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HDU 1724 Ellipse [辛普森积分]

2024-09-02 17:41:43   211人阅读

Ellipse

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1868    Accepted Submission(s): 792

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:

技术分享


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 技术分享, 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
6.283 3.142
可以计算椭圆面积了哈哈
注意eps小一点
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;typedef long long ll;const double eps=1e-10;double a,b,l,r;inline double F(double x){    return sqrt((a*a-x*x)/(a*a))*b;}inline double cal(double l,double r){    return (F(l)+F(r)+4*F((l+r)/2))*(r-l)/6;}double Simpson(double l,double r,double now){    double mid=(l+r)/2,p=cal(l,mid),q=cal(mid,r);    if(abs(now-p-q)<eps) return now;    else return Simpson(l,mid,p)+Simpson(mid,r,q);}int main(int argc, const char * argv[]) {    int T;scanf("%d",&T);    while(T--){        scanf("%lf%lf%lf%lf",&a,&b,&l,&r);        l=max(l,-a);r=min(r,a);        printf("%.3f\n",2*Simpson(l,r,cal(l,r)));    }    return 0;}

 

 

HDU 1724 Ellipse [辛普森积分]


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