hdu-1742Ellipse(自适应辛普森)

Ellipse

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2418    Accepted Submission(s): 1076

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

#include<map>
#include<queue>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#define inf 0x3f3f3f
#define ll long long
#define maxn 100005
using namespace std;
double a1,b1;
double F(double x)      //这里自定义函数
{
    return b1 * sqrt(1.0 - (x * x) / (a1 * a1));
}
double simpson(double a,double b)   //返回区间[a,b]的三点辛普森值
{
    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 = simpson(a,c), R = simpson(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,simpson(a,b));
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double l,r;
        double eps=1e-5;
        cin>>a1>>b1>>l>>r;
        double ans=asr(l,r,eps);
        ans*=2.0;
        printf("%.3lf
",ans);
    }
}