//javada  ,in java

import java.util.Scanner;
public class ONETHOUSANDTWO {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int s1 = 0, s2 = 0, i = 0;
        if ((m * n - 1) % 2 == 0) {
            System.out.println();
        }

        while(i==0)
        {
            if(n%2==0||m%2==0)
            {
                if(n%2==0&&m%2==0)
                {
                    if(n>m)n=n/2;
                    else m=m/2;
                }
                else if(n%2==0)n=n/2;
                else m=m/2;
                s1+=n*m;
            }
            else
            {
                s1+=1;
                s2+=n*m-1;
                n=0;m=0;i=1;
            }

            //------------------------------------------------
            if(n%2==0||m%2==0)
            {
                if(n%2==0&&m%2==0)
                {
                    if(n>m)n=n/2;
                    else m=m/2;
                }
                else if(n%2==0)n=n/2;
                else m=m/2;
                s2+=n*m;
            }
            else
            {
                s2+=1;
                s1+=n*m-1;
                n=0;m=0;i=1;
            }

        }
        if (s1>=s2) System.out.println("[:=[first]");
        else System.out.println("[second]=:]");
    }
}

Thank you very much!)
You saved hours of my life!

Is it possible at all?

That's the thought. It is conjectured that however youcut it during the process, the total steps are always the same, but I didn't prove it. Now Iget it. Thank you.

Спасибо за пояснение;) а каким же образом люди в топе выходят за 0.05 секунд? Вон первое место в А+B аж за 0.015 работает! Наверняка люди обходят стороной медленные методы и операции. Может вы знаете ресурс где можно почитать об этом? Поиск в гугле ни привёл ни к чем

и a=(m*n-1)%2;