結果
| 問題 |
No.2074 Product is Square ?
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2022-09-14 12:42:38 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 518 ms / 2,000 ms |
| コード長 | 2,932 bytes |
| コンパイル時間 | 424 ms |
| コンパイル使用メモリ | 82,220 KB |
| 実行使用メモリ | 71,936 KB |
| 最終ジャッジ日時 | 2024-12-14 23:56:19 |
| 合計ジャッジ時間 | 10,004 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 1 |
| other | AC * 33 |
ソースコード
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import log,gcd
input = lambda :sys.stdin.readline()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
def isPrimeMR(n):
if n==1:
return 0
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
if n in L:
return 1
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
from math import gcd
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def isqrt(a):
ok = 0
ng = 10**9+1
while ng-ok>1:
mid = (ok+ng)//2
if mid*mid <= a:
ok = mid
else:
ng = mid
return ok
def solve(N,_A):
A = _A.copy()
for i in range(N):
for j in range(i+1,N):
g = gcd(A[i],A[j])
A[i] //= g
A[j] //= g
for a in A:
if isqrt(a)**2 !=a:
return False
return True
def big_int(N,A):
p = 1
for a in A:
p *= a
ok = 0
ng = 10**(18*N) + 1
while ng-ok>1:
mid = (ok+ng)//2
if mid*mid <= p:
ok = mid
else:
ng = mid
return ok*ok == p
for _ in range(int(input())):
N = int(input())
A = li()
print("Yes" if solve(N,A) else "No")