POJ 1730 Perfect Pth Powers (枚举||分解质因子)

Perfect Pth Powers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16638		Accepted: 3771

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

Source

Waterloo local 2004.01.31

 

 

从31往前枚举(因为数据最大不会超过32-bit (int))，用pow将要求的数字求出来，再进行验证，这里要注意精度问题。

这道题比较坑的是会有负数，对于负数的话，只能开奇数次方。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
using namespace std;
int main()
{
    //freopen("in.txt","r",stdin);
    int n,x,y;
    while(scanf("%d",&n)&&n)
    {
        if(n>0)
        {
            for(int i=31;i>=1;i--)
            {
                x=(int)(pow(n*1.0,1.0/i)+0.5);
                y=(int)(pow(x*1.0,1.0*i)+0.5);
                if(n==y)
                {
                    printf("%d
",i);
                    break;
                }
            }
        }
        else
        {
            n=-n;
            for(int i=31;i>=1;i-=2)
            {
                x=(int)(pow(n*1.0,1.0/i)+0.5);
                y=(int)(pow(x*1.0,1.0*i)+0.5);
                if(n==y)
                {
                    printf("%d
",i);
                    break;
                }
            }
        }
    }
    return 0;
}

还可以用分解质因子的方法来做

所求的答案就是所给的数的所有质因子的指数的最大公约数

大牛是这么做的

http://www.cnblogs.com/Lyush/archive/2012/07/14/2591872.html