Codeoforces 558 B. Duff in Love 【 Codeforces Round #326 (Div. 2)】

B. Duff in Love

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Duff is in love with lovely numbers! A positive integer x is called lovely if
and only if there is no such positive integer a > 1 such that a2 is
a divisor of x.



Malek has a number store! In his store, he has only divisors of positive integer n (and he has all of them). As a birthday present,
Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.

Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.

Input

The first and only line of input contains one integer, n (1 ≤ n ≤ 1012).

Output

Print the answer in one line.

Sample test(s)

input

10

output

10

input

12

output

6

Note

In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.

In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 22,
so 12 is not lovely, while 6 is indeedlovely.

题目大意：
给你一个数 n【注意 n 的范围，要用long long】，然后找它所有因子数中没有 平方数 的最大值
样例解释：
第二个样例：12的因子有1 2 3 4 6 12，又因为12中有 4 = 2*2，所以12舍去，就剩下 6 了，

解题思路:
说穿了，这其实就是一个所有的素因子乘起来的问题，所以我们只需要对 n 进行素因子分解就 ok 了，
上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;

#define MM(a) memset(a,0,sizeof(a))

typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e7+5;
const int mod = 1000000007;
const double eps = 1e-7;
bool prime[maxn];
LL p[maxn],k;
///素数筛选
void isprime()
{
    k = 0;
    MM(prime);
    for(LL i=2; i<maxn; i++)
    {
        if(!prime[i])
        {
            p[k++] = i;
            for(LL j=i*i; j<maxn; j+=i)
                prime[j] = 1;
        }
    }
}
LL fac[maxn], num[maxn], cnt;
///分解素因子算法
void Dec(LL x)
{
    MM(num);
    cnt = 0;
    for(LL i=0; p[i]*p[i]<=x&&i<k; i++)
    {
        if(x%p[i] == 0)
        {
            fac[cnt] = p[i];
            while(x%p[i] == 0)
            {
                num[cnt]++;
                x /= p[i];
            }
            cnt++;
        }
    }
    if(x > 1)
    {
        fac[cnt] = x;
        num[cnt++] = 1;
    }
}
int main()
{
    LL n;
    isprime();
    while(cin>>n)
    {
        Dec(n);
        LL ans = 1;
        for(int i=0; i<cnt; i++)
            ans *= fac[i];
        cout<<ans<<endl;
    }
    return 0;
}