
| 98,676
| 0
| 197,352
|
No
|
output
| 1
| 98,676
| 0
| 197,353
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
|
instruction
| 0
| 98,808
| 0
| 197,616
|
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
def fn(s,a):
if len(s)==1:
if s==a:
return 0
else:
return 1
a1=0
a2=0
l=len(s)//2
t=s[:l]
p=s[l:]
for i in t:
if i!=a:
a1+=1
for i in p:
if i!=a:
a2+=1
return min(a1+fn(p,chr(ord(a)+1)),a2+fn(t,chr(ord(a)+1)))
for _ in range(int(input())):
n=int(input())
a=input()
print(fn(a,'a'))
```
|
output
| 1
| 98,808
| 0
| 197,617
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
|
instruction
| 0
| 98,809
| 0
| 197,618
|
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
LETRAS = "abcdefghijklmnopqrstuvwxyz"
def string_boa(indice, string):
letras = LETRAS[indice]
if len(string) == 1:
if string != letras:
return 1
else:
return 0
direita = 0
esquerda = 0
for l in string[:len(string)//2]:
if l != letras:
direita += 1
for l in string[len(string)//2:]:
if l != letras:
esquerda += 1
lado_direito = direita + string_boa(indice + 1, string[len(string)//2:])
lado_esquerdo = esquerda + string_boa(indice + 1, string[:len(string)//2:])
return min(lado_direito , lado_esquerdo)
entrada = int(input())
for i in range(entrada):
n = int(input())
string = input()
print(string_boa(0, string))
```
|
output
| 1
| 98,809
| 0
| 197,619
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
|
instruction
| 0
| 98,810
| 0
| 197,620
|
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
"""
Author: Q.E.D
Time: 2020-07-17 10:11:52
"""
def count(s, c):
c2 = chr(ord(c) + 1)
if len(s) == 1:
return 1 - int(s[0] == c)
else:
n = len(s)
h = n // 2
x1 = sum(1 for a in s[:h] if a != c)
x2 = sum(1 for a in s[h:] if a != c)
y1 = count(s[h:], c2)
y2 = count(s[:h], c2)
return min(x1 + y1, x2 + y2)
T = int(input())
for _ in range(T):
n = int(input())
s = input()
ans = count(s, 'a')
print(ans)
```
|
output
| 1
| 98,810
| 0
| 197,621
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
|
instruction
| 0
| 98,811
| 0
| 197,622
|
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
import sys
from math import log
def num(l, r, d):
mid = (l+r)//2
if d>=k:
return 0 if list_s[l] == n2a[d] else 1
count_l = len([0 for i in range(l, mid+1) if list_s[i] != n2a[d]])
count_r = len([0 for i in range(mid+1, r+1) if list_s[i] != n2a[d]])
return min(num(l, mid, d+1)+count_r, num(mid+1, r, d+1)+count_l)
a2n = {chr(97+i):i for i in range(26)}
n2a = {i:chr(97+i) for i in range(26)}
T=int(sys.stdin.readline())
for _ in range(T):
n = int(sys.stdin.readline()) # 0<=k<=17
list_s = sys.stdin.readline().strip()
k = int(log(n, 2))
print(num(0, n-1, 0))
```
|
output
| 1
| 98,811
| 0
| 197,623
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
|
instruction
| 0
| 98,812
| 0
| 197,624
|
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
import sys
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s); sys.stdout.write('\n')
def wi(n): sys.stdout.write(str(n)); sys.stdout.write('\n')
def wia(a, sep=' '): sys.stdout.write(sep.join([str(x) for x in a])); sys.stdout.write('\n')
def solve(n, s, c):
if n == 1:
return 0 if s[0] == c else 1
mid = n // 2
c1 = s[:mid].count(c)
c2 = s[mid:].count(c)
return min(mid - c1 + solve(mid, s[mid:], chr(ord(c) + 1)), mid - c2 + solve(mid, s[:mid], chr(ord(c) + 1)))
def main():
for _ in range(ri()):
n = ri()
s = rs()
wi(solve(n, s, 'a'))
if __name__ == '__main__':
main()
```
|
output
| 1
| 98,812
| 0
| 197,625
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
|
instruction
| 0
| 98,813
| 0
| 197,626
|
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
import sys, threading
sys.setrecursionlimit(10**8)
threading.stack_size(10**8) # new thread will get stack of such size
def main():
def findMinCost(s,targetChar):
if len(s)==1:
return int(s[0]!=targetChar)
mid=len(s)//2
leftCost=sum(c!=targetChar for c in s[:mid])+findMinCost(s[mid:],chr(ord(targetChar)+1))
rightCost=sum(c!=targetChar for c in s[mid:])+findMinCost(s[:mid],chr(ord(targetChar)+1))
return min(leftCost,rightCost)
t=int(input())
for _ in range(t):
_,s=int(input()),input()
print(findMinCost(s,"a"))
threading.Thread(target=main).start()
```
|
output
| 1
| 98,813
| 0
| 197,627
|
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