solution1（通过10%）

#include<stdio.h>
#include<math.h>
typedef long long LL;
int isPrime(LL n){
	LL sqr = (int)sqrt(1.0 * n);
	for(int i = 2; i <= sqr; i++){
		if(n % i == 0) return 0;
	}
	return 1;
}
int main(){
	int t;
	LL a;
	scanf("%d", &t);
	while(t--){
		LL f[100] = {0}, sqr, num = 0, odd = 0, even = 0, flag = 0;
		scanf("%lld", &a);
		if(isPrime(a)) flag = 1;
		else{
			sqr = (int)sqrt(1.0*a);
			for(int i = 2; i <= sqr; i++){
				if(a % i == 0){
					while(a % i == 0){
						a /= i;
						f[num]++;
					}
          if(f[num] == 1) {
            flag = 1;
            break;
          }
          else if(f[num] % 2 == 0) even++;
          else odd++;				
					num++;
				}
				if(odd > 2 || (odd == 2 && even > 0)){
					flag = 1;
					break;
				}
			}
			if(a > 1) flag = 1;
		}
 		if(flag) printf("no\n");
		else printf("yes\n");
	} 
	return 0;
}

solution2

1 <= a_i <= 10¹⁸,
奇数次幂可以分解为偶数次幂*奇数次幂
则能够拆分的数最后一定能分解为下列形式：

• a²*b³
  ==》至多有五次方，而上界10¹⁸开五次方大约是4000左右

#include<stdio.h>
#include<math.h>
int main(){
	printf("%lld", (long long)pow(1e18, 0.2));//3981
	return 0;
}

列出4000以内的素数，看是否有只出现一次的质因数，若有则一定不能拆分，木有人和它配对不满足指数大于等于2
分解后a等于1或者是大于4000的因子，这样一番分解后a不会太大了，直接判断是否能够分解为平方或者立方的形式融入其中即可

#include<stdio.h>
#include<math.h>
typedef long long LL;
LL p[4005] = {0}, prime[4005], n = 0;
int check(LL a){
	LL s, num;
	for(int i = 1; i <= n && prime[i] * prime[i] <= a; i++){
		num = 0;
		if(a % prime[i] == 0){
			while(a % prime[i] == 0){
				a /= prime[i];
				num++;
			}
	        if(num == 1) return 0;
		}
	}
	s = sqrt(1.0 * a);
	while(s*s <= a){
		if(s*s == a) return 1;
		s++;
	}
	s = pow(a, 1.0 * 1 / 3);
	while(s*s*s <= a){
		if(s*s*s == a) return 1;
		s++;
	} 
	return 0;
}
int main(){
	int t;
	LL a;
	scanf("%d", &t);
	for(int i = 2; i <= 4000; i++){
		if(!p[i]){
			prime[++n] = i;
			for(int j = i * i; j <= 4000; j += i){
				p[j] = 1;
			}
		}
	}
	while(t--){
		scanf("%lld", &a);
 		if(check(a)) printf("yes\n");
		else printf("no\n");
	} 
	return 0;
}