Jan’s LightOJ :: 1354 - IP Checking Solution

1354 - IP Checking

#include <stdio.h>
int DecBin(int n)
{
        int multiplier = 1,result = 0;
        while(n>0)
        {

        result+=n%2*multiplier;
        multiplier*=10;
        n/=2;

        }
        return result;


}
int main()
{
        int T,da,db,dc,dd,ba,bb,bc,bd;
        scanf("%d",&T);
        int i;
        for(i =0; i<T;i++)
        {
                scanf("%d.%d.%d.%d",&da,&db,&dc,&dd);
                scanf("%d.%d.%d.%d",&ba,&bb,&bc,&bd);

                if(DecBin(da)==ba&&DecBin(db)==bb&&DecBin(dc)==bc&&DecBin(dd)==bd)
                printf("Case %d: Yes\n",i+1);
                else
                printf("Case %d: No\n",i+1);

        }
        return 0;


}

Jan’s LightOJ :: 1042 - Secret Origins Solution

1042 - Secret Origins

#include <stdio.h>
int oneCount(int n)
    {
    int count = 0;
    long long t = 1;
    for(t = 1; t<=n; t*=2) {
            if((n&(t))!=0) {
                    count++;
                    }

            }
    return count;


    }
long long int next(long long int num)
    {
    long long res,t;
    for(t = 1; t<=num; t*=2) {
            if((num&(t))!=0) {
                    res = num+t;
                    break;

                    }

            }
    int diff = oneCount(num)-oneCount(res);
    int i;
    for(i = 0; i<diff; i++) {
            res+=(1<<i);

            }
    return res;



    }
int main()
    {
    int T,i,N,j;
    scanf("%d",&T);
    for(i =0; i<T; i++) {
            scanf("%d",&N);
            printf("Case %d: %lld\n",i+1,next(N));

            }
    return 0;



    }

Jan’s LightOJ :: 1006 - Hex-a-bonacci Solution

1006 - Hex-a-bonacci

    #include <cstdio>
    #include <bits/stdc++.h>

    using namespace std;
    long long data[100005];
    long long a, b, c, d, e, f;
    long long fn( long long n )
    {
        if( n == 0 ) return a;
        else
        if( n == 1 ) return b;
        else
        if( n == 2 ) return c;
        else
        if( n == 3 ) return d;
        else
        if( n == 4 ) return e;
        else
        if( n == 5 ) return f;
        else
        if(data[n]!=-1) return data[n];
        else
        {

            data[n] = ( fn(n-1) + fn(n-2) + fn(n-3) + fn(n-4) + fn(n-5) + fn(n-6) )%10000007;

            return data[n];
        }
    }
    int main()
    {
        long long n, caseno = 0, cases;
        scanf("%lld", &cases);
        while( cases-- )
        {
        memset(data,-1,sizeof(data));
            scanf("%lld %lld %lld %lld %lld %lld %lld", &a, &b, &c, &d, &e, &f, &n);
            printf("Case %lld: %lld\n", ++caseno, fn(n) % 10000007);
        }
        return 0;
    }

Jan’s LightOJ :: 1000 - Greetings from LightOJ Solution

1000 - Greetings from LightOJ

#include <stdio.h>

int main()
{
    int n,i, a, b, sum;

    scanf("%d",&n);
    for (i = 0; i< n; i++)
    {
        scanf("%d %d",&a,&b);
        sum = a + b;
         printf("Case %d: %d\n",i+1,sum);

    }



    return 0;
}

Jan’s LightOJ :: 1069 - Lift Solution

1069 - Lift

    #include <stdio.h>

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

        int i;
        int a,b;
        for(i =0; i<T;i++)
        {
            scanf("%d%d",&a,&b);
            if(b==a)
            printf("Case %d: %d\n",i+1,(19+b*4));
            else if(a<b)
            printf("Case %d: %d\n",i+1,(19+b*4));
            else if(a>b)

            printf("Case %d: %d\n",i+1,(19+a*4+(a-b)*4));



        }

        return 0;
    }

Jan’s LightOJ :: 1107 - How Cow Solution

1107 - How Cow

#include <stdio.h>

int main()
{
    int T,M;
    scanf("%d",&T);
    int i,j;
    for(i =0; i<T; i++)
    {
        int x1,x2,y1,y2,a,b;

        printf("Case %d:\n",i+1);
        scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
        scanf("%d",&M);
        for(j=0; j<M; j++)
        {
            scanf("%d%d",&a,&b);

            if((a>=x1&&a<=x2)&&(b>=y1&&b<=y2))
            {
                printf("Yes\n");
            }
            else
                printf("No\n");


        }
    }
    return 0;

}

Jan’s LightOJ :: 1297 - Largest Box Solution

1297 - Largest Box

#include <stdio.h>
#include <math.h>

int main()
{
        int T,i;
        double L,W,x;
        scanf("%d",&T);
        for(i =0; i<T; i++)
        {
                scanf("%lf%lf",&L,&W);
                x = (W+L-sqrt(W*W+L*L-W*L))/6;
                printf("Case %d: %lf\n",i+1,(W-2*x)*(L-2*x)*x);


        }
        return 0;


}

Jan’s LightOJ :: 1053 - Higher Math Solution

1053 - Higher Math

#include <stdio.h>
#include <math.h>
int main(){
int input,i=1;

    unsigned a,o,h;
    scanf("%d",&input);

    while(input-->0)
    {
        scanf("%d%d%d",&a,&o,&h);
        if(sqrt(a*a+o*o)==h)
        {
            printf("Case %d: yes\n",i++);
        }
        else if(sqrt(h*h+o*o)==a)
        {
            printf("Case %d: yes\n",i++);
        }
        else if(sqrt(a*a+h*h)==o)
        {
            printf("Case %d: yes\n",i++);
        }
        else
            printf("Case %d: no\n",i++);



    }

    return 0;
}

Jan’s LightOJ :: 1116 - Ekka Dokka Solution

1116 - Ekka Dokka

#include <stdio.h>
#include <math.h>

int main()
    {
    int T,x,i,j,m;
    long int w;
    scanf("%d",&T);
    for(x =0; x<T; x++) {

            scanf("%ld",&w);
            if(w%2!=0)
                printf("Case %d: Impossible\n",x+1);
            else {
                    m = 1;
                    while(w%2==0) {
                            m*=2;
                            w/=2;

                            }
                            printf("Case %d: %ld %d\n",x+1,w,m);



                    }





            }
    return 0;
    }

Jan’s LightOJ :: 1001 - Opposite Task Solution

1001 - Opposite Task 

#include <stdio.h>
int oneCount(int n)
    {
    int count = 0;
    long long t = 1;
    for(t = 1; t<=n; t*=2) {
            if((n&(t))!=0) {
                    count++;
                    }

            }
    return count;


    }
long long int next(long long int num)
    {
    long long res,t;
    for(t = 1; t<=num; t*=2) {
            if((num&(t))!=0) {
                    res = num+t;
                    break;

                    }

            }
    int diff = oneCount(num)-oneCount(res);
    int i;
    for(i = 0; i<diff; i++) {
            res+=(1<<i);

            }
    return res;



    }
int main()
    {
    int T,i,N,j;
    scanf("%d",&T);
    for(i =0; i<T; i++) {
            scanf("%d",&N);
            printf("Case %d: %lld\n",i+1,next(N));

            }
    return 0;



    }

Jan’s LightOJ :: 1216 - Juice in the Glass Solution

1216 - Juice in the Glass

    #include <stdio.h>
    #include <math.h>
    #define PI acos(-1)
    int main()
        {
        int T,i,r1,r2,h,p;
        double R,R2,V;
        scanf("%d",&T);

        for(i =0; i<T; i++) {
                scanf("%d%d%d%d",&r1,&r2,&h,&p);
                R = r2+ (r1-r2)*((double)(p)/h);
                V = ((PI*p)*(R*R+R*r2+r2*r2))/3.0;
                printf("Case %d: %lf\n",i+1,V);


                }
        return 0;


        }

Jan’s LightOJ :: 1305 - Area of a Parallelogram Solution

1305 - Area of a Parallelogram

#include <stdio.h>
#include <math.h>
int main()
{
        int i,T,Dx,Dy,Ax,Ay,Bx,By,Cx,Cy,height,base;
        int area;

        scanf("%d",&T);
        for(i  =0 ;i<T; i++)
        {
                scanf("%d%d%d%d%d%d",&Ax,&Ay,&Bx,&By,&Cx,&Cy);
                Dx = Ax+Cx-Bx;
                Dy = Ay-By+Cy;
                height = Cy-By;
                base = Bx-Ax;
                area = ((Ax*By)+(Bx*Cy)+(Cx*Dy)+(Dx*Ay))-((Ay*Bx)+(By*Cx)+(Cy*Dx)+(Dy*Ax));


                printf("Case %d: %d %d %d\n",i+1,Dx,Dy,abs(area/2));


        }

        return 0;


}

Jan’s LightOJ :: 1385 - Kingdom Division Solution

1385 - Kingdom Division

    #include <stdio.h>
    #include <math.h>
    int main()
        {
        int T,i;
        double a,b,c,result,div;
        scanf("%d",&T);
        for(i = 0; i<T; i++) {
                scanf("%lf%lf%lf",&a,&b,&c);
                div = ((b * b)-(a * c))/b;
                if((b*b-a*c)/b<=0)
                    printf("Case %d: -1\n",i+1);
                else {
                        result = (a*c)*(b+c+a+b)/div;
                        printf("Case %d: %.8lf\n",i+1,result/b);
                        }

                }


        return 0;
        }

Jan’s LightOJ :: 1225 - Palindromic Numbers (II) Solution

1225 - Palindromic Numbers (II)

    #include 

    int main()
    {
        int T;
        scanf("%d",&T);
        int n;
        int reverse=0;
        int in;
        int i;
        for(i =0; i        {
            scanf("%d",&n);
            in = n;
            while(n!=0)
            {
                reverse = reverse*10;
                reverse = reverse + n%10;

                n = n/10;
            }
            if(in==0)
            printf("Case %d: Yes\n",i+1);
            else

            if(in==reverse)
            printf("Case %d: Yes\n",i+1);
            else
            printf("Case %d: No\n",i+1);

            reverse = 0;


        }
    return 0;
    }

Jan’s LightOJ :: 1433 - Minimum Arc Distance Solution

1433 - Minimum Arc Distance

#include <stdio.h>
#include <math.h>
int main()
{
        int i,T, Ox,Oy,Ax,Ay,Bx,By;
        double angle,s,AB,OB,OA;
        scanf("%d",&T);

        for(i =0; i<T;i++)
        {
                scanf("%d%d%d%d%d%d",&Ox,&Oy,&Ax,&Ay,&Bx,&By);
                AB = sqrt(pow(Ax-Bx,2)+pow(Ay-By,2));
                OB = sqrt(pow(Ox-Bx,2)+pow(Oy-By,2));
                OA= OB;
                angle = acos((OB*OB + OA*OA-AB*AB)/(2*OB*OA));
                s = OB*angle;
                printf("Case %d: %f\n",i+1,s);



        }
        return 0;




}

Jan’s LightOJ :: 1294 - Positive Negative Sign Solution

1294 - Positive Negative Sign

#include <stdio.h>

int main()
{
        int T,i;
        long long n,m;
        scanf("%d",&T);
        for(i =0; i<T;i++)
        {
                scanf("%ld%ld",&n,&m);
                printf("Case %d: %lld\n",i+1,(m*(n/2)));

        }
        return 0;

}

Jan’s LightOJ :: 1022 - Circle in Square Solution

1022 - Circle in Square

#include <stdio.h>
#include <math.h>
int main()
    {
    int T,i;
    double r;
    scanf("%d",&T);
    for(i =0; i<T; i++) {
            scanf("%lf",&r);
            printf("Case %d: %.2f\n",i+1,((2*r)*(2*r)-((2*acos(0.0))*r*r))+1e-9);
            }
    return 0;
    }

Jan’s LightOJ :: 1331 - Agent J Solution

1331 - Agent J

#include <stdio.h>
#include <math.h>
const double EPS = 1e-9;
int main()
    {
    int i,T;
    scanf("%d",&T);
    double r1,r2,r3,a,b,c,s,area,A,B,C,result;
    for(i =0; i<T; i++) {
            scanf("%lf%lf%lf",&r1,&r2,&r3);
            a = r1+r2;
            b = r2+r3;
            c = r3 + r1;
            s = (a+b+c)*0.5;
            area = sqrt(s*(s-a)*(s-b)*(s-c));
            A = acos((a*a+c*c-b*b)/(2.0*a*c));
            B = acos((a*a+b*b-c*c)/(2.0*a*b));
            C = acos((b*b+c*c-a*a)/(2.0*b*c));
            result = area-A*r1*r1*0.5-B*r2*r2*0.5-C*r3*r3*0.5;
            printf("Case %d: %.8lf\n",i+1,result+EPS);
            }
    return 0;

    }

Jan’s LightOJ :: 1275 - Internet Service Providers Solution

1275 - Internet Service Providers

    #include <stdio.h>

    int main()
        {
        long long N,C;
        int T,i,ans;
        scanf("%d",&T);
        for(i =0; i<T; i++
           ) {
                scanf("%lld%lld",&N,&C);
                if(N==0)
                    ans = 0;
                else
                    ans = C/(2*N);

                            int res1 = ans*C-ans*ans*N,res2=(ans+1)*C-(ans+1)*(ans+1)*N;
                            res1>=res2?1:(ans+=1);
                if(N==0) ans = 0;
                printf("Case %d: %d\n",i+1,ans);
                }


        return 0;
        }

Jan’s LightOJ :: 1387 - Setu Solution

1387 - Setu

#include <stdio.h>

int main()
    {
    int i,j,y,input,sum=0,donation;
    scanf("%d",&input);
    char step[6];

    for(i = 0; i<input; i++) {
            scanf("%d",&j);

                    printf("Case %d:\n",i+1);
            for(y = 0; y<j; y++) {
                    scanf("%s",step);
                    if(step[0]=='d') {
                            scanf("%d",&donation);
                            sum+=donation;

                            }
                    if(step[0]=='r')
                        printf("%d\n",sum);



                    }
                    sum = 0;

            }
    return 0;

    }

Jan’s LightOJ :: 1020 - A Childhood Game Solution

1020 - A Childhood Game
Solution:

    #include <stdio.h>
    int main()
        {
        int T,i,n;
        char name[5];
        scanf("%d",&T);

        for(i =0; i<T; i++) {
                scanf("%d%s",&n,name);
                if(name[0]=='A') {
                        if(n%3==1)
                            printf("Case %d: Bob\n",i+1);
                        else
                            printf("Case %d: Alice\n",i+1);
                        }
                else if(name[0]=='B') {
                        if(n%3==0)
                            printf("Case %d: Alice\n",i+1);
                        else
                            printf("Case %d: Bob\n",i+1);



                        }

                }

        return 0;
        }

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;
    }

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;


    }