acd LCM Challenge(求1~n的随意三个数的最大公倍数)

Problem Description

Some days ago, I learned the concept of LCM (least common multiple). I‘ve played with it for several times and I want to make a big number with it.

But I also don‘t want to use many numbers, so I‘ll choose three positive integers (they don‘t have to be distinct) which are not greater thann. Can you help me to find the maximum possible least common multiple
of these three integers?

Input

The first line contains an integer n (1?≤?n?≤?10^6) — the n mentioned in the statement.

Output

Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.

Sample Input

9

Sample Output

504

仅仅要这三个数中有两个数是奇数一个是偶数，最小公倍数就是这三个数的积。

#include<stdio.h>
int main()
{
    long long  LCM,n;
    while(scanf("%lld",&n)>0)
    {
        if(n==1)LCM=1;
        if(n==2)LCM=2;
        if(n>2)
        {
            if(n%2)LCM=n*(n-1)*(n-2);
            else
            {
                if(n*(n-1)*(n-2)/2<n*(n-1)*(n-3))
                    LCM=n*(n-1)*(n-3);
                else LCM=n*(n-1)*(n-2)/2;
            }
        }
        printf("%lld\n",LCM);
    }
}

acd LCM Challenge(求1~n的随意三个数的最大公倍数)