Code Forces 833 A The Meaningless Game (thinking, math)

Code Forces 833 A The Meaningless Game

Subject

There are two people playing the game, each round gives a natural number k, the winning person multiplies k ^ 2, the losing person multiplies k, gives the score of the last two people, asks whether two people can reach this score

I have to spit out a long English title. Only one sentence is translated ...

solution

It ’s also good to think that the
product is cubed to determine whether it is a factor of two numbers.
If it is, it is obviously not true,
otherwise it will be Yes.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#define int long long
using namespace std;

inline int read(){
    int x = 0, w = 1;
    char ch = getchar();
    for(; ch > '9' || ch < '0'; ch = getchar()) if(ch == '-') w = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
    return x * w;
}

signed main(){
    int n = read();
    while(n--){
        int a = read(), b = read();
        int tmp = a * b;
        // int awsl = pow(tmp, (1.0 / 3)) + 0.5;
        int awsl = cbrt((double)a*(double)b);//在网上找到了这个开三次方的函数，啧啧
        if(awsl * awsl * awsl != tmp || a % awsl || b % awsl) cout << "No" << endl;
        else cout << "Yes" << endl;
    }
    return 0;
}