辛普森自适应方程- HDU-1724

最新推荐文章于 2019-03-20 21:10:43 发布

七分好月

最新推荐文章于 2019-03-20 21:10:43 发布

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本文链接：https://blog.youkuaiyun.com/qq_42506224/article/details/84061524

本文介绍了一种使用自适应辛普森递归过程计算椭圆与垂直于X轴直线交集面积的方法。通过输入椭圆参数a、b及两条直线的位置，算法能够精确计算出交集的蓝色区域面积。

B - Ellipse

HDU - 1724

Math is important!! Many students failed in 2+2’s mathematical test, so let's AC this problem to mourn for our lost youth..
Look this sample picture:

A ellipses in the plane and center in point O. the L,R lines will be vertical through the X-axis. The problem is calculating the blue intersection area. But calculating the intersection area is dull, so I have turn to you, a talent of programmer. Your task is tell me the result of calculations.(defined PI=3.14159265 , The area of an ellipse A=PI*a*b )

Input

Input may contain multiple test cases. The first line is a positive integer N, denoting the number of test cases below. One case One line. The line will consist of a pair of integers a and b, denoting the ellipse equation , A pair of integers l and r, mean the L is (l, 0) and R is (r, 0). (-a <= l <= r <= a).

Output

For each case, output one line containing a float, the area of the intersection, accurate to three decimals after the decimal point.

Sample Input

2
2 1 -2 2
2 1 0 2

Sample Output

6.283
3.142

#include<stdio.h>
#include<math.h> 

double a,b,l,r;

double F(double x) {
	return sqrt(b*b-(b*b*x*x)/(a*a));
}

double xps(double a,double b) {
	
    double c = a + (b - a) / 2.0;
  return (F(a) + 4 * F(c) + F(b)) * (b - a) / 6.0;

}

double asr(double a,double b,double eps,double A) { //自适应辛普森递归过程
	double c = a + (b - a) / 2.0;                   //A为区间[a,b]的三点辛普森值
	double L = xps(a,c), R = xps(c,b);
	if(fabs(L + R - A) <= 15 * eps) return L + R + (L + R - A) / 15.0;
	return asr(a,c,eps/2.0,L) + asr(c,b,eps/2.0,R);
}

double asr(double a,double b,double eps) {  //自适应辛普森主过程
	return asr(a,b,eps,xps(a,b));
}

int main() {
	int t;
	scanf("%d",&t);
	while(t--) {
		
		scanf("%lf%lf%lf%lf",&a,&b,&l,&r);
		
    	printf("%.3f\n",2*asr(l,r,0.0001));
		
	}
	return 0;
}