#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;

int main() {
    int x1, y1, x2, y2, x0, y0, l;

    cin >> x1 >> y1 >> x2 >> y2;
    cin >> x0 >> y0 >> l;
    cout << fixed << setprecision(2); // <|ENG: Set 2 symbols after dot |> <|RUS: stavim 2 simvola dlia tochki |>
    //<|ENG: get all sides |>
    //<|RUS: poluchaem vse storonui |>
    double firstEdge = sqrt(pow(x1 - x0, 2) + pow(y1 - y0, 2));
    double secondEdge = sqrt(pow(x2 - x0, 2) + pow(y2 - y0, 2));
    double field = sqrt(pow(x2 - x1, 2) + pow(y2 - y1, 2));
    double ans2 = max(firstEdge, secondEdge) - l; // <|ENG: Second answer - max length |> <|RUS: Vtoroi otvet = prosto maximum do krainih tochek |>

    double ans1;
    //<|ENG: get the squares of the sides|>
    //<|RUS: Vosvodim v kvadrat storonui |>
    double a = firstEdge * firstEdge;
    double b = secondEdge * secondEdge;
    double c = field * field;
    //<|ENG: checking if there are obtuse angles on the side that is the pineapple field |>
    //<|RUS: Proverka est li typie ygli vosle toi storoni kotoraiy iavliaetsia polem c ananasami |>
    if (a+c < b || b+c < a) {
        ans1 = min(firstEdge, secondEdge) - l;
    }
    else {
        //<|ENG: get half the perimeter of triangle|>
        //<|RUS: poluchaem poly perimetr treygolnika |>
        double p = (firstEdge + secondEdge + field) / 2;
        //<|ENG: get square of triangle|>
        //<|RUS: poluchaem ploshad treygolnika |>
        double S = sqrt(p * (p - firstEdge) * (p - secondEdge) * (p - field));

        //<|ENG: If the area is zero and the field with pineapples does not exist (test: 2 2 2 2 3 3 1)|>
        //<|RUS: Esli ploshad ravna 0 i polia c ananasami kak bu net (test: 2 2 2 2 3 3 1)|>
        if (S == 0 && field == 0) {
            ans1 = min(firstEdge, secondEdge) - l;
        }
        //<|ENG: If the area is zero and more than one side is not zero (test: -5 0 5 0 1 0 1)|>
        //<|RUS: Esli ploshad ravna 0 i ni odna storona ne ravna 0 (test:  -5 0 5 0 1 0 1)|>
        else if (S == 0 && firstEdge != 0 && secondEdge != 0 && field != 0) {
            ans1 =  0;
        }
        else {
            //<|ENG: Default. We just look for the height using the formula|>
            //<|RUS: Obichni slychai. Prosto po formule S = a*h/2 (test:  -5 0 5 0 1 0 1)|>
            double d = 2 * S / field;
            ans1 = d - l;
        }
    }
    //<|ENG: If, after subtracting the rope, the length becomes negative, then do 0|>
    //<|RUS: Esli sle vichetania verevki otvet stal menihe nylia to delayem 0|>
    ans1 = ans1 < 0 ? 0 : ans1;
    ans2 = ans2 < 0 ? 0 : ans2;

    //<|ENG: Next just cout "round(ans1 * 100) / 100" and "round(ans2 * 100) / 100"|>
    //<|RUS: Teper prosto vivedi "round(ans1 * 100) / 100" i "round(ans2 * 100) / 100"|>

    //!!!!!!!!!!!!!!!!!!!
    //cout << round(ans1 * 100) / 100 << endl;
    //cout << round(ans2 * 100) / 100;
    return 0;

    //<|ENG: Thanks a lot for your time here!|>
    //<|RUS: Spasibo chto pochitali i uia vam pomog (navernoe)|>
    // :)
}

#include <iostream>
#include <cmath>

using namespace std;

int main(){
    double x1 = 0, y1 = 0, x2 = 0, y2 = 0;
    double px = 0, py = 0;
    double length;

    double A = 0;
    double B = -1;
    double C = 0;
    double slope = 0;

    cin>>x1>>y1>>x2>>y2>>px>>py>>length;

    // get the equation of pineapple line in the form Ax + By + c = 0

    if((x2 - x1) != 0){
        slope = (y2 - y1) / (x2 - x1);
        A = slope;
        C = y2 - slope * x2;
    }

    if((x2 - x1) == 0){
        A = 1;
        B = 0;
        C = -x1;
    }

    // now since we have Ax + By + C = 0
    // we can find the shortest distance between the point and the line

    double short_dist = fabs(A*px + B*py + C) / sqrt(pow(A, 2) + pow(B, 2));
    double dist_x1y1 = fabs(sqrt(pow(px-x1, 2) + pow(py-y1, 2)));
    double dist_x2y2 = fabs(sqrt(pow(px-x2, 2) + pow(py-y2, 2)));

    // once we have calculated the shorted distance, we need to
    // find out the point lying on that pineapple line which is
    // closest to the peg

    double slope_new = 0;
    double B_new = 0;
    double C_new = 0;
    double A_new = 0;
    bool status = true;

    if((x2 - x1) == 0){
        slope_new = 0; // the other line has slope of inifinity, so this is parallel
        B_new = 1;
        A_new = 0;
        C_new = -py;
        status = false;
    }

    if((y2 - y1) == 0){
        B_new = 0;
        A_new = 1;
        C_new = -px;
        status = false;
    }

    if(status){
        slope_new = -1.0/slope;
        C_new = py - slope_new*px;
        A_new = slope_new;
        B_new = -1;
    }

    // calculating the point of intersection

    double inter_X = (-C*B_new - B*(-C_new)) / (A*B_new - B*A_new);
    double inter_Y = (A*(-C_new) - (-C*A_new))/(A*B_new - B*A_new);

    // now check whether this point lies between the given points or not ?
    // to do that we need to check whether both of the above calculated
    // coordinates lie between the range of x1, y1 and x2, y2

    double min_x_coor = min(x1, x2);
    double min_y_coor = min(y1, y2);
    double max_x_coor = max(x1, x2);
    double max_y_coor = max(y1, y2);

    if(inter_X >= min_x_coor && inter_X <= max_x_coor && inter_Y >= min_y_coor && inter_Y <= max_y_coor){
        // this means, the point lies on the line
        double one_pineapple_distance = short_dist - length;
        if(one_pineapple_distance <= 0){
            cout<<"0.00"<<endl;
        }
        else
            printf("%.2f\n", one_pineapple_distance);

        // now to eat all the pineapples
        double larger_distance = max(dist_x1y1, dist_x2y2);
        if(larger_distance - length <= 0){
            cout<<"0.00"<<endl;
        }
        else
            printf("%.2f\n", larger_distance - length);
    }

    // or else the perpendicuar point does not lies on the line
    else{
        // first shorter one
        double shorter_distance = min(dist_x1y1, dist_x2y2);
        double larger_distance = max(dist_x1y1, dist_x2y2);

        double pineapple_1 = shorter_distance - length;
        if(pineapple_1 <= 0){
            cout<<"0.00"<<endl;
        }
        else
            printf("%.2f\n", pineapple_1);

        double pineapple_all = larger_distance - length;
        if(pineapple_all <= 0){
            cout<<"0.00"<<endl;
        }
        else
            printf("%.2f\n", pineapple_all);
    }
    return 0;
}

435 981 301 -384 -1 345 0

Correct answer:
371.78
789.08

Damn, i used min(xa, xb) <= x_h <= max(xa, xb) and got WA5, then I wrote min(xa, xb) <= x_h && x_h <= max(xa, xb) and got AC

#include<bits/stdc++.h>
using namespace std;
int main()
{
    double a,b,c,d,m,n,l,p,t,s,k;
    cin>>a>>b>>c>>d>>m>>n>>l;
    p=sqrt((a-m)*(a-m)+(b-n)*(b-n));
    t=sqrt((c-m)*(c-m)+(d-n)*(d-n));
    s=max(p,t);
    if(s>l){s-=l;
    }
    else{ s=0;}
    k=min(p,t);
    if(k<=l){
        k=0;
    }
    else if(p==t){
        a=(a+c)/2,b=(b+d)/2;
        k=sqrt((a-m)*(a-m)+(b-n)*(b-n));
        if(k<=l){
        k=0;
    }
    else{
        k-=l;
    }
    }
    else{
    while(abs(p-t)>.001){
        if(p<t){
            c=(a+c)/2,d=(b+d)/2;
            t=sqrt((c-m)*(c-m)+(d-n)*(d-n));
        }
        else{
            a=(a+c)/2,b=(b+d)/2;
            p=sqrt((a-m)*(a-m)+(b-n)*(b-n));
        }
    }
    k=min(p,t);
        if(k<=l){
        k=0;
    }
    else{
        k-=l;
    }
    }
    cout<<fixed<<setprecision(2)<<k<<endl;
    cout<<fixed<<setprecision(2)<<s<<endl;
}

#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
#include <iomanip>
#include <set>
using namespace std;
int main()
{
    int x1, x2, y1, y2, x3, y3, l;
    cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3 >> l;
    // h = 2/s3 в€љp(p-s3)(p-s1)(p-s2),
    int a = y1 - y2, b = x2 - x1, c = x1 * y2 - x2 * y1;
    double s1 = sqrt((x1 - x3) * (x1 - x3) + (y1 - y3) * (y1 - y3)),
        s2 = sqrt((x2 - x3) * (x2 - x3) + (y2 - y3) * (y2 - y3)),
        s3 = sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));
    double p = s1 + s2 + s3; p /= 2;
    double ans = (2 / s3 * sqrt(p * (p - s3) * (p - s1) * (p - s2)));
    if (max(s1, s2) * max(s1, s2) > min(s1, s2) * min(s1, s2) + s3 * s3) {
        ans = min(sqrt((x1 - x3) * (x1 - x3) + (y1 - y3) * (y1 - y3)), sqrt((x2 - x3) * (x2 - x3) + (y2 - y3) * (y2 - y3)));
    }
    if (s3 == 0.0) {
        ans = sqrt((x1 - x3) * (x1 - x3) + (y1 - y3) *(y1 - y3));
    }
    printf("%0.2f\n", max(0.0, ans - l));
    printf("%0.2f\n", max(0.0, max(sqrt((x1 - x3) * (x1 - x3) + (y1 - y3) * (y1 - y3)), sqrt((x2 - x3) * (x2 - x3) + (y2 - y3) * (y2 - y3))) - l));
}

why wa#17?
help pls

Edited by author 23.11.2018 08:05

I tried different tasks and always got the right answer.

You can solve it using :
(1) canonical straight line formula for perpendicular length ;
(2) cosine theorem to know whether your perpendicular length is the answer to the first question or not;
(3) Just formula for Euclidian distance and maximum function to answer the second question.

type
 Ta=record
 x,y:extended;
end;

var a,b,c:Ta;
    ab,ac,bc,p,s,h,m:extended;
    l:extended;

    function max(a,b:extended):extended;
    begin
     if a-b>0 then max:=a else max:=b;
    end;

begin
 readln(a.x,a.y,b.x,b.y);
 readln(c.x,c.y,l);

 ac:=sqrt(sqr(a.x-c.x)+sqr(a.y-c.y));
 ab:=sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
 bc:=sqrt(sqr(b.x-c.x)+sqr(b.y-c.y));
 p:=(ab+ac+bc)/2;
 s:=sqrt(p*(p-ac)*(p-ab)*(p-bc));

 if ab=0 then h:=ac else h:=2*s/ab;

 if h-l>0 then          writeln(h-l:0:2)          else writeln('0.00');

m:=max(ac,bc);
 if m-l>0 then writeln(m-l:0:2) else writeln('0.00');
end.

Edited by author 03.08.2008 16:37

import java.util.Scanner;


public class Problem_1348 {
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int ax,ay,bx,by,cx,cy,l;
        double ac,bc,ab,x,min,A,B,C,h;
        x=h=0;
        ax=in.nextInt();
        ay=in.nextInt();
        bx=in.nextInt();
        by=in.nextInt();
        cx=in.nextInt();
        cy=in.nextInt();
        l=in.nextInt();

        ab= Math.sqrt( (bx-ax)*(bx-ax)+(by-ay)*(by-ay) );
        ac =Math.sqrt( (cx-ax)*(cx-ax)+ (cy-ay)*(cy-ay) );
        bc= Math.sqrt( (cx-bx)*(cx-bx) + (cy-by)*(cy-by) );
        A=ac; B=bc; C=ab;

        x=(A*A-B*B+C*C)/(2*C);
       h= Math.sqrt((A*A-x*x));
         min=h;


          if(A>B){
            if(min-l<0)System.out.printf("%.2f",0.00); else System.out.printf("%.2f",min-l);
            System.out.println();
            if(A-l<0)System.out.printf("%.2f",0.00);else System.out.printf("%.2f",A-l);
        }else{
            if(min-l<0)System.out.printf("%.2f",0.00); else System.out.printf("%.2f",min-l);
            System.out.println();
            if(A-l<0)System.out.printf("%.2f",0.00);else System.out.printf("%.2f",B-l);
        }

    }
 }

Edited by author 06.05.2014 16:57

What is the test #21 ?

#include <iostream>
#include <cmath>
#include <stdio.h>
#include <algorithm>
#include <cstdio>

using namespace std;

double dist(int x1, int y1, int x2, int y2)
{
     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}



int main()
{
     double mlen,rlen,dmin,dmax,dtemp,dab_bc_ca[3];
     //mlen is the perpendicular distance from C to AB
     //rlen is Rope Length
     //dab_bc_ca is length of AB, BC and CA
     int x1,y1,x2,y2,cx,cy;
     cin>>x1>>y1>>x2>>y2>>cx>>cy>>rlen;

     if (x1==x2 & y1==y2) //when A and B are same points
     {
          dtemp=dist(x1,y1,cx,cy);
          if (dtemp>rlen)
          dmin=dist(x1,y1,cx,cy)-rlen;
          else dmin=0.00;
          printf("%.2f\n%.2f",dmin,dmin);

     }
     else if (x1!=x2 or y1!=y2)
     {
     mlen=cx*(y1-y2)+cy*(x2-x1)+y1*(x1-x2)-x1*(y1-y2);
     mlen=abs(mlen)/dist(x1,y1,x2,y2);

     if (mlen==0)//Check the position of C, inside AB or Not?
     {
          if (dist(x1,y1,cx,cy)+dist(x2,y2,cx,cy)==dist(x1,y1,x2,y2)) printf("0.00\n");
          else {
                    dtemp=min(dist(cx,cy,x1,y1),dist(cx,cy,x2,y2));
                    if (rlen>=dtemp) printf("0.00\n");
                    else printf("%.2f\n",dtemp-rlen);
               }
          dmax=max(dist(x1,y1,cx,cy),dist(x2,y2,cx,cy));
          if (rlen>=dmax) cout<<"0.00";
          else printf("%.2f",dmax-rlen);
     }
     else //C is outside of AB
     {
          dab_bc_ca[0]=dist(x1,y1,x2,y2); //AB
          dab_bc_ca[1]=dist(x2,y2,cx,cy); //BC
          dab_bc_ca[2]=dist(x1,y1,cx,cy); //CA
          sort(dab_bc_ca+0,dab_bc_ca+3);
          if (dab_bc_ca[2]*dab_bc_ca[2]>dab_bc_ca[1]*dab_bc_ca[1]+dab_bc_ca[0]*dab_bc_ca[0])
          {
               //For Obtuse Angle Triangle
               dab_bc_ca[2]=dist(x1,y1,cx,cy); //CA
               dab_bc_ca[1]=dist(x2,y2,cx,cy); //CB
               dtemp=max(dab_bc_ca[2],dab_bc_ca[1]);
               //Rope length is biggest
               if (rlen>=dtemp) printf("0.00\n0.00");
               else
               {
                    dtemp=min(dist(x1,y1,cx,cy),dist(x2,y2,cx,cy));
                    if (rlen>=dtemp) printf("0.00\n");
                    else printf("%.2f\n",dtemp-rlen);
                    dtemp=max(dist(x1,y1,cx,cy),dist(x2,y2,cx,cy));
                    printf("%.2f",dtemp-rlen);
               }

          }

          else
          {
               //for Acute angle Triangle
               if (rlen>=mlen) cout<<"0.00\n"<<endl;
               else printf("%.2f\n",mlen-rlen);
               dtemp=max(dist(x1,y1,cx,cy),dist(x2,y2,cx,cy));
               if (rlen>=dtemp) cout<<"0.00";
               else printf("%.2f",dtemp-rlen);
          }
     }
     }
     return 0;
}

you need to write answer with two digits after point, so don't use cout!
use printf("%.2f\n",  );
and don't forget #include <cstdio>

i've been finding this mistake for 2 days!!

This is my code
   #include <iostream>
   #include <stdio.h>
   #include <math.h>
   using namespace std;
long n,x[201],y[201];

   int main()
  {double xa,xb,xc,ya,yb,yc,l,a,b,c,x,h;
   cin>>xa>>ya>>xb>>yb>>xc>>yc>>l;
  b=sqrt((xc-xa)*(xc-xa)+(yc-ya)*(yc-ya));
  a=sqrt((xc-xb)*(xc-xb)+(yc-yb)*(yc-yb));
  c=sqrt((xa-xb)*(xa-xb)+(ya-yb)*(ya-yb));
  if(c!=0){
  x=(c*c+b*b-a*a)/(2.0*c);
  h=sqrt(b*b-x*x);
  if(b!=0&&a!=0){
  if((b*b+c*c-a*a)/(2.0*b*c)<=0) h=b;
  if((a*a+c*c-b*b)/(2.0*a*c)<=0) h=a;}

  if(h<=l) h=0; else h=h-l;}
  else h=a;
  printf("%0.2f\n",h);

  if(a>=b) {if(a<=l) h=0; else h=a-l;}
  else {if(b<=l) h=0; else h=b-l;}
  printf("%0.2f\n",h);


      return 0;
   }

Please give me some test

Try this test(Sorry for my poor English):
-2 -1 -4 -3
2 -1 1

correct output data:
3.00
5.32

I got accepted, but my program dont work on test:
1 1 3 3 2 2
1

correct answer - 0 0.41
my program write - 0.41 0.41

please rejudge

Edited by author 21.08.2008 16:27

Edited by author 12.07.2009 05:14

Oh no WA#25! Please help me!

var
  a1,a2,a3,h,s,aa,bb:real;
  a:array[1..3,1..2] of integer;
  i,l:integer;


procedure readdata;
begin
  readln(a[1,1],a[1,2],a[2,1],a[2,2]);
  readln(a[3,1],a[3,2],l);
end;

procedure writedata;
begin
  if aa>=0 then
  writeln(aa:0:2) else writeln('0.00');
  if bb>=0 then
  writeln(bb:0:2) else writeln('0.00');
end;

function d(x1,y1,x2,y2:integer):real;
begin
  d:=sqrt(sqr(x1-x2)+sqr(y1-y2));
end;

procedure square;
var
  p:real;
begin
  a1:=d(a[1,1],a[1,2],a[2,1],a[2,2]);
  a2:=d(a[3,1],a[3,2],a[2,1],a[2,2]);
  a3:=d(a[1,1],a[1,2],a[3,1],a[3,2]);
  p:=(a1+a2+a3)/2;
  s:=sqrt(p*(p-a1)*(p-a2)*(p-a2));
end;

begin
  readdata;
  square;
  if (a[1,1]=a[2,1]) and (a[1,2]=a[2,2]) then
  begin
    aa:=d(a[1,1],a[1,2],a[3,1],a[3,2]);
    bb:=aa;
  end  else
  begin
  if (a1*a1+a2*a2<a3*a3) or (a2*a2+a3*a3<a1*a1) or (a1*a1+a3*a3<a2*a2) then
  begin
    aa:=d(a[1,1],a[1,2],a[3,1],a[3,2]);
    if d(a[2,1],a[2,2],a[3,1],a[3,2])<aa
    then aa:=d(a[2,1],a[2,2],a[3,1],a[3,2]);
  end  else
  begin
    h:=d(a[1,1],a[1,2],a[2,1],a[2,2]);
    aa:=s/(0.5*h);
  end;

  bb:=d(a[3,1],a[3,2],a[2,1],a[2,2]);
  if d(a[3,1],a[3,2],a[1,1],a[1,2])>bb then
  bb:=d(a[3,1],a[3,2],a[1,1],a[1,2]);
  end;
  aa:=aa-l;
  bb:=bb-l;
  writedata;
end.

Show me test 13 please.

#include <cstdio>
#include <cmath>
#include <algorithm>

struct point
{
    int x, y;
};

double distance(const point P1, const point P2)
{
    return sqrt((double)(P1.x - P2.x)*(P1.x - P2.x) + (double)(P1.y - P2.y)*(P1.y - P2.y));
}

bool operator ==(const point P1, const point P2)
{
    if (P1.x == P2.x && P1.y == P2.y) return 1;
                                      return 0;
}

int main()
{
    point A, B, C;
    unsigned int L;
    scanf("%d%d%d%d%d%d%d", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y, &L);

    const double pi = 3.14159265;
    const double epsilon = 0.00000001;
    double alpha, beta, gamma;
    double a, b, c;
    a = distance(B, C);
    b = distance(A, C);
    c = distance(A, B);
    alpha = acos((b*b + c*c - a*a)/(2*b*c));
    beta = acos((a*a + c*c - b*b)/(2*a*c));
    gamma = acos((a*a + b*b - c*c)/(2*a*b));
    if (A == B)
    {
        printf("1 is working\n");
        printf("a is %.2lf\nb is %.2lf\nc is %.2lf\n", a , b , c);
        printf("alpha is %.2lf\nbeta is %.2lf\ngamma is %.2lf\n",  alpha, beta, gamma);
        if (L >= a) printf("0.00\n0.00\n");
        else printf("%.2lf\n%.2lf\n", a - L, b - L);
        return 0;
    }

    if(abs(alpha + beta - pi) < epsilon)
    {
        printf("2 is working\n");
        printf("a is %.2lf\nb is %.2lf\nc is %.2lf\n", a , b , c);
        printf("alpha is %.2lf\nbeta is %.2lf\ngamma is %.2lf\n",  alpha, beta, gamma);
        double nl1 = std::min(a, b);
        double nl2 = std::max(a, b);
        if (L >= nl1) printf("0.00\n");
        else printf("%.2lf\n", nl1 - L);
        if (L >= nl2) printf("0.00\n");
        else printf("%.2lf\n", nl2 - L);
        return 0;
    }
    if (abs(alpha - pi/2) < epsilon)
    {
        printf("3 is working\n");
        printf("a is %.2lf\nb is %.2lf\nc is %.2lf\n", a , b , c);
        printf("alpha is %.2lf\nbeta is %.2lf\ngamma is %.2lf\n",  alpha, beta, gamma);
        double nl1 = b;
        double nl2 = std::max(a, b);
        if (L >= nl1) printf("0.00\n");
        else printf("%.2lf\n", nl1 - L);
        if (L >= nl2) printf("0.00\n");
        else printf("%.2lf\n", nl2 - L);
        return 0;
    }
    if (abs(beta - pi/2) < epsilon)
    {
        printf("4 is working\n");
        printf("a is %.2lf\nb is %.2lf\nc is %.2lf\n", a , b , c);
        printf("alpha is %.2lf\nbeta is %.2lf\ngamma is %.2lf\n",  alpha, beta, gamma);
        double nl1 = a;
        double nl2 = std::max(a, b);
        if (L >= nl1) printf("0.00\n");
        else printf("%.2lf\n", nl1 - L);
        if (L >= nl2) printf("0.00\n");
        else printf("%.2lf\n", nl2 - L);
        return 0;
    }
    if(alpha > pi/2)
    {
        printf("5 is working\n");
        printf("a is %.2lf\nb is %.2lf\nc is %.2lf\n", a , b , c);
        printf("alpha is %.2lf\nbeta is %.2lf\ngamma is %.2lf\n",  alpha, beta, gamma);
        double nl1 = b;
        double nl2 = a;
        if (L >= nl1) printf("0.00\n");
        else printf("%.2lf\n", nl1 - L);
        if (L >= nl2) printf("0.00\n");
        else printf("%.2lf\n", nl2 - L);
        return 0;
    }
    if(beta > pi/2)
    {
        printf("6 is working\n");
        printf("a is %.2lf\nb is %.2lf\nc is %.2lf\n", a , b , c);
        printf("alpha is %.2lf\nbeta is %.2lf\ngamma is %.2lf\n",  alpha, beta, gamma);
        double nl1 = a;
        double nl2 = b;
        if (L >= nl1) printf("0.00\n");
        else printf("%.2lf\n", nl1 - L);
        if (L >= nl2) printf("0.00\n");
        else printf("%.2lf\n", nl2 - L);
        return 0;
    }

    printf("7 is working\n");
    printf("a is %.2lf\nb is %.2lf\nc is %.2lf\n", a , b , c);
    printf("alpha is %.2lf\nbeta is %.2lf\ngamma is %.2lf\n",  alpha, beta, gamma);
    double nl1 = b * sin(alpha);
    double nl2 = std::max(a, b);
    if (L >= nl1) printf("0.00\n");
    else printf("%.2lf\n", nl1 - L);
    if (L >= nl2) printf("0.00\n");
    else printf("%.2lf\n", nl2 - L);
    return 0;
}

I have tested all the inputs that was given in disscutions and the output for all of them were correct.
When I submit my solution, it gives WA#3.
Please give me test 3.