Codeforces Round #326 (Div. 2) B

最新推荐文章于 2024-10-14 09:52:52 发布

G_rass

最新推荐文章于 2024-10-14 09:52:52 发布

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本文介绍了一个数学问题：如何从一个给定整数n的所有因数中找到最大的可爱数。可爱数定义为不被任何大于1的整数的平方整除的正整数。通过分解n的质因数并利用这些质因数的独特性质来解决问题。

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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.

$\text{[math]}$

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

Output

Input

Output

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 indeed lovely.

Find all prime divisors of n. Assume they arep₁, p₂, ..., p_k (in $\text{[math]}$ ). If answer isa, then we know that for each 1 ≤ i ≤ k, obviously a is not divisible byp_i² (and all greater powers ofp_i). Soa ≤ p₁ × p₂ × ... × p_k. And we know thatp₁ × p₂ × ... × p_k is itself lovely. So,answer isp₁ × p₂ × ... × p_k

Time complexity: $\text{[math]}$

大致思路是所有质因子相乘

比较简洁的代码

#include <iostream>
#include <cstring>
#include <stdio.h>
using namespace std;
int main()
{
    long long n;
    while(scanf("%I64d",&n)!=EOF)
    {
        long long  i;
        for(i=2;i*i<=n;i++)
        {
            if(n%(i*i)==0)
            {
                n/=i;i=1;
            }
        }
        printf("%I64d\n",n);
    }
}