CF833 A The Meaningless Game

题干

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 kk 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 k^2 , 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 nn 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.

输入格式

In the first string, the number of games nn (1<=n<=350000)(1<=n<=350000) is given.

Each game is represented by a pair of scores aa , bb (1<=a,b<=10^{9})(1<=a,b<=109) – the results of Slastyona and Pushok, correspondingly.

输出格式

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).

翻译

Slastyona和她的忠实狗狗普什克正在玩一个毫无意义但是很有趣的游戏。游戏包括多个回合。

它的规则非常简单：先选择一个自然数k。然后，谁说（或吠）的比另一个快就会赢得一局。胜利者的得分在那之后会乘以k的平方，而输了的人的得分就只能乘以k。比赛开始时，Slastyona和PurSok都有一个初始分数。不幸的是，Slastyona丢失了记事本，那里记录了他们玩过的N个游戏的历史。她设法回忆了每一场比赛的最终结果，但是记忆都很模糊。帮助Slastyona验证它们的正确性，或者，换句话说，对于每一对给定的分数，确定游戏是否能够完成这样的结果。

样例

​ 输入：

6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000

​ 输出：

Yes
Yes
Yes
No
No
Yes

Tips

First game might have been consisted of one round, in which the number 22 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 55 , and in the second one, Pushok would have barked the number 33 .

瞎掰掰 ≧ ﹏ ≦

首先，这是一道很简单的思维题（然而最近脑子煞笔了的我依然卡了半天...），所以不要被不知所云的题干给吓唬到。

我们把花里胡哨的题干给去掉，真正的问题其实就是让我们判断每一组数据的合法性吧。这题的要点（其实也算不上难想，可我这个傻缺就是没看出来QwQ）就是我们不能分别考虑两个数，我们要把两个数之间的合法关系规律给找出来，也就是我们把两个数合起来...

初始分数是未知的，也就是任意的。改变一共有两种，乘k或者乘k^2, 问题是我们对于每轮谁乘k,谁乘k^2 ,并不清楚。而这时我们应该能反应过来，虽然这两个数各自的变化我们不知道，但他们的乘积的变化是一定的，也就是乘上了k^3，那么问题就解决了：只要两个最终分数的乘积是一个完全立方数，那么这组数据就是合法的（前提是这个完全立方数的立方根是这两个数的因数，例如1和8，虽然相乘是2的完全立方，但2并不是1的因数，所以并不合法）。

剩下的就是敲代码了...我采用了纯离线操作...因为a，b的最大值都是10^9, 所以乘积的最大值是10^18，开数组显然开不下，所以用了map...

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<map>
using namespace std;
const int maxn=350000+10;
#define ll long long
unsigned long long a[maxn],b[maxn],tmp;
int main(){
    int n;
    cin>>n;
    map<unsigned long long,int> num;
    for(ll i=1;i<=1000000;i++){//10^18是10^6的立方
        tmp=i*i*i;
        num[tmp]=i;
    }
    for(int i=1;i<=n;i++)
        scanf("%llu %llu",&a[i],&b[i]);
    for(int i=1;i<=n;i++){
        //cout<<a[i]<<" "<<b[i]<<endl;
        if(a[i]==1&&b[i]==1){
            printf("Yes\n");
            continue;
        }
        unsigned long long tmpp=a[i]*b[i];
        if(num[tmpp]>0){
            unsigned long long t=num[tmpp];
            if(a[i]%t==0&&b[i]%t==0){
                printf("Yes\n");
                continue;
            }
            else printf("No\n");
        }
        else printf("No\n");
    }
    return 0;
}