POJ 1730 Perfect Pth Powers

Perfect Pth Powers

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 17339 Accepted: 3974

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

171073741824250

Sample Output

1302

题目大意：要求出一个完美的平方数，完美的平方数是这样的，n = b^p的当P最大的时候才是一个完美的平方数

解题思路：将P从31到1遍历枚举，使用POW函数将他求出来，不过有一点问题就是精度问题，当用POW（125，1/3），直接取整的时候是4，所以需要在后面加一个0.1，也就是（int）（POW(125,1/3)+0.1）.这样就可以求出结果了，还有要注意的是，他给出的N可能是负数，所以要对负数特殊处理，也就是说当N是负数的时候，P不可能是偶数，只能是奇数。

#include <cstdio>#include <cmath>using namespace std;//typedef long long ll;//ll x;int x;int main(){    while (scanf("%d", &x), x){        if (x > 0){            for (int i = 31; i>= 1; i--){                int a = (int)(pow(x * 1.0, 1.0 / i) + 0.1);                int b = (int)(pow(a * 1.0, 1.0 * i) + 0.1);                if (x == b){                    printf("%d\n", i);                    break;                }            }        }        else{            x = -x;            for (int i = 31; i>= 1; i -= 2){                int a = (int)(pow(x * 1.0, 1.0 / i) + 0.1);                int b = (int)(pow(a * 1.0, 1.0 * i) + 0.1);                if (x == b){                    printf("%d\n", i);                    break;                }            }        }    }    return 0;}