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

假设选了五个数，分别是k1,k2,k3,k4,k5
a = k1*k2*k2*k3*k4*k4*k5
b = k1*k1*k2*k3*k3*k4*k5*k5
a*b=k1^3 * k2^3 * k3^3 * k4^3 *k5^3
a,b<=10^9,a*b<=10^18，a*b开三次根号就是10^6
在(1,10^6)内二分，找到mid^3 = a*b,并且a%mid=0,b%mid=0

#include <stdio.h>typedef long long LL;const LL MAXN = 1e6;LL a,b;bool flag;int main(){    int n;    LL mul;    scanf("%d",&n);    while(n--)    {        flag = true;        scanf("%I64d %I64d",&a,&b);        mul = a*b;        LL l = 1;        LL r = MAXN;        LL mid = (l+r) >> 1;        while(l <= r)        {            if(mid*mid*mid > mul)                r = mid-1;            else if(mid*mid*mid < mul)                l = mid+1;            else if(a%mid == 0 && b%mid == 0)            {                printf("Yes\n");                flag = false;                break;            }            else                break;            mid = (l+r) >> 1;        }        if(flag)            printf("No\n");    }    return 0;}