結果

問題 No.2074 Product is Square ?
ユーザー ygd.
提出日時 2022-09-16 23:18:59
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 3,393 bytes
コンパイル時間 489 ms
コンパイル使用メモリ 82,432 KB
実行使用メモリ 79,940 KB
最終ジャッジ日時 2024-12-21 22:57:04
合計ジャッジ時間 30,118 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 25 WA * 1 RE * 2 TLE * 5
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

import sys
#input = sys.stdin.readline
#input = sys.stdin.buffer.readline #文字列はダメ
#sys.setrecursionlimit(1000000)
import math
#import bisect
#import itertools
#import random
#from heapq import heapify, heappop, heappush
from collections import defaultdict 
#from collections import deque
#import copy #DeepCopy: hoge = [_[:] for _ in hogehoge]
#from functools import lru_cache
#@lru_cache(maxsize=None)
#MOD = pow(10,9) + 7
#MOD = 998244353
#dx = [1,0,-1,0]
#dy = [0,1,0,-1]
#dx8 = [1,1,0,-1,-1,-1,0,1]
#dy8 = [0,1,1,1,0,-1,-1,-1]
def gcd(a, b):
    while b:
        a, b = b, a % b
    return a
 
 
def isPrimeMR(n):
    d = n - 1
    d = d // (d & -d)
    L = [2]
    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
#無断使用すいません
def findFactorRho(n):
    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): #これを実行
    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
#高速素因数分解を用いた約数全列挙
import itertools
def factorization_fast(x):
    P = primeFactor(x)
    base = []
    L = []
    for p in P:
        base.append(p)
        temp = [i for i in range(P[p]+1)]
        L.append(temp)
    ret = []
    for T in itertools.product(*L):
        val = 1
        for i,t in enumerate(T):
            val *= pow(base[i],t)
        ret.append(val)
    ret.sort() #必要ならソート
    return ret
def main():
    T = int(input())
    for _ in range(T):
        N = int(input())
        A = list(map(int,input().split()))
        dic = defaultdict(int)
        for a in A:
            PL = primeFactor(a)
            for p,num in PL.items():
                dic[p] ^= num%2
        
        if max(dic.values()) == 0:
            print('Yes')
        else:
            print('No')
if __name__ == '__main__':
    main()
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