Mysterious Bacteria （唯一分解++欧拉素数筛）

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

3

17

1073741824

25

Sample Output

Case 1: 1

Case 2: 30

Case 3: 2

 题目大意：

给你一个数x = b^p,求p的最大值

 

x = p1^x1*p2^x2*p3^x3*...*ps^xs

开始我以为是找x1、x2、... 、xs中的最大值，后来发现想错了，x = b^p, x只有一个因子的p次幂构成

如果x = 12 = 2^2*3^1，要让x = b^p,及12应该是12 = 12^1

所以p = gcd(x1, x2, x3, ... , xs);

比如：24 = 2^3*3^1，p应该是gcd(3, 1) = 1,即24 = 24^1

         324 = 3^4*2^2,p应该是gcd(4, 2) = 2,即324 = 18^2

 

本题有一个坑，就是x可能为负数，如果x为负数的话，x = b^q, q必须使奇数，所以将x转化为正数求得的解如果是偶数的话必须将其一直除2转化为奇数

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
bool IsPrime[1000010];    
int Prim[1000010],num = 0;    
void  init(){    
    int  j;    
    for(int i = 2; i <= maxn; i ++){    
        if(!IsPrime[i])    
            Prim[num ++] = i;    
        for(j  = 0; j < num; j ++){    
            if(i * Prim[j] > maxn)    
                break;    
            IsPrime[i * Prim[j]] = true;    
            if(i % Prim[j] == 0)    
                break;    
        }    
    }//printf("%d
",num);    
}
int gcd(int a,int b){
	return !b?a:gcd(b,a%b);
}
int main()
{
	init();
	int t,ncase = 1;
	scanf("%d",&t);
	while(t--){
		bool falg = false;
		ll n;
		scanf("%lld",&n);
		if(n<0) {
			falg = true;
			n = -n;
		}
		int ans = 0;
		for(int i = 0;i<num && Prim[i]<=n ; i++){
			//if(n == 1) break;
			int s = 0;
			if(n%Prim[i] == 0){
				while(!(n%Prim[i])){
					s++; n /= Prim[i];
				}
				if(!ans) ans = s;
				else ans = gcd(ans,s);
			}
			
		}
		if(n!=1) ans = 1;
		if(falg){
			while(!(ans&1)){
				ans >>= 1;
			}
		}
		printf("Case %d: %d
",ncase++,ans);
	}
	return 0;
}