The long path of Xiao Sha (graph theory, Euler diagram)

https://ac.nowcoder.com/acm/contest/23477/E

I drew the shortest path according to the drawing , Is to draw every three adjacent edges into the shortest , It can be found that the shortest path is always $n-1$, Then according to the definition , Eulatus is either all even points , Or there are two singularities , You can find , When n In an odd number of , Every point of the complete graph is an even point , So you can go all the way ; When it's even , Every point is a singularity , At least a few edges need to be deleted so that the graph has only two singularities , That's subtracting $n/2-1$. Then the total number of sides of a complete graph is $n\ast (n-1)/2$, From the formula of the first term plus the last term .

#include<bits/stdc++.h>
#define int long long
using namespace std;
signed main()
{
    
	int n;
	cin >> n;
	cout << n - 1 << " ";
	if (n % 2 == 0) cout << (1 + n - 1) * (n - 1) / 2 - (n / 2 - 1);
	else cout << (1 + n - 1) * (n - 1) / 2;
	return 0;
}