HDU 1714 RedField 一重积分

题目链接：点击打开链接

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;
}