message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 17,242 | 22 | 34,484 |
Tags: dp, math
Correct Solution:
```
s, x = map(int, input().split(' '))
if (s-x)%2 or s < x:
print(0)
else:
c = bin((s-x)//2)[2:][::-1]
t = bin(x)[2:][::-1]
for i in range(len(t)):
if t[i] == '1' and i < len(c) and c[i] == '1':
print(0)
exit(0)
print(pow(2, bin(x)[2:].count('1'))-(2 if s==x else 0))
``` | output | 1 | 17,242 | 22 | 34,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 17,243 | 22 | 34,486 |
Tags: dp, math
Correct Solution:
```
import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math
alphabets = list('abcdefghijklmnopqrstuvwxyz')
def isPrime(x):
for i in range(2,x):
if i*i>x:
break
if (x%i==0):
return False
return True
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def si():
return input()
def prefix_sum(arr):
r = [0] * (len(arr)+1)
for i, el in enumerate(arr):
r[i+1] = r[i] + el
return r
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def ii():
return int(input())
def li():
return list(map(int,input().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
def power_set(L):
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred
def fastPlrintNextLines(a):
# 12
# 3
# 1
#like this
#a is list of strings
print('\n'.join(map(str,a)))
def sortByFirstAndSecond(A):
A = sorted(A,key = lambda x:x[0])
A = sorted(A,key = lambda x:x[1])
return list(A)
#__________________________TEMPLATE__________________OVER_______________________________________________________
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
else:
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def solve():
s,x = li()
andval = (s-x) /2
# print(andval)
if andval!=int(andval) or s<x or int(andval)&x:
print(0)
else:
w = andval
andval=int(andval)
andval = list(bin(andval)[2:]);x = list(bin(x)[2:])
s=andval[:]
if len(x)>len(s):
s = ['0']*(len(x)-len(s))+s
else:
x = ['0']*(len(s)-len(x))+x
andval=s[:]
ans=0
for i in range(len(s)-1,-1,-1):
if x[i]=='1':
ans+=1
# print(andval,x)
ans=2**ans
if w==0:
ans-=2
print(ans)
# s = list(bin(s)[1:]);x = list(bin(x)[1:])
# if len(x)>len(s):
# s = [0]*(len(x)-len(s))+s
# else:
# x = [0]*(len(s)-len(x))+x
# dp = [[0,0]]*(len(s))
# for i in range(len(s)-1,-1,-1):
# if s[i]=='0' and x[i]=='0':
# if i<len(s)-1:
# dp[i][0]=dp[i+1][0]+dp[i+1][1]
# else:
# dp[i][0]=1
# continue
# if s[i]=='1' and x[i]=='1':
t = 1
# t = int(input())
for _ in range(t):
solve()
``` | output | 1 | 17,243 | 22 | 34,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 17,244 | 22 | 34,488 |
Tags: dp, math
Correct Solution:
```
a, b = [int(x) for x in input().split()]
c = (a-b) / 2
if c < 0 or not c.is_integer() or int(c) & b:
print(0)
exit(0)
t = 0
while b:
t += b & 1
b >>= 1
t = 1 << t
if c == 0:
t -= 2
print(t)
``` | output | 1 | 17,244 | 22 | 34,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 17,245 | 22 | 34,490 |
Tags: dp, math
Correct Solution:
```
'''
___ ____
____ _____ _____/ (_)_ ______ ____ _____/ / /_ __ ______ ___ __
/ __ `/ __ `/ __ / / / / / __ \/ __ `/ __ / __ \/ / / / __ `/ / / /
/ /_/ / /_/ / /_/ / / /_/ / /_/ / /_/ / /_/ / / / / /_/ / /_/ / /_/ /
\__,_/\__,_/\__,_/_/\__,_/ .___/\__,_/\__,_/_/ /_/\__, /\__,_/\__, /
/_/ /____/ /____/
'''
import math
from collections import deque
import os.path
from math import gcd, floor, ceil
from collections import *
import sys
mod = 1000000007
INF = float('inf')
def st(): return list(sys.stdin.readline().strip())
def li(): return list(map(int, sys.stdin.readline().split()))
def mp(): return map(int, sys.stdin.readline().split())
def inp(): return int(sys.stdin.readline())
def pr(n): return sys.stdout.write(str(n)+"\n")
def prl(n): return sys.stdout.write(str(n)+" ")
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
s, x = mp()
if x > s:
pr(0)
else:
a = (s-x)
if a & 1:
pr(0)
else:
b = 0
a //= 2
if x & a != 0:
pr(0)
else:
while x:
b += x & 1
x //= 2
ans = pow(2, b)
if a == 0:
ans -= 2
pr(ans)
``` | output | 1 | 17,245 | 22 | 34,491 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 17,246 | 22 | 34,492 |
Tags: dp, math
Correct Solution:
```
s, x = map(int, input().split())
print(0 if s < x or (s - x) & (2 * x + 1) else 2 ** bin(x).count('1') - 2 * (s == x))
# Made By Mostafa_Khaled
``` | output | 1 | 17,246 | 22 | 34,493 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 17,247 | 22 | 34,494 |
Tags: dp, math
Correct Solution:
```
suma,xor=map(int,input().split())
mitad=round((suma-xor)/2)
if(mitad<0 or mitad*2+xor!=suma or (mitad&xor)!=0):
print(0)
else:
print(2**(bin(xor).count("1"))-2*(suma==xor))
``` | output | 1 | 17,247 | 22 | 34,495 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 17,248 | 22 | 34,496 |
Tags: dp, math
Correct Solution:
```
s, xx = map(int, input().split())
if s - xx & 1 or s < xx:
exit(print(0))
y = bin(s - xx)[2:-1]
x = bin(xx)[2:]
ans = 1
if len(y) < len(x):
y = '0' * (len(x) - len(y)) + y
else:
x = '0' * (len(y) - len(x)) + x
for i in range(len(x)):
if x[i] == '1' and y[i] == '1':
exit(print(0))
if x[i] == '1' and y[i] == '0':
ans <<= 1
print(ans - 2 * (s == xx))
``` | output | 1 | 17,248 | 22 | 34,497 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 17,249 | 22 | 34,498 |
Tags: dp, math
Correct Solution:
```
s, x = map(int, input().split())
a, b = 1, 0
for i in range(50):
c, d = s & (1 << i), x & (1 << i)
# print(i, a, b, c, d)
if c == d:
if c:
a, b = 2 * a, 0
else:
a, b = a, a
else:
if c:
a, b = b, b
else:
a, b = 0, 2 * b
if s == x:
a -= 2
print(a)
``` | output | 1 | 17,249 | 22 | 34,499 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha and Grisha like studying sets of positive integers.
One day Grisha has written a set A containing n different integers ai on a blackboard. Now he asks Masha to create a set B containing n different integers bj such that all n2 integers that can be obtained by summing up ai and bj for all possible pairs of i and j are different.
Both Masha and Grisha don't like big numbers, so all numbers in A are from 1 to 106, and all numbers in B must also be in the same range.
Help Masha to create the set B that satisfies Grisha's requirement.
Input
Input data contains multiple test cases. The first line contains an integer t — the number of test cases (1 ≤ t ≤ 100).
Each test case is described in the following way: the first line of the description contains one integer n — the number of elements in A (1 ≤ n ≤ 100).
The second line contains n integers ai — the elements of A (1 ≤ ai ≤ 106).
Output
For each test first print the answer:
* NO, if Masha's task is impossible to solve, there is no way to create the required set B.
* YES, if there is the way to create the required set. In this case the second line must contain n different positive integers bj — elements of B (1 ≤ bj ≤ 106). If there are several possible sets, output any of them.
Example
Input
3
3
1 10 100
1
1
2
2 4
Output
YES
1 2 3
YES
1
YES
1 2 | instruction | 0 | 17,352 | 22 | 34,704 |
Tags: brute force, constructive algorithms
Correct Solution:
```
from random import randint
h = [-1] * 2000001
for t in range(int(input())):
n, A = int(input()), list(map(int, input().split()))
B = []
while len(B) < n:
b = randint(1, 1000000)
if all(h[a + b] != t for a in A):
for a in A: h[a + b] = t
B.append(b)
print('YES\n' + ' '.join(map(str, B)))
``` | output | 1 | 17,352 | 22 | 34,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha and Grisha like studying sets of positive integers.
One day Grisha has written a set A containing n different integers ai on a blackboard. Now he asks Masha to create a set B containing n different integers bj such that all n2 integers that can be obtained by summing up ai and bj for all possible pairs of i and j are different.
Both Masha and Grisha don't like big numbers, so all numbers in A are from 1 to 106, and all numbers in B must also be in the same range.
Help Masha to create the set B that satisfies Grisha's requirement.
Input
Input data contains multiple test cases. The first line contains an integer t — the number of test cases (1 ≤ t ≤ 100).
Each test case is described in the following way: the first line of the description contains one integer n — the number of elements in A (1 ≤ n ≤ 100).
The second line contains n integers ai — the elements of A (1 ≤ ai ≤ 106).
Output
For each test first print the answer:
* NO, if Masha's task is impossible to solve, there is no way to create the required set B.
* YES, if there is the way to create the required set. In this case the second line must contain n different positive integers bj — elements of B (1 ≤ bj ≤ 106). If there are several possible sets, output any of them.
Example
Input
3
3
1 10 100
1
1
2
2 4
Output
YES
1 2 3
YES
1
YES
1 2 | instruction | 0 | 17,353 | 22 | 34,706 |
Tags: brute force, constructive algorithms
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int, input().split()))
a.sort()
b=[]
b.append(1)
t=10**6
used=[0]*(t+1)
used[1]=1
can_be=2
diffs=[]
for i in range(n):
for j in range(i+1,n):
diffs.append(a[j]-a[i])
for i in range(1,n):
for j in diffs:
used[min(b[i-1]+j,t)]=1
while(used[can_be]):
can_be+=1
b.append(can_be)
used[can_be]=1
if len(b)==n:
print('YES')
for i in b:
print(i,end=' ')
print('')
else:
print('NO')
``` | output | 1 | 17,353 | 22 | 34,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha and Grisha like studying sets of positive integers.
One day Grisha has written a set A containing n different integers ai on a blackboard. Now he asks Masha to create a set B containing n different integers bj such that all n2 integers that can be obtained by summing up ai and bj for all possible pairs of i and j are different.
Both Masha and Grisha don't like big numbers, so all numbers in A are from 1 to 106, and all numbers in B must also be in the same range.
Help Masha to create the set B that satisfies Grisha's requirement.
Input
Input data contains multiple test cases. The first line contains an integer t — the number of test cases (1 ≤ t ≤ 100).
Each test case is described in the following way: the first line of the description contains one integer n — the number of elements in A (1 ≤ n ≤ 100).
The second line contains n integers ai — the elements of A (1 ≤ ai ≤ 106).
Output
For each test first print the answer:
* NO, if Masha's task is impossible to solve, there is no way to create the required set B.
* YES, if there is the way to create the required set. In this case the second line must contain n different positive integers bj — elements of B (1 ≤ bj ≤ 106). If there are several possible sets, output any of them.
Example
Input
3
3
1 10 100
1
1
2
2 4
Output
YES
1 2 3
YES
1
YES
1 2 | instruction | 0 | 17,354 | 22 | 34,708 |
Tags: brute force, constructive algorithms
Correct Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect
#?############################################################
def isPrime(x):
for i in range(2, x):
if i*i > x:
break
if (x % i == 0):
return False
return True
#?############################################################
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den, p - 2, p)) % p
#?############################################################
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
#?############################################################
def power(x, y, p):
ab = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
ab = (ab * x) % p
y = y >> 1
x = (x * x) % p
return ab
#?############################################################
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#?############################################################
def digits(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
#?############################################################
def ceil(n, x):
if (n % x == 0):
return n//x
return n//x+1
#?############################################################
def mapin():
return [int(x) for x in input().split()]
#?############################################################
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
bb = [-1]*(2000009)
t = int(input())
for j in range(t):
n = int(input())
a = mapin()
a.sort()
ab = []
v = 1
while len(ab) < n:
fl = True
for i in a:
if bb[i + v] == j:
fl =False
break
if not fl:
v += 1
continue
for i in a:
bb[i + v] = j
ab.append(v)
v += 1
print("YES")
print(*ab)
``` | output | 1 | 17,354 | 22 | 34,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha and Grisha like studying sets of positive integers.
One day Grisha has written a set A containing n different integers ai on a blackboard. Now he asks Masha to create a set B containing n different integers bj such that all n2 integers that can be obtained by summing up ai and bj for all possible pairs of i and j are different.
Both Masha and Grisha don't like big numbers, so all numbers in A are from 1 to 106, and all numbers in B must also be in the same range.
Help Masha to create the set B that satisfies Grisha's requirement.
Input
Input data contains multiple test cases. The first line contains an integer t — the number of test cases (1 ≤ t ≤ 100).
Each test case is described in the following way: the first line of the description contains one integer n — the number of elements in A (1 ≤ n ≤ 100).
The second line contains n integers ai — the elements of A (1 ≤ ai ≤ 106).
Output
For each test first print the answer:
* NO, if Masha's task is impossible to solve, there is no way to create the required set B.
* YES, if there is the way to create the required set. In this case the second line must contain n different positive integers bj — elements of B (1 ≤ bj ≤ 106). If there are several possible sets, output any of them.
Example
Input
3
3
1 10 100
1
1
2
2 4
Output
YES
1 2 3
YES
1
YES
1 2 | instruction | 0 | 17,355 | 22 | 34,710 |
Tags: brute force, constructive algorithms
Correct Solution:
```
visited = [-1] * (2 * 10 ** 6 + 1)
t = int(input())
for i in range(t):
n, A = int(input()), list(map(int, input().split()))
A.sort()
res = []
v = 1
while len(res) < n:
flag = True
for a in A:
if visited[a + v] == i:
flag = False
break
if not flag:
v += 1
continue
for a in A:
visited[a + v] = i
res.append(v)
v += 1
print("YES\n" + ' '.join(map(str,res)))
``` | output | 1 | 17,355 | 22 | 34,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha and Grisha like studying sets of positive integers.
One day Grisha has written a set A containing n different integers ai on a blackboard. Now he asks Masha to create a set B containing n different integers bj such that all n2 integers that can be obtained by summing up ai and bj for all possible pairs of i and j are different.
Both Masha and Grisha don't like big numbers, so all numbers in A are from 1 to 106, and all numbers in B must also be in the same range.
Help Masha to create the set B that satisfies Grisha's requirement.
Input
Input data contains multiple test cases. The first line contains an integer t — the number of test cases (1 ≤ t ≤ 100).
Each test case is described in the following way: the first line of the description contains one integer n — the number of elements in A (1 ≤ n ≤ 100).
The second line contains n integers ai — the elements of A (1 ≤ ai ≤ 106).
Output
For each test first print the answer:
* NO, if Masha's task is impossible to solve, there is no way to create the required set B.
* YES, if there is the way to create the required set. In this case the second line must contain n different positive integers bj — elements of B (1 ≤ bj ≤ 106). If there are several possible sets, output any of them.
Example
Input
3
3
1 10 100
1
1
2
2 4
Output
YES
1 2 3
YES
1
YES
1 2 | instruction | 0 | 17,356 | 22 | 34,712 |
Tags: brute force, constructive algorithms
Correct Solution:
```
from random import randint
visited = [-1] * (2 * 10 ** 6 + 1)
t = int(input())
for i in range(t):
n, A = int(input()), list(map(int, input().split()))
res = []
while len(res) < n:
v = randint(1, 10**6)
flag = True
for a in A:
if visited[a + v] == i:
flag = False
break
if flag:
for a in A:
visited[a + v] = i
res.append(v)
#if all(visited[a + v] != i for a in A):
# for a in A:
# visited[a + v] = i
# res.append(v)
print("YES\n" + ' '.join(map(str,res)))
``` | output | 1 | 17,356 | 22 | 34,713 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha and Grisha like studying sets of positive integers.
One day Grisha has written a set A containing n different integers ai on a blackboard. Now he asks Masha to create a set B containing n different integers bj such that all n2 integers that can be obtained by summing up ai and bj for all possible pairs of i and j are different.
Both Masha and Grisha don't like big numbers, so all numbers in A are from 1 to 106, and all numbers in B must also be in the same range.
Help Masha to create the set B that satisfies Grisha's requirement.
Input
Input data contains multiple test cases. The first line contains an integer t — the number of test cases (1 ≤ t ≤ 100).
Each test case is described in the following way: the first line of the description contains one integer n — the number of elements in A (1 ≤ n ≤ 100).
The second line contains n integers ai — the elements of A (1 ≤ ai ≤ 106).
Output
For each test first print the answer:
* NO, if Masha's task is impossible to solve, there is no way to create the required set B.
* YES, if there is the way to create the required set. In this case the second line must contain n different positive integers bj — elements of B (1 ≤ bj ≤ 106). If there are several possible sets, output any of them.
Example
Input
3
3
1 10 100
1
1
2
2 4
Output
YES
1 2 3
YES
1
YES
1 2 | instruction | 0 | 17,357 | 22 | 34,714 |
Tags: brute force, constructive algorithms
Correct Solution:
```
from random import randint
def solve():
n, aa = int(input()), list(map(int, input().split()))
bb, ab = set(), set()
while True:
b = randint(1, 1000000)
for a in aa:
if a + b in ab:
break
else:
bb.add(b)
if len(bb) == n:
break
for a in aa:
ab.add(a + b)
print("YES")
print(' '.join(map(str, bb)))
def main():
for _ in range(int(input())):
solve()
if __name__ == '__main__':
main()
``` | output | 1 | 17,357 | 22 | 34,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha and Grisha like studying sets of positive integers.
One day Grisha has written a set A containing n different integers ai on a blackboard. Now he asks Masha to create a set B containing n different integers bj such that all n2 integers that can be obtained by summing up ai and bj for all possible pairs of i and j are different.
Both Masha and Grisha don't like big numbers, so all numbers in A are from 1 to 106, and all numbers in B must also be in the same range.
Help Masha to create the set B that satisfies Grisha's requirement.
Input
Input data contains multiple test cases. The first line contains an integer t — the number of test cases (1 ≤ t ≤ 100).
Each test case is described in the following way: the first line of the description contains one integer n — the number of elements in A (1 ≤ n ≤ 100).
The second line contains n integers ai — the elements of A (1 ≤ ai ≤ 106).
Output
For each test first print the answer:
* NO, if Masha's task is impossible to solve, there is no way to create the required set B.
* YES, if there is the way to create the required set. In this case the second line must contain n different positive integers bj — elements of B (1 ≤ bj ≤ 106). If there are several possible sets, output any of them.
Example
Input
3
3
1 10 100
1
1
2
2 4
Output
YES
1 2 3
YES
1
YES
1 2 | instruction | 0 | 17,358 | 22 | 34,716 |
Tags: brute force, constructive algorithms
Correct Solution:
```
d = [-1] * 1000001
for t in range(int(input())):
n, a = int(input()), list(map(int, input().split()))
a.sort()
for i in range(n):
for j in range(i + 1, n): d[a[j] - a[i]] = t
i = 1
while any(d[i * j] == t for j in range(1, n)): i += 1
print("YES\n" + ' '.join(str(j * i + 1) for j in range(n)))
``` | output | 1 | 17,358 | 22 | 34,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Masha and Grisha like studying sets of positive integers.
One day Grisha has written a set A containing n different integers ai on a blackboard. Now he asks Masha to create a set B containing n different integers bj such that all n2 integers that can be obtained by summing up ai and bj for all possible pairs of i and j are different.
Both Masha and Grisha don't like big numbers, so all numbers in A are from 1 to 106, and all numbers in B must also be in the same range.
Help Masha to create the set B that satisfies Grisha's requirement.
Input
Input data contains multiple test cases. The first line contains an integer t — the number of test cases (1 ≤ t ≤ 100).
Each test case is described in the following way: the first line of the description contains one integer n — the number of elements in A (1 ≤ n ≤ 100).
The second line contains n integers ai — the elements of A (1 ≤ ai ≤ 106).
Output
For each test first print the answer:
* NO, if Masha's task is impossible to solve, there is no way to create the required set B.
* YES, if there is the way to create the required set. In this case the second line must contain n different positive integers bj — elements of B (1 ≤ bj ≤ 106). If there are several possible sets, output any of them.
Example
Input
3
3
1 10 100
1
1
2
2 4
Output
YES
1 2 3
YES
1
YES
1 2 | instruction | 0 | 17,359 | 22 | 34,718 |
Tags: brute force, constructive algorithms
Correct Solution:
```
from random import randint
visited = [-1] * (2 * 10 ** 6 + 1)
t = int(input())
for i in range(t):
n, A = int(input()), list(map(int, input().split()))
res = []
while len(res) < n:
v = randint(1, 10**6)
if all(visited[a + v] != i for a in A):
for a in A:
visited[a + v] = i
res.append(v)
print("YES\n" + ' '.join(map(str,res)))
``` | output | 1 | 17,359 | 22 | 34,719 |
Provide a correct Python 3 solution for this coding contest problem.
Remainder of Big Integers
Given two integers $A$ and $B$, compute the remainder of $\frac{A}{B}$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the remainder in a line.
Constraints
* $0 \leq A, B \leq 10^{1000}$
* $B \ne 0$
Sample Input 1
5 8
Sample Output 1
5
Sample Input 2
100 25
Sample Output 2
0
Example
Input
5 8
Output
5 | instruction | 0 | 17,677 | 22 | 35,354 |
"Correct Solution:
```
a,b=input().split(" ")
a=int(a)
b=int(b)
print(a%b)
``` | output | 1 | 17,677 | 22 | 35,355 |
Provide a correct Python 3 solution for this coding contest problem.
Remainder of Big Integers
Given two integers $A$ and $B$, compute the remainder of $\frac{A}{B}$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the remainder in a line.
Constraints
* $0 \leq A, B \leq 10^{1000}$
* $B \ne 0$
Sample Input 1
5 8
Sample Output 1
5
Sample Input 2
100 25
Sample Output 2
0
Example
Input
5 8
Output
5 | instruction | 0 | 17,678 | 22 | 35,356 |
"Correct Solution:
```
if __name__ == "__main__":
a, b = list(map(lambda x: int(x), input().split()))
print(a % b)
``` | output | 1 | 17,678 | 22 | 35,357 |
Provide a correct Python 3 solution for this coding contest problem.
Remainder of Big Integers
Given two integers $A$ and $B$, compute the remainder of $\frac{A}{B}$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the remainder in a line.
Constraints
* $0 \leq A, B \leq 10^{1000}$
* $B \ne 0$
Sample Input 1
5 8
Sample Output 1
5
Sample Input 2
100 25
Sample Output 2
0
Example
Input
5 8
Output
5 | instruction | 0 | 17,679 | 22 | 35,358 |
"Correct Solution:
```
def main():
a,b = map(int, input().split())
print(a%b)
main()
``` | output | 1 | 17,679 | 22 | 35,359 |
Provide a correct Python 3 solution for this coding contest problem.
Remainder of Big Integers
Given two integers $A$ and $B$, compute the remainder of $\frac{A}{B}$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the remainder in a line.
Constraints
* $0 \leq A, B \leq 10^{1000}$
* $B \ne 0$
Sample Input 1
5 8
Sample Output 1
5
Sample Input 2
100 25
Sample Output 2
0
Example
Input
5 8
Output
5 | instruction | 0 | 17,680 | 22 | 35,360 |
"Correct Solution:
```
a, b = input().split()
print(int(a) % int(b))
``` | output | 1 | 17,680 | 22 | 35,361 |
Provide a correct Python 3 solution for this coding contest problem.
Remainder of Big Integers
Given two integers $A$ and $B$, compute the remainder of $\frac{A}{B}$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the remainder in a line.
Constraints
* $0 \leq A, B \leq 10^{1000}$
* $B \ne 0$
Sample Input 1
5 8
Sample Output 1
5
Sample Input 2
100 25
Sample Output 2
0
Example
Input
5 8
Output
5 | instruction | 0 | 17,681 | 22 | 35,362 |
"Correct Solution:
```
a, b=input().split(' ')
a=int(a)
b=int(b)
print(a-a//b*b)
``` | output | 1 | 17,681 | 22 | 35,363 |
Provide a correct Python 3 solution for this coding contest problem.
Remainder of Big Integers
Given two integers $A$ and $B$, compute the remainder of $\frac{A}{B}$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the remainder in a line.
Constraints
* $0 \leq A, B \leq 10^{1000}$
* $B \ne 0$
Sample Input 1
5 8
Sample Output 1
5
Sample Input 2
100 25
Sample Output 2
0
Example
Input
5 8
Output
5 | instruction | 0 | 17,682 | 22 | 35,364 |
"Correct Solution:
```
print(eval(input().replace(" ","%")))
``` | output | 1 | 17,682 | 22 | 35,365 |
Provide a correct Python 3 solution for this coding contest problem.
Remainder of Big Integers
Given two integers $A$ and $B$, compute the remainder of $\frac{A}{B}$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the remainder in a line.
Constraints
* $0 \leq A, B \leq 10^{1000}$
* $B \ne 0$
Sample Input 1
5 8
Sample Output 1
5
Sample Input 2
100 25
Sample Output 2
0
Example
Input
5 8
Output
5 | instruction | 0 | 17,683 | 22 | 35,366 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Big Integers - Remainder of Big Integers
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=NTL_2_E&lang=jp
"""
import sys
def main(args):
A, B = map(int, input().split())
print(A % B)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 17,683 | 22 | 35,367 |
Provide a correct Python 3 solution for this coding contest problem.
Remainder of Big Integers
Given two integers $A$ and $B$, compute the remainder of $\frac{A}{B}$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the remainder in a line.
Constraints
* $0 \leq A, B \leq 10^{1000}$
* $B \ne 0$
Sample Input 1
5 8
Sample Output 1
5
Sample Input 2
100 25
Sample Output 2
0
Example
Input
5 8
Output
5 | instruction | 0 | 17,684 | 22 | 35,368 |
"Correct Solution:
```
A,B = map(int,input().split())
print(A % B)
``` | output | 1 | 17,684 | 22 | 35,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Remainder of Big Integers
Given two integers $A$ and $B$, compute the remainder of $\frac{A}{B}$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the remainder in a line.
Constraints
* $0 \leq A, B \leq 10^{1000}$
* $B \ne 0$
Sample Input 1
5 8
Sample Output 1
5
Sample Input 2
100 25
Sample Output 2
0
Example
Input
5 8
Output
5
Submitted Solution:
```
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
return [I() for i in range(n)]
def LIR(n):
return [LI() for i in range(n)]
def SR(n):
return [S() for i in range(n)]
def LSR(n):
return [LS() for i in range(n)]
sys.setrecursionlimit(1000000)
mod = 1000000007
#1_A
"""
import math
n = int(input())
m = n
ans = []
i = 2
k = math.sqrt(n)
while i <= k:
if n%i == 0:
n = n//i
ans.append(i)
i -= 1
if n == 1:
break
i += 1
if len(ans) == 0 or n != 1:
ans.append(n)
print(m,end = ": ")
for i in range(len(ans)-1):
print(ans[i], end = " ")
print(ans[-1])
"""
#1_B
"""
m,n = map(int, input().split())
b = bin(n)[2:]
num = [m]
for i in range(len(b)-1):
num.append(num[-1]**2%1000000007)
ans = 1
for i in range(len(b)):
ans *= num[i] if b[len(b)-1-i] == "1" else 1
ans %= 1000000007
print(ans)
"""
#1_C
"""
import math
l = int(input())
a = list(map(int, input().split()))
dict = {}
for n in a:
ans = []
i = 2
k = math.sqrt(n)
dic = {}
while i <= k:
if n%i == 0:
n = n//i
ans.append(i)
i -= 1
if n == 1:
break
i += 1
if len(ans) == 0 or n != 1:
ans.append(n)
for i in ans:
if i in dic:
dic[i] += 1
else:
dic[i] = 1
for i in dic.keys():
if i in dict:
dict[i] = max(dict[i], dic[i])
else:
dict[i] = dic[i]
sum = 1
for x,y in dict.items():
sum *= x**y
print(sum)
"""
#1_D
"""
import math
n = int(input())
m = n
ans = []
i = 2
k = math.sqrt(n)
dic = {}
while i <= k:
if n%i == 0:
n = n//i
ans.append(i)
i -= 1
if n == 1:
break
i += 1
if len(ans) == 0 or n != 1:
ans.append(n)
ans = list(set(ans))
sum = m
for i in ans:
sum *= (1-1/i)
print(int(sum))
"""
#1_E
"""
def gcd(a,b):
if a==0:
return b
return gcd(b%a,a)
a,b = LI()
g = gcd(a,b)
a//=g
b//=g
if b == 1:
print(0,1)
quit()
if b < 4:
f = b-1
else:
p = []
x = 2
m = b
while x**2 <= b:
if not m%x:
p.append(x)
while not m%x:
m//=x
x += 1
if m != 1:p.append(m)
f = b
for i in p:
f *= (1-1/i)
f = int(f)
x_ = pow(a,f-1,b)
y_ = (1-a*x_)//b
ans = [x_,y_,abs(x_)+abs(y_)]
for i in range(-10000,10000):
s = -b*i+x_
t = a*i+y_
m = abs(s)+abs(t)
if m < ans[2]:
ans = [s,t,m]
elif m == ans[2] and s <= t:
ans = [s,t,m]
print(ans[0],ans[1])
"""
#2
#2_A
"""
a,b = LI()
print(a+b)
"""
#2_B
"""
a,b = LI()
print(a-b)
"""
#2_C
"""
a,b = LI()
print(a*b)
"""
#2_D
"""
a,b = LI()
if a*b < 0:
print(-(-a//b))
else:
print(a//b)
"""
#2_E
a,b = LI()
print(a%b)
``` | instruction | 0 | 17,688 | 22 | 35,376 |
Yes | output | 1 | 17,688 | 22 | 35,377 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.
Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.
Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.
Input
The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.
Output
The only line should contain your answer modulo 10^9+7.
Examples
Input
2
Output
1
Input
3
Output
4
Input
6
Output
120
Note
Consider the second example: these are the permutations of length 3:
* [1,2,3], f(p)=1.
* [1,3,2], f(p)=1.
* [2,1,3], f(p)=2.
* [2,3,1], f(p)=2.
* [3,1,2], f(p)=2.
* [3,2,1], f(p)=2.
The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2. | instruction | 0 | 17,763 | 22 | 35,526 |
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
p=10**9+7
import math
def r(l):
x=1
for m in l:
x=x*m%p
return x
n=int(input())
a,k,x,t=[],int(math.log2(n)),n,0
while x>0:
a.append(x-x//2)
x//=2
b=[n//(3*2**i)-n//(6*2**i) for i in range(k+1)]
d=[n//2**i-n//(3*2**i) for i in range(k+1)]
y=r(i for i in range(2,n+1))
s=k if n<3*2**(k-1) else 0
for j in range(s,k+1):
e=[a[i] for i in range(j)]+[d[j]]+[b[i] for i in range(j,k)]
x=y*r(e)%p
f=r(sum(e[:i+1]) for i in range(k+1))
while f>1:
x*=p//f+1
f=f*(p//f+1)%p
t+=x%p
print(t%p)
``` | output | 1 | 17,763 | 22 | 35,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.
Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.
Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.
Input
The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.
Output
The only line should contain your answer modulo 10^9+7.
Examples
Input
2
Output
1
Input
3
Output
4
Input
6
Output
120
Note
Consider the second example: these are the permutations of length 3:
* [1,2,3], f(p)=1.
* [1,3,2], f(p)=1.
* [2,1,3], f(p)=2.
* [2,3,1], f(p)=2.
* [3,1,2], f(p)=2.
* [3,2,1], f(p)=2.
The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2. | instruction | 0 | 17,764 | 22 | 35,528 |
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
p=10**9+7
import math
def inv(k,p):
prod=1
while k>1:
prod*=(p//k+1)
k=(k*(p//k+1))%p
return prod%p
n=int(input())
a=[]
k=int(math.log2(n))
x=n
while x>0:
y=x//2
a.append(x-y)
x=y
c=[sum(a[i:]) for i in range(k+1)]
b=[n//(3*2**i)-n//(6*2**i) for i in range(k+1)]
d=[n//2**i-n//(3*2**i) for i in range(k+1)]
facs=[1]*(n+1)
for i in range(2,n+1):
facs[i]=(i*facs[i-1])%p
if n<3*(2**(k-1)):
start=k
else:
start=0
tot=0
for j in range(start,k+1):
prod=1
for i in range(j,k):
prod*=b[i]
prod*=d[j]
for i in range(j):
prod*=a[i]
prod%=p
prod*=facs[n]
e=[a[i] for i in range(j)]+[d[j]]+[b[i] for i in range(j,k)]
f=[sum(e[:i+1]) for i in range(k+1)]
g=1
for guy in f:
g*=guy
prod*=inv(g,p)
prod%=p
tot+=prod
print(tot%p)
``` | output | 1 | 17,764 | 22 | 35,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.
Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.
Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.
Input
The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.
Output
The only line should contain your answer modulo 10^9+7.
Examples
Input
2
Output
1
Input
3
Output
4
Input
6
Output
120
Note
Consider the second example: these are the permutations of length 3:
* [1,2,3], f(p)=1.
* [1,3,2], f(p)=1.
* [2,1,3], f(p)=2.
* [2,3,1], f(p)=2.
* [3,1,2], f(p)=2.
* [3,2,1], f(p)=2.
The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2. | instruction | 0 | 17,765 | 22 | 35,530 |
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
import math
p=10**9+7
n=int(input())
k=int(math.log2(n))
f=[[n//(2**i*3**j) for j in range(3)] for i in range(k+1)]
old=[[0,0,0] for i in range(k+1)]
old[k][0]=1
if n>=3*2**(k-1):
old[k-1][1]=1
m=n//2+2
for i in range(2,m):
dp=[[0,0,0] for i in range(k+1)]
for j in range(k):
for l in range(2):
dp[j][l]=(old[j][l]*(f[j][l]-i+1)+old[j+1][l]*(f[j][l]-f[j+1][l])+old[j][l+1]*(f[j][l]-f[j][l+1]))%p
old=dp
curr=old[0][0]
for i in range(1,n-m+2):
curr*=i
curr%=p
print(curr)
``` | output | 1 | 17,765 | 22 | 35,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.
Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.
Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.
Input
The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.
Output
The only line should contain your answer modulo 10^9+7.
Examples
Input
2
Output
1
Input
3
Output
4
Input
6
Output
120
Note
Consider the second example: these are the permutations of length 3:
* [1,2,3], f(p)=1.
* [1,3,2], f(p)=1.
* [2,1,3], f(p)=2.
* [2,3,1], f(p)=2.
* [3,1,2], f(p)=2.
* [3,2,1], f(p)=2.
The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2. | instruction | 0 | 17,766 | 22 | 35,532 |
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
p=10**9+7
import math
def prod(l):
x=1
for m in l:
x*=m
return x
n=int(input())
a=[]
k=int(math.log2(n))
x=n
while x>0:
y=x//2
a.append(x-y)
x=y
c=[sum(a[i:]) for i in range(k+1)]
b=[n//(3*2**i)-n//(6*2**i) for i in range(k+1)]
d=[n//2**i-n//(3*2**i) for i in range(k+1)]
facs=[1]*(n+1)
for i in range(2,n+1):
facs[i]=(i*facs[i-1])%p
start=k if n<3*(2**(k-1)) else 0
tot=0
for j in range(start,k+1):
e=[a[i] for i in range(j)]+[d[j]]+[b[i] for i in range(j,k)]
x=(facs[n]*prod(e))%p
f=prod([sum(e[:i+1]) for i in range(k+1)])
while f>1:
x*=p//f+1
f=(f*(p//f+1))%p
tot+=x%p
print(tot%p)
``` | output | 1 | 17,766 | 22 | 35,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.
Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.
Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.
Input
The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.
Output
The only line should contain your answer modulo 10^9+7.
Examples
Input
2
Output
1
Input
3
Output
4
Input
6
Output
120
Note
Consider the second example: these are the permutations of length 3:
* [1,2,3], f(p)=1.
* [1,3,2], f(p)=1.
* [2,1,3], f(p)=2.
* [2,3,1], f(p)=2.
* [3,1,2], f(p)=2.
* [3,2,1], f(p)=2.
The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2. | instruction | 0 | 17,767 | 22 | 35,534 |
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
p=10**9+7
import math
def r(l):
x=1
for m in l:
x=x*m%p
return x
n=int(input())
a,k,x,t=[],int(math.log2(n)),n,0
while x>0:
a.append(x-x//2)
x//=2
b=[n//(3*2**i)-n//(6*2**i) for i in range(k+1)]
d=[n//2**i-n//(3*2**i) for i in range(k+1)]
y=r([i for i in range(2,n+1)])
s=k if n<3*2**(k-1) else 0
for j in range(s,k+1):
e=[a[i] for i in range(j)]+[d[j]]+[b[i] for i in range(j,k)]
x=y*r(e)%p
f=r([sum(e[:i+1]) for i in range(k+1)])
while f>1:
x*=p//f+1
f=f*(p//f+1)%p
t+=x%p
print(t%p)
``` | output | 1 | 17,767 | 22 | 35,535 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.
Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.
Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.
Input
The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.
Output
The only line should contain your answer modulo 10^9+7.
Examples
Input
2
Output
1
Input
3
Output
4
Input
6
Output
120
Note
Consider the second example: these are the permutations of length 3:
* [1,2,3], f(p)=1.
* [1,3,2], f(p)=1.
* [2,1,3], f(p)=2.
* [2,3,1], f(p)=2.
* [3,1,2], f(p)=2.
* [3,2,1], f(p)=2.
The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2. | instruction | 0 | 17,768 | 22 | 35,536 |
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
def perm(a,b):
ret = 1
for i in range(a, a - b, -1):
ret = (ret * i)%1000000007
return ret
num = int(input())
num2 = num
c = 0
while num2:
c += 1
num2 >>= 1
pow2 = 1 << (c-1)
array = [pow2]
for i in range(c-1):
array.append(array[-1]>>1)
def calc(arr):
data = []
#sm = 0
av = num
for i in range(c):
data.append((num // arr[i] - (num - av), av))
av -= data[-1][0]
#print(data)
ans = 1
for d in data:
ans = (ans * (d[0] * perm(d[1] - 1, d[0] - 1)))%1000000007
return ans
answer = calc(array)
if num >= pow2 // 2 * 3:
for i in range(1,c):
dat = [1]
for j in range(1, c):
if j == i:
dat = [3*dat[0]] + dat
else:
dat = [2*dat[0]] + dat
#print(dat)
a = calc(dat)
answer += a
print(answer%1000000007)
``` | output | 1 | 17,768 | 22 | 35,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.
Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.
Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.
Input
The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.
Output
The only line should contain your answer modulo 10^9+7.
Examples
Input
2
Output
1
Input
3
Output
4
Input
6
Output
120
Note
Consider the second example: these are the permutations of length 3:
* [1,2,3], f(p)=1.
* [1,3,2], f(p)=1.
* [2,1,3], f(p)=2.
* [2,3,1], f(p)=2.
* [3,1,2], f(p)=2.
* [3,2,1], f(p)=2.
The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2. | instruction | 0 | 17,769 | 22 | 35,538 |
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
p=10**9+7
import math
def prod(l):
x=1
for m in l:
x=x*m%p
return x
n=int(input())
a,k,x,t=[],int(math.log2(n)),n,0
while x>0:
a.append(x-x//2)
x//=2
c=[sum(a[i:]) for i in range(k+1)]
b=[n//(3*2**i)-n//(6*2**i) for i in range(k+1)]
d=[n//2**i-n//(3*2**i) for i in range(k+1)]
y=prod([i for i in range(2,n+1)])
s=k if n<3*(2**(k-1)) else 0
for j in range(s,k+1):
e=[a[i] for i in range(j)]+[d[j]]+[b[i] for i in range(j,k)]
x=(y*prod(e))%p
f=prod([sum(e[:i+1]) for i in range(k+1)])
while f>1:
x*=p//f+1
f=(f*(p//f+1))%p
t+=x%p
print(t%p)
``` | output | 1 | 17,769 | 22 | 35,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.
Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.
Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.
Input
The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.
Output
The only line should contain your answer modulo 10^9+7.
Examples
Input
2
Output
1
Input
3
Output
4
Input
6
Output
120
Note
Consider the second example: these are the permutations of length 3:
* [1,2,3], f(p)=1.
* [1,3,2], f(p)=1.
* [2,1,3], f(p)=2.
* [2,3,1], f(p)=2.
* [3,1,2], f(p)=2.
* [3,2,1], f(p)=2.
The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2.
Submitted Solution:
```
p=10**9+7
import math
def inv(k,p):
prod=1
while k>1:
prod*=(p//k+1)
k=(k*(p//k+1))%p
return prod%p
n=int(input())
a=[]
k=int(math.log2(n))
x=n
while x>0:
y=x//2
a.append(x-y)
x=y
c=[sum(a[i:]) for i in range(k+1)]
b=[n//(3*2**i)-n//(6*2**i) for i in range(k+1)]
d=[n//2**i-n//(3*2**i) for i in range(k+1)]
facs=[1]*(n+1)
for i in range(2,n+1):
facs[i]=(i*facs[i-1])%p
if n<3*(2**(k-1)):
start=k
else:
start=0
tot=0
for j in range(start,k+1):
prod=1
for i in range(j,k):
prod*=b[i]
prod*=d[j]
for i in range(j):
prod*=a[i]
prod%=p
prod*=facs[n]
e=[a[i] for i in range(j)]+[d[j]]+[b[i] for i in range(j,k)]
f=[sum(e[:i+1]) for i in range(k+1)]
g=1
for guy in f:
g*=guy
prod*=inv(g,p)
prod%=p
tot+=prod
print(prod)
``` | instruction | 0 | 17,770 | 22 | 35,540 |
No | output | 1 | 17,770 | 22 | 35,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.
Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.
Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.
Input
The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.
Output
The only line should contain your answer modulo 10^9+7.
Examples
Input
2
Output
1
Input
3
Output
4
Input
6
Output
120
Note
Consider the second example: these are the permutations of length 3:
* [1,2,3], f(p)=1.
* [1,3,2], f(p)=1.
* [2,1,3], f(p)=2.
* [2,3,1], f(p)=2.
* [3,1,2], f(p)=2.
* [3,2,1], f(p)=2.
The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2.
Submitted Solution:
```
n=int(input())
ans=1
m=10**9+7
for i in range(2,n):
ans*=i
ans%=m
print(ans)
``` | instruction | 0 | 17,771 | 22 | 35,542 |
No | output | 1 | 17,771 | 22 | 35,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.
Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.
Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.
Input
The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.
Output
The only line should contain your answer modulo 10^9+7.
Examples
Input
2
Output
1
Input
3
Output
4
Input
6
Output
120
Note
Consider the second example: these are the permutations of length 3:
* [1,2,3], f(p)=1.
* [1,3,2], f(p)=1.
* [2,1,3], f(p)=2.
* [2,3,1], f(p)=2.
* [3,1,2], f(p)=2.
* [3,2,1], f(p)=2.
The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2.
Submitted Solution:
```
p=10**9+7
def inv(k,p):
prod=1
while k>1:
prod*=(p//k+1)
k=(k*(p//k+1))%p
return prod%p
n=int(input())
a=[]
x=n
while x>0:
y=x//2
a.append(x-y)
x=y
frontsum=[sum(a[:i]) for i in range(1,len(a)+1)]
prod=1
for guy in a:
prod*=guy
prod%=p
ind=0
for i in range(1,n):
if frontsum[ind]!=i:
prod*=i
prod%=p
else:
ind+=1
if n<3*(2**(len(a)-2)):
print(prod)
else:
backsum=[sum(a[i:]) for i in range(len(a))]
spec=[guy//3 for guy in backsum]
for i in range(len(a)-2,-1,-1):
spec[i]-=spec[i+1]
unspec=[a[i]-spec[i] for i in range(len(a))]
unsum=[sum(unspec[i:]) for i in range(len(a))]
fracs=[(spec[i],unsum[i+1]) for i in range(len(a)-1)]
fracs=fracs[::-1]
invs=[inv(guy[1],p) for guy in fracs]
curr=prod
for i in range(len(a)-1):
curr=(curr*fracs[i][0]*invs[i])%p
prod+=curr
print(prod%p)
``` | instruction | 0 | 17,772 | 22 | 35,544 |
No | output | 1 | 17,772 | 22 | 35,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.
Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.
Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.
Input
The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.
Output
The only line should contain your answer modulo 10^9+7.
Examples
Input
2
Output
1
Input
3
Output
4
Input
6
Output
120
Note
Consider the second example: these are the permutations of length 3:
* [1,2,3], f(p)=1.
* [1,3,2], f(p)=1.
* [2,1,3], f(p)=2.
* [2,3,1], f(p)=2.
* [3,1,2], f(p)=2.
* [3,2,1], f(p)=2.
The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2.
Submitted Solution:
```
def frac(n):
res = 1
for i in range(1, n+1):
res = res * i
res %= MOD
return res
n = int(input())
MOD = 10**9 + 7
size_2 = 1
cnt_2 = 0
while size_2 <= n:
size_2 *= 2
cnt_2 += 1
cnt_2 = cnt_2 - 1
cnt_2_li = [0]*cnt_2
for i in range(cnt_2):
cnt_2_li[i] = n // (2)**(cnt_2-i)
ans = 1
for i in range(len(cnt_2_li)):
ans *= (cnt_2_li[i] - i)
ans %= MOD
ans *= frac(n-len(cnt_2_li))
if 3 * 2**(cnt_2 - 1) > n:
print(ans%MOD)
if 3 * 2**(cnt_2 - 1) <= n:
cnt_3 = cnt_2
cnt_3_li = [0]*cnt_3
for i in range(cnt_3):
cnt_3_li[i] = n // ((2)**(cnt_2-i-1) * 3)
ans = 0
for i in range(len(cnt_2_li)+1):
flag = False
tmp_ans = 1
for j in range(len(cnt_2_li)):
if i <= j:
flag = True
if flag:
tmp_ans *= (cnt_2_li[j] - j)
else:
tmp_ans *= (cnt_3_li[j] - j)
ans += tmp_ans
ans %= MOD
ans *= frac(n-len(cnt_2_li))
print(ans%MOD)
``` | instruction | 0 | 17,773 | 22 | 35,546 |
No | output | 1 | 17,773 | 22 | 35,547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0. | instruction | 0 | 17,811 | 22 | 35,622 |
Tags: math, number theory
Correct Solution:
```
from math import gcd
# from operator import mul
# from collections import defaultdict
# from itertools import count
# from functools import reduce
# primes_cache, prime_jumps = [], defaultdict(list)
# def primes():
# prime = 1
# for i in count():
# if i < len(primes_cache): prime = primes_cache[i]
# else:
# prime += 1
# while prime in prime_jumps:
# for skip in prime_jumps[prime]:
# prime_jumps[prime + skip] += [skip]
# del prime_jumps[prime]
# prime += 1
# prime_jumps[prime + prime] += [prime]
# primes_cache.append(prime)
# yield prime
# def factorize(n):
# for prime in primes():
# if prime > n: return
# exponent = 0
# while n % prime == 0:
# exponent, n = exponent + 1, n / prime
# if exponent != 0:
# yield prime, exponent
# def totient(n):
# return int(reduce(mul, (1 - 1.0 / p for p, exp in factorize(n)), n))
def phi(n) :
result = n # Initialize result as n
# Consider all prime factors
# of n and for every prime
# factor p, multiply result with (1 - 1 / p)
p = 2
while(p * p<= n) :
# Check if p is a prime factor.
if (n % p == 0) :
# If yes, then update n and result
while (n % p == 0) :
n = n // p
result = result * (1.0 - (1.0 / (float) (p)))
p = p + 1
# If n has a prime factor
# greater than sqrt(n)
# (There can be at-most one
# such prime factor)
if (n > 1) :
result = result * (1.0 - (1.0 / (float)(n)))
return (int)(result)
def solve(a,m):
k = gcd(a,m)
return phi(m//k)
t = int(input())
for _ in range(t):
a, m = map(int, input().split())
print(solve(a,m))
``` | output | 1 | 17,811 | 22 | 35,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0. | instruction | 0 | 17,812 | 22 | 35,624 |
Tags: math, number theory
Correct Solution:
```
from math import gcd
def Fenjie(n):
k = {}
if (n == 1):
return {}
a = 2
while (n >= 2):
if (a*a > n):
if (n in k):
k[n] += 1
else:
k[n] = 1
break
b = n%a
if (b == 0):
if (a in k):
k[a] += 1
else:
k[a] = 1
n = n//a
else:
a += 1
return k
def Euler(n):
if (n == 1):
return 1
k = Fenjie(n)
m = n
for i in k:
m = m // i * (i-1)
return m
t = int(input())
for _ in range(t):
a, b = map(int, input().split())
b = b//gcd(a,b)
print(Euler(b))
``` | output | 1 | 17,812 | 22 | 35,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0. | instruction | 0 | 17,813 | 22 | 35,626 |
Tags: math, number theory
Correct Solution:
```
import sys, math
for ii in range(int(input())):
a,m = map(int,input().split())
g = math.gcd(a,m)
m=int(m/g)
ans = m
for i in range(2,int(math.sqrt(m))+1):
if m%i==0:
ans-=(ans/i)
while m%i==0:
m=int(m/i)
if m>1:
ans-=ans/m
print(int(ans))
#3SUPE
#WWK73
#7PC7Z
#OBRPO
#T9RJQ
``` | output | 1 | 17,813 | 22 | 35,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0. | instruction | 0 | 17,814 | 22 | 35,628 |
Tags: math, number theory
Correct Solution:
```
primes=[True]*1000001
primes[0]=False
primes[1]=False
for i in range(2, 100001):
if primes[i]:
for j in range(i*2, 100001, i):
primes[j]=False
pL=[]
for i in range(2, 100001):
if primes[i]:pL.append(i)
def fact(n):
L=[]
for i in pL:
if n%i==0:
while n%i==0:
L.append(i)
n//=i
if n!=1:L.append(n)
return L
def gcd(a,b):
while b:
a,b=b,a%b
return a
for i in ' '*int(input()):
a,m=map(int,input().split())
g=gcd(a,m)
aa=a//g
mm=m//g
L=fact(mm)
M=[]
for i in L:
if i not in M:M.append(i)
for i in M:
mm*=(i-1)
mm//=i
print(mm)
``` | output | 1 | 17,814 | 22 | 35,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0. | instruction | 0 | 17,815 | 22 | 35,630 |
Tags: math, number theory
Correct Solution:
```
from math import gcd
def phi(n) :
res=n
p=2
while(p*p<=n):
if(n%p==0):
while(n%p==0) :
n//=p
res//=p
res*=(p-1)
p+=1
if(n>1):
res//=n
res*=(n-1)
return(res)
t=int(input())
for _ in range(t):
a,m=map(int,input().split())
g=gcd(a,m)
m//=g
print(phi(m))
``` | output | 1 | 17,815 | 22 | 35,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0. | instruction | 0 | 17,816 | 22 | 35,632 |
Tags: math, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
import math
t=int(input())
for test in range(t):
a,m=map(int,input().split())
GCD=math.gcd(a,m)
x=m//GCD
L=int(math.sqrt(x))
FACT=dict()
for i in range(2,L+2):
while x%i==0:
FACT[i]=FACT.get(i,0)+1
x=x//i
if x!=1:
FACT[x]=FACT.get(x,0)+1
ANS=m//GCD
for f in FACT:
ANS=ANS*(f-1)//f
print(ANS)
``` | output | 1 | 17,816 | 22 | 35,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0. | instruction | 0 | 17,817 | 22 | 35,634 |
Tags: math, number theory
Correct Solution:
```
import math
def Phi_Euler(val) :
ans = val
i = 2
while i*i <= val :
if val%i == 0 :
ans -= ans//i
while val%i == 0 : val //= i
i += 1
if val > 1 : ans -= ans//val
return ans
t = int(input())
while t > 0 :
a, m = map(int, input().split())
print(Phi_Euler(m//math.gcd(a, m)))
t -= 1
``` | output | 1 | 17,817 | 22 | 35,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0. | instruction | 0 | 17,818 | 22 | 35,636 |
Tags: math, number theory
Correct Solution:
```
def gcd(x,y):
while(y):
x,y=y,x%y
return x
def totient(x):
tot=x
n=x
i=1
prime_factors=[]
q=2
while n>1 and q*q<=n:
if n%q==0:
prime_factors.append(q)
while(n%q==0):
n//=q
q+=1
if n>1:
prime_factors.append(n)
for p in prime_factors:
tot=(tot/p)*(p-1)
return int(tot)
for _ in range(int(input())):
a,m=map(int,input().split())
g=gcd(m,a)
ans=0
count=0
prime_factors=[]
m1=(m)//g
ans=totient(m1)
print(ans)
``` | output | 1 | 17,818 | 22 | 35,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0.
Submitted Solution:
```
from sys import stdin
from collections import deque
mod = 10**9 + 7
import sys
sys.setrecursionlimit(10**5)
from queue import PriorityQueue
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
# n,k = map(int,input().split())
#
# l = list(map(int,input().split()))
t = int(input())
for _ in range(t):
a,b = map(int,input().split())
g = gcd(a,b)
m = b//g
b//=g
l = []
for i in range(2,int(sqrt(m)) + 1):
if b%i==0:
l.append(i)
while(b%i==0):
b//=i
if b>1:
l.append(b)
ans = 0
for i in l:
m*=(i-1)
m//=i
print(m)
``` | instruction | 0 | 17,819 | 22 | 35,638 |
Yes | output | 1 | 17,819 | 22 | 35,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0.
Submitted Solution:
```
from math import gcd
def euler_phi(n):
res=n
for x in range(2,int(n**.5)+2):
if n%x==0:
res=(res//x)*(x-1)
while n%x==0:
n//=x
if n!=1:
res=(res//n)*(n-1)
return res
for _ in range(int(input())):
a,m=map(int,input().split())
g=gcd(a,m)
a//=g
m//=g
res=euler_phi(m)
print(res)
``` | instruction | 0 | 17,820 | 22 | 35,640 |
Yes | output | 1 | 17,820 | 22 | 35,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0.
Submitted Solution:
```
from math import *
from bisect import *
def phi(n):
# Initialize result as n
result = n;
# Consider all prime factors
# of n and subtract their
# multiples from result
p = 2;
while(p * p <= n):
# Check if p is a
# prime factor.
if (n % p == 0):
# If yes, then
# update n and result
while (n % p == 0):
n = int(n / p);
result -= int(result / p);
p += 1;
# If n has a prime factor
# greater than sqrt(n)
# (There can be at-most
# one such prime factor)
if (n > 1):
result -= int(result / n);
return result;
test = int(input())
for i in range(test):
a, m = input().split()
a = int(a)
m = int(m)
d = gcd(a, m)
num = m//d
ans = phi(num)
print(ans)
``` | instruction | 0 | 17,821 | 22 | 35,642 |
Yes | output | 1 | 17,821 | 22 | 35,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0.
Submitted Solution:
```
# Design_by_JOKER
import math
import cmath
from decimal import * # su dung voi so thuc
from fractions import * # su dung voi phan so
# getcontext().prec = x # lay x-1 chu so sau giay phay ( thuoc decimal)
# Decimal('12.3') la 12.3 nhung Decimal(12.3) la 12.30000000012
# Fraction(a) # tra ra phan so bang a (Fraction('1.23') la 123/100 Fraction(1.23) la so khac (thuoc Fraction)
# a = complex(c, d) a = c + d(i) (c = a.real, d = a.imag)
# a.capitalize() bien ki tu dau cua a(string) thanh chu hoa, a.lower() bien a thanh chu thuong, tuong tu voi a.upper()
# a.swapcase() doi nguoc hoa thuong, a.title() bien chu hoa sau dau cach, a.replace('a', 'b', slg)
# a.join['a', 'b', 'c'] = 'a'a'b'a'c, a.strip('a') bo dau va cuoi ki tu 'a'(rstrip, lstrip)
# a.split('a', slg = -1) cat theo ki tu 'a' slg lan(rsplit(), lsplit()), a.count('aa', dau = 0, cuoi= len(a)) dem slg
# a.startswith('a', dau = 0, cuoi = len(a)) co bat dau bang 'a' ko(tuong tu endswith())
# a.find("aa") vi tri dau tien xuat hien (rfind())
# input = open(".inp", mode='r') a = input.readline()
# out = open(".out", mode='w')
T = int(input())
while T > 0:
T -= 1
a, m = map(int, input().split())
m //= math.gcd(a, m)
ans = m
aa = int(math.sqrt(m))
for x in range(2, max(3, aa+1)):
if m <= 1:
break
if m % x == 0:
while m % x == 0:
m //= x
ans = (ans * (x - 1)) // x
if m > 1:
ans = (ans * (m-1)) // m
print(ans)
``` | instruction | 0 | 17,822 | 22 | 35,644 |
Yes | output | 1 | 17,822 | 22 | 35,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0.
Submitted Solution:
```
from sys import stdin
from collections import deque
mod = 10**9 + 7
import sys
sys.setrecursionlimit(10**5)
from queue import PriorityQueue
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
# n,k = map(int,input().split())
#
# l = list(map(int,input().split()))
t = int(input())
for _ in range(t):
a,b = map(int,input().split())
g = gcd(a,b)
m = b//g
l = []
for i in range(2,int(sqrt(m)) + 1):
if b%i==0:
l.append(i)
while(b%i==0):
b//=i
if b>1:
l.append(b)
ans = 0
for i in l:
m*=(i-1)
m//=i
print(m)
``` | instruction | 0 | 17,823 | 22 | 35,646 |
No | output | 1 | 17,823 | 22 | 35,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).
Note: \gcd(a, b) is the greatest common divisor of a and b.
Input
The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).
Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.
Example
Input
3
4 9
5 10
42 9999999967
Output
6
1
9999999966
Note
In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].
In the second test case the only appropriate x is 0.
Submitted Solution:
```
import math
def divisors(n):
res = []
d = 1
while d * d <= n:
if n % d == 0:
res.append(d)
d = d + 1
if n != 1:
res.append(n)
return res
def u(n):
if n == 1:
return 1
d = 2
divs = 0
count = 0
while d * d <= n:
if n % d == 0:
n = n // d
divs += 1
if count == 1:
return 0
else:
count = 1
else:
d = d + 1
count = 0
if n > 1:
if n == d and count == 1:
return 0
divs += 1
if divs % 2 == 0:
return 1
else:
return -1
t = int(input())
for i in range(t):
am = list(map(int, input().split()))
a = am[0]
m = am[1]
g = math.gcd(a, m)
divs = divisors(m // g)
sum = 0
for div in divs:
sum += u(div) * (m // g // div)
print(sum)
``` | instruction | 0 | 17,824 | 22 | 35,648 |
No | output | 1 | 17,824 | 22 | 35,649 |
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