Codeforces 235A. LCM Challenge

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 than n. 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 ≤ 106) — 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.

先处理边界情况1和2

然后n是奇数，自然想到n*(n-1)*(n-2)最大

n是偶数的情况,n和n-1肯定互质，但是n-2就不一定了，所以往下暴力找一个数，使三者互质。但是这样的结果不一定有(n-1)*(n-2)*(n-3)大，所以比较一下

#include <iostream>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;


LL gcd(LL a,LL b){
  if (b == 0) return a;
  return gcd(b,a % b);
}
LL ans = 1,n,total;
int main()
{
    // freopen("test.in","r",stdin);
    cin >> n;
    if (n == 2){
      cout << 2; return 0;
    }
    if (n == 1){
      cout << 1; return 0;
    }
    if (n % 2){
      cout << n * (n-1) * (n-2);
    }
    else {
      ans = n * (n-1);
      LL now = n-2;
      while (gcd(now,n) != 1 || gcd(now,n-1) != 1){
        now --;
      }
      ans *= now;
      cout << max(ans,(n-1)*(n-2)*(n-3));
    }

    return 0;
}