hdu 1724 Ellipse（Simpson积分法）

Ellipse

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1152    Accepted Submission(s): 435

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

22 1 -2 22 1 0 2

 

Sample Output

6.2833.142
 
 
 
Simpson积分法：http://zh.wikipedia.org/wiki/%E8%BE%9B%E6%99%AE%E6%A3%AE%E7%A9%8D%E5%88%86%E6%B3%95
 
 
AC代码：
#include <iostream>#include <cmath>#include <cstdlib>#include <cstring>#include <cstdio>#include <queue>#include <ctime>#include <algorithm>#define ll long longusing namespace std;const int INF = 1e9;const int maxn = 505;const double eps = 1e-8;double a, b;double F(double x){    return 2.0 * b * sqrt(1.0 - x * x / (a * a));}double simpson(double l, double r){    double mid = (l + r) / 2.0;    return (r - l) * (F(l) + 4.0 * F(mid) + F(r)) / 6.0;}double work(double l, double r, double sum){    double mid = (l + r) / 2.0;    double ls = simpson(l, mid);    double rs = simpson(mid, r);    if(abs(ls + rs - sum) <= eps) return sum;    return work(l, mid, ls) + work(mid, r, rs);}int main(){    int t;    double l, r;    scanf("%d", &t);    while(t--)    {        scanf("%lf%lf%lf%lf", &a, &b, &l, &r);        double ans = work(l, r, simpson(l, r));        printf("%.3f\n", ans);    }    return 0;}