Jan’s LightOJ :: 1178 - Trapezium Solution

1178 - Trapezium
Solution:

#include <stdio.h>
#include <math.h>
int main()
    {
    int T,i;
    double a,b,c,d,s,h,w,area,result;
    scanf("%d",&T);

    for(i =0; i<T; i++) {
            scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
            s = (d+b+a-c)/2;
            area = sqrt(s*(s-d)*(s-b)*(s-a+c));
                        h = (2*area)/(fabs(a-c));
                        result = ((a+c)*h)/2;
                        printf("Case %d: %lf\n",i+1,result);



            }

    return 0;


    }

Jan’s LightOJ :: 1043 - Triangle Partitioning Solution

1043 - Triangle Partitioning

#include <stdio.h>
#include <math.h>
int main()
{
        int T;
        scanf("%d",&T);
        float AB,AC,BC,ratio;
        int i;
        for(i =0; i<T; i++)
        {
                scanf("%f%f%f%f",&AB,&AC,&BC,&ratio);
                printf("Case %d: %f\n",i+1,(sqrt(ratio)/sqrt(ratio+1))*AB);



        }

return 0;

}

Jan’s LightOJ :: 1015 – Brush(I) Solution

1015 Brush(I)

Solution:

    #include <stdio.h>

    int main()
    {
        int T,N;
        scanf("%d\n",&T);

        int i,j,dust;
        unsigned long long int sum = 0;
        for(i =0; i<T;i++)
        {
            scanf("%d",&N);
            for(j=0;j<N;j++)
            {
                scanf("\n%d",&dust);
                if(dust>0)
                sum+=dust;


            }
            printf("Case %d: %lld\n",i+1,sum);
            sum =0;



        }

        return 0;
    }