LightOJ

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 = 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

#include<stack>
#include<queue>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<map>
#include<iostream>
#include<string.h>
#include<algorithm>
#define maxn 1000005
#define maxm 10000005
#define MAXN 100005
#define MAXM 10005
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
bool vis[maxm+5];
ll tot;
ll p[maxn];
void get_prime(){
     tot=0;
    mem(vis,true);
    for(ll i=2;i<=maxm;i++){
        if(vis[i]){
            p[tot++]=i;
            for(ll j=i*i;j<=maxm;j+=i)vis[j]=false;
        }
    }
}
int main(){
    get_prime();
    int t,test=0;
    scanf("%d",&t);
    while(t--){
     ll n;scanf("%lld",&n);
     ll m=n;
     n=abs(n);
     ll x=0,ans=-1;
     while(p[x]<=n&&x<tot){
        ll y=0;
        while(n%p[x]==0){y++;n/=p[x];}
        x++;
        if(ans==-1)ans=y;
        else
            ans=__gcd(ans,y);
     }
     if(n>1)ans=__gcd(ans,(ll)1);
     if(m<0){
        while(ans%2==0)ans/=2;
     }
     else if(m==1||m==0)
        ans=1;
        printf("Case %d: %lld
",++test,ans);
    }
}