POJ 1730 Perfect Pth Powers

M - Perfect Pth Powers

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

Submit Status

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


此题解题思路:由于给定的数比较大，一一枚举计算量比较大，所以需要从另一个方面出发！由于给定的数不会超过某个数的32次方，那么我就可以从32次方往下枚举，以减少计算量，不过过程中需要再次优化！见代码中的解释
另外需要注意负数的情况！！！
 
解题代码：

// File Name:b.cpp
// Author: sheng
// Created Time: 2013年04月02日 星期二 20时24分39秒

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;

typedef long long LL;
const double INF = 1e-7;//防止精度问题而导致的出错

int main()
{
    int i, flag;
    LL n, temp;
    while (~scanf ("%lld", &n) && n)
    {
        flag = 0;
        if (n < 0)
        {
            flag = 1;
            n = -n;
        }
        for (i = 32; i >=1; i --)//从32次方开始枚举
        {
            temp = (LL)(pow(n, 1./i) + INF);//计算n的 i 方跟，一次到位
            if (pow(temp, i) == n && (!flag || (flag & i)))//判断temp的i次方是否与n==，并对负数情况进行处理
                break;
        }
        printf ("%d\n", i);
    }
    return 0;
}