poj 1730Perfect Pth Powers（分解质因数）

Problem Description

We say that x is a perfect square if, for some integer b, x = b2. Similarly, x is a perfect cube if, for some integer b, x = b3. More generally, x is a perfect pth power if, for some integer b, x = bp. Given an integer x you are to determine the largest p such that x is a perfect pth power.

Input

Each test case is given by a line of input containing x. The value of x will have magnitude at least 2 and be within the range of a (32-bit) int in C, C++, and Java. A line containing 0 follows the last test case.

Output

For each test case, output a line giving the largest integer p such that x is a perfect p<i>th</i> power.

Sample Input

17 1073741824 25 0

Sample Output

1 30 2

题意：给出一个整数x，把x写成x=a^p，求p最大是多少？

思路:把x分解质因数，x = a1^b1 * a2^b2 … ak^bk，则最终结果为b1,b2,…bk的最大公约数。注意x有可能是负数。

如果x是负数，则要把求得的答案一直除以2，直到结果是一个奇数，因为一个数的偶数次方不可能是负数。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
ll p[1001],num[1001];
ll cnt;
void fenjie(ll n)
{
    memset(p,0,sizeof(p));
    memset(num,0,sizeof(num));
    cnt=0;
    for(ll i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {p[++cnt]=i;
        num[cnt]=0;
        while(n%i==0)
        {
        num[cnt]++;
        n/=i;
    }
}


    }
    if(n!=1)
    {
        p[++cnt]=n;
        num[cnt]=1;
    }
}
ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    ll x;
    while(~scanf("%lld",&x))
    {

        if(x==0)
        break;
        ll ans=0;
        int flag=0;
        if(x<0)
        {
            x=-x;
            flag=1;
        }
        fenjie(x);
        for(ll i=1;i<=cnt;i++)
        {
        ans=gcd(ans,num[i]);}

        if(flag)
        {while(ans%2==0)
        ans/=2;}
        printf("%lld\n",ans);


    }
    return 0;
}