Codeforces 834C The Meaningless Game【思维】

C. The Meaningless Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 theround. After that, the winner's score is multiplied byk², and the loser's score is multiplied byk. 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 scoresa, 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).

Example

Input

62 475 458 816 16247 9941000000000 1000000

Output

YesYesYesNoNoYes

Note

First game might have been consisted of one round, in which the number2 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 number5, and in the second one, Pushok would have barked the number3.

题目大意：

现在有两个人玩游戏，一开始两个人的分数都是1，每一轮游戏两个人会共同挑选一个数K，游戏获得胜利的选手将自己的分数*k^2，输了的选手将自己的分数*k.

现在有n个查询，每个查询表示两个人最终的分数，问是否有一种游戏过程，使得这个最终分数是可能出现的。

思路：

这个题就素一个煞笔题啊。

一个人获得k^2的倍数，另一个人获得k的倍数。

那么两个人相乘的分数就相当于增长了k^3的倍数。

我们知道，k^3*k^3=(2k)^3

我们还知道，k^3*q^3=(kq)^3

那么问题就是在问你这两个人的分数相乘是不是三次方数。

如果是，就可能是答案，同时，我们已知乘积的三次根号的结果，如果a%这个结果==0&&b%这个结果==0.那么表示数字a和数字b确实可以得到这样的结果。

否则就是No。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<map>using namespace std;#define ll __int64int main(){    map<ll ,int>s;    for(ll i=1; i<=1000000; i++)    {        ll tmp=i*i*i;        s[tmp]=i;    }    int t;    scanf("%d",&t);    while(t--)    {        ll a,b;        scanf("%I64d%I64d",&a,&b);        if(a==1&&b==1)        {            printf("Yes\n");            continue;        }        ll c=a*b;        if(s[c]>0)        {            ll tmpp=s[c];            if(a%tmpp==0&&b%tmpp==0)printf("Yes\n");            else printf("No\n");        }        else        {            printf("No\n");        }    }}