POJ 1730 Perfect Pth Powers（暴力枚举）

题目链接：

https://cn.vjudge.net/problem/POJ-1730

题目描述：

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

17
1073741824
25
0

Sample Output

1
30
2

 /*
 题意描述
 输入x，计算并输出满足x=b的p次方中最大的p

 解题思路
 因为x的大小为2到2的31次方，故可采取枚举次方的方法，计算出b，再计算出b的p次方为temp，看temp和x是否相等即可。
 另外需要注意的是x可能为负数，首先需要将负数变为正数，另外枚举的时候只能枚举奇数次方，因为偶数次方不能的到负数。
 另外是用pow开方和乘方时注意精度问题，比如4.999999直接取整误差很大，加上0.1即可避免此类问题。
 */
 #include<cstdio>
 #include<cmath>

 int main()
 {
     int x;
     while(scanf("%d",&x),x != ){
         if(x > ){
             for(int i=;i>=;i--){
                 int b=(int)(pow(x*1.0,1.0/(i*1.0)) + 0.1);
                 int temp=(int)(pow(b*1.0,i*1.0) + 0.1);
                 if(x == temp){
                     printf("%d\n",i);
                     break;
                 }
             }
         }
         else{
             x *= -;
             for(int i=;i>=;i-=){
                 int b=(int)(pow(x*1.0,1.0/(i*1.0)) + 0.1);
                 int temp=(int)(pow(b*1.0,i*1.0) + 0.1);
                 if(x == temp){
                     printf("%d\n",i);
                     break;
                 }
             }
         }
     }
     return ;
 }