HDU 1714 RedField 一重积分

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给定椭圆的标准方程，椭圆外一点(x,y) 且 x>=a , abs(y)>=b

求阴影面积

先求出直线与椭圆的交点(x1, y1),

然后积分即可。

#include <stdio.h>#include <iostream>#include <algorithm>#include <sstream>#include <stdlib.h> #include <iomanip>  #include <string.h>#include <limits.h>#include <vector>#include <string>#include <time.h>#include <math.h>#include <queue>#include <stack>#include <set>#include <map>const double eps = 1e-8;const double pi = acos(-1.0);template <class T>inline bool rd(T &ret) {char c; int sgn;if (c = getchar(), c == EOF) return 0;while (c != '-' && (c<'0' || c>'9')) c = getchar();sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');ret *= sgn;return 1;}template <class T>inline void pt(T x) {if (x <0) { putchar('-'); x = -x; }if (x>9) pt(x / 10);putchar(x % 10 + '0');}using namespace std;typedef long long ll;double a, b, x, y;int main(){int T; rd(T);while (T--){cin >> a >> b >> x >> y;y = abs(y);double k = y / x;double x1 = a*b / sqrt(b*b + a*a*k*k);double y1 = k * x1;double ans =  - x1*sqrt(a*a - x1*x1) / 2 + a*a*acos(x1 / a) / 2;ans *= b / a;ans += x1*y1 / 2;cout << fixed << setprecision(2) << ans << endl;}return 0;}

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