Perfect Pth Powers解题报告

最新推荐文章于 2019-08-03 10:20:25 发布

xieqinghuang

最新推荐文章于 2019-08-03 10:20:25 发布

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分类专栏：数论文章标签： integer output input each c

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本文介绍了一种算法，用于确定给定整数作为完美幂次的最大指数。通过循环计算不同指数下的根，判断整数是否可以表示为某个基数的完美幂次，并返回最大的可能指数。

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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

Sample Output

1
30
2
#include<stdio.h>
#include<math.h>
int main()
{
	double n,b;
	long x,y;
	int i;
	while(scanf("%lf",&n)!=EOF)
	{
		if(n==0)
			break;
		if(n>0)
		{
			for(i=32;i>=0;i--)
			{
				b=pow(n,1.0/i);
				x=b;    //类型转换为long
				y=x+1;  //
				if(b-x<=1e-12||y-b<=1e-12)
				{
					printf("%d\n",i);
					break;
				}
			}
		}
		if(n<0)
		{
			n=-n;  //if n《0，先转化为正数先，然后i 只能为奇数，查找i。
			for(i=31;i>=1;i-=2)
			{
				b=pow(n,1.0/i);
				x=b;      //类型转换为long。
				y=x+1;    //
				if(b-x<=1e-12||y-b<=1e-12)
				{
					printf("%d\n",i);
					break;
				}
			}
		}
	}
	return 0;
}