Wednesday 8 June 2016

Solution of URI 2059 :: Odd, Even or Cheating

   , ,  1 comment

Before seeing the solution make sure that you tried enough. Don’t paste the whole code, just find out the logic. If you stuck in trouble, just inform me on comment.

/**Bismillahir Rahmanir Rahim.**/

#include <stdio.h>
int main()
{
    int p, j1, j2, r, a, win, sum;
    scanf("%d %d %d %d %d", &p, &j1, &j2, &r, &a);
    sum = j1 + j2;
    if((sum%2==0 && p==1) || (sum%2==1 && p==0)) win = 1;
    else win = 2;
    if((r==1 && a==0) || (r==0 && a==1)) win = 1;
    else if(r==1 && a==1) win=2;
    printf("Jogador %d ganha!\n", win);
    return 0;
}

1 comment:

  1. Anonymous31 March 2018 at 02:27

    What's wrong in this code?
    (Wrong answer (10%))
    #include
    int main(){
    int p, j1, j2, r, a, x;
    scanf("%d %d %d %d %d", &p, &j1, &j2, &r, &a);
    x= (j1+j2) %2;
    if (x==0) {
    if ((p==1) || (p==0 && r==1 && a==0))
    printf("Jogador 1 ganha!\n");
    else printf("Jogador 2 ganha!\n");
    }
    else if (x!=0) {
    if ((p==0) || (p==1 && r==1 && a==0))
    printf("Jogador 1 ganha!\n");

    else printf("Jogador 2 ganha!\n");
    }
    return 0;
    }

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