SDNU 1464. The maximum least common multiple (thinking)

Description

Problem Description Given a positive integer N, Q ~ from any 1 N elect three numbers, their least common multiple can be up to much.

Input

Enter a positive integer N (1 <= N <= 10 ^ 6)

Output

Output An integer that represents you find the least common multiple of

Sample Input

9

Sample Output

504 
ideas: the Internet, said the problem in a o (n ^ 3) will be T, are false! I o (log n) approach is also a T! inhuman. Later, the method can only be used mathematics. 
          If n is odd, the (maximum least common multiple) = n * (n-1 ) * (n-2) 
          if n is even and, if n is a multiple of 3, then the (maximum least common multiple) = (n-1) * (n-2) * (n -3) 
                                   or (greatest common divisor) = n * (n-1 ) * (n-3)

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define ll long long
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;

ll n, sum;

int main()
{
    scanf("%lld", &n);
    if(n <= 2)printf("%lld\n", n);
    else if(n%2 != 0)
    {
        sum = n*(n-1)*(n-2);
        printf("%lld\n", sum);
    }
    else if(n%3 == 0)
    {
        sum = (n-1)*(n-2)*(n-3);
        printf("%lld\n", sum);
    }
    else
    {
        sum = n*(n-1)*(n-3);
        printf("%lld\n", sum);
    }
    return 0;
}