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 =  ;
 int np[N >> ], p[N >> ],ans[N]= {};
 void init()
 {
     int n = N;
     p[] = , p[] = ;
     for (int i = ; i <= n; i++, i++)
     {
         if (np[i >> ] & ( << ((i >> ) & ))) continue;
         p[++p[]] = i;
         for (int j = (i << ) + i; j <= n; j += (i << ))
         {
             np[j >> ] |= ( << ((j >> ) & ));
         }
     }
 }
 int gcd(int x,int y)
 {
     if(y==)return x;
     return gcd(y,x%y);
 }
 int fun(long long x)
 {
     int i,ans=,an=;
     bool flag=;
     if(x<)flag=,x=-x;;
     for(i=; x>=p[i]&&i<=p[]; i++)
     {
         if(x%p[i]==)
         {
             ans=;
             while(x%p[i]==)ans++,x/=p[i];
             if(an==)an=ans;
             else
             {
                 an=gcd(ans,an);
             }
         }
     }
     if(x!=)
     {
         an=;
     }
     while(flag&&(an%==))an>>=;
     if(an==)return ;
     return an;
 }
 int main()
 {
     init();
     long long n;
     while(~scanf("%I64d",&n),n)
     {
         printf("%d\n",fun(n));
     }
 }