题目：
Dr. Mob has just discovered a Deathly Bacteria. He named it RC-01. RC-01 has a very strange reproduction system. RC-01 lives exactly xdays. Now RC-01 produces exactly p new deadly Bacteria where x = bp (where b, p are integers). More generally, x is a perfect pth power. Given the lifetime x of a mother RC-01 you are to determine the maximum number of new RC-01 which can be produced by the mother RC-01.
Input
Input starts with an integer T (≤ 50), denoting the number of test cases.
Each case starts with a line containing an integer x. You can assume that x will have magnitude at least 2 and be within the range of a 32 bit signed integer.
Output
For each case, print the case number and the largest integer p such that x is a perfect pth power.
Sample Input
3
17
1073741824
25
Sample Output
Case 1: 1
Case 2: 30
Case 3: 2
题目大意：
给定x,寻找满足 x=b^p（b,p为整数）的p的最大值。
解题思路：
原以为挺简单的一道题，死活过不了，看了题解才明白，一是没有考虑负数，而是我的暴力毫无技巧，而这是一道思维题。。
例：给你一个数x = b^p,求p的最大值
x = p1^x1*p2^x2*p3^x3*...*ps^xs【数论--唯一分解定理】
如果x = 12 = 2^2*3^1，要让x = b^p,及12应该是12 = 12^1
所以p = gcd(x1, x2, x3, ... , xs);
比如：24 = 2^3*3^1，p应该是gcd(3, 1) = 1,即24 = 24^1
324 = 3^4*2^2，p应该是gcd(4, 2) = 2,即324 = 18^2
本坑：x可能为负数，如果x为负数的话，x = b^q, q 必须使奇数，所以将x转化为正数求得的解如果是偶数的话，必须将其一直除2转化为奇数，x= -16 = 16 = 2^4=4^2=16^1= -16 ^1。
实现代码：
#include<bits/stdc++.h>
using namespace std;
typedef int Int;
#define int long long
set<int>s;
void get_fac(int n){
for(int i=2;i*i<=n;i++){
int ans=0;
if(n%i==0){ //每一位的上角标
while(n%i==0) n/=i,ans++;
s.insert(ans);
}
}
if(n>1) s.insert(1);//未除尽的数，上角标默认为1
}
int gcd(int a,int b) //求最大公因子
{
return b==0?a:gcd(b,a%b);
}
Int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
for(int ca=1;ca<=t;ca++)
{
cout<<"Case "<<ca<<": ";
int n;
cin>>n;
s.clear();
get_fac(abs(n));
if(s.size()==1){
int ans=*s.begin();
if(n<0){
while(ans%2==0) ans/=2;
cout<<ans<<endl;
}
else cout<<ans<<endl;
}
else{
int ans=*s.begin();
set<int>::iterator it;
for(it=s.begin();it!=s.end();it++) ans=gcd(ans,*it);
if(n<0) cout<<1<<endl;
else cout<<ans<<endl;
}
}
cin.get(),cin.get();
return 0;
}