codeforces 588 B Duff in Love

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最新推荐文章于 2023-12-10 21:11:55 发布

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分类专栏： ACM 文章标签：分解公因数

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本文介绍了一种算法，用于找出给定数及其因子中最大的Lovely数。Lovely数定义为不能被任何大于1的完全平方数整除的正整数。通过分解质因数并结合不重复的因子来解决问题。

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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 
        
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 <iostream>  
#include <cstdio>  
#include <climits>  
#include <cstring>  
#include <cstdlib>  
#include <cmath>  
#include <vector>  
#include <queue>  
#include <algorithm>  
#define esp 1e-6  
#define inf 0x0f0f0f0f  
#define LL long long    
using namespace std;  
  
/************************************************  
  
Desiner:hl  
time:2016/02/06
Exe.Time:31 ms
Exe.Memory:0 KB  
 
题目链接：http://codeforces.com/contest/588/problem/B
 
题意： 给你一个数。问这个数和他的因子中最大的不会被n^2(n=2,3,4,5....)整除的数是哪个 
 
题解：分解质因数 然后把不重复的乘起来就好了,注意要long long 没用坑了几把。。。 还有别忘记优化。 
************************************************/   

int main(){
	int i,j,k,l,m,n;
	long long ans=1,p;
	scanf("%I64d",&p);
	long long c=p;
	long long a[1001];
	a[0]=1;
	j=1;
	for(i=2;i<=c;i++){
		while(c%i==0){
			if(a[j-1]!=i){
				a[j++]=i;
			}
			c=c/i;
		}
		//优化，如果大于c的 
		if(i>sqrt(c)){
			a[j++]=c;
			break;
		} 

	}
	for(i=0;i<j;i++){
		ans=ans*a[i];
	}
	printf("%I64d\n",ans);
}