POJ 1730 Perfect Pth Powers

最新推荐文章于 2020-02-05 22:56:30 发布

转载最新推荐文章于 2020-02-05 22:56:30 发布 · 102 阅读

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原文链接：http://www.cnblogs.com/l609929321/p/6763561.html

本文介绍了一个算法，用于确定一个整数是否为某个整数的完美p次幂，并找出最大的p值。通过开方和比较的方法，该算法可以有效处理正数和负数情况。

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Perfect Pth Powers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17685		Accepted: 4068

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

Source

Waterloo local 2004.01.31

#include<stdio.h>
#include<math.h>

int main()
{
    int n;
    while(~scanf("%d",&n) && n)
    {
        if(n > 0)
        {
            for(int i = 31; i >= 1; i--)
            {
                int t = (int)(pow(n*1.0,1.0/i) + 0.1);//先开最大次方
                int x = (int)(pow(t*1.0,1.0*i) + 0.1);
                if(n == x)//判断重新求的数与原数相等就是完美次数
                {
                    printf("%d\n",i);
                    break;
                }
            }

        }
        else
        {
            n = -n;//考虑负数的情况
            for(int i = 31; i >= 1; i-=2)
            {
                int t = (int)(pow(n*1.0,1.0/i) + 0.1);
                int x = (int)(pow(t*1.0,1.0*i) + 0.1);
                if(n == x)
                {
                    printf("%d\n",i);
                    break;
                }
            }
        }
    }
    return 0;
}

http://blog.youkuaiyun.com/u013647282/article/details/37816089

大神做的

转载于:https://www.cnblogs.com/l609929321/p/6763561.html