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.
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.
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).
6 2 4 75 45 8 8 16 16 247 994 1000000000 1000000
Yes Yes Yes No No Yes
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
题目大意：有两个人玩游戏，每个人的初始分数是1，他们每次喊一个数k，然后赢的人分数乘上k的平方，输的人分数乘k，给出两个数，问是不是这两人玩游戏所得到的最后分数
可以发现每一个过程的分数的乘积都是一个三次方，所以最后能得到最后的分数相乘是可以开三次方的，而且a和b对这个三次方根取余==0（如2,32就只能满足前面那个），只要满足这两个就对了
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#include<iostream>
#include<cstdio>
using namespace std;
typedef long long ll;
int main()
{
int test;
cin>>test;
while(test--)
{
ll a,b;
//cin>>a>>b;
scanf("%lld%lld",&a,&b);
ll res=a*b;
ll left=1,right=1000000;
ll mid;
while(left<=right)
{
mid=(left+right)/2;
ll tmp=mid*mid*mid;
if(tmp<res)
{
left=mid+1;
}
else if(tmp>res)
{
right=mid-1;
}
else break;
}
if(mid*mid*mid==res&&a%mid==0&&b%mid==0)
{
puts("Yes");
//cout<<"Yes"<<endl;
}
else
{
puts("No");
//cout<<"No"<<endl;
}
}
return 0;
}