UVA 10622 Perfect P-th Powers

Posted 2020-10-03

https://vjudge.net/problem/UVA-10622

 

将n分解质因数，指数的gcd就是答案

如果n是负数，将答案除2至奇数

原理：（a*b）^p=a^p*b^p

 

 

#include<cmath>
#include<cstdio>
#include<algorithm>
#define N 65550
using namespace std;
int gcd(int a,int b) { return !b ? a : gcd(b,a%b); }
int main()
{
    int m,sum,ans;
    long long n,nn;
    while(scanf("%lld",&nn)!=EOF)
    {
        if(!nn) return 0;
        n=abs(nn);
        m=sqrt(n);
        ans=0; 
        for(int i=2;i<=m;i++)
        {
            sum=0;
            if(n%i==0)
                 while(n%i==0) sum++,n/=i;
            ans=gcd(ans,sum);
        }
        if(!ans) 
        {
            printf("1\n");
            continue;
        }
        if(nn<0) 
        while(!(ans&1)) ans>>=1;
        printf("%d\n",ans);
    }
    
}