J - Perfect Pth Powers Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

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

題意：輸入一個數x，若為0則結束。否則，我們知道這個數x必定存在相應的a，b，使得x等於a的b次方；這樣的a，b有至少一對。你所要做的就是求出最大的b。

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
	long long n;     //int不能滿足要求了...
	while(scanf("%lld",&n)!=EOF)
	{
		if(n==0) break;
		long long  m;
		int i;
		int ans=1;
		if(n<0)       //n可能為為負數
		{	m=-n;i=3;}    //負數當然得要是某個負數的奇次冪咯
		else
		{	m=n;i=2;}
		double eps=1e-12;   //精度很重要的,不能太小
		double p;
		for(;i<=32;)
		{
			p=pow(m*1.0,1.0/i);    //pow(int,int)提交報錯....改為float
			if(fabs(int(p+eps)-p)<eps)    //x等於a的b次方,b是整數,a也要是整數哦.
				ans=i;
			if(n<0)
				i+=2;     //負數當然得要是某個負數的奇次冪咯
			else
				i++;  
		}
		printf("%d\n",ans);
	}
	return 0;
}