Saturday 6 February 2016

Solution of URI 1066 :: Even, Odd, Positive and Negative

   , ,  3 comments

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 a,b,c1=0,c2=0,c3=0,c4=0;
    for(a=1;a<=5;a++)
    {
        scanf("%d", &b);
        if(b%2==0) c1++;
        if(b%2==1||b%2==-1) c2++;
        if(b>0) c3++;
        if(b<0) c4++;
    }
    printf("%d valor(es) par(es)\n",c1);
    printf("%d valor(es) impar(es)\n",c2);
    printf("%d valor(es) positivo(s)\n",c3);
    printf("%d valor(es) negativo(s)\n",c4);

    return 0;
}

3 comments:

  1. Unknown3 September 2017 at 18:13

    #include
    int main ()
    {
    int a,b,par= 0,impar=0,positivo=0, negativo=0;

    for (a=1;a<=5;a++)
    {
    scanf ("%d", &b);
    if (b%2==0)
    {
    par++;
    }
    else
    {
    impar++;
    }
    if (b>0)
    {
    positivo++;
    }
    else if (b==0)
    {
    positivo = positivo;
    }
    else
    {
    negativo++;
    }
    }
    printf ("%d valor(es) par(es)\n%d valor(es) impar(es)\n%d valor(es) positivo(s)\n%d valor(es) negativo(s)\n", par,impar,positivo,negativo);
    return 0;
    }

    ReplyDelete
    Replies
    1. Unknown16 December 2018 at 22:00

      #include
      int main()
      {
      int a,b,c1=0,c2=0,c3=0,c4=0;
      for (a=0; a<5; a++)
      {
      scanf("%d",&b);
      if(b%2 == 0)
      c1++;
      if(b%2 == 1 || b%2 == -1)
      c2++;
      if(b>0)
      c3++;
      if(b<0)
      c4++;
      }
      printf("%d valor(es) par(es)\n",c1);
      printf("%d valor(es) impar(es)\n",c2);
      printf("%d valor(es) positivo(s)\n",c3);
      printf("%d valor(es) negativo(s)\n",c4);

      return 0;
      }

      THE RIGHT IS!

      Delete
  2. Savio9 August 2019 at 23:15

    #include



    int main(){

    int n1,n2,n3,n4,n5;
    int vetor[5];
    int i,x;
    int par,impar,pos,neg;
    scanf("%d\n",&n1);
    scanf("%d\n",&n2);
    scanf("%d\n",&n3);
    scanf("%d\n",&n4);
    scanf("%d\n",&n5);
    vetor[0]=n1;
    vetor[1]=n2;
    vetor[2]=n3;
    vetor[3]=n4;
    vetor[4]=n5;
    x=0;
    pos=0;
    neg=0;
    impar=0;
    par=0;
    for(i=0;i<5;i++){
    x=vetor[i];
    //printf("%d",x);
    if (x>0){
    pos+=1;
    }
    if(x<0){
    neg+=1;
    }
    if(x%2==0||x==0){
    par+=1;
    }
    if (x%2!=0){
    impar+=1;
    }
    }
    printf("%d valor(es) par(es)\n",par);
    printf("%d valor(es) impar(es)\n",impar);
    printf("%d valor(es) positivo(s)\n",pos);
    printf("%d valor(es) negativos(s)\n",neg);
    return 0;
    }

    wrong answer 5%, i tried and cannot see the correct answer

    ReplyDelete

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