1929. Teddybears are not for Everyone (Timus) (combination+reading questions)

http://acm.timus.ru/problem.aspx?space=1&num=1929

combination problems. 排列组合问题。

According to the problems, it is assumed that Holden is chosen and there are two more open positions. And based on this combination, constraints are needed to be satisfied that there must be at least one Teddyhater in each group.  Totally

n: people, m: haters and n%3==0

k = n%3

case 1: m < k: impossible 0

case 2: m == k

Holden is a hater: (Holden)(non-hater)(non-hater) = (n-m)(n-m-1)/2

Holden is not a hater: (Holden)(Hater)(non-hater) = (m)(n-m-1)

case 3: m == k+1 : one more haters

Holden is hater: (Holden)(non-hater)(non-hater) + (Holden)(hater)(non-hater)

Holden is not a hater: (Holden)(Hater)(non-hater) + (Holden)(Hater)(hater)

case 4: else : two more or three more

Holden is hater: (Holden)(non-hater)(non-hater) + (Holden)(hater)(non-hater) + (Holden)(hater)(hater) = (Holden)(others)(others)

Holden is not a hater: (Holden)(Hater)(non-hater) + (Holden)(Hater)(hater)

import java.util.Scanner;
 
public class timus1929 {
    //https://github.com/fanofxiaofeng/timus/blob/master/1929/main.py -- reference
    //http://acm.timus.ru/problem.aspx?space=1&num=1929 -- problem
    // Holden is here and there two more spots left
    public static void main(String[] args) {
        Scanner in  =new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        int k = n/3; //groups
        int tag = 0;//Hole is not haters
        for(int i = 0; i<m; i++){
            if(in.nextInt()==1)
                tag = 1;
        }
        int res = 1;
        if(m<k){
            System.out.println(0);
            return;
        }
        if(tag==0){
            if(m==k) res = m*(n-m-1);
            //else res = m*(n-m-1) + m*(m-1)/2;
            else if(m==k+1) res = m*(n-m-1) + m*(m-1)/2;
            else res = (n-1)*(n-1-1)/2 - (n-m-1)*(n-m-2)/2;
        }else {//Holden is haters
            if(m==k) res = (n-k)*(n-k-1)/2;
            //else res = (n-m)*(n-m-1)/2 + (m-1)*(n-m);
            else if(m==k+1) res = (n-m)*(n-m-1)/2 + (m-1)*(n-m);
            else res = (n-m)*(n-m-1)/2 + (m-1)*(n-m) + (m-1)*(m-2)/2; //res = (n-1)*(n-1-1)/2;//1: Holden and// res = (n-m)*(n-m-1)/2 + (m-1)*(n-m) + (m-1)(m-2)/2
        }
 
        //int res = combination(4,0);
        System.out.println(res);
    }
}

It is hard to figure out the meaning of the problems

tai tm nan le!!!!