HDU Ellipse(simpson积分)

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2677    Accepted Submission(s): 1208

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

 

 

Recommend

lcy   |   We have carefully selected several similar problems for you:       

 

 

直接强上simpso积分就行

稍微有点卡精度

 

#include<cstdio>
#include<cmath>
int N;
double a, b, L, R;
double f(double x) {
    return 2 * b * sqrt(1.0 - (x * x) / (a * a));
}
double sim(double l, double r) {
    return (f(l) + f(r) + 4.0 * f((l + r) / 2.0)) * (r - l) / 6.0;
}
double asr(double l, double r, double eps, double ans) {
    double mid = (l + r) / 2;
    double la = sim(l, mid), ra = sim(mid, r);
    if(fabs(la + ra - ans) < eps) return la + ra;
    return asr(l, mid, eps / 2, sim(l, mid)) + asr(mid, r, eps / 2, sim(mid, r));
}
main() {
    scanf("%d", &N);
    while(N--) {
        scanf("%lf %lf %lf %lf", &a, &b, &L, &R);
        printf("%.3lf\n", asr(L, R, 1e-5, sim(L, R)));
    }
}