Codeforces Round #172 (Div. 2) C. Rectangle Puzzle 数学题几何

C. Rectangle Puzzle

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/281/problem/C

Description

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 Oxaxis, 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 Input

1 1 45

Sample Output

0.828427125

HINT

题意

给你一个中心在原点的矩形，这个矩形的边平行于坐标轴，然后将这个矩形旋转k°，然后问你旋转之后的矩形和原来的矩形相交的面积是多少

题解：

解方程，有两种情况一种是上图给你的，一种是大正方形减去两个矩形的那种

分别把公式列出来，然后解就好了。。

高中几何题，蛋疼。。

代码

#include<iostream>
#include<math.h>
#include<stdio.h>
using namespace std;
const double pi = acos(-1.0);
int main()
{
    double w,h,t;cin>>w>>h>>t;
    if(w<h)swap(w,h);
    if(t>=90)t=180-t;
    t=t*pi/180;
    double x1,y1,x2,y2;
    y2 = (h-tan(t)*w/(1+1.0/cos(t)))/(1+1/cos(t)-tan(t)*tan(t)/(1+1/cos(t)));
    x2 = tan(t)*y2;
    y1 = (w-y2*tan(t))/(1+1/cos(t));
    x1 = y1*tan(t);

    //cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
    //cout<<tan(t)<<endl;
    if(tan(t/2)<h/w)
    {
        double ans = w*h - x1*y1 - x2*y2;
        printf("%.16f
",ans);
    }
    else
    {
        double x = h / sin(t);
        printf("%.10f
",w*h-(w-x)*h);
    }

}