HDU 1724 Ellipse 自适应simpson函数模板题（二）

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Ellipse

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2012 Accepted Submission(s): 857

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

Author

威士忌

Source

HZIEE 2007 Programming Contest

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模板如图：

模板理解：a和b分别代表所求积分的上下限。eps的值一般为： 1e-6

同时还要加上一个函数double F（double x）；这个函数就是你要求的函数（这题要求出x关于y的函数，所以要化简一下）

上代码：

#include<cstdio>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>using namespace std;const double eps=1e-6;///eps的值double a,b,l,r;double F(double x ){    return b*sqrt(1-x*x/(a*a));///对应题目所需求解的方程                               ///求出x关于y的方程}double simpson(double l,double r){    double c=l+(r-l)/2;    return (F(l) +4*F(c)+F(r))*(r-l)/6;}double asr(double l,double r,double eps, double A){    double c=l+(b-l)/2;    double L=simpson(l,c),R=simpson(c,r);    if(fabs(L+R-A) <= 15*eps )        return L+R+(L+R-A)/15;    return asr(l,c,eps/2,L) + asr(c,r,eps/2,R);}double asr(double l,double r,double eps){    return asr(l,r,eps,simpson(l,r));}int main(){    int t;    cin>>t;    while(t--)    {        scanf("%lf%lf%lf%lf",&a,&b,&l,&r);        printf("%.3lf\n",2*asr(l,r,eps));///传入上下限和eps                                         ///此题不要忘记把最终结果乘2    }    return 0;}

要努力啊QAQ

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