Saturday, 30 April 2016

Solution of URI 1186 :: Below the Secundary Diagonal


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()
{
    double a=0.0, M[12][12];
    char T[2];
    int C,x,y,z,p=11;
    scanf("%s", &T);
    for(x=0;x<=11;x++)
    {
        for(y=0; y<=11; y++)
            scanf("%lf", &M[x][y]);
    }
    for(z=1; z<=11;z++)
    {
        for(C=11; C>=p;C--)
            a+=M[z][C];
        p--;
    }
    if(T[0]=='S')
        printf("%.1lf\n",a);
    else if(T[0]=='M')
    {
        a=a/66.0;
        printf("%.1lf\n",a);
    }
    return 0;
}

2 comments:

  1. #include
    int main()
    {

    int i,j,d=11;
    float a[12][12],s=0.0,m=0.0,x=0.0;
    char c;
    scanf("%c",&c);
    for(i=0; i<=11; i++)
    {
    for(j=0; j<=11; j++)
    {
    scanf("%f",&a[i][j]);
    if( i>=1 && j>d )
    {
    s=s+a[i][j];
    m++;
    }

    }
    d--;
    }
    if(c=='S')
    {
    printf("%.1f\n",s);
    }
    else if(c=='M')
    {
    x=s/m;
    printf("%.1f\n",x);
    }
    return 0;
    }

    ReplyDelete
  2. This comment has been removed by the author.

    ReplyDelete

Note: only a member of this blog may post a comment.