(Codeforces Round #426 (Div. 2)) C.The Meaningless Game

题目

http://codeforces.com/contest/834/problem/C

题解

先用GCD求出x,y的最小公倍数z

---

再求得p=x/z，q=y/z;

---

若x%p*p == 0 && y%q*q==0且z/p*q可开1/3次方为整数，则为Yes，否则为No

---

注意点

---

scanf(“%lld%lld”,&x,&y);//cin,cout影响输入速度会t

LL c =ceil( pow(a*1.0, 1.0/3)),d =ceil(pow(b*1.0, 1.0/3));//调整精度

---

AC代码

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define maxn 50000 + 10#define For(a,b)  for(int i = a; i < b; i++)typedef long long LL;using namespace std;int a[maxn];LL gcd(LL a,LL b){        if(a%b == 0) return b;        else return gcd(b,a%b);}int main(){    int n;    scanf("%d",&n);    while(n--)    {            LL  x,y,z;            LL  p,q;            scanf("%I64d%I64d",&x,&y);            z = gcd(x,y);            p = x/z;            q = y/z;       if(x % (p*p) == 0 && y % (q*q) == 0)          {               LL  a = x/(p*p*q),b = y/(q*q*p);               LL c =ceil( pow(a*1.0, 1.0/3)),d =ceil(pow(b*1.0, 1.0/3));                  if(c*c*c == a  && d*d*d == b)                       printf("Yes\n");                 else   printf("No\n");          }       else printf("No\n");    }      return 0;        }