light oj 1220 Mysterious Bacteria

Mysterious Bacteria  

Time limit

500 ms

Memory limit

32768 kB

OS

Linux

Dr. Mob has just discovered a Deathly Bacteria. He named it RC-01. RC-01 has a very strange reproduction system. RC-01 lives exactlyx days. Now RC-01 produces exactlyp new deadly Bacteria wherex = b^p (whereb, p are integers). More generally,x is a perfectp^th power. Given the lifetimex 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 integerx. You can assume thatx 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 integerp such thatx is a perfect p^th power.

Sample Input

3

17

1073741824

25

Sample Output

Case 1: 1

Case 2: 30

Case 3: 2

题意：给一个数字n，求能使p^k等于n的最大整数k,n没说是正数还是负数

根据唯一分解定理，每一个数都可以分解成素数乘积组成的因式，求最大的整数k实际就是求因式指数的最大公约数，但是要注意的是n可能小于0，但n小于0是就要将答案中含2的因数全部去掉，因为p^k(k为偶数)>=0的

#include<iostream>#include<algorithm>#include<cmath>#include<cstdio>#include<queue>#include<cstring>#include<string>#include<map>#include<set>using namespace std;typedef long long ll;#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)#define sf scanf#define pf printf#define mem(i,a) memset(i,a,sizeof i)#define pi acos(-1.0)#define eps 1e-10const int Max=1e6+7;int p[Max+1],cou,n,T;ll m;bool vis[Max+1];void prime()//素数打表{    cou=0;    mem(vis,0);    for(int i=2;i<=Max;i++)    {        if(!vis[i])p[cou++]=i;        for(int j=0;j<cou&&(ll)i*p[j]<=Max;j++)        {            vis[i*p[j]]=1;            if(!(i%p[j]))break;        }    }}int GCD(int x,int y)//求最大公约数{    return !y?x:GCD(y,x%y);}int main(){    prime();    sf("%d",&T);    for1(i,T)    {        sf("%lld",&m);        int tag=1;        if(m<0)tag=0,m=-m;        int ans=0;        for(int j=0;j<cou&&p[j]<=sqrt(m);j++)//唯一分解        {            int num=0;            if(!(m%p[j]))                while(!(m%p[j]))                    num++,m/=p[j];            if(!ans)ans=num;//对ans赋初值            else ans=GCD(ans,num);//对分解的因式的次数求GCD        }        if(m>1)ans=1;        if(!tag)//n小于0的情况要将偶数次幂除去            while(!(ans%2))                ans>>=1;        pf("Case %d: %d\n",i,ans);    }    return 0;}