1 <= N <= 106。
规律题:
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
long long n, number;
cin>>n;
if( n <= 2)
{
cout<<2;
}
else if(n % 2)
{
number = n * (n - 1) * (n - 2);
cout<<number;
}
else
{
if( n % 3 == 0)
{
number = (n - 1) * (n - 2) * (n - 3) ;
}
else number = n * (n - 1) * (n - 3);
cout<<number;
}
return 0;
}原文地址:http://blog.csdn.net/u014492609/article/details/44726957