Blue Bridge Cup practice system-algorithm training min Max common multiple

Problem description
A positive integer n is known, asking for three numbers from 1~n, and the maximum number of least common multiple they can have.

Input format
Enter a positive integer n.

Output format
Output An integer that represents the least common multiple you found.
Sample input
9
Sample output
504
Data size and conventions
1 <= N <= 106.

Analysis:

The meaning of this question is to find three numbers within the range of the 1~n, so that their least common multiple in this range is the largest combination.

Next, let's say one conclusion: two contiguous natural numbers greater than 1 are bound to coprime.

And for the 1~n range, it must be N (n-1) * (n-2) The product of the largest, if the three number is 22 coprime that is the best.

If n is an odd number, then N, n-1, n-2 must be 22 coprime, if there is some entanglement, then we analyze under what circumstances there may be a public factor. n is an odd number, then the n,n-1,n-2 must be two odd plus one even case. Male factor 2 Direct pass because there is only one even number. Assuming that one of the remaining n,n-2 is divisible by 3, then the number of common factors must be n or n-2 plus minus 3 to obtain the condition. For this reason, the product of n,n-1,n-2 is not only the largest, but also a certain 22 coprime.

If n is an even number, continue to analyze N (n-1) * (n-2), so that N and N-2 must have a common factor of 2, then change to the N (n-1) * (n-3). And then think about it, no ah, if the even number itself can be divisible by 3, then the formula N (n-1) * (n-3) is not set up, and the n-3 has a common factor of 3, and then think about it, the formula becomes (n-1) * (n-2) * (n-3), two odd clip a even situation.

Code

#include <iostream>using namespacestd; intMain () {Long LongN, ans;  while(Cin >>N) {if(N <=2) {ans=N; }           Else if(n%2) {ans= n * (n-1) * (N-2); }          Else {              if(n%3) ans = n * (n1) * (n3); ElseAns= (n1) * (n2) * (n3); } cout<< ans <<Endl; }      return 0; }   

Blue Bridge Cup practice system-algorithm training min Max common multiple