Friday 25 March 2016

Solution of LIGHT OJ 1022 :: Circle in Square


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>
#include <math.h>
#define PI 2*acos(0.0)
int main()
{
    int a,b;
    double c,d,e,f,g,h,r;
    scanf("%d", &b);
    for(a=1; a<=b; a++)
    {
        scanf("%lf", &r);
        c=((r+r)*(r+r))-(PI*r*r);
        printf("Case %d: %.2lf\n",a,c);
    }
    return 0;
}

2 comments:

  1. #include
    #include
    #define pi 2*acos(0.0)
    int main()
    {
    int T,i;
    scanf("%d",&T);
    for(i=1;i<=T;i++)
    {
    float area,r;
    scanf("%f",&r);
    area=(4*r*r)-(pi*r*r);
    printf("Case %d: %.2f\n",i,area);
    }
    return 0;
    }

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