Mysterious Bacteria(唯一分解定理)

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 x days. Now RC-01 produces exactly p new deadly Bacteria where x = b^p (where b, p are integers). More generally, x is a perfect p^th 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

题意： 给x（可为负），求满足x=b^p,p的最大值

思路： 由唯一分解定理，一个整数x=p1 ^a1 *p2^a2*p3^a3…，即对于一个x，我们先把它分解为n个质数相乘,则所求p= gcd(a1,a2,a3,a4,a5,…,an),如果x为负，指数就不能为偶数（负数的偶次方为正数），如果算出答案为偶，就除2到为奇

AC代码：

#include<stdio.h>
#include<string.h>
#define ll long long
#include<algorithm>
using namespace std;
const int N=110000;
ll p[N],book[N],l,n;
void init()
{ll i,j;l=0;memset(book,0,sizeof(book));for(i=2;i<N;i++){if(!book[i]){p[l++]=i;for(j=0;i*p[j]<=N&&j<l;j++){book[i*p[j]]=1;if(i%p[j]==0)break;}}}
}
int gcd(ll a,ll b)
{if(b==0)return a;elsereturn gcd(b,a%b);
}
int main()
{ll t,c=1;scanf("%lld",&t);init();while(t--){int sum=0,f=0;scanf("%lld",&n);if(n<0){f=1;n=-n;}for(ll i=0;i<l&&p[i]*p[i]<=n;i++){if(n%p[i]==0){int cnt=0;while(n%p[i]==0){n/=p[i];cnt++;}if(cnt==0)sum=cnt;elsesum=gcd(sum,cnt);}}if(n!=1)sum=1;if(f){while(sum%2==0)sum/=2;}printf("Case %lld: %lld\n",c++,sum);}return 0;
}

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