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
1 /*
2 题意描述
3 输入x，计算并输出满足x=b的p次方中最大的p
4
5 解题思路
6 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。
7 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。
8 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。
9 */
10 #include<cstdio>
11 #include<cmath>
12
13 int main()
14 {
15 int x;
16 while(scanf("%d",&x),x != 0){
17 if(x > 0){
18 for(int i=31;i>=1;i--){
19 int b=(int)(pow(x*1.0,1.0/(i*1.0)) + 0.1);
20 int temp=(int)(pow(b*1.0,i*1.0) + 0.1);
21 if(x == temp){
22 printf("%d\n",i);
23 break;
24 }
25 }
26 }
27 else{
28 x *= -1;
29 for(int i=31;i>=1;i-=2){
30 int b=(int)(pow(x*1.0,1.0/(i*1.0)) + 0.1);
31 int temp=(int)(pow(b*1.0,i*1.0) + 0.1);
32 if(x == temp){
33 printf("%d\n",i);
34 break;
35 }
36 }
37 }
38 }
39 return 0;
40 }