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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601
Submitted Solution:
```
import sys
stdin = sys.stdin
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip() # ignore trailing spaces
# 最高次は等しくないといけないので、
# y % x = y-xになる
# yがxを包含していればok
mod = 1000000007
def ca(x, y):
ct = 0
for i in range(x, y+1):
for j in range(i, y+1):
if j % i == (j^i):
ct += 1
return ct
ans = 0
l, r = na()
dp = [0] * 4
for i in range(60, -1, -1):
ndp = [0] * 4
base = 1<<i
if base <= r and l < base*2:
ptn = 0
if base+1 <= l:
ptn += 1
if r < base*2-1:
ptn += 2
ndp[ptn] += 1
if l>>i != r>>i:
if (l>>i&1) == 0 and (r>>i&1) == 1:
ndp[1] += dp[3]
ndp[2] += dp[3]
if ((l>>i)&(r>>i)) == (l>>i):
ndp[3] += dp[3]
if (l >> i & 1) == 0:
ndp[0] += dp[1]
ndp[1] += dp[1]
ndp[1] += dp[1]
if (r >> i & 1) == 1:
ndp[0] += dp[2]
ndp[2] += dp[2]
ndp[2] += dp[2]
ndp[0] += dp[0] * 3
for k in range(4):
ndp[k] %= mod
dp = ndp
print(sum(dp) % mod)
``` | instruction | 0 | 69,622 | 22 | 139,244 |
Yes | output | 1 | 69,622 | 22 | 139,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601
Submitted Solution:
```
def count(lb, rb):
assert lb[0] == '1'
assert rb[0] == '1'
assert len(lb) == len(rb)
dp = [1, 0, 0, 0]
for lc, rc in zip(lb[1:], rb[1:]):
ndp = [dp[0], 0, 0, 0]
if rc == '0':
ndp[1] += dp[1]
if lc == '1':
ndp[0] = 0
else:
ndp[1] += dp[1] * 2
ndp[2] += dp[1]
if lc == '0':
ndp[1] += dp[0]
ndp[3] += dp[0]
if lc == '0':
ndp[2] += dp[3]
ndp[3] += dp[3] * 2
else:
ndp[3] += dp[3]
ndp[2] += dp[2] * 3
dp = ndp
return sum(dp)
l, r = map(int, input().split())
lb = bin(l)[2:]
rb = bin(r)[2:]
ld = len(lb)
rd = len(rb)
ans = 0
MOD = 10 ** 9 + 7
for d in range(ld, rd + 1):
tlb = lb if d == ld else '1' + '0' * (d - 1)
trb = rb if d == rd else '1' * d
ans = (ans + count(tlb, trb)) % MOD
print(ans)
``` | instruction | 0 | 69,623 | 22 | 139,246 |
Yes | output | 1 | 69,623 | 22 | 139,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601
Submitted Solution:
```
# coding: utf-8
# Your code here!
import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline #文字列入力のときは注意
#n = int(input())
l,r = [int(i) for i in readline().split()]
def make_dp(init, size):
res = "[{}]*{}".format(init,size[-1])
for i in reversed(size[:-1]):
res = "[{} for _ in [0]*{}]".format(res,i)
return eval(res)
MOD = 10**9 + 7
R = bin(r)[2:]
L = bin(l)[2:]
L = "0"*(len(R)-len(L)) + L
dp = make_dp(0,(len(R)+1,2,2,2))
dp[0][0][0][0] = 1
for i, (r,l) in enumerate(zip(R,L)): #i桁目からi+1桁目に遷移
ri = int(r)
li = int(l)
for is_less in range(2):
for is_more in range(2):
for is_nonzero in range(2):
for dl in range(0 if is_more else li,2):
for dr in range(dl, 2 if is_less else ri+1): # d: i+1桁目の数字
dp[i+1][is_nonzero or dr != 0][is_more or li < dl][is_less or dr < ri] += dp[i][is_nonzero][is_more][is_less]*(dr==dl or is_nonzero)
dp[i+1][is_nonzero or dr != 0][is_more or li < dl][is_less or dr < ri] %= MOD
#print(dp[-1])
ans = 0
for i in range(2):
for j in range(2):
for k in range(2):
ans += dp[-1][i][j][k]
print(ans%MOD)
``` | instruction | 0 | 69,624 | 22 | 139,248 |
Yes | output | 1 | 69,624 | 22 | 139,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601
Submitted Solution:
```
p = 10**9+7
l,r = map(int, input().split())
if l==r:
print(1)
else:
ret = [min(i, r-i+1)%p for i in range(l,r+1)]
print(sum(ret)%p)
``` | instruction | 0 | 69,625 | 22 | 139,250 |
No | output | 1 | 69,625 | 22 | 139,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601
Submitted Solution:
```
# -*- coding: utf-8 -*-
l, r = map(int, input().split())
count = 0
for x in range(l, r + 1):
for y in range(l, r + 1):
if (x % y) == (x ^ y):
count += 1
print(count % (10**9 + 7))
``` | instruction | 0 | 69,626 | 22 | 139,252 |
No | output | 1 | 69,626 | 22 | 139,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601
Submitted Solution:
```
from itertools import product
L, R = map(int, input().split())
MOD = 10**9 + 7
dp = [[[0, 0] for _ in range(2)] for _ in range(2)]
pp = [[[0, 0] for _ in range(2)] for _ in range(2)]
dp[0][0][0] = 1
for d in range(60, -1, -1):
pp, dp = dp, pp
dp = [[[0, 0] for _ in range(2)] for _ in range(2)]
lb = L>>d & 1
rb = R>>d & 1
if d == 1:
pass
ans = 0
for lrs in product((0,1), repeat=3):
l, r, s = lrs
for xy in product((0,1), repeat=2):
nl, nr, ns = l, r, s
x, y = xy
if x and not y:
continue
if not s and x != y:
continue
if x and y:
ns = 1
if not l and not x and lb:
continue
if x and not lb:
nl = 1
if not r and y and not rb:
continue
if not y and rb:
nr = 1
dp[nl][nr][ns] += pp[l][r][s]
dp[nl][nr][ns] %= MOD
print(sum(dp[l][r][s] for l in (0,1) for r in (0,1) for s in (0,1)))
``` | instruction | 0 | 69,627 | 22 | 139,254 |
No | output | 1 | 69,627 | 22 | 139,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601
Submitted Solution:
```
L,R=input().split()
l=int(L)
r=int(R)
pa=0
for i in range(l,r+1):
for j in range(l,i+1):
if i%j==i^j:
pa+=1
print(pa%(10**9+7))
``` | instruction | 0 | 69,628 | 22 | 139,256 |
No | output | 1 | 69,628 | 22 | 139,257 |
Provide a correct Python 3 solution for this coding contest problem.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7 | instruction | 0 | 69,792 | 22 | 139,584 |
"Correct Solution:
```
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=NTL_1_A&lang=ja
def prime_factors(n): # 戻り値はiterable type
i = 2
while i * i <= n:
if n % i:
i += 1
else:
n //= i
yield i
if n > 1:
yield n
N = int(input())
P = list(prime_factors(N)) # 素因数のリスト
print(str(N) + ":", *P)
``` | output | 1 | 69,792 | 22 | 139,585 |
Provide a correct Python 3 solution for this coding contest problem.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7 | instruction | 0 | 69,793 | 22 | 139,586 |
"Correct Solution:
```
n=int(input())
A=[]
B=n
while n%2==0:
n//=2
A.append(2)
while n%3==0:
n//=3
A.append(3)
while n%5==0:
n//=5
A.append(5)
while n%7==0:
n//=7
A.append(7)
if n>=10000000:
A.append(n)
else:
for j in range(3,n+1,2):
while n%j==0:
n//=j
A.append(j)
print(B,end="")
print(':',*A)
``` | output | 1 | 69,793 | 22 | 139,587 |
Provide a correct Python 3 solution for this coding contest problem.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7 | instruction | 0 | 69,794 | 22 | 139,588 |
"Correct Solution:
```
def factorization(n):
L=[]
temp=n
print(n,end="")
print(":",end="")
for i in range(2,int(n**(1/2)//1)+1):
if temp%i==0:
c=0
while temp%i==0:
c+=1
temp//=i
L.append([i,c])
for j in range(c):
print("",i,end="")
if temp!=1:
print("",temp,end="")
print()
n=int(input())
factorization(n)
``` | output | 1 | 69,794 | 22 | 139,589 |
Provide a correct Python 3 solution for this coding contest problem.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7 | instruction | 0 | 69,795 | 22 | 139,590 |
"Correct Solution:
```
from math import sqrt
n = int(input())
def prime_factors(N):
a = []
while N % 2 == 0:
a.append(2)
N //= 2
f = 3
while f * f <= N:
if N % f == 0:
a.append(f)
N //= f
else:
f += 2
if N != 1:
a.append(N)
return a
factors = prime_factors(n)
print("{}:".format(n), *factors)
``` | output | 1 | 69,795 | 22 | 139,591 |
Provide a correct Python 3 solution for this coding contest problem.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7 | instruction | 0 | 69,796 | 22 | 139,592 |
"Correct Solution:
```
n=int(input())
def prime_factorize(n):
a = []
while n % 2 == 0:
a.append(2)
n //= 2
f = 3
while f * f <= n:
if n % f == 0:
a.append(f)
n //= f
else:
f += 2
if n != 1:
a.append(n)
return a
b=prime_factorize(n)
print(f'{n}: ',end="")
print(*b)
``` | output | 1 | 69,796 | 22 | 139,593 |
Provide a correct Python 3 solution for this coding contest problem.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7 | instruction | 0 | 69,797 | 22 | 139,594 |
"Correct Solution:
```
def prime_factor(n):
ass = []
for i in range(2,int(n**0.5)+1):
while n % i==0:
ass.append(i)
n = n//i
if n != 1:
ass.append(n)
return ass
a = int(input())
print(str(a)+': ',end='')
print(*prime_factor(a))
``` | output | 1 | 69,797 | 22 | 139,595 |
Provide a correct Python 3 solution for this coding contest problem.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7 | instruction | 0 | 69,798 | 22 | 139,596 |
"Correct Solution:
```
n=int(input())
n1=n
l=[]
def isprime(n):
if n==2:
return True
elif n==1:
return False
else:
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
for i in range(2,int(n**0.5)+1):
while n1%i==0:
l.append(i)
n1=n1//i
if isprime(n1):
l.append(n1)
break
w=str(n)+":"
for i in l:
w+=" "+str(i)
print(w)
``` | output | 1 | 69,798 | 22 | 139,597 |
Provide a correct Python 3 solution for this coding contest problem.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7 | instruction | 0 | 69,799 | 22 | 139,598 |
"Correct Solution:
```
n = int(input())
m = int(n**0.5)+2
is_prime = [1]*m
is_prime[0] = is_prime[1] = 0
for p in range(2,m):
for q in range(2*p,m,p):
is_prime[q] = 0
primes = [p for p, c in enumerate(is_prime) if c]
print(str(n)+': ', end='')
ans = []
for p in primes:
while n % p == 0:
ans.append(p)
n //= p
if n > 1:
ans.append(n)
print(*ans)
``` | output | 1 | 69,799 | 22 | 139,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7
Submitted Solution:
```
def f(n):
a=[]
while n%2==0:
a.append(2)
n=n//2
f=3
while f*f<=n:
if n%f==0:
a.append(f)
n=n//f
else:
f+=2
if n!=1:
a.append(n)
return(a)
n=int(input())
a=f(n)
b=[str(i) for i in a]
A=" ".join(b)
print(f"{n}: {A}")
``` | instruction | 0 | 69,800 | 22 | 139,600 |
Yes | output | 1 | 69,800 | 22 | 139,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7
Submitted Solution:
```
n = int(input())
F = {}
tmp = n
i = 2
while i**2 <= tmp:
cnt = 0
while tmp % i == 0:
cnt += 1
tmp //= i
if cnt > 0:
F[i] = cnt
i += 1
if tmp != 1 or F == {}:
F[tmp] = 1
G = []
for p in F:
for i in range(F[p]):
G.append(str(p))
G = ' '.join(G)
print(f'{n}: {G}')
``` | instruction | 0 | 69,801 | 22 | 139,602 |
Yes | output | 1 | 69,801 | 22 | 139,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7
Submitted Solution:
```
def factorize(n):
i = 2
m = n
factors = []
while i*i <= m:
if m%i==0:
m //= i
factors.append(i)
else:
i += 1
if m > 1:
factors.append(m)
return factors
n = int(input())
primes = factorize(n)
print("{0}: {1}".format(n, " ".join(list(map(str,primes)))))
``` | instruction | 0 | 69,802 | 22 | 139,604 |
Yes | output | 1 | 69,802 | 22 | 139,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7
Submitted Solution:
```
p=lambda x:print(x,end=' ')
n=input()
p(n+':')
n=int(n);s=n**.5
while n%2==0 and n>3:
p(2)
n//=2
d=3
while s>d and n>d:
if n%d==0:
p(d)
n//=d
else:d+=2
print(n)
``` | instruction | 0 | 69,803 | 22 | 139,606 |
Yes | output | 1 | 69,803 | 22 | 139,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7
Submitted Solution:
```
# -*- coding:utf-8 -*-
n = int(input())
print(str(n)+':',end='')
array = []
count = 0
while True:
for i in range(2,n+1):
if n%i == 0 and i != n:
array.append(i)
n = int(n/i)
break
if i == n:
count += 1
break
if count == 15:
break
array.append(n)
for i in range(len(array)):
print(' ',array[i],sep = '',end='')
print('')
``` | instruction | 0 | 69,804 | 22 | 139,608 |
No | output | 1 | 69,804 | 22 | 139,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7
Submitted Solution:
```
import math
def generate_prime(max_number):
prime_list = []
num = 1
if max_number == 1:
return 0
elif max_number >= 2:
yield 2
# count up to max number
for i in range(3,max_number,2):
num += 2
# judge: 1
if num == 1:
continue
# judge: 2
if num == 2:
yield 2
continue
if num % 2 == 0:
continue
if num > int(max_number / 2):
yield max_number
break
k = 1
ok = ''
while k < num:
k += 2
if num % k == 0:
break
else:
ok = 'X'
if ok == 'X':
yield num
def main():
n = 1000000#int(input())
if n >= 2:
print(str(n)+':',end='')
for factor in generate_prime(n):
end = ''
while True:
# if n == factor:
# print('',factor,end='')
# break
if n % factor == 0:
n = int(n / factor)
print('',factor,end='')
if n == 1:
end = 'X'
break
else:
break
if end == 'X':
break
print()
if __name__ == '__main__':
main()
``` | instruction | 0 | 69,805 | 22 | 139,610 |
No | output | 1 | 69,805 | 22 | 139,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7
Submitted Solution:
```
try:
n = int(input())
except:
exit
max_len = n
work = [1 for i in range(0,max_len+1)]
prime = []
soinsu = []
for i in range(2,max_len+1):
if (work[i] == 1 ):
prime.append(i)
j = i * 2
while ( j <= max_len ):
work[j] = 0
j += i
t = n
while (t > 1):
for i in prime:
if (t % i == 0 ):
soinsu.append(i)
t = t // i
soinsu.sort()
print (n,":",end ="",sep ="")
for i in soinsu:
print (" ",i, end ="",sep="")
print ("\n")
``` | instruction | 0 | 69,806 | 22 | 139,612 |
No | output | 1 | 69,806 | 22 | 139,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. Print a space before each prime factor.
Examples
Input
12
Output
12: 2 2 3
Input
126
Output
126: 2 3 3 7
Submitted Solution:
```
a=int(input())
n=a
i=2
fc=[]
while i*i<=n:
while n%i==0:
n//=i
fc.append(i)
i+=1
if n>1:
fc.append(n)
print(str(n)+": "+" ".join(str(b) for b in fc))
``` | instruction | 0 | 69,807 | 22 | 139,614 |
No | output | 1 | 69,807 | 22 | 139,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct positive integers. Your task is to find the size of it's largest k-multiple free subset.
Input
The first line of the input contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109). The next line contains a list of n distinct positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).
All the numbers in the lines are separated by single spaces.
Output
On the only line of the output print the size of the largest k-multiple free subset of {a1, a2, ..., an}.
Examples
Input
6 2
2 3 6 5 4 10
Output
3
Note
In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}.
Submitted Solution:
```
import sys
from math import *
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
n, k = mints()
if k == 1:
print(n)
exit(0)
m = dict()
for i in mints():
c = 0
x = i
while x % k == 0:
x //= k
c += 1
if x not in m:
m[x] = [0]*34
m[x][c] += 1
r = 0
for i in m:
f = True
for j in m[i]:
if f and j != 0:
r += 1
f = False
else:
f = True
print(r)
``` | instruction | 0 | 70,139 | 22 | 140,278 |
Yes | output | 1 | 70,139 | 22 | 140,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3 | instruction | 0 | 70,475 | 22 | 140,950 |
Tags: implementation, math
Correct Solution:
```
print('\n'.join(('F','Inf')[pow(b,64,q)*p%q>0]+'inite'for p,q,b
in(map(int,input().split())for _ in[0]*int(input()))))
``` | output | 1 | 70,475 | 22 | 140,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3 | instruction | 0 | 70,476 | 22 | 140,952 |
Tags: implementation, math
Correct Solution:
```
import sys
def main():
n = int(input())
ans = []
for i in range(n):
p, q, b = map(int, input().split(" "))
t = pow(b, 111, q)
if p * t % q == 0:
ans.append("Finite")
else:
ans.append("Infinite")
print("\n".join(ans))
main()
``` | output | 1 | 70,476 | 22 | 140,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3 | instruction | 0 | 70,477 | 22 | 140,954 |
Tags: implementation, math
Correct Solution:
```
# python3
import sys
from fractions import gcd
def is_finite(p, q, b):
return not p * pow(b, 64, q) % q
input()
print("\n".join("Finite" if is_finite(*map(int, line.split())) else "Infinite"
for line in sys.stdin.readlines()))
``` | output | 1 | 70,477 | 22 | 140,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3 | instruction | 0 | 70,478 | 22 | 140,956 |
Tags: implementation, math
Correct Solution:
```
import sys
input()
out = []
for line in sys.stdin:
p,q,b=[int(x) for x in line.split()]
out.append(['Finite','Infinite'][bool(p*pow(b,60,q)%q)])
print('\n'.join(out))
``` | output | 1 | 70,478 | 22 | 140,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3 | instruction | 0 | 70,479 | 22 | 140,958 |
Tags: implementation, math
Correct Solution:
```
import sys
import math
ini = lambda: int(sys.stdin.readline())
inl = lambda: [int(x) for x in sys.stdin.readline().split()]
def solve():
p, q, b = inl()
if p == 0:
return True
g = math.gcd(p, q)
p //= g
q //= g
if q == 1:
return True
for i in range(6):
b = b * b % q
if b == 0:
return True
return False
n = ini()
for i in range(n):
print(["Infinite", "Finite"][solve()])
``` | output | 1 | 70,479 | 22 | 140,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3 | instruction | 0 | 70,480 | 22 | 140,960 |
Tags: implementation, math
Correct Solution:
```
input()
print('\n'.join(map(lambda x: (lambda p, q, b: 'Infinite' if p * pow(b, 60, q) % q else 'Finite')(*x), map(lambda l:map(int, l.split()), __import__('sys').stdin.readlines()))))
``` | output | 1 | 70,480 | 22 | 140,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3 | instruction | 0 | 70,481 | 22 | 140,962 |
Tags: implementation, math
Correct Solution:
```
n=int(input())
s=''
for i in range(n):
p,q,b=map(int,input().split())
for i in range(6):
b=(b*b)%q
if((p*b)%q):
s+='Infinite\n'
else:
s+='Finite\n'
print(s)
``` | output | 1 | 70,481 | 22 | 140,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3 | instruction | 0 | 70,482 | 22 | 140,964 |
Tags: implementation, math
Correct Solution:
```
from sys import stdin, stdout
n=int(stdin.readline())
s=''
for i in range(n):
p,q,b=map(int,input().split())
for i in range(6):
b=(b*b)%q
if((p*b)%q):
s+='Infinite\n'
else:
s+='Finite\n'
print(s)
``` | output | 1 | 70,482 | 22 | 140,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3
Submitted Solution:
```
sy = 'Finite'
sn = 'Infinite'
n = int(input())
result = ['' for i in range(n)]
for i in range(n):
p, q, b = list(map(int, input().split()))
if p * pow(b, 60, q) % q:
result[i] = sn
else:
result[i] = sy
print('\n'.join(result))
``` | instruction | 0 | 70,483 | 22 | 140,966 |
Yes | output | 1 | 70,483 | 22 | 140,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3
Submitted Solution:
```
n = int(input())
ans = []
while n:
n += -1
p, q, b = map(int, input().split())
if p * pow(b, 62, q) % q: ans.append("Infinite")
else: ans.append("Finite")
for _ in ans: print(_)
``` | instruction | 0 | 70,484 | 22 | 140,968 |
Yes | output | 1 | 70,484 | 22 | 140,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3
Submitted Solution:
```
from math import gcd
import os
from io import BytesIO
input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
def vvod(): return int(input())
def vvod1(): return map(int, input().split())
n = vvod()
for i in range(n):
p, q, k = vvod1()
dele = gcd(p, q)
q = q // dele
dele = gcd(k, q)
while dele != 1:
while q % dele == 0:
q = q // dele
dele = gcd(k, q)
if q == 1:
print("Finite")
else:
print("Infinite")
``` | instruction | 0 | 70,486 | 22 | 140,972 |
Yes | output | 1 | 70,486 | 22 | 140,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3
Submitted Solution:
```
N = int(input())
def gcd(a,b):
r0 = a
r1 = b
while r1 != 0:
r0,r1 = r1,r0%r1
return r0
for _ in range(N):
p,q,b = [int(x) for x in input().split()]
g = gcd(p,q)
q //= g
if gcd(b,q) == 1: print("Infinite")
else: print("Finite")
``` | instruction | 0 | 70,487 | 22 | 140,974 |
No | output | 1 | 70,487 | 22 | 140,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3
Submitted Solution:
```
from math import gcd
norm = "Finite"
nenorm = "Infinite"
def factors(n):
factors = [1, 2]
d = 2
while d * d <= n:
if n % d == 0:
factors.append(d)
n//=d
else:
d += 1
if n > 1:
factors.append(n)
else:
break
return factors
for i in range(int(input())):
p, q, b = map(int, input().split(' '))
g = gcd(q, p)
q //= g
q_factors = set(factors(q))
b_factors = set(factors(b))
for factor in q_factors:
if factor not in b_factors:
print(nenorm)
break
else:
print(norm)
# while q != 1:
# j = gcd(q, b)
# if j==1:
# print(nenorm)
# break
# q //= j
# else:
# print(norm)
``` | instruction | 0 | 70,488 | 22 | 140,976 |
No | output | 1 | 70,488 | 22 | 140,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given several queries. Each query consists of three integers p, q and b. You need to answer whether the result of p/q in notation with base b is a finite fraction.
A fraction in notation with base b is finite if it contains finite number of numerals after the decimal point. It is also possible that a fraction has zero numerals after the decimal point.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of queries.
Next n lines contain queries, one per line. Each line contains three integers p, q, and b (0 ≤ p ≤ 10^{18}, 1 ≤ q ≤ 10^{18}, 2 ≤ b ≤ 10^{18}). All numbers are given in notation with base 10.
Output
For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.
Examples
Input
2
6 12 10
4 3 10
Output
Finite
Infinite
Input
4
1 1 2
9 36 2
4 12 3
3 5 4
Output
Finite
Finite
Finite
Infinite
Note
6/12 = 1/2 = 0,5_{10}
4/3 = 1,(3)_{10}
9/36 = 1/4 = 0,01_2
4/12 = 1/3 = 0,1_3
Submitted Solution:
```
from fractions import gcd
TC = int(input())
for T in range(TC):
p, q, b = map(int, input().split())
g = gcd(p, q)
p //= g
q //= g
seen = set()
while(True):
seen.add(p)
p %= q
p *= b
if(p == 0 or p in seen):
break
print(len(seen))
print(["Finite", "Infinite"][bool(p)])
``` | instruction | 0 | 70,490 | 22 | 140,980 |
No | output | 1 | 70,490 | 22 | 140,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Janusz is a businessman. He owns a company "Januszex", which produces games for teenagers. Last hit of Januszex was a cool one-person game "Make it one". The player is given a sequence of n integers a_i.
It is allowed to select any subset of them, and the score is equal to the greatest common divisor of selected elements. The goal is to take as little elements as it is possible, getting the score 1. Now Janusz wonders, for given sequence, how much elements should the player choose?
Input
The first line contains an only integer n (1 ≤ n ≤ 300 000) — the number of integers in the sequence.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000).
Output
If there is no subset of the given sequence with gcd equal to 1, output -1.
Otherwise, output exactly one integer — the size of the smallest subset with gcd equal to 1.
Examples
Input
3
10 6 15
Output
3
Input
3
2 4 6
Output
-1
Input
7
30 60 21 42 70 15 30
Output
3
Note
In the first example, selecting a subset of all numbers gives a gcd of 1 and for all smaller subsets the gcd is greater than 1.
In the second example, for all subsets of numbers the gcd is at least 2.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
k=-1
i=2
d1,d0=False,False
while not d1 and not d0 and i <= min(a):
d1,d0=False,False
for j in range(n):
if a[j]%i==1: d1=True
elif a[j]%i==0: d0=True
i+=1
if d1 and d0: k=i
print(k)
``` | instruction | 0 | 70,741 | 22 | 141,482 |
No | output | 1 | 70,741 | 22 | 141,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Janusz is a businessman. He owns a company "Januszex", which produces games for teenagers. Last hit of Januszex was a cool one-person game "Make it one". The player is given a sequence of n integers a_i.
It is allowed to select any subset of them, and the score is equal to the greatest common divisor of selected elements. The goal is to take as little elements as it is possible, getting the score 1. Now Janusz wonders, for given sequence, how much elements should the player choose?
Input
The first line contains an only integer n (1 ≤ n ≤ 300 000) — the number of integers in the sequence.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000).
Output
If there is no subset of the given sequence with gcd equal to 1, output -1.
Otherwise, output exactly one integer — the size of the smallest subset with gcd equal to 1.
Examples
Input
3
10 6 15
Output
3
Input
3
2 4 6
Output
-1
Input
7
30 60 21 42 70 15 30
Output
3
Note
In the first example, selecting a subset of all numbers gives a gcd of 1 and for all smaller subsets the gcd is greater than 1.
In the second example, for all subsets of numbers the gcd is at least 2.
Submitted Solution:
```
from sys import *
maxn = 3 * 10 ** 5 + 5
fre = [0 for i in range(maxn)]
isprime = [1 for i in range(maxn)]
prime = []
divi = [0 for i in range(maxn)]
fact = [1] * 10
def nCr(n, r):
if n < r:
return 0
if n == r:
return 1
pro = 1
for i in range(r):
pro *= (n - i)
pro //= fact[r]
return pro
n = int(stdin.readline())
arr = list(map(int, stdin.readline().split()))
for i in arr:
if i is 1:
print(1)
exit()
fre[i] += 1
divi[1] = n
for i in range(2, maxn):
if isprime[i] is 1:
prime.append(i)
for j in range(1, maxn):
if i * j >= maxn:
break
isprime[i * j] = 0
divi[i] += fre[i * j]
for i in range(1, 10):
fact[i] = fact[i - 1] * i
mobius = [0 for i in range(maxn)]
for i in range(1, maxn):
mobius[i] = 1
for p in prime:
if p * p >= maxn:
break
x = p * p
for j in range(x, maxn, x):
mobius[j] = 0
for p in prime:
for j in range(p, maxn, p):
mobius[j] *= -1
for r in range(3, 4):
coprime = 0
for d in range(1, maxn):
ncr = nCr(divi[d], r)
coprime += mobius[d] * ncr
if coprime > 0:
print(r)
exit()
print(-1)
``` | instruction | 0 | 70,742 | 22 | 141,484 |
No | output | 1 | 70,742 | 22 | 141,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Janusz is a businessman. He owns a company "Januszex", which produces games for teenagers. Last hit of Januszex was a cool one-person game "Make it one". The player is given a sequence of n integers a_i.
It is allowed to select any subset of them, and the score is equal to the greatest common divisor of selected elements. The goal is to take as little elements as it is possible, getting the score 1. Now Janusz wonders, for given sequence, how much elements should the player choose?
Input
The first line contains an only integer n (1 ≤ n ≤ 300 000) — the number of integers in the sequence.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000).
Output
If there is no subset of the given sequence with gcd equal to 1, output -1.
Otherwise, output exactly one integer — the size of the smallest subset with gcd equal to 1.
Examples
Input
3
10 6 15
Output
3
Input
3
2 4 6
Output
-1
Input
7
30 60 21 42 70 15 30
Output
3
Note
In the first example, selecting a subset of all numbers gives a gcd of 1 and for all smaller subsets the gcd is greater than 1.
In the second example, for all subsets of numbers the gcd is at least 2.
Submitted Solution:
```
def isprime(x):
if x > 3 and x % 2 == 0 or x <= 1:
return False
for i in range(3, int(x ** 0.5) + 1, 2):
if x % i == 0:
return False
return True
def delit(x):
d=set()
for i in range(2,x//2+1):
if x%i==0 and isprime(i): d.add(i)
return d
k=0
n = int(input())
a = list(map(int,input().split()))
vsedel = set()
otdel = set()
for ch in a:
chdel=delit(ch)
dobdel=set()
dobdel.update(chdel)
dobdel.difference_update(otdel)
vsedel.update(dobdel)
if len(vsedel.difference(chdel))>0:
otdel.update(vsedel.difference(chdel))
vsedel.intersection_update(chdel)
k+=1
if len(vsedel)>0 and not(vsedel in delit(a[0])):k+=1
if k:print(k)
else:print(-1)
``` | instruction | 0 | 70,743 | 22 | 141,486 |
No | output | 1 | 70,743 | 22 | 141,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Janusz is a businessman. He owns a company "Januszex", which produces games for teenagers. Last hit of Januszex was a cool one-person game "Make it one". The player is given a sequence of n integers a_i.
It is allowed to select any subset of them, and the score is equal to the greatest common divisor of selected elements. The goal is to take as little elements as it is possible, getting the score 1. Now Janusz wonders, for given sequence, how much elements should the player choose?
Input
The first line contains an only integer n (1 ≤ n ≤ 300 000) — the number of integers in the sequence.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000).
Output
If there is no subset of the given sequence with gcd equal to 1, output -1.
Otherwise, output exactly one integer — the size of the smallest subset with gcd equal to 1.
Examples
Input
3
10 6 15
Output
3
Input
3
2 4 6
Output
-1
Input
7
30 60 21 42 70 15 30
Output
3
Note
In the first example, selecting a subset of all numbers gives a gcd of 1 and for all smaller subsets the gcd is greater than 1.
In the second example, for all subsets of numbers the gcd is at least 2.
Submitted Solution:
```
from sys import *
from collections import *
from math import *
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def lsi():
x = li()
x.pop()
return x
def si(): return stdin.readline()
fre = [0] * (4 * 10 ** 5 + 5)
divi = [0] * (4 * 10 ** 5 + 5)
isprime = [0] * (4 * 10 ** 5 + 5)
prime = []
def seive(n):
for i in range(n):
isprime[i] = True
for i in range(2, n):
if isprime[i] is True:
prime.append(i)
for j in range(1, n):
if i * j >= n:
break
isprime[i * j] = False
divi[i] += fre[i * j]
mobius = [0] * (4 * 10 ** 5 + 5)
def calc_mobius(n):
for i in range(1, n):
mobius[i] = 1
for p in prime:
if p * p >= n:
break
x = p * p
for j in range(x, n, x):
mobius[j] = 0
for p in prime:
for j in range(p, n, p):
mobius[j] *= -1
fact = [1] * 10
def calc_fact():
fact[0] = 1
for i in range(1, 10):
fact[i] = i * fact[i - 1]
def nCr(n, r):
pro = 1
for i in range(r):
pro *= (n - i)
pro /= fact[r]
return pro
def count_coprime(r):
coprime = 0
for d in range(1, 3 * 10 ** 5 + 1):
ncr = nCr(divi[d], r)
# print("HERE {}={}, {}={}".format(d, divi[d], r, ncr))
coprime += mobius[d] * ncr
return coprime;
if __name__ == "__main__":
n = int(stdin.readline())
arr = list(map(int, stdin.readline().split()))
for i in arr:
if i is 1:
print(i)
exit()
fre[i] += 1
divi[1] = n
seive(3 * 10 ** 5 + 1)
calc_mobius(3 * 10 ** 5 + 1)
calc_fact()
for r in range(1, 10):
coprime = count_coprime(r)
# print("{}={}".format(r, coprime))
if coprime > 0:
exit(print(r))
print(-1)
``` | instruction | 0 | 70,744 | 22 | 141,488 |
No | output | 1 | 70,744 | 22 | 141,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you have received two positive integer numbers x and y. You forgot them, but you remembered a shuffled list containing all divisors of x (including 1 and x) and all divisors of y (including 1 and y). If d is a divisor of both numbers x and y at the same time, there are two occurrences of d in the list.
For example, if x=4 and y=6 then the given list can be any permutation of the list [1, 2, 4, 1, 2, 3, 6]. Some of the possible lists are: [1, 1, 2, 4, 6, 3, 2], [4, 6, 1, 1, 2, 3, 2] or [1, 6, 3, 2, 4, 1, 2].
Your problem is to restore suitable positive integer numbers x and y that would yield the same list of divisors (possibly in different order).
It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers x and y.
Input
The first line contains one integer n (2 ≤ n ≤ 128) — the number of divisors of x and y.
The second line of the input contains n integers d_1, d_2, ..., d_n (1 ≤ d_i ≤ 10^4), where d_i is either divisor of x or divisor of y. If a number is divisor of both numbers x and y then there are two copies of this number in the list.
Output
Print two positive integer numbers x and y — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.
Example
Input
10
10 2 8 1 2 4 1 20 4 5
Output
20 8 | instruction | 0 | 70,777 | 22 | 141,554 |
Tags: brute force, greedy, math, number theory
Correct Solution:
```
def factors(number):
factors=[]
for i in range(1,number+1):
if number%i==0:
factors.append(i)
return factors
n=int(input())
fac=input().split()
fac=list(map(int,fac))
number=max(fac)
for x in factors(number):
fac.remove(x)
number2=max(fac)
print(number,number2)
``` | output | 1 | 70,777 | 22 | 141,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you have received two positive integer numbers x and y. You forgot them, but you remembered a shuffled list containing all divisors of x (including 1 and x) and all divisors of y (including 1 and y). If d is a divisor of both numbers x and y at the same time, there are two occurrences of d in the list.
For example, if x=4 and y=6 then the given list can be any permutation of the list [1, 2, 4, 1, 2, 3, 6]. Some of the possible lists are: [1, 1, 2, 4, 6, 3, 2], [4, 6, 1, 1, 2, 3, 2] or [1, 6, 3, 2, 4, 1, 2].
Your problem is to restore suitable positive integer numbers x and y that would yield the same list of divisors (possibly in different order).
It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers x and y.
Input
The first line contains one integer n (2 ≤ n ≤ 128) — the number of divisors of x and y.
The second line of the input contains n integers d_1, d_2, ..., d_n (1 ≤ d_i ≤ 10^4), where d_i is either divisor of x or divisor of y. If a number is divisor of both numbers x and y then there are two copies of this number in the list.
Output
Print two positive integer numbers x and y — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.
Example
Input
10
10 2 8 1 2 4 1 20 4 5
Output
20 8 | instruction | 0 | 70,778 | 22 | 141,556 |
Tags: brute force, greedy, math, number theory
Correct Solution:
```
n = int(input())
d = list(map(int, input().split()))
x = max(d)
if d.count(x) == 2:
print(x, x)
else:
setd = set(d)
notx = []
y = []
for i in setd:
if x % i != 0:
notx.append(i)
if len(notx) != 0:
print(x, max(notx))
else:
for i in setd:
if d.count(i) == 2:
y.append(i)
print(x, max(y))
``` | output | 1 | 70,778 | 22 | 141,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you have received two positive integer numbers x and y. You forgot them, but you remembered a shuffled list containing all divisors of x (including 1 and x) and all divisors of y (including 1 and y). If d is a divisor of both numbers x and y at the same time, there are two occurrences of d in the list.
For example, if x=4 and y=6 then the given list can be any permutation of the list [1, 2, 4, 1, 2, 3, 6]. Some of the possible lists are: [1, 1, 2, 4, 6, 3, 2], [4, 6, 1, 1, 2, 3, 2] or [1, 6, 3, 2, 4, 1, 2].
Your problem is to restore suitable positive integer numbers x and y that would yield the same list of divisors (possibly in different order).
It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers x and y.
Input
The first line contains one integer n (2 ≤ n ≤ 128) — the number of divisors of x and y.
The second line of the input contains n integers d_1, d_2, ..., d_n (1 ≤ d_i ≤ 10^4), where d_i is either divisor of x or divisor of y. If a number is divisor of both numbers x and y then there are two copies of this number in the list.
Output
Print two positive integer numbers x and y — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.
Example
Input
10
10 2 8 1 2 4 1 20 4 5
Output
20 8 | instruction | 0 | 70,779 | 22 | 141,558 |
Tags: brute force, greedy, math, number theory
Correct Solution:
```
import math
n = eval(input())
array = list(map(int, input().split()))
array.sort()
x = array[-1]
arr = []
max1 = -1
for i in array:
if(x%i!=0 or (x%i==0 and array.count(i)==2)):
max1 = max(max1, i)
y = max1
print(x, y)
``` | output | 1 | 70,779 | 22 | 141,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you have received two positive integer numbers x and y. You forgot them, but you remembered a shuffled list containing all divisors of x (including 1 and x) and all divisors of y (including 1 and y). If d is a divisor of both numbers x and y at the same time, there are two occurrences of d in the list.
For example, if x=4 and y=6 then the given list can be any permutation of the list [1, 2, 4, 1, 2, 3, 6]. Some of the possible lists are: [1, 1, 2, 4, 6, 3, 2], [4, 6, 1, 1, 2, 3, 2] or [1, 6, 3, 2, 4, 1, 2].
Your problem is to restore suitable positive integer numbers x and y that would yield the same list of divisors (possibly in different order).
It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers x and y.
Input
The first line contains one integer n (2 ≤ n ≤ 128) — the number of divisors of x and y.
The second line of the input contains n integers d_1, d_2, ..., d_n (1 ≤ d_i ≤ 10^4), where d_i is either divisor of x or divisor of y. If a number is divisor of both numbers x and y then there are two copies of this number in the list.
Output
Print two positive integer numbers x and y — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.
Example
Input
10
10 2 8 1 2 4 1 20 4 5
Output
20 8 | instruction | 0 | 70,780 | 22 | 141,560 |
Tags: brute force, greedy, math, number theory
Correct Solution:
```
n=int(input())
d=input().split()
a=[0 for j in range(10**4+1)]
for i in range(n):
d[i]=int(d[i])
a[d[i]]+=1
d.sort(reverse=True)
x=d[0]
for j in range(1,x+1):
if(x%j==0):
a[j]-=1
for l in range(10**4+1):
if(a[l]==1):
y=l
print(x,y)
``` | output | 1 | 70,780 | 22 | 141,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you have received two positive integer numbers x and y. You forgot them, but you remembered a shuffled list containing all divisors of x (including 1 and x) and all divisors of y (including 1 and y). If d is a divisor of both numbers x and y at the same time, there are two occurrences of d in the list.
For example, if x=4 and y=6 then the given list can be any permutation of the list [1, 2, 4, 1, 2, 3, 6]. Some of the possible lists are: [1, 1, 2, 4, 6, 3, 2], [4, 6, 1, 1, 2, 3, 2] or [1, 6, 3, 2, 4, 1, 2].
Your problem is to restore suitable positive integer numbers x and y that would yield the same list of divisors (possibly in different order).
It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers x and y.
Input
The first line contains one integer n (2 ≤ n ≤ 128) — the number of divisors of x and y.
The second line of the input contains n integers d_1, d_2, ..., d_n (1 ≤ d_i ≤ 10^4), where d_i is either divisor of x or divisor of y. If a number is divisor of both numbers x and y then there are two copies of this number in the list.
Output
Print two positive integer numbers x and y — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.
Example
Input
10
10 2 8 1 2 4 1 20 4 5
Output
20 8 | instruction | 0 | 70,781 | 22 | 141,562 |
Tags: brute force, greedy, math, number theory
Correct Solution:
```
n = int(input())
a = [int(j) for j in input().split()]
x = max(a)
y=1
for i in range(1,x+1):
if x%i == 0:
a.remove(i)
y=max(a)
print(x,y)
# print(a)
``` | output | 1 | 70,781 | 22 | 141,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you have received two positive integer numbers x and y. You forgot them, but you remembered a shuffled list containing all divisors of x (including 1 and x) and all divisors of y (including 1 and y). If d is a divisor of both numbers x and y at the same time, there are two occurrences of d in the list.
For example, if x=4 and y=6 then the given list can be any permutation of the list [1, 2, 4, 1, 2, 3, 6]. Some of the possible lists are: [1, 1, 2, 4, 6, 3, 2], [4, 6, 1, 1, 2, 3, 2] or [1, 6, 3, 2, 4, 1, 2].
Your problem is to restore suitable positive integer numbers x and y that would yield the same list of divisors (possibly in different order).
It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers x and y.
Input
The first line contains one integer n (2 ≤ n ≤ 128) — the number of divisors of x and y.
The second line of the input contains n integers d_1, d_2, ..., d_n (1 ≤ d_i ≤ 10^4), where d_i is either divisor of x or divisor of y. If a number is divisor of both numbers x and y then there are two copies of this number in the list.
Output
Print two positive integer numbers x and y — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.
Example
Input
10
10 2 8 1 2 4 1 20 4 5
Output
20 8 | instruction | 0 | 70,782 | 22 | 141,564 |
Tags: brute force, greedy, math, number theory
Correct Solution:
```
n = int(input())
lst = [int(j) for j in input().split(' ') if j!='' and j!=' ']
lst.sort()
num1 = lst[-1]
print(num1,end=' ')
i = 1
while i<=num1:
if num1%i==0:
lst.remove(i)
i+=1
print(lst[-1])
``` | output | 1 | 70,782 | 22 | 141,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you have received two positive integer numbers x and y. You forgot them, but you remembered a shuffled list containing all divisors of x (including 1 and x) and all divisors of y (including 1 and y). If d is a divisor of both numbers x and y at the same time, there are two occurrences of d in the list.
For example, if x=4 and y=6 then the given list can be any permutation of the list [1, 2, 4, 1, 2, 3, 6]. Some of the possible lists are: [1, 1, 2, 4, 6, 3, 2], [4, 6, 1, 1, 2, 3, 2] or [1, 6, 3, 2, 4, 1, 2].
Your problem is to restore suitable positive integer numbers x and y that would yield the same list of divisors (possibly in different order).
It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers x and y.
Input
The first line contains one integer n (2 ≤ n ≤ 128) — the number of divisors of x and y.
The second line of the input contains n integers d_1, d_2, ..., d_n (1 ≤ d_i ≤ 10^4), where d_i is either divisor of x or divisor of y. If a number is divisor of both numbers x and y then there are two copies of this number in the list.
Output
Print two positive integer numbers x and y — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.
Example
Input
10
10 2 8 1 2 4 1 20 4 5
Output
20 8 | instruction | 0 | 70,783 | 22 | 141,566 |
Tags: brute force, greedy, math, number theory
Correct Solution:
```
def ls(l,n):
# l = [l[i] for i in range(0,n)]
desc = sorted(l,reverse=True)
n1 = desc[0]
l1 = []
l2 = []
for i in range(1,len(desc)):
if (n1 % desc[i])or desc.count(desc[i])>1:
l1.append(desc[i])
else:
l2.append(desc[i])
# print(n1)
if len(l1)==0:
print(n1,l2[0])
else:
print(n1,l1[0])
# return
# print(l1,l2)
n = int(input())
# print(n)
ls(list(map(int, input().rstrip().split())),n)
``` | output | 1 | 70,783 | 22 | 141,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently you have received two positive integer numbers x and y. You forgot them, but you remembered a shuffled list containing all divisors of x (including 1 and x) and all divisors of y (including 1 and y). If d is a divisor of both numbers x and y at the same time, there are two occurrences of d in the list.
For example, if x=4 and y=6 then the given list can be any permutation of the list [1, 2, 4, 1, 2, 3, 6]. Some of the possible lists are: [1, 1, 2, 4, 6, 3, 2], [4, 6, 1, 1, 2, 3, 2] or [1, 6, 3, 2, 4, 1, 2].
Your problem is to restore suitable positive integer numbers x and y that would yield the same list of divisors (possibly in different order).
It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers x and y.
Input
The first line contains one integer n (2 ≤ n ≤ 128) — the number of divisors of x and y.
The second line of the input contains n integers d_1, d_2, ..., d_n (1 ≤ d_i ≤ 10^4), where d_i is either divisor of x or divisor of y. If a number is divisor of both numbers x and y then there are two copies of this number in the list.
Output
Print two positive integer numbers x and y — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.
Example
Input
10
10 2 8 1 2 4 1 20 4 5
Output
20 8 | instruction | 0 | 70,784 | 22 | 141,568 |
Tags: brute force, greedy, math, number theory
Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
X = max(A)
div = set()
for i in range(1, int(X ** (1/2)) + 2):
if X % i == 0:
div.add(i)
div.add(X // i)
f = []
for a in A:
if a in div:
div.remove(a)
else:
f += [a]
Y = max(f)
print(X, end=' ')
print(Y)
``` | output | 1 | 70,784 | 22 | 141,569 |
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