codeforces B. Duff in Love

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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。可爱数是指不能被任何大于1的整数的平方整除的数。文章通过一个简洁的C++代码示例展示了如何高效地解决这个问题。

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.
Examples
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.

解析：如果该数是完全平方数的倍数，那么让它除以这个i，就可以保证他是其他因子的乘积（最大的），也不是完全平方数的倍数

#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rep1(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=1e6+100;
ll arr[N];
int main()
{
    ios::sync_with_stdio(false);
    ll n;
    cin>>n;
    for(ll i=2;i*i<=n;i++)
    {
        while(n%(i*i)==0)
            n/=i;
    }
    cout<<n<<endl;
    return 0;
}