J - Perfect Pth Powers解题报告（张宇）

最新推荐文章于 2019-08-03 10:20:25 发布

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本文介绍了一个算法问题，即确定一个给定整数作为完美幂的最大指数。通过使用C++实现，该程序能够处理正负整数，并找到使该数成为某个整数的幂的最大指数。

J - Perfect Pth Powers

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1730

Description

We say that x is a perfect square if, for some integer b, x = b ². Similarly, x is a perfect cube if, for some integer b, x = b ³. 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 p th 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 th power.

Sample Input

Sample Output

1
30
2

题意：输入一个数x，若为0则结束。否则，我们知道这个数x必定存在相应的a，b，使得x等于a的b次方；这样的a，b有至少一对。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不能满足要求了...
	while(scanf("%lld",&n)!=EOF)
	{
		if(n==0) break;
		long long  m;
		int i;
		int ans=1;
		if(n<0)       //n可能为为负数
		{	m=-n;i=3;}    //负数当然得要是某个负数的奇次幂咯
		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)提交报错....改为float
			if(fabs(int(p+eps)-p)<eps)    //x等于a的b次方,b是整数,a也要是整数哦.
				ans=i;
			if(n<0)
				i+=2;     //负数当然得要是某个负数的奇次幂咯
			else
				i++;  
		}
		printf("%d\n",ans);
	}
	return 0;
}