lightoj 1220

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

Output for Sample Input

3

17

1073741824

25

Case 1: 1

Case 2: 30

Case 3: 2

题意：让求x = b^p的最大的p。

这个x是这个细菌能活的天数，而且还能是负的？！有点神奇

就是把x用算数基本定理化成质因数的幂乘积的形式，然后找出所有幂次的最大公约数，就是答案了。

要注意是如果x是负数的话，那么幂次就不能是偶数了，因为一个数的偶数次方肯定是正的嘛，就要把幂次变成奇数再求最大公约数。

#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define LL long longusing namespace std;const int maxn = 1e6 + 10;int f[maxn],p[maxn],cnt = 0;void prime(){    LL i,j;    f[1] = 1;    for(i=2;i<maxn;i++)    {        if(f[i])            continue;        p[cnt++] = i;        for(j=i*i;j<maxn;j+=i)        {            f[j] = 1;        }    }}int gcd(int a,int b){    if(b == 0)        return a;    else        return gcd(b,a%b);}int main(void){    int T,i,j;    LL n;    prime();    scanf("%d",&T);    int cas = 1;    while(T--)    {        scanf("%lld",&n);        int flag = 0;        if(n < 0)        {            flag = 1;            n = -n;        }        i = 0;        int g = 0;        while(p[i]*p[i] <= n && i < cnt)        {            int t = 0;            while(n % p[i] == 0)            {                n /= p[i];                t++;            }            i++;            if(t == 0)                continue;            if(flag == 1)            {                while(t % 2 == 0)                {                    t /= 2;                }            }            g = gcd(g,t);        }        if(n > 1)            g = 1;        printf("Case %d: %d\n",cas++,g);    }    return 0;}