hdu 1724 辛普森积分公式

Ellipse

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2425 Accepted Submission(s): 1081

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
题意：裸的辛普森积分公式，他只能用来积连续的优美的曲线。
有一个收获就是double型a*a/(b*b)比a*a/b/b快了100ms，大概是乘法比除法快吧。

#include<bits/stdc++.h>using namespace std;double eps = 1e-10;int a,b,l,r;double f(double x){    return b*sqrt(1-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 solve(double l,double r){    double mid =(l+r)/2;    double res = simpson(l,r);    if(fabs(res - simpson(l,mid)-simpson(mid,r))< eps){        return res;    }    else return solve(l,mid)+solve(mid,r);}int main(){    int T;    cin >> T;    while(T--){        scanf("%d %d %d %d",&a,&b,&l,&r);        printf("%.3f\n",2*solve(l,r));    }    return 0;}