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;
    }

Jan’s LightOJ :: 1182 – Parity Solution

1182 - Parity Solutoin
 

    #include <stdio.h>
    int oneCount(int n)
    {
            int count = 0;
            while(n>0)
            {

                    if(n%2==1) count++;
                    n/=2;
            }
          return count;

    }

    int main()
    {
            int T;
            unsigned long long int n;
            scanf("%d",&T);
            int i;
            for(i =0; i<T;i++)
            {
                    scanf("%lld",&n);
                    if(oneCount(n)%2==0)
                    printf("Case %d: even\n",i+1);
                    else
                    printf("Case %d: odd\n",i+1);

            }
    return 0;


    }