Mysterious Bacteria 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

给一个数x（可能是负数）求出 使得 x = a ^ b 的 b 的最大值

算术基本定理，x 可以分解成不同的素因子的幂的积，那么找出所有素因子的幂的最小公约数，即为结果。
如果x 为负数，那么先求出 - x 的结果ans，如果 ans 为偶数，说明当前答案不行，应继续将它/2 直到找到一个奇数为止。

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<queue>#include<algorithm>#define ll long long#define inf 0x3f3f3f3f#define maxn 100007  //1e5 + 7using namespace std;bool vis[maxn];int prime[10010];int num[1000];int cnt = 0;void find_prime(){    vis[0] = 1;    vis[1] = 1;    for(int i = 2; i < maxn; i ++)    {        if(vis[i] == 0)        {            prime[cnt++] = i;            for(int j = i+i; j < maxn; j += i)            {                vis[j] = 1;            }        }    }}int gcd(int a,int b){    if( a < b)    {        a = a ^ b;        b = b ^ a;        a = a ^ b;    }    return b == 0 ? a : gcd( b ,a % b);}int main(){    int t;    long long n;    find_prime();    scanf("%d",&t);    for(int i = 1; i <= t; i ++)    {        scanf("%lld",&n);        int flag = 0;        if(n < 0)            {                n = -n;                flag = 1;            }        if( n == 1 || n == -1)   // 1 和 -1 特判下             {                printf("Case %d: 1\n",i);                continue;               }        int j = 0;        int count = 0;        memset(num,0,sizeof(num));        while( j < cnt && prime[j] <= n)        {            if( n % prime[j] == 0)            {                while( n % prime[j] == 0)                {                    num[count] ++;                    n /= prime[j];                                  //  printf(" %d " ,num[count]);                }                count ++;            }            j ++;        }        if( n > 1)   //此时说明有个存在一个比较大的素数（该素数不在prime[] 数组中）         {            // 而且有一个，所以最小公约数为 1，答案为 1 ；             printf("Case %d: 1\n",i);            continue;        }    //  printf(" __%d__ ",count);        if(count == 1)            {                   if(flag == 1 && num[0] % 2 == 0 )                    {                        int tmp = num[0];                    //  printf("%d ",tmp);                        while(tmp % 2 == 0)   // n 为负数，且所求答案为偶数，不断除2 直到得到奇数答案                         {                            tmp /= 2;                        }                        num[0] = tmp;                    }                printf("Case %d: %d\n",i,num[0]);                continue;            }        int tmp = gcd(num[0],num[1]);        for(int j = 3; j < count; j ++)            tmp = gcd(num[j],tmp);        if( flag == 1 && tmp % 2 == 0)  // n 为负数，同上         {                while( tmp % 2 == 0 )                {                    tmp  /= 2;                }        }        printf("Case %d: %d\n",i,tmp);    }    return 0;}