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
分解 质因子，如果是正数，直接输出因子的gcd-p,否则如果是奇数，分解--x,输出因子gcd，如果偶数，不断转化为奇数，因为负数的偶数次幂不可能是负数
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 1000005
using namespace std;
#define ll long long
ll num[maxn];
ll p[maxn];
bool vis[maxn];
int flag;
int k;
ll gcd(ll a,ll b)
{
return b==0?a:gcd(b,a%b);
}
void sp()
{
k=0;
memset(vis,0,sizeof(vis));
vis[0]=vis[1]=1;
for(int i=2;i<maxn;i++)
{
if(!vis[i])
p[k++]=i;
for(int j=0;j<k&&i*p[j]<maxn;j++)
{
vis[i*p[j]]=1;
if(i%p[j]==0)
break;
}
}
}
ll getf(ll x)
{
ll sum=0;
for(int i=0;i<k&&p[i]*p[i]<=x;i++)
{
ll res=0;
while(x%p[i]==0&&x)
{
x/=p[i];
res++;
}
if(flag)
while(res%2==0&&res)
res/=2;
if(res&&!sum)
sum=res;
else if(res)
sum=gcd(res,sum);
}
if(x>1)
sum=1;
return sum;
}
int main()
{
int t;
scanf("%d",&t);
int w=0;
sp();
while(t--)
{ ll x;
w++;
scanf("%lld",&x);
flag=0;
if(x<0)
{
x=-x;
flag=1;
}
printf("Case %d: %lld\n",w,getf(x));
}
return 0;
}