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Codeforces Round #172 (Div. 2) Rectangle Puzzle

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C. Rectangle Puzzle

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two rectangles on a plane. The centers of both rectangles are located in the origin of coordinates (meaning the center of the rectangle’s symmetry). The first rectangle’s sides are parallel to the coordinate axes: the length of the side that is parallel to the Ox_axis, equals _w, the length of the side that is parallel to the Oy axis, equals h. The second rectangle can be obtained by rotating the first rectangle relative to the origin of coordinates by angle α.     Your task is to find the area of the region which belongs to both given rectangles. This region is shaded in the picture.

Input

The first line contains three integers w, h, α (1 ≤ w, h ≤ 106; 0 ≤ α ≤ 180). Angle α is given in degrees.

Output

In a single line print a real number — the area of the region which belongs to both given rectangles. The answer will be considered correct if its relative or absolute error doesn’t exceed 10 - 6.

Sample test(s)

input

1 1 45

output

0.828427125

input

6 4 30

output

19.668384925

Note

The second sample has been drawn on the picture above.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iomanip>

using namespace std;

const double PI=acos(-1);

int main()
{
    double w,h,a;
    double ans;
    cin>>w>>h>>a;

    cout<<setiosflags(ios::fixed)<<setprecision(8);
    if(a>90)a=180-a;
    a*=PI/180;

    if(fabs(a/PI*180-90)<1e-5)
    {
        //ans=w*h;
        if(w>h)ans=h*h;
        else ans=w*w;
        cout<<ans;
    }
    else if(a>=atan(2*h/w) && w>h/sin(a)+h/tan(a))
    {
        ans=h*h/sin(a);
        cout<<ans;
    }
    else if(a>=atan(2*w/h) && h>w/sin(a)+w/tan(a))
    {
        ans=w*w/sin(a);
        cout<<ans;
    }
    else
    {
        ans=w*h;


        /*
        double a11=1.0/sin(a)/cos(a)+tan(a)-1.0/tan(a),
        a12=2/cos(a),
        a21=2/cos(a),
        a22=1.0/sin(a)/cos(a)+tan(a)-1.0/tan(a),
        b1=w/cos(a)+h*tan(a)-w,
        b2=w*tan(a)+h/cos(a)-h;
        double x1=(b1*a22-b2*a12)/(a11*a22-a12*a21),x2=(a11*b2-a21*b1)/(a11*a22-a12*a21);
        ans-=x2*x2/tan(a)+(w-x2-x1/sin(a))*(w-x2-x1/sin(a))*tan(a);
        */
        ans=(sin(a/2)*h*h - 2*cos(a/2)*h*w + sin(a/2)*w*w)/(2*cos(a/2) - 4*cos(a/2)*cos(a/2)*cos(a/2));
        cout<<ans;
    }
    //system("pause");
}

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