acd LCM Challenge(求1~n的随意三个数的最大公倍数)

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?

9

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);
    }
}