Rectangle Puzzle CodeForces - 281C （几何）

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.

Examples
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.

思路：

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2) { ans = ans * a % MOD; } a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
typedef long double ld;
ld w, h, a;
const ld pi = acos(-1);
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    cin >> w >> h >> a;
    if (h > w) {
        swap(h, w);
    }
    a = min(a, 180.0 - a);
    a /= 180;
    a *= pi;
    ld z = (w - (h * sin(a)) / (1.0 + cos(a))) / (1.0 + cos(a) - sin(a) * sin(a) / (1 + cos(a)));
    ld v = (h - z * sin(a)) / (1.0 + cos(a));
    if (v > 0.0) {
        ld x = z * cos(a);
        ld y = z * sin(a);
        ld ans = w * h;
        ans -= x * y;
        x = v * cos(a);
        y = v * sin(a);
        ans -= x * y;
        cout << fixed << setprecision(7) << ans << endl;
    } else {
        v = h / sin(a);
        ld x = v * cos(a / 2.0);
        ld y = v * sin(a / 2.0);
        ld ans = x * y * 2.0;
        cout << fixed << setprecision(7) << ans << endl;
    }

    return 0;
}

inline void getInt(int *p)
{
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    } else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}