C. The Meaningless Game

题目链接

题意

给你两个数，开始都为1,然后每轮可以任选一个k，一边可以乘以$k$，另一边乘以$k^2$,然后问你最终是否可以得到所给的两个数a,b;

思路

$a×b = t^3$,二分是否存在$t$,如果不存在肯定不可以，如果存在，那么要保证a中有t中所有的因子，b中也是，那么就是$ amodt==0 and bmodt == 0$,因为$t = k1k2...kn$,那么$a$中要么是$k1^2$要么是$k1$,同理$b$

代码

#include<bits/stdc++.h>
#define N 100005
using namespace std;
typedef long long LL;
LL gcd(LL n,LL m);
int main(void)
{int T;scanf("%d",&T);while(T--){LL a,b;scanf("%lld %lld",&a,&b);LL c = a*b;LL l = 0,r = 1000000;LL id;while(l <= r){LL mid = (l+r)/(LL)2;if(mid*mid*mid <= a*b){l = mid + 1;id = mid;}else r = mid - 1;}if(id*id*id != a*b){printf("No\n");}else{   //aprintf("%d\n",id);if(!(a%id)&&!(b%id))printf("Yes\n");else printf("No\n");}}return 0;
}
LL gcd(LL n,LL m)
{if(m == 0)return n;else return gcd(m,n%m);
}

转载于:https://www.cnblogs.com/zzuli2sjy/p/7266051.html

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C. The Meaningless Game

C. The Meaningless Game

题意

思路

代码

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