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 pthpower.
Sample Input
3
17
1073741824
25
Sample Output
Case 1: 1
Case 2: 30
Case 3: 2
分析：
n = p1^x1 * p2^x2 * p3^x3 ……
故找到 gcd（x1,x2,x3……）即可；
特别注意n为负,即只能为次幂！
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
int prime[maxn],cnt = 0;
bool vis[maxn];
void is_prime()
{
memset(prime,0,sizeof prime);
memset(vis,0,sizeof vis);
for(int i = 2; i < maxn; i++)
{
if(!vis[i])
{
prime[cnt++] = i;
for(int j = 2*i; j < maxn; j += i)
vis[j] = 1;
}
}
}
int gcd(int a,int b)
{
if(a%b == 0)
return b;
return gcd(b,a%b);
}
int main()
{
is_prime();
int T,Case = 1;
scanf("%d",&T);
while(T--)
{
ll n;
scanf("%lld",&n);
bool judge = true;
if(n < 0)
{
n = -n;
judge = false;
}
int sum = 0;
for(int i = 0; i < cnt && prime[i]*prime[i] <= n; i++)
{
if(n%prime[i] == 0)
{
int cou = 0;
while(n%prime[i] == 0)
{
cou++;
n /= prime[i];
}
if(sum == 0) //凑到幂一样就行了
sum = cou;
else
sum = gcd(sum,cou);
}
}
if(n > 1)
sum = gcd(sum,1);
if(!judge)
while(sum%2 == 0)
sum /= 2;
printf("Case %d: %d\n",Case++,sum);
}
return 0;
}