The Meaningless Game（思维）

Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.

The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner’s score is multiplied by k2, and the loser’s score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.

Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.

Input
In the first string, the number of games n (1 ≤ n ≤ 350000) is given.
Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109) – the results of Slastyona and Pushok, correspondingly.

Output
For each pair of scores, answer “Yes” if it’s possible for a game to finish with given score, and “No” otherwise.
You can output each letter in arbitrary case (upper or lower).

有两个人玩游戏，每轮给出一个自然数k，赢得人乘k^2，输得人乘k，给出最后两个人的分数，问两个人能否达到这个分数

解题思路：将两个人的分数相乘然后开立方即可;
a,b一开始都是1，当经过几轮后，我们得到a，b的值，a*b=k1^3 * k2^3……
因此我们有：

#include<stdio.h>
#include<math.h>
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		int k=cbrt((double)a*(double)b);//运用cbrt函数来开三次方
		int aa=a/k,bb=b/k;
		if(aa*aa*bb==a&&bb*bb*aa==b)
		printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}