椭圆面积公式
Ellipse
Time limit: 1000/1000 MS (java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1061 Accepted Submission(s): 388
Problem Description
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
Author
威士忌
Source
HZIEE 2007 Programming Contest
| Run ID | Submit Time | Judge Status | Pro.ID | Exe.Time | Exe.Memory | Code Len. | Language | Author |
| 10006793 | 2014-01-21 21:22:02 | Accepted | 1724 | 468MS | 316K | 628 B | C++ |
思路：
这道题目的意思是：求出所给椭圆的蓝色阴影部分面积。由于阴影部分是不规则图形，所以，可以考虑用数值分析的办法－
辛普森公式或者牛顿－科特斯公式求积分法，更多的公式见数值分析教材，辛普森的公式如下，为了增加精度，可以用二分法
迭代细化。
这是核心公式，具体实现见代码：（这种办法的效率真不怎么样，跑了400ms）
#include <iOStream>
#include <math.h>
using namespace std;
const double esp = 1e-10;
double a,b;
double f(double x)
{
return b * sqrt(1.0-(x*x)/(a*a));
}
double simpson(double l,double r)
{
return (f(l)+4*f((l+r)/2.0)+f(r))/6.0*(r-l);
}
double integral(double l,double r)
{
double mid = (l+r)/2.0;
double res = simpson(l,r);
if (fabs(res-simpson(l,mid)-simpson(mid,r)) < esp)
return res;
else
return integral(l,mid) + integral(mid,r);
}
int main()
{
int T;
double l,r;
cin >> T;
while (T--)
{
cin >> a >> b >> l >> r;
printf("%.3lf\n",2*integral(l,r));
}
return 0;
}
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