Perfect Pth Powers poj1730

Perfect Pth Powers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16383		Accepted: 3712

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 = b^p. 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 pth power.

Sample Input

17
1073741824
25
0

Sample Output

1
30
2

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#define MAXN 100010
#define INF 0x7fffffff
using namespace std;
const int N = 77000 ;
int np[N >> 6], p[N >> 2],ans[N]= {0};
void init()
{
    int n = N;
    p[0] = 1, p[1] = 2;
    for (int i = 3; i <= n; i++, i++)
    {
        if (np[i >> 6] & (1 << ((i >> 1) & 31))) continue;
        p[++p[0]] = i;
        for (int j = (i << 1) + i; j <= n; j += (i << 1))
        {
            np[j >> 6] |= (1 << ((j >> 1) & 31));
        }
    }
}
int gcd(int x,int y)
{
    if(y==0)return x;
    return gcd(y,x%y);
}
int fun(long long x)
{
    int i,ans=0,an=0;
    bool flag=0;
    if(x<0)flag=1,x=-x;;
    for(i=1; x>=p[i]&&i<=p[0]; i++)
    {
        if(x%p[i]==0)
        {
            ans=0;
            while(x%p[i]==0)ans++,x/=p[i];
            if(an==0)an=ans;
            else
            {
                an=gcd(ans,an);
            }
        }
    }
    if(x!=1)
    {
        an=1;
    }
    while(flag&&(an%2==0))an>>=1;
    if(an==0)return 1;
    return an;
}
int main()
{
    init();
    long long n;
    while(~scanf("%I64d",&n),n)
    {
        printf("%d
",fun(n));
    }
}