【Codeforces Round 326 (Div 2)B】【质因数分解】Duff in Love n的最大因子使其不为平方数倍数

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本文介绍了一个算法问题，即从给定正整数n的所有因子中找出最大的“可爱”数。这里“可爱”的定义是该数不能被任何大于1的整数的平方整除。文章通过质因数分解的方法给出了一个O(sqrt(n))的时间复杂度解决方案。

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
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.

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<iostream>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=0,M=0,Z=1e9+7,ms63=1061109567;
int casenum,casei;
int main()
{
	LL n;
	while(~scanf("%lld",&n))
	{
		int top=sqrt(n);
		LL ans=1;
		for(int i=2;i<=top;i++)if(n%i==0)
		{
			ans*=i;
			while(n%i==0)n/=i;
		}
		if(n)ans*=n;
		printf("%lld\n",ans);
	}
	return 0;
}
/*
【题意】
给你一个n（[1,1e12]），
我们想要找到n中的最大的一个因子，使得这个因子不是任何平方数的倍数。

【类型】
质因数分解

【分析】
不是任何平方数的倍数，等价于不是任何素数的倍数。
不是任何素数的倍数，等价于任何素数这个数都最多只含有一个。
于是我们对n做素数拆分，所有其因子素数的单次乘积结果，就是答案啦。

【时间复杂度&&优化】
O（sqrt(n)）

*/