Mysterious Bacteria 素因子分解（唯一分解定理）

题意：（题目在最后）

给你一个整数n,让你求满足b^p=x的最大的p,例如72=2^3*3^2;结果是1；因为只有72^1=72;

216=2^3*3^3=6^3,结果是3；

利用 唯一分解定理x=p1^e1*p2^e2*......pi^ei,存下每一个指数，再求最大公约数，得到答案；

注意：两个点

1: n有可能是负的，这时候先把n变成正数求，因为只有b^奇数才可能为负数，（n=-4时，-2^2=4,显然不可能=n）

所以要把所有的指数不断除2转化成奇数（n=-64时，-2^6=(-2^2)^3=-4^3,结果是3）

2：素数打表用long long ,不然平方时会炸。

#include<cstdio>#include<cstring>using namespace std;typedef long long ll;const int N=100005;int tot=0,d,a[100];ll ans[N/10];bool vis[N];void Prime(){    memset(vis,true,sizeof(vis));    for(int i=2;i<N;i++)    {        if(vis[i]) ans[++tot]=i;        for(int j=1;(j<=tot)&&(i*ans[j]<N);j++)        {            vis[i*ans[j]]=false;            if(i%ans[j]==0) break;        }    }}int gcd(int a,int b) { return b?gcd(b,a%b):a;}void sbreak(ll x){    d=0;    memset(a,0,sizeof(a));    for(int i=1;ans[i]*ans[i]<=x;i++)    {        int cnt=0;        if(x%ans[i]==0) d++;        while(x%ans[i]==0)        {            x/=ans[i];            a[d]=++cnt;        }    }    if(x>1) {d=1;a[1]=1;}}int main(){    Prime();    int t,cas=0;    scanf("%d",&t);    while(t--)    {        ll x;         bool tag=false;        scanf("%lld",&x);        if(x<0) {x=-x;tag=true;}        sbreak(x);        int t=a[1];        if(tag)        {            for(int i=1;i<=d;i++)            {                while(a[i]%2==0)                    a[i]/=2;                t=gcd(t,a[i]);            }        }        else            for(int i=1;i<=d;i++)               t=gcd(t,a[i]);        printf("Case %d: %d\n",++cas,t);    }    return 0;}

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 p^th power.

Sample Input

3

17

1073741824

25

Sample Output

Case 1: 1

Case 2: 30

Case 3: 2