message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains two integers N, M (1 โค N โค 1024, 2 โค M โค 16), separated by a single space.
Output
Output "YES" or "NO".
Examples
Input
2 3
Output
YES
Input
3 2
Output
NO
Input
33 16
Output
YES
Input
26 5
Output
NO
Submitted Solution:
```
import random
a=input()
if a=="2 3":
print("YES")
elif a=="3 2":
print("NO")
elif a=="33 16":
print("YES")
elif a=="26 5":
print("NO")
else:
if random.randint(0,1)==1:
print("YES")
else:
print("NO")
#a1
``` | instruction | 0 | 15,318 | 5 | 30,636 |
No | output | 1 | 15,318 | 5 | 30,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains two integers N, M (1 โค N โค 1024, 2 โค M โค 16), separated by a single space.
Output
Output "YES" or "NO".
Examples
Input
2 3
Output
YES
Input
3 2
Output
NO
Input
33 16
Output
YES
Input
26 5
Output
NO
Submitted Solution:
```
import math
n,m=list(map(int,input().split()))
x=math.log2(n)
y=m
x=int(x)
if y<3*x:
print("NO")
else:
print("YES")
``` | instruction | 0 | 15,319 | 5 | 30,638 |
No | output | 1 | 15,319 | 5 | 30,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains two integers N, M (1 โค N โค 1024, 2 โค M โค 16), separated by a single space.
Output
Output "YES" or "NO".
Examples
Input
2 3
Output
YES
Input
3 2
Output
NO
Input
33 16
Output
YES
Input
26 5
Output
NO
Submitted Solution:
```
import random
a=input()
if a=="2 3":
print("YES")
elif a=="3 2":
print("NO")
elif a=="33 16":
print("YES")
elif a=="26 5":
print("NO")
else:
if random.randint(0,1)==1:
print("YES")
else:
print("NO")
#a19
``` | instruction | 0 | 15,320 | 5 | 30,640 |
No | output | 1 | 15,320 | 5 | 30,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For an array b of length m we define the function f as
f(b) = \begin{cases} b[1] & if m = 1 \\\ f(b[1] โ b[2],b[2] โ b[3],...,b[m-1] โ b[m]) & otherwise, \end{cases}
where โ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
For example, f(1,2,4,8)=f(1โ2,2โ4,4โ8)=f(3,6,12)=f(3โ6,6โ12)=f(5,10)=f(5โ10)=f(15)=15
You are given an array a and a few queries. Each query is represented as two integers l and r. The answer is the maximum value of f on all continuous subsegments of the array a_l, a_{l+1}, โฆ, a_r.
Input
The first line contains a single integer n (1 โค n โค 5000) โ the length of a.
The second line contains n integers a_1, a_2, ..., a_n (0 โค a_i โค 2^{30}-1) โ the elements of the array.
The third line contains a single integer q (1 โค q โค 100 000) โ the number of queries.
Each of the next q lines contains a query represented as two integers l, r (1 โค l โค r โค n).
Output
Print q lines โ the answers for the queries.
Examples
Input
3
8 4 1
2
2 3
1 2
Output
5
12
Input
6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2
Output
60
30
12
3
Note
In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.
In second sample, optimal segment for first query are [3,6], for second query โ [2,5], for third โ [3,4], for fourth โ [1,2].
Submitted Solution:
```
n = int(input())
values = list(map(int, input().split()))
queries = int(input())
dp = [[0] * 5009 for i in range(5009)]
for i in range(n):
dp[0][i] = values[i]
for i in range(1, n): # 0 is already populated
for j in range(n-i+1):
top = dp[i-1][j]
right = dp[i-1][j+1]
dp[i][j] = top ^ right
for i in range(1, n):
for j in range(n-i+1):
top = dp[i-1][j]
right = dp[i-1][j+1]
dp[i][j] = max(right, max(dp[i][j], top))
for i in range(queries):
left, right = map(int, input().split())
last_row = (right - 1) - (left - 1)
last_column = (left - 1)
print(dp[last_row][last_column])
``` | instruction | 0 | 15,691 | 5 | 31,382 |
Yes | output | 1 | 15,691 | 5 | 31,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For an array b of length m we define the function f as
f(b) = \begin{cases} b[1] & if m = 1 \\\ f(b[1] โ b[2],b[2] โ b[3],...,b[m-1] โ b[m]) & otherwise, \end{cases}
where โ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
For example, f(1,2,4,8)=f(1โ2,2โ4,4โ8)=f(3,6,12)=f(3โ6,6โ12)=f(5,10)=f(5โ10)=f(15)=15
You are given an array a and a few queries. Each query is represented as two integers l and r. The answer is the maximum value of f on all continuous subsegments of the array a_l, a_{l+1}, โฆ, a_r.
Input
The first line contains a single integer n (1 โค n โค 5000) โ the length of a.
The second line contains n integers a_1, a_2, ..., a_n (0 โค a_i โค 2^{30}-1) โ the elements of the array.
The third line contains a single integer q (1 โค q โค 100 000) โ the number of queries.
Each of the next q lines contains a query represented as two integers l, r (1 โค l โค r โค n).
Output
Print q lines โ the answers for the queries.
Examples
Input
3
8 4 1
2
2 3
1 2
Output
5
12
Input
6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2
Output
60
30
12
3
Note
In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.
In second sample, optimal segment for first query are [3,6], for second query โ [2,5], for third โ [3,4], for fourth โ [1,2].
Submitted Solution:
```
n = int(input())
arr = list(map(int, input().split()))
aux = [[0] * n for i in range(n)]
for i in range(n):
aux[0][i] = arr[i]
for i in range(1, n):
for j in range(n-i):
aux[i][j] = aux[i-1][j] ^ aux[i-1][j+1]
for i in range(1, n):
for j in range(n-i):
aux[i][j] = max(aux[i][j], aux[i-1][j], aux[i-1][j+1])
q = int(input())
for i in range(q):
buscaEsquerda, buscaDireita = map(int, input().split())
buscaEsquerda, buscaDireita = buscaEsquerda-1, buscaDireita-1
print(aux[buscaDireita-buscaEsquerda][buscaEsquerda])
``` | instruction | 0 | 15,692 | 5 | 31,384 |
Yes | output | 1 | 15,692 | 5 | 31,385 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For an array b of length m we define the function f as
f(b) = \begin{cases} b[1] & if m = 1 \\\ f(b[1] โ b[2],b[2] โ b[3],...,b[m-1] โ b[m]) & otherwise, \end{cases}
where โ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
For example, f(1,2,4,8)=f(1โ2,2โ4,4โ8)=f(3,6,12)=f(3โ6,6โ12)=f(5,10)=f(5โ10)=f(15)=15
You are given an array a and a few queries. Each query is represented as two integers l and r. The answer is the maximum value of f on all continuous subsegments of the array a_l, a_{l+1}, โฆ, a_r.
Input
The first line contains a single integer n (1 โค n โค 5000) โ the length of a.
The second line contains n integers a_1, a_2, ..., a_n (0 โค a_i โค 2^{30}-1) โ the elements of the array.
The third line contains a single integer q (1 โค q โค 100 000) โ the number of queries.
Each of the next q lines contains a query represented as two integers l, r (1 โค l โค r โค n).
Output
Print q lines โ the answers for the queries.
Examples
Input
3
8 4 1
2
2 3
1 2
Output
5
12
Input
6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2
Output
60
30
12
3
Note
In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.
In second sample, optimal segment for first query are [3,6], for second query โ [2,5], for third โ [3,4], for fourth โ [1,2].
Submitted Solution:
```
n = int(input())
matrix = [[0] * n for i in range(n)]
inpt = list(map(int, input().split()))
for i in range(n):
matrix[i][i] = inpt[i]
for i in range(n - 1, -1, -1):
for j in range(i + 1, n):
a = matrix[i][j - 1]
b = matrix[i + 1][j]
xor = a ^ b
matrix[i][j] = xor
for i in range(n - 1, -1, -1):
for j in range(i + 1, n):
a = matrix[i][j - 1]
b = matrix[i + 1][j]
matrix[i][j] = max(matrix[i][j], a, b)
m = int(input())
for i in range(m):
query = list(map(int, input().split()))
print(matrix[query[0] - 1][query[1] - 1])
``` | instruction | 0 | 15,693 | 5 | 31,386 |
Yes | output | 1 | 15,693 | 5 | 31,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For an array b of length m we define the function f as
f(b) = \begin{cases} b[1] & if m = 1 \\\ f(b[1] โ b[2],b[2] โ b[3],...,b[m-1] โ b[m]) & otherwise, \end{cases}
where โ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
For example, f(1,2,4,8)=f(1โ2,2โ4,4โ8)=f(3,6,12)=f(3โ6,6โ12)=f(5,10)=f(5โ10)=f(15)=15
You are given an array a and a few queries. Each query is represented as two integers l and r. The answer is the maximum value of f on all continuous subsegments of the array a_l, a_{l+1}, โฆ, a_r.
Input
The first line contains a single integer n (1 โค n โค 5000) โ the length of a.
The second line contains n integers a_1, a_2, ..., a_n (0 โค a_i โค 2^{30}-1) โ the elements of the array.
The third line contains a single integer q (1 โค q โค 100 000) โ the number of queries.
Each of the next q lines contains a query represented as two integers l, r (1 โค l โค r โค n).
Output
Print q lines โ the answers for the queries.
Examples
Input
3
8 4 1
2
2 3
1 2
Output
5
12
Input
6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2
Output
60
30
12
3
Note
In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.
In second sample, optimal segment for first query are [3,6], for second query โ [2,5], for third โ [3,4], for fourth โ [1,2].
Submitted Solution:
```
import bisect
import os, sys, atexit,threading
from io import BytesIO, StringIO
input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
_OUTPUT_BUFFER = StringIO()
sys.stdout = _OUTPUT_BUFFER
@atexit.register
def write():
sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
def calculate(array):
n = len(array)
finalarray = []
finalarray.append(array)
finalarray.append([])
while (n!=1):
for x in range(n-1):
finalarray[-1].append(finalarray[-2][x]^finalarray[-2][x+1])
finalarray.append([])
n-=1
return finalarray
def solve():
n = int(input())
array = [0]
array.extend(list(map(int,input().split())))
subArrays = []
for x in range(n+1):
subArrays.append([0]*(n+1))
for x in range(1,n):
value = array[x]
subArrays[1][x] = max(subArrays[1][x],value)
subArrays[x][x] = max(subArrays[x][x],value)
value = array[x]^array[x+1]
subArrays[1][x+1] = max(subArrays[1][x+1],value)
subArrays[x][x+1] = max(subArrays[x][x+1],value)
subArrays[1][n] = max(subArrays[1][n],array[n])
subArrays[n][n] = max(subArrays[n][n],array[n])
finalarray = calculate(array)
# print (finalarray,len(finalarray))
for x in range(1,n+1):
for y in range(x+2,n+1):
# print (y-x+1,x,finalarray[y-x+1])
value = finalarray[y-x][x]
subArrays[1][y] = max(subArrays[1][y],value)
subArrays[x][y] = max(subArrays[x][y],value)
# print (subArrays)
for x in range(1,n+1):
for y in range(2,n+1):
subArrays[x][y] = max(subArrays[x][y],subArrays[x][y-1])
# print (subArrays)
for y in range(1,n+1):
for x in range(n-1,0,-1):
subArrays[x][y] = max(subArrays[x][y],subArrays[x+1][y])
# print (subArrays)
q = int(input())
for _ in range(q):
l, r = map(int,input().split())
print (subArrays[l][r])
try:
solve()
except Exception as e:
print (e)
``` | instruction | 0 | 15,694 | 5 | 31,388 |
Yes | output | 1 | 15,694 | 5 | 31,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For an array b of length m we define the function f as
f(b) = \begin{cases} b[1] & if m = 1 \\\ f(b[1] โ b[2],b[2] โ b[3],...,b[m-1] โ b[m]) & otherwise, \end{cases}
where โ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
For example, f(1,2,4,8)=f(1โ2,2โ4,4โ8)=f(3,6,12)=f(3โ6,6โ12)=f(5,10)=f(5โ10)=f(15)=15
You are given an array a and a few queries. Each query is represented as two integers l and r. The answer is the maximum value of f on all continuous subsegments of the array a_l, a_{l+1}, โฆ, a_r.
Input
The first line contains a single integer n (1 โค n โค 5000) โ the length of a.
The second line contains n integers a_1, a_2, ..., a_n (0 โค a_i โค 2^{30}-1) โ the elements of the array.
The third line contains a single integer q (1 โค q โค 100 000) โ the number of queries.
Each of the next q lines contains a query represented as two integers l, r (1 โค l โค r โค n).
Output
Print q lines โ the answers for the queries.
Examples
Input
3
8 4 1
2
2 3
1 2
Output
5
12
Input
6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2
Output
60
30
12
3
Note
In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.
In second sample, optimal segment for first query are [3,6], for second query โ [2,5], for third โ [3,4], for fourth โ [1,2].
Submitted Solution:
```
read=lambda : map(int,input().split())
n=int(input())
a=list(read())
f=[[0]*5000 for _ in range(5000)]
for i in range(n):
f[i][i]=a[i]
for l in range(2,n+1):
for i in range(n):
j=i+l-1
if j>=n:
break
f[i][j]=max(f[i+1][j],max(f[i][j-1],f[i+1][j]^f[i][j-1]))
q=int(input())
while q>0:
q-=1
l,r=read()
l,r=l-1,r-1
print(f[l][r])
``` | instruction | 0 | 15,695 | 5 | 31,390 |
No | output | 1 | 15,695 | 5 | 31,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For an array b of length m we define the function f as
f(b) = \begin{cases} b[1] & if m = 1 \\\ f(b[1] โ b[2],b[2] โ b[3],...,b[m-1] โ b[m]) & otherwise, \end{cases}
where โ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
For example, f(1,2,4,8)=f(1โ2,2โ4,4โ8)=f(3,6,12)=f(3โ6,6โ12)=f(5,10)=f(5โ10)=f(15)=15
You are given an array a and a few queries. Each query is represented as two integers l and r. The answer is the maximum value of f on all continuous subsegments of the array a_l, a_{l+1}, โฆ, a_r.
Input
The first line contains a single integer n (1 โค n โค 5000) โ the length of a.
The second line contains n integers a_1, a_2, ..., a_n (0 โค a_i โค 2^{30}-1) โ the elements of the array.
The third line contains a single integer q (1 โค q โค 100 000) โ the number of queries.
Each of the next q lines contains a query represented as two integers l, r (1 โค l โค r โค n).
Output
Print q lines โ the answers for the queries.
Examples
Input
3
8 4 1
2
2 3
1 2
Output
5
12
Input
6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2
Output
60
30
12
3
Note
In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.
In second sample, optimal segment for first query are [3,6], for second query โ [2,5], for third โ [3,4], for fourth โ [1,2].
Submitted Solution:
```
import sys
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def main():
n = int(input())
A = list(map(int, input().split()))
dp = [[0]*(n+1) for i in range(n+1)]
for i in range(n):
dp[i][i+1] = A[i]
for d in range(2, n+1):
for i in range(n+1-d):
j = i + d
for k in range(i+1, j):
dp[i][j] = max(dp[i][j], max(dp[i][k], dp[k][j], dp[i][k]^dp[k][j]))
q = int(input())
for i in range(q):
l, r = map(int, input().split())
l, r = l-1, r-1
print(dp[l][r+1])
``` | instruction | 0 | 15,696 | 5 | 31,392 |
No | output | 1 | 15,696 | 5 | 31,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For an array b of length m we define the function f as
f(b) = \begin{cases} b[1] & if m = 1 \\\ f(b[1] โ b[2],b[2] โ b[3],...,b[m-1] โ b[m]) & otherwise, \end{cases}
where โ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
For example, f(1,2,4,8)=f(1โ2,2โ4,4โ8)=f(3,6,12)=f(3โ6,6โ12)=f(5,10)=f(5โ10)=f(15)=15
You are given an array a and a few queries. Each query is represented as two integers l and r. The answer is the maximum value of f on all continuous subsegments of the array a_l, a_{l+1}, โฆ, a_r.
Input
The first line contains a single integer n (1 โค n โค 5000) โ the length of a.
The second line contains n integers a_1, a_2, ..., a_n (0 โค a_i โค 2^{30}-1) โ the elements of the array.
The third line contains a single integer q (1 โค q โค 100 000) โ the number of queries.
Each of the next q lines contains a query represented as two integers l, r (1 โค l โค r โค n).
Output
Print q lines โ the answers for the queries.
Examples
Input
3
8 4 1
2
2 3
1 2
Output
5
12
Input
6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2
Output
60
30
12
3
Note
In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.
In second sample, optimal segment for first query are [3,6], for second query โ [2,5], for third โ [3,4], for fourth โ [1,2].
Submitted Solution:
```
import sys
import math
from collections import defaultdict
n=int(sys.stdin.readline())
arr=list(map(int,sys.stdin.readline().split()))
q=int(sys.stdin.readline())
pre=[0]
x,y,z=0,0,0
odd=[0]
even=[0]
for i in range(n):
if i%2==0:
y^=arr[i]
else:
z^=arr[i]
even.append(z)
odd.append(y)
x^=arr[i]
pre.append(x)
#print(pre,'pre')
#print(odd,'odd')
#print(even,'even')
l=[[0 for _ in range(n)] for x in range(n)]
for i in range(1,n+1):
for j in range(i,n+1):
if(j-i+1)%2==0:
a=(j-i+1)//2
if a%2==0 or a==1:
l[i-1][j-1]=pre[j]^pre[i-1]
else:
l[i-1][j-1]=arr[i-1]^arr[i]^arr[j-2]^arr[j-1]
else:
if ((j-i)//2)%2==0:
l[i-1][j-1]=arr[i-1]^arr[j-1]
continue
if i%2!=0:
l[i-1][j-1]=odd[j]^odd[i-1]
else:
l[i-1][j-1]=even[j]^even[i-1]
#print(l,'l')
dp=[[0 for _ in range(n+1)] for x in range(n+1)]
for i in range(n,0,-1):
for j in range(i,n+1):
dp[i][j]=l[i-1][j-1]
if i+1<n:
dp[i][j]=max(dp[i][j],dp[i+1][j])
if j-1>=0:
dp[i][j]=max(dp[i][j],dp[i][j-1])
#dp[i][j]=max(l[i-1][j-1],dp[i][j-1],dp[i+1][j])
#print(dp,'dp')
for _ in range(q):
i,j=map(int,sys.stdin.readline().split())
print(dp[i][j])
``` | instruction | 0 | 15,697 | 5 | 31,394 |
No | output | 1 | 15,697 | 5 | 31,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For an array b of length m we define the function f as
f(b) = \begin{cases} b[1] & if m = 1 \\\ f(b[1] โ b[2],b[2] โ b[3],...,b[m-1] โ b[m]) & otherwise, \end{cases}
where โ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
For example, f(1,2,4,8)=f(1โ2,2โ4,4โ8)=f(3,6,12)=f(3โ6,6โ12)=f(5,10)=f(5โ10)=f(15)=15
You are given an array a and a few queries. Each query is represented as two integers l and r. The answer is the maximum value of f on all continuous subsegments of the array a_l, a_{l+1}, โฆ, a_r.
Input
The first line contains a single integer n (1 โค n โค 5000) โ the length of a.
The second line contains n integers a_1, a_2, ..., a_n (0 โค a_i โค 2^{30}-1) โ the elements of the array.
The third line contains a single integer q (1 โค q โค 100 000) โ the number of queries.
Each of the next q lines contains a query represented as two integers l, r (1 โค l โค r โค n).
Output
Print q lines โ the answers for the queries.
Examples
Input
3
8 4 1
2
2 3
1 2
Output
5
12
Input
6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2
Output
60
30
12
3
Note
In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.
In second sample, optimal segment for first query are [3,6], for second query โ [2,5], for third โ [3,4], for fourth โ [1,2].
Submitted Solution:
```
def f(array):
newArray = []
for i in range(len(array) - 1):
currentNum = array[i]
nextNum = array[i +1]
newArray.append(currentNum ^ nextNum)
return newArray
def getMax(array, a ,b):
maxNum = array[a]
for i in range(a + 1, b + 1):
if( array[i] > maxNum):
maxNum = array[i]
return maxNum
arrayLenght = int(input())
textArray = input().split(' ')
array = []
for i in textArray:
array.append(int(i))
processed = [array]
while len(array) > 1:
print(array)
array = f(array)
processed.append(array)
print(array)
queriesNum = int(input())
for i in range(queriesNum):
query = input().split(' ')
queryA = int(query[0])
queryB = int(query[1])
querySize = queryB - queryA + 1
maxNum = getMax(processed[0], queryA - 1 , queryB - 1)
for i in range(1, querySize):
actualMax = getMax(processed[i], queryA - 1, queryB - i - 1)
if(actualMax > maxNum):
maxNum = actualMax
print(maxNum)
``` | instruction | 0 | 15,698 | 5 | 31,396 |
No | output | 1 | 15,698 | 5 | 31,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider placing N flags on a line. Flags are numbered through 1 to N.
Flag i can be placed on the coordinate X_i or Y_i. For any two different flags, the distance between them should be at least D.
Decide whether it is possible to place all N flags. If it is possible, print such a configulation.
Constraints
* 1 \leq N \leq 1000
* 0 \leq D \leq 10^9
* 0 \leq X_i < Y_i \leq 10^9
* All values in Input are integer.
Input
Input is given from Standard Input in the following format:
N D
X_1 Y_1
X_2 Y_2
\vdots
X_N Y_N
Output
Print `No` if it is impossible to place N flags.
If it is possible, print `Yes` first. After that, print N lines. i-th line of them should contain the coodinate of flag i.
Examples
Input
3 2
1 4
2 5
0 6
Output
Yes
4
2
0
Input
3 3
1 4
2 5
0 6
Output
No
Submitted Solution:
```
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")
n,d = list(map(int, input().split()))
xy = [tuple(map(int, input().split())) for _ in range(n)]
class SAT2:
def __init__(self, n):
self.n = n
self.ns = [[] for _ in range(2*n)]
def add_clause(self, i, f, j, g):
"""(xi==f) and (xj==g)
"""
n = self.n
u0,u1 = (i,i+n) if not f else (i+n,i)
v0,v1 = (j,j+n) if not g else (j+n,j)
self.ns[u1].append(v0)
self.ns[v1].append(u0)
def solve(self):
"""ๅผท้ฃ็ตๆๅๅ่งฃ
ใใใญใธใซใซใฝใผใ้ ใฎ้้ ใซ่ฟใ
"""
preorder = {}
lowlink = {}
seen = [False]*(2*self.n)
scc_queue = []
i = 0 # Preorder counter
# ans = [None]*(2*self.n)
count = 1
for source in range(n):
if seen[source]:
continue
queue = [source]
while queue:
v = queue[-1]
if v not in preorder:
i = i + 1
preorder[v] = i
done = True
for w in self.ns[v]:
if w not in preorder:
queue.append(w)
done = False
break
if done:
lowlink[v] = preorder[v]
for w in self.ns[v]:
if seen[w]:
continue
if preorder[w] > preorder[v]:
lowlink[v] = min([lowlink[v], lowlink[w]])
else:
lowlink[v] = min([lowlink[v], preorder[w]])
queue.pop()
if lowlink[v] == preorder[v]:
scc = {v}
while scc_queue and preorder[scc_queue[-1]] > preorder[v]:
k = scc_queue.pop()
scc.add(k)
for v in scc:
seen[v] = count
count += 1
else:
scc_queue.append(v)
ans = [None]*self.n
for i in range(self.n):
if seen[i]==seen[i+self.n]:
return None
elif seen[i]>seen[i+self.n]:
ans[i] = True
else:
ans[i] = False
return ans
solver = SAT2(n)
for i in range(n):
for j in range(i+1,n):
if abs(xy[i][0]-xy[j][0])<d:
solver.add_clause(i,False,j,False)
if abs(xy[i][0]-xy[j][1])<d:
solver.add_clause(i,False,j,True)
if abs(xy[i][1]-xy[j][0])<d:
solver.add_clause(i,True,j,False)
if abs(xy[i][1]-xy[j][1])<d:
solver.add_clause(i,True,j,True)
ans = solver.solve()
if ans is not None:
print("Yes")
write("\n".join(map(str, [xy[i][0 if ans[i] else 1] for i in range(n)])))
else:
print("No")
``` | instruction | 0 | 15,713 | 5 | 31,426 |
No | output | 1 | 15,713 | 5 | 31,427 |
Provide a correct Python 3 solution for this coding contest problem.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9 | instruction | 0 | 15,715 | 5 | 31,430 |
"Correct Solution:
```
A,B,N=map(int,input().split())
k = min(B-1,N)
print(int(A*k/B))
``` | output | 1 | 15,715 | 5 | 31,431 |
Provide a correct Python 3 solution for this coding contest problem.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9 | instruction | 0 | 15,716 | 5 | 31,432 |
"Correct Solution:
```
a, b, n = [int(x) for x in input().split()]
y = min(b-1, n)
print((a*y)//b)
``` | output | 1 | 15,716 | 5 | 31,433 |
Provide a correct Python 3 solution for this coding contest problem.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9 | instruction | 0 | 15,717 | 5 | 31,434 |
"Correct Solution:
```
a,b,n=map(int,input().split())
k=min(n,b-1)
print(int(a*k/b))
``` | output | 1 | 15,717 | 5 | 31,435 |
Provide a correct Python 3 solution for this coding contest problem.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9 | instruction | 0 | 15,718 | 5 | 31,436 |
"Correct Solution:
```
A, B, N = list(map(int, input().split()))
x = min(N, B-1)
ans = A*x//B
print(ans)
``` | output | 1 | 15,718 | 5 | 31,437 |
Provide a correct Python 3 solution for this coding contest problem.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9 | instruction | 0 | 15,719 | 5 | 31,438 |
"Correct Solution:
```
a,b,c=map(int,input().split())
x=min(b-1,c)
print((a*x)//b - a*(x//b))
``` | output | 1 | 15,719 | 5 | 31,439 |
Provide a correct Python 3 solution for this coding contest problem.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9 | instruction | 0 | 15,720 | 5 | 31,440 |
"Correct Solution:
```
A, B, N = map(int,input().split())
max = int(A*min(B-1,N)/B)
print(max)
``` | output | 1 | 15,720 | 5 | 31,441 |
Provide a correct Python 3 solution for this coding contest problem.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9 | instruction | 0 | 15,721 | 5 | 31,442 |
"Correct Solution:
```
a, b, n = map(int,input().split())
x = min(b -1, n)
ans = int(a * x/ b)
print(ans)
``` | output | 1 | 15,721 | 5 | 31,443 |
Provide a correct Python 3 solution for this coding contest problem.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9 | instruction | 0 | 15,722 | 5 | 31,444 |
"Correct Solution:
```
A, B, N = map(int,input().split())
m = min(B-1,N)
print(int(A*(m/B)))
``` | output | 1 | 15,722 | 5 | 31,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9
Submitted Solution:
```
a,b,n=map(int,input().split())
x=min(n,b-1)
print(int((a*x/b)//1))
``` | instruction | 0 | 15,723 | 5 | 31,446 |
Yes | output | 1 | 15,723 | 5 | 31,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9
Submitted Solution:
```
a,b,n=map(int,input().split())
print((a*min(b-1,n)//b))
``` | instruction | 0 | 15,724 | 5 | 31,448 |
Yes | output | 1 | 15,724 | 5 | 31,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9
Submitted Solution:
```
A,B,N = map(int,input().split())
n = min(B-1,N)
ans = (A*n)//B
print(ans)
``` | instruction | 0 | 15,725 | 5 | 31,450 |
Yes | output | 1 | 15,725 | 5 | 31,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9
Submitted Solution:
```
#abc165d
a,b,n=map(int,input().split())
c=min(b-1,n)
print(a*c//b)
``` | instruction | 0 | 15,726 | 5 | 31,452 |
Yes | output | 1 | 15,726 | 5 | 31,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9
Submitted Solution:
```
import math
a, b, n = map(int, input().split())
def f(x):
return (a * x) // b - a * (x // b)
left, right = 0, n
while right > left + 10:
mid1 = (right * 2 + left) / 3
mid2 = (right + left * 2) / 3
if f(mid1) <= f(mid2):
# ไธ้ใไธใใ๏ผๆๅฐๅคใใจใxใฏใใใกใใไธใใฎๆฐใ ใช๏ผ
right = mid2
else:
# ไธ้ใไธใใ๏ผๆๅฐๅคใใจใxใฏใใใกใใไธใใฎๆฐใ ใช๏ผ
left = mid1
listalt = []
for i in range(int(left), int(right) + 1):
listalt.append(f(i))
listalt.append(f(0))
listalt.append(f(n))
print(max(listalt))
``` | instruction | 0 | 15,727 | 5 | 31,454 |
No | output | 1 | 15,727 | 5 | 31,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9
Submitted Solution:
```
import math
A,B,N=map(int,input().split())
han=0
kot=0
i=0
if B-1>=N:
print(math.floor(A*N/B)-A*math.floor(N/B))
i=1
else:
for a in range(B-1,N,2*B):
if han<(math.floor(A*a/B)-A*math.floor(a/B)):
han=(math.floor(A*a/B)-A*math.floor(a/B))
kot=a
if i==0:
print(han)
``` | instruction | 0 | 15,728 | 5 | 31,456 |
No | output | 1 | 15,728 | 5 | 31,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9
Submitted Solution:
```
from math import floor
A,B,N = map(int,input().split())
ans = 0
if B <= N:
for x in range(B-1,N+1,B):
ans = max(ans, floor(A*x/B) - A*floor(x/B))
else:
ans = floor((A*N)/B) - A*floor(N/B)
print(ans)
``` | instruction | 0 | 15,729 | 5 | 31,458 |
No | output | 1 | 15,729 | 5 | 31,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers A, B, and N.
Find the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N.
Here floor(t) denotes the greatest integer not greater than the real number t.
Constraints
* 1 โค A โค 10^{6}
* 1 โค B โค 10^{12}
* 1 โค N โค 10^{12}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B N
Output
Print the maximum possible value of floor(Ax/B) - A ร floor(x/B) for a non-negative integer x not greater than N, as an integer.
Examples
Input
5 7 4
Output
2
Input
11 10 9
Output
9
Submitted Solution:
```
import math
A,B,N=map(int,input().split())
list=0
for i in range(1,N+1):
if i>B:
list=math.floor(A*i/B)
else:
c=(math.floor((A*i)/B))-(A*(math.floor(i/B)))
if list<c:
list=c
print(list)
``` | instruction | 0 | 15,730 | 5 | 31,460 |
No | output | 1 | 15,730 | 5 | 31,461 |
Provide a correct Python 3 solution for this coding contest problem.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0 | instruction | 0 | 15,751 | 5 | 31,502 |
"Correct Solution:
```
import sys,queue,math,copy,itertools,bisect,collections,heapq
def main():
MOD = 2**61-1
SI = lambda : sys.stdin.readline().rstrip()
s = SI()
t = SI()
lens = len(s)
lent = len(t)
def cx(x):
return ord(x) - ord('a') + 1
hsh = 0
for x in t:
hsh = (hsh * 26 + cx(x)) % MOD
n = (lens+lent) * 2 + 1
dp = [0] * n
h = 0
ans = 0
for i in range(n):
if i >= lent:
h -= cx(s[(i - lent)%lens]) * pow(26,lent-1,MOD)
h = (h * 26 + cx(s[i % lens])) % MOD
if h == hsh:
if i < lent:
dp[i] = 1
else:
dp[i] = dp[i-lent] + 1
ans = max(ans,dp[i])
if n - ans * lent < lent * 2:
print(-1)
else:
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 15,751 | 5 | 31,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
Submitted Solution:
```
from collections import defaultdict
#ๆณใจๅบๆฐ่จญๅฎ
mod1 = 10 ** 9 + 7
base1 = 1007
def getlist():
return list(map(int, input().split()))
class UnionFind:
def __init__(self, n):
self.par = [i for i in range(n + 1)]
self.rank = [0] * (n + 1)
self.size = [1] * (n + 1)
self.judge = "No"
def find(self, x):
if self.par[x] == x:
return x
else:
self.par[x] = self.find(self.par[x])
return self.par[x]
def same_check(self, x, y):
return self.find(x) == self.find(y)
def union(self, x, y):
x = self.find(x); y = self.find(y)
if self.rank[x] < self.rank[y]:
if self.same_check(x, y) != True:
self.size[y] += self.size[x]
self.size[x] = 0
else:
self.judge = "Yes"
self.par[x] = y
else:
if self.same_check(x, y) != True:
self.size[x] += self.size[y]
self.size[y] = 0
else:
self.judge = "Yes"
self.par[y] = x
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
def siz(self, x):
x = self.find(x)
return self.size[x]
#NใฎๆๅญใใใชใๆๅญๅAใฎMๆๅญใฎใใใทใฅๅคใ่จ็ฎ
def rollingHash(N, M, A):
mul1 = 1
for i in range(M):
mul1 = (mul1 * base1) % mod1
val1 = 0
for i in range(M):
val1 = val1 * base1 + A[i]
val1 %= mod1
hashList1 = [None] * (N - M + 1)
hashList1[0] = val1
for i in range(N - M):
val1 = (val1 * base1 - A[i] * mul1 + A[i + M]) % mod1
hashList1[i + 1] = val1
return hashList1
#ๅฆ็ๅ
ๅฎน
def main():
s = list(input())
t = list(input())
N = len(s)
M = len(t)
#sใฎใปใใ้ทใ่ชฟๆด
if N < M:
var = int(M // N) + 1
s = s * var
N = N * var
s = s + s
for i in range(2 * N):
s[i] = ord(s[i]) - 97
for i in range(M):
t[i] = ord(t[i]) - 97
sHash1 = rollingHash(2 * N, M, s)
tHash1 = rollingHash(M, M, t)
tHash1 = tHash1[0]
value = "No"
UF = UnionFind(N)
for i in range(N):
j = (i + M) % N
if sHash1[i] == tHash1:
value = "Yes"
if sHash1[j] == tHash1:
UF.union(i, j)
if value == "No":
print(0)
return
if UF.judge == "Yes":
print(-1)
return
for i in range(N):
UF.par[i] = UF.find(i)
ans = 0
for i in range(N):
ans = max(ans, UF.size[i])
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 15,755 | 5 | 31,510 |
Yes | output | 1 | 15,755 | 5 | 31,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
Submitted Solution:
```
class RollingHash():
def __init__(self,s):
self.length=len(s)
self.base1=1009; self.base2=1013
self.mod1=10**9+7; self.mod2=10**9+9
self.hash1=[0]*(self.length+1); self.hash2=[0]*(self.length+1)
self.pow1=[1]*(self.length+1); self.pow2=[1]*(self.length+1)
for i in range(self.length):
self.hash1[i+1]=(self.hash1[i]+ord(s[i]))*self.base1%self.mod1
self.hash2[i+1]=(self.hash2[i]+ord(s[i]))*self.base2%self.mod2
self.pow1[i+1]=self.pow1[i]*self.base1%self.mod1
self.pow2[i+1]=self.pow2[i]*self.base2%self.mod2
def get(self,l,r):
h1=((self.hash1[r]-self.hash1[l]*self.pow1[r-l])%self.mod1+self.mod1)%self.mod1
h2=((self.hash2[r]-self.hash2[l]*self.pow2[r-l])%self.mod2+self.mod2)%self.mod2
return (h1,h2)
def solve(s,t):
ls=len(s); lt=len(t)
RHs=RollingHash(s*2)
RHt=RollingHash(t)
Judge=[False]*ls
B=RHt.get(0,lt)
for i in range(ls):
if RHs.get(i,i+lt)==B:
Judge[i]=True
ret=0
Visited=[-1]*ls
for i in range(ls) :
if Judge[i] and Visited[i]==-1:
idx=i
cnt=0
while Judge[idx]:
if Visited[idx]!=-1:
cnt+=Visited[idx]
break
cnt+=1
Visited[idx]=1
idx=(idx+lt)%ls
if idx==i:
return -1
Visited[i]=cnt
ret=max(ret,cnt)
return ret
s=input(); t=input()
s*=(len(t)+len(s)-1)//len(s)
print(solve(s,t))
``` | instruction | 0 | 15,756 | 5 | 31,512 |
Yes | output | 1 | 15,756 | 5 | 31,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
Submitted Solution:
```
import collections
class KMP():
def __init__(self, pattern):
self.pattern = pattern
self.n = len(pattern)
self.create_k_table()
def create_k_table(self):
ktable = [-1]*(self.n+1)
j = -1
for i in range(self.n):
while j >= 0 and self.pattern[i] != self.pattern[j]:
j = ktable[j]
j += 1
if i+1 < self.n and self.pattern[i+1] == self.pattern[j]:
ktable[i+1] = ktable[j]
else:
ktable[i+1] = j
self.ktable = ktable
def match(self,s):
n = len(s)
j = 0
ret = [0]*n
for i in range(n):
while j >= 0 and (j == self.n or s[i] != self.pattern[j]):
j = self.ktable[j]
j += 1
if j == self.n:
ret[(i-self.n+1)%n] = 1
return ret
def main():
s = input()
t = input()
n = len(s)
m = len(t)
while len(s) < m + n:
s += s
a = KMP(t)
b = a.match(s)
edges = [[] for i in range(n)]
indegree = [0] * n
for i in range(n):
k = (i+m)%n
if b[i]:
edges[i].append(k)
indegree[k] += 1
def topSort():
q = collections.deque([i for i in range(n) if indegree[i] == 0])
while q:
cur = q.popleft()
res.append(cur)
for nei in edges[cur]:
indegree[nei] -= 1
if indegree[nei] == 0:
q.append(nei)
return len(res) == n
res = []
dp = [0] * n
flag = topSort()
if flag:
for k in range(n):
i = res[k]
for j in edges[i]:
dp[j] = max(dp[j], dp[i] + 1)
ans = max(dp)
else:
ans = -1
return ans
if __name__ == "__main__":
print(main())
``` | instruction | 0 | 15,757 | 5 | 31,514 |
Yes | output | 1 | 15,757 | 5 | 31,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
Submitted Solution:
```
class KMP():
def __init__(self, pattern):
self.pattern = pattern
self.n = len(pattern)
self.create_k_table()
def create_k_table(self):
ktable = [-1]*(self.n+1)
j = -1
for i in range(self.n):
while j >= 0 and self.pattern[i] != self.pattern[j]:
j = ktable[j]
j += 1
if i+1 < self.n and self.pattern[i+1] == self.pattern[j]:
ktable[i+1] = ktable[j]
else:
ktable[i+1] = j
self.ktable = ktable
def match(self,s):
n = len(s)
j = 0
ret = [0]*n
for i in range(n):
while j >= 0 and (j == self.n or s[i] != self.pattern[j]):
j = self.ktable[j]
j += 1
if j == self.n:
ret[(i-self.n+1)%n] = 1
return ret
def main():
s = input()
t = input()
n = len(s)
m = len(t)
k = 1 - (-(m-1)//n)
si = s*k
a = KMP(t)
b = a.match(si)
# edges = [[] for i in range(n)]
# for i in range(n):
# if b[i]:
# edges[i].append((i+m)%n)
#
# def dfs(node):
# if visited[node] == 1:
# return True
# if visited[node] == 2:
# return False
# visited[node] = 1
# for nei in edges[node]:
# if dfs(nei):
# return True
# visited[node] = 2
# res.append(node)
# return False
# visited = [-1] * n
# res = []
# for i in range(n):
# if dfs(i):
# return -1
# dp = [0] * n
# ans = 0
# for i in range(n-1, -1, -1):
# t = res[i]
# for c in edges[t]:
# dp[c] = max(dp[c], dp[t] + 1)
# ans = max(ans, dp[c])
# return ans
visited = [0]*n
loop = False
ans = 0
for i in range(n):
if visited[i]:
continue
visited[i] = 1
cur = i
right = 0
while b[cur]:
nxt = (cur + m) % n
if visited[nxt]:
loop = True
break
visited[nxt] = 1
cur = nxt
right += 1
cur = i
left = 0
while b[(cur-m)%n]:
nxt = (cur-m) % n
if visited[nxt]:
loop = True
break
visited[nxt] = 1
cur = nxt
left += 1
if not loop:
ans = max(ans, right+left)
else:
ans = -1
break
return ans
if __name__ == "__main__":
print(main())
``` | instruction | 0 | 15,758 | 5 | 31,516 |
Yes | output | 1 | 15,758 | 5 | 31,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
Submitted Solution:
```
s = input()
t = input()
def z_algorithm(s):
a = [0] * len(s)
i = 1
j = 0
a[0] = len(s)
while i < len(s):
while i + j < len(s) and s[j] == s[i+j]:
j += 1
a[i] = j
if j == 0:
i += 1
continue
k = 1
while i + k < len(s) and k + a[k] < j:
a[i+k] = a[k]
k += 1
i += k
j -= k
return a
def solve(i, li):
ans = 0
while True:
if visited[i]:
break
if i < 0 or len(li) <= i:
break
if li[i] < len(t):
visited[i] = True
break
if li[i] >= len(t):
visited[i] = True
ans += 1
i += len(t)
return ans
#sใไผธใฐใ
new_s = ""
while True:
new_s += s
if len(new_s) > len(t):
s = new_s
break
s = s*3
li = z_algorithm(t + s)[len(t):]
visited = [False] * len(li)
ans1 = 0
for i in range(len(li)):
ans1 = max(ans1, solve(i, li))
'''
s += s
li = z_algorithm(t + s)[len(t):]
visited = [False] * len(li)
ans2 = 0
for i in range(len(li)):
ans2 = max(ans2, solve(i, li))
if ans1 == ans2:
print(ans1)
else:
print(-1)
'''
print(ans1)
``` | instruction | 0 | 15,759 | 5 | 31,518 |
No | output | 1 | 15,759 | 5 | 31,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
Submitted Solution:
```
import sys,queue,math,copy,itertools,bisect,collections,heapq
def main():
MOD = 2**61-1
SI = lambda : sys.stdin.readline().rstrip()
s = SI()
t = SI()
s = s * 3
lens = len(s)
lent = len(t)
def cx(x):
return ord(x) - ord('a') + 1
hash = 0
for x in t:
hash = (hash * 26 + cx(x)) % MOD
cnt = 0
f = False
h = 0
last_i = 0
ans = 0
for i in range(lens):
if i >= lent:
h -= cx(s[i-lent]) * pow(26,lent-1,MOD)
h = (h * 26 + cx(s[i])) % MOD
if h == hash:
cnt += 1
ans = max(ans,cnt)
last_i = i
else:
if i - last_i >= lent and last_i > 0:
cnt = 0
f = True
if f:
print(ans)
else:
print(-1)
if __name__ == '__main__':
main()
``` | instruction | 0 | 15,760 | 5 | 31,520 |
No | output | 1 | 15,760 | 5 | 31,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
Submitted Solution:
```
S = input()
W = input()
M = len(S)
S *= 6 * (len(W)//len(S)+1)
N = len(W)
def primeFactor(N):
i, n, ret, d, sq = 2, N, {}, 2, 99
while i <= sq:
k = 0
while n % i == 0: n, k, ret[i] = n//i, k+1, k+1
if k > 0 or i == 97: sq = int(n**(1/2)+0.5)
if i < 4: i = i * 2 - 1
else: i, d = i+d, d^6
if n > 1: ret[n] = 1
return ret
def divisors(N):
pf = primeFactor(N)
ret = [1]
for p in pf:
ret_prev = ret
ret = []
for i in range(pf[p]+1):
for r in ret_prev:
ret.append(r * (p ** i))
return sorted(ret)
D = divisors(N)
p = 1
for d in D:
if W[:-d] == W[d:]:
W = W[:d]
p = N//d
N = d
break
W = W[:d]
T = [-1] * (len(W)+1)
ii = 2
jj = 0
T[0] = -1
T[1] = 0
while ii < len(W):
if W[ii - 1] == W[jj]:
T[ii] = jj + 1
ii += 1
jj += 1
elif jj > 0:
jj = T[jj]
else:
T[ii] = 0
ii += 1
def KMP(i0):
ret = -1
m = 0
i = 0
while m + i < len(S) - i0:
if W[i] == S[m + i + i0]:
i += 1
if i == len(W):
return m
else:
m = m + i - T[i]
if i > 0:
i = T[i]
return len(S) - i0
i = 0
c = 0
cmax = 0
s = 0
while s < M:
# print("i =", i)
k = KMP(i)
if k + i == len(S):
break
elif k == len(W) - 1:
c += 1
else:
s = i
c = 1
cmax = max(cmax, c)
i += k + 1
if s < M:
print(-1)
else:
print(cmax//p)
``` | instruction | 0 | 15,761 | 5 | 31,522 |
No | output | 1 | 15,761 | 5 | 31,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
Submitted Solution:
```
import sys
import random
s = input()
t = input()
sl = len(s)
tl = len(t)
l = 0
r = 1000000//tl + 2
r2 = r
s *= 999999//sl + 1
n = len(s)
base1 = 1007
mod1 = 10**9+7
modTank1 = [3000012541,3000012553,3000012563,3000012649,3000012683,3000012709]
mod1 = modTank1[random.randint(0,5)]
base1 = 1007
mod1 = 10**9+7
hash1 = [0]*(n+1)
power1 = [1]*(n+1)
for i,e in enumerate(s):
hash1[i+1] = (hash1[i]*base1 + ord(e))%mod1
power1[i+1] = (power1[i]*base1)%mod1
def rolling_hash(i, j):
return (hash1[j]-hash1[i]*power1[j-i]%mod1)%mod1
while r - l > 1:
m = l + (r-l)//2
lt = len(t)*m
hash2 = [0]*(lt+1)
power2 = [1]*(lt+1)
for i,e in enumerate(t*m):
hash2[i+1] = (hash2[i]*base1 + ord(e))%mod1
power2[i+1] = (power2[i]*base1)%mod1
r_hash = hash2[-1]%mod1
lt = len(t)*m
flag = 0
for i in range(len(s)-lt+1):
if r_hash == rolling_hash(i, i+lt):
flag = 1
break
if flag == 1:
l = m
else:
r = m
if r == r2:
print(-1)
else:
print(l)
``` | instruction | 0 | 15,762 | 5 | 31,524 |
No | output | 1 | 15,762 | 5 | 31,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integers N,\ A and B. Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists.
* P_0=A
* P_{2^N-1}=B
* For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.
Constraints
* 1 \leq N \leq 17
* 0 \leq A \leq 2^N-1
* 0 \leq B \leq 2^N-1
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there is no permutation that satisfies the conditions, print `NO`.
If there is such a permutation, print `YES` in the first line. Then, print (P_0,\ P_1,\ ...\ P_{2^N-1}) in the second line, with spaces in between. If there are multiple solutions, any of them is accepted.
Examples
Input
2 1 3
Output
YES
1 0 2 3
Input
3 2 1
Output
NO
Submitted Solution:
```
import sys
stdin = sys.stdin
sys.setrecursionlimit(10**5)
def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x)-1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())
# i-bit็ฎใจj-bit็ฎใๅ
ฅใๆฟใใ
def swap_digit(num:int, i:int, j:int):
if num & (1<<i) and not (num & (1<<j)):
return num - (1<<i) + (1<<j)
elif not (num & (1<<i)) and num & (1<<j):
return num + (1<<i) - (1<<j)
else:
return num
# 2ใคใฎๆฐใฎbitใ็ฐใชใๆกใ็นๅฎใใ
def different_digit(n: int, a: int, b: int):
ret = n-1
for digit in range(n-1, -1, -1):
if (a^b) & (1<<digit):
return digit
return ret
# ไธไฝbitใใๅๅธฐ็ใซๆฑบใใ
def rec(n: int, a: int, b: int):
if n == 1:
return [a,b]
dd = different_digit(n,a,b)
a = swap_digit(a, n-1, dd)
b = swap_digit(b, n-1, dd)
na = a & ((1<<(n-1)) - 1)
nb = b & ((1<<(n-1)) - 1)
first = rec(n-1, na, na^1)
latte = rec(n-1, na^1, nb)
if a & (1<<(n-1)):
ret = list(map(lambda x: x + (1<<(n-1)), first)) + latte
else:
ret = first + list(map(lambda x: x + (1<<(n-1)), latte))
return [swap_digit(reti, n-1, dd) for reti in ret]
n,a,b = li()
if bin(a).count('1') % 2 == bin(b).count('1') % 2:
print("NO")
else:
print("YES")
print(*rec(n,a,b))
``` | instruction | 0 | 15,771 | 5 | 31,542 |
Yes | output | 1 | 15,771 | 5 | 31,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integers N,\ A and B. Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists.
* P_0=A
* P_{2^N-1}=B
* For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.
Constraints
* 1 \leq N \leq 17
* 0 \leq A \leq 2^N-1
* 0 \leq B \leq 2^N-1
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there is no permutation that satisfies the conditions, print `NO`.
If there is such a permutation, print `YES` in the first line. Then, print (P_0,\ P_1,\ ...\ P_{2^N-1}) in the second line, with spaces in between. If there are multiple solutions, any of them is accepted.
Examples
Input
2 1 3
Output
YES
1 0 2 3
Input
3 2 1
Output
NO
Submitted Solution:
```
from sys import setrecursionlimit
setrecursionlimit(10 ** 9)
n, a, b = [int(i) for i in input().split()]
if bin(a ^ b).count('1') % 2 == 0:
print('NO')
exit()
def dfs(i, a, b):
if i == 1:
return [a, b]
d = (a ^ b) & -(a ^ b)
ad = ((a & (~d ^ d - 1)) >> 1) + (a & d - 1)
bd = ((b & (~d ^ d - 1)) >> 1) + (b & d - 1)
c = ad ^ 1
res1 = dfs(i - 1, ad, c)
res2 = dfs(i - 1, c, bd)
ans1 = [((r & ~(d - 1)) << 1) + (r & d - 1) + (d & a) for r in res1]
ans2 = [((r & ~(d - 1)) << 1) + (r & d - 1) + (d & b) for r in res2]
return ans1 + ans2
print('YES')
print(*dfs(n, a, b))
``` | instruction | 0 | 15,772 | 5 | 31,544 |
Yes | output | 1 | 15,772 | 5 | 31,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integers N,\ A and B. Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists.
* P_0=A
* P_{2^N-1}=B
* For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.
Constraints
* 1 \leq N \leq 17
* 0 \leq A \leq 2^N-1
* 0 \leq B \leq 2^N-1
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there is no permutation that satisfies the conditions, print `NO`.
If there is such a permutation, print `YES` in the first line. Then, print (P_0,\ P_1,\ ...\ P_{2^N-1}) in the second line, with spaces in between. If there are multiple solutions, any of them is accepted.
Examples
Input
2 1 3
Output
YES
1 0 2 3
Input
3 2 1
Output
NO
Submitted Solution:
```
d = ((0, 0), (0, 1), (1, 1), (1, 0))
n, a, b = map(int, input().split())
c = 0
p = a ^ b
z, o = [], []
for i in range(n):
if (p >> i) & 1:
o.append(i)
else:
z.append(i)
if len(o) % 2 == 0:
print('NO')
exit(0)
print('YES')
ans = [0] * pow(2, n)
if n % 2:
i = o.pop()
for j in range(pow(2, n - 1), pow(2, n)):
ans[j] += pow(2, i)
else:
i = o.pop()
j = z.pop()
for k in range(4):
tmp = pow(2, n) // 4
for l in range(tmp):
ans[k * tmp + l] += pow(2, i) * d[k][0] + pow(2, j) * d[k][1]
t = 1
while o:
tmp = pow(2, n) // t // 8
i = o.pop()
j = o.pop()
idx = 0
for l in range(tmp):
if l == 0:
for p in range(4):
for q in range(t):
ans[idx] += d[p][0] * \
pow(2, i) + d[p][1] * pow(2, j)
idx += 1
for p in range(4):
for q in range(t):
ans[idx] += d[p - 1][0] * \
pow(2, i) + d[p - 1][1] * pow(2, j)
idx += 1
else:
for p in range(4):
for q in range(t):
ans[idx] += d[p - 2][0] * \
pow(2, i) + d[p - 2][1] * pow(2, j)
idx += 1
for p in range(4):
for q in range(t):
ans[idx] += d[1 - p][0] * \
pow(2, i) + d[1 - p][1] * pow(2, j)
idx += 1
t *= 4
while z:
tmp = pow(2, n) // t // 8
i = z.pop()
j = z.pop()
idx = 0
for l in range(tmp):
for p in range(4):
for q in range(t):
ans[idx] += d[p][0] * \
pow(2, i) + d[p][1] * pow(2, j)
idx += 1
for p in range(4):
for q in range(t):
ans[idx] += d[3 - p][0] * \
pow(2, i) + d[3 - p][1] * pow(2, j)
idx += 1
t *= 4
print(' '.join(map(lambda x: str(x ^ a), ans)))
# for i in range(pow(2, n) - 1):
# print(ans[i + 1] - ans[i], end=' ')
``` | instruction | 0 | 15,773 | 5 | 31,546 |
Yes | output | 1 | 15,773 | 5 | 31,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integers N,\ A and B. Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists.
* P_0=A
* P_{2^N-1}=B
* For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.
Constraints
* 1 \leq N \leq 17
* 0 \leq A \leq 2^N-1
* 0 \leq B \leq 2^N-1
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there is no permutation that satisfies the conditions, print `NO`.
If there is such a permutation, print `YES` in the first line. Then, print (P_0,\ P_1,\ ...\ P_{2^N-1}) in the second line, with spaces in between. If there are multiple solutions, any of them is accepted.
Examples
Input
2 1 3
Output
YES
1 0 2 3
Input
3 2 1
Output
NO
Submitted Solution:
```
def extract(a, i):
return (a>>(i+1) << i) | (a&((1<<i)-1))
def space(a, i):
return ((a>>i) << (i+1)) | (a&((1<<i)-1))
def compose(n, a, b):
if n==1:
return [a, b]
for i in range(n):
if (a>>i&1) ^ (b>>i&1):
x = i
a_bool = (a>>i & 1) << i
b_bool = a_bool ^ (1 << i)
a_dash = extract(a, i)
b_dash = extract(b, i)
c = a_dash ^ 1
break
Q = compose(n-1, a_dash, c)
R = compose(n-1, c, b_dash)
n_Q = [space(i, x)|a_bool for i in Q]
n_R = [space(i, x)|b_bool for i in R]
return n_Q + n_R
n, a, b = map(int, input().split())
cnt = 0
c = a^b
for i in range(n):
cnt += c>>i & 1
if cnt&1:
print("YES")
print(*compose(n, a, b))
else:
print("NO")
``` | instruction | 0 | 15,774 | 5 | 31,548 |
Yes | output | 1 | 15,774 | 5 | 31,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integers N,\ A and B. Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists.
* P_0=A
* P_{2^N-1}=B
* For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.
Constraints
* 1 \leq N \leq 17
* 0 \leq A \leq 2^N-1
* 0 \leq B \leq 2^N-1
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there is no permutation that satisfies the conditions, print `NO`.
If there is such a permutation, print `YES` in the first line. Then, print (P_0,\ P_1,\ ...\ P_{2^N-1}) in the second line, with spaces in between. If there are multiple solutions, any of them is accepted.
Examples
Input
2 1 3
Output
YES
1 0 2 3
Input
3 2 1
Output
NO
Submitted Solution:
```
N, A, B = list(map(int, input().split(' ')))
num = format(A^B, 'b').count('1')
if num >= 2**N - 1 or num % 2 == 0:
print('NO')
else:
C = A^B
c = format(A^B, 'b')
p = list()
p.append(A)
for i in range(len(c)):
if int(c[len(c)-1-i]) == 1:
tmp = p[-1] ^ (C & (1 << i))
p.append(tmp)
for i in range(2**N-len(p)):
p.append(p[-2])
print('YES')
for i in range(len(p)):
print(p[i], end=' ')
``` | instruction | 0 | 15,775 | 5 | 31,550 |
No | output | 1 | 15,775 | 5 | 31,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integers N,\ A and B. Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists.
* P_0=A
* P_{2^N-1}=B
* For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.
Constraints
* 1 \leq N \leq 17
* 0 \leq A \leq 2^N-1
* 0 \leq B \leq 2^N-1
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there is no permutation that satisfies the conditions, print `NO`.
If there is such a permutation, print `YES` in the first line. Then, print (P_0,\ P_1,\ ...\ P_{2^N-1}) in the second line, with spaces in between. If there are multiple solutions, any of them is accepted.
Examples
Input
2 1 3
Output
YES
1 0 2 3
Input
3 2 1
Output
NO
Submitted Solution:
```
N, A, B = map(int, input().split())
a_str = bin(A)[2:]
b_str = bin(B)[2:]
a_bin = int(a_str)
b_bin = int(b_str)
a = [0] * (N+1)
b = [0] * (N+1)
c = [A] + [0]*(2**N-2) + [B]
# for i in range(len(a_str),2,-1):
# a.append(int(a_str[i-1]))
# for i in range(len(b_str),2,-1):
# b.append(int(b_str[i-1]))
cnta = 0
cntb = 0
while (a_bin):
a[cnta] = a_bin%10
a_bin = a_bin//10
cnta += 1
while (b_bin):
b[cntb] = b_bin%10
b_bin = b_bin//10
cntb += 1
# print(cnta,cntb)
d = A^B
h = 0
while(d):
d &= d-1
h += 1
# print(h)
if (2**N-h)%2 == 0 or h > 2**N-2:
print('NO', end='')
exit()
else:
print('YES')
j = 1
C = A
for i in range(max(cnta, cntb)):
if a[i] != b[i]:
if a[i] == 1:
C -= 2**i
c[j] = C
else:
C += 2**i
c[j] = C
j += 1
for i in range(j,2**N-1,2):
c[j],c[j+1] = c[j-2],c[j-1]
for i in range(2**N):
print(c[i], end=' ')
``` | instruction | 0 | 15,776 | 5 | 31,552 |
No | output | 1 | 15,776 | 5 | 31,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integers N,\ A and B. Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists.
* P_0=A
* P_{2^N-1}=B
* For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.
Constraints
* 1 \leq N \leq 17
* 0 \leq A \leq 2^N-1
* 0 \leq B \leq 2^N-1
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there is no permutation that satisfies the conditions, print `NO`.
If there is such a permutation, print `YES` in the first line. Then, print (P_0,\ P_1,\ ...\ P_{2^N-1}) in the second line, with spaces in between. If there are multiple solutions, any of them is accepted.
Examples
Input
2 1 3
Output
YES
1 0 2 3
Input
3 2 1
Output
NO
Submitted Solution:
```
N, A, B = map(int, input().split())
AB = A ^ B
if bin(AB)[2:].count('1') != 1:
print('NO')
else:
print('YES')
C = [0] * (1 << N)
C[0] = str(A)
pos = 0
while AB % 2 == 0:
pos += 1
AB >>= 1
for i in range(1, 1 << N):
k = 0
b = i
while b % 2 == 0:
k += 1
b >>= 1
A ^= 1 << ((pos + k + 1) % N)
C[i] = str(A)
print(' '.join(C))
``` | instruction | 0 | 15,777 | 5 | 31,554 |
No | output | 1 | 15,777 | 5 | 31,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integers N,\ A and B. Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists.
* P_0=A
* P_{2^N-1}=B
* For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.
Constraints
* 1 \leq N \leq 17
* 0 \leq A \leq 2^N-1
* 0 \leq B \leq 2^N-1
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
If there is no permutation that satisfies the conditions, print `NO`.
If there is such a permutation, print `YES` in the first line. Then, print (P_0,\ P_1,\ ...\ P_{2^N-1}) in the second line, with spaces in between. If there are multiple solutions, any of them is accepted.
Examples
Input
2 1 3
Output
YES
1 0 2 3
Input
3 2 1
Output
NO
Submitted Solution:
```
N,A,B=[int(x) for x in input().split()]
SA=bin(A)[2:]
SB=bin(B)[2:]
mx=max(len(SA),len(SB))
SA=("0"*mx+SA)[-mx:]
SB=("0"*mx+SB)[-mx:]
diff=0
L=[]
for i in range(len(SA)):
if SA[i]!=SB[i]:
L.append(i)
diff+=1
if diff%2==0:
print('NO')
else:
out=str(int(SA,2))+" "
print('YES')
for i in L:
if SA[i]=="1":
SA=SA[:i]+"0"+SA[i+1:]
else:
SA=SA[:i]+"1"+SA[i+1:]
out+=str(int(SA,2))+" "
for j in range(2**N-i-2):
if SA[0]=="0":
SA="1"+SA[1:]
else:
SA="0"+SA[1:]
out+=str(int(SA,2))+" "
print(out.strip())
"""
for t in out.split():
print(("0000"+bin(int(t))[2:])[-3:])
"""
``` | instruction | 0 | 15,778 | 5 | 31,556 |
No | output | 1 | 15,778 | 5 | 31,557 |
Provide a correct Python 3 solution for this coding contest problem.
Kurohashi has never participated in AtCoder Beginner Contest (ABC).
The next ABC to be held is ABC N (the N-th ABC ever held). Kurohashi wants to make his debut in some ABC x such that all the digits of x in base ten are the same.
What is the earliest ABC where Kurohashi can make his debut?
Constraints
* 100 \leq N \leq 999
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If the earliest ABC where Kurohashi can make his debut is ABC n, print n.
Examples
Input
111
Output
111
Input
112
Output
222
Input
750
Output
777 | instruction | 0 | 15,779 | 5 | 31,558 |
"Correct Solution:
```
n = int(input())
ct = 0
while ct < n:
ct += 111
print(ct)
``` | output | 1 | 15,779 | 5 | 31,559 |
Provide a correct Python 3 solution for this coding contest problem.
Kurohashi has never participated in AtCoder Beginner Contest (ABC).
The next ABC to be held is ABC N (the N-th ABC ever held). Kurohashi wants to make his debut in some ABC x such that all the digits of x in base ten are the same.
What is the earliest ABC where Kurohashi can make his debut?
Constraints
* 100 \leq N \leq 999
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If the earliest ABC where Kurohashi can make his debut is ABC n, print n.
Examples
Input
111
Output
111
Input
112
Output
222
Input
750
Output
777 | instruction | 0 | 15,780 | 5 | 31,560 |
"Correct Solution:
```
print(int(111*(1+int((int(input())-1)/111))))
``` | output | 1 | 15,780 | 5 | 31,561 |
Provide a correct Python 3 solution for this coding contest problem.
Kurohashi has never participated in AtCoder Beginner Contest (ABC).
The next ABC to be held is ABC N (the N-th ABC ever held). Kurohashi wants to make his debut in some ABC x such that all the digits of x in base ten are the same.
What is the earliest ABC where Kurohashi can make his debut?
Constraints
* 100 \leq N \leq 999
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If the earliest ABC where Kurohashi can make his debut is ABC n, print n.
Examples
Input
111
Output
111
Input
112
Output
222
Input
750
Output
777 | instruction | 0 | 15,781 | 5 | 31,562 |
"Correct Solution:
```
import math
N = int(input())
r = math.ceil(N/111.0)*111
print(r)
``` | output | 1 | 15,781 | 5 | 31,563 |
Provide a correct Python 3 solution for this coding contest problem.
Kurohashi has never participated in AtCoder Beginner Contest (ABC).
The next ABC to be held is ABC N (the N-th ABC ever held). Kurohashi wants to make his debut in some ABC x such that all the digits of x in base ten are the same.
What is the earliest ABC where Kurohashi can make his debut?
Constraints
* 100 \leq N \leq 999
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If the earliest ABC where Kurohashi can make his debut is ABC n, print n.
Examples
Input
111
Output
111
Input
112
Output
222
Input
750
Output
777 | instruction | 0 | 15,782 | 5 | 31,564 |
"Correct Solution:
```
print(str((int(input())-1)//111+1)*3)
``` | output | 1 | 15,782 | 5 | 31,565 |
Provide a correct Python 3 solution for this coding contest problem.
Kurohashi has never participated in AtCoder Beginner Contest (ABC).
The next ABC to be held is ABC N (the N-th ABC ever held). Kurohashi wants to make his debut in some ABC x such that all the digits of x in base ten are the same.
What is the earliest ABC where Kurohashi can make his debut?
Constraints
* 100 \leq N \leq 999
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
Output
If the earliest ABC where Kurohashi can make his debut is ABC n, print n.
Examples
Input
111
Output
111
Input
112
Output
222
Input
750
Output
777 | instruction | 0 | 15,783 | 5 | 31,566 |
"Correct Solution:
```
a = int(input())
while len(set(str(a)))!=1:
a+=1
print(a)
``` | output | 1 | 15,783 | 5 | 31,567 |
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