HDOJ1724-椭圆

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

HZIEE 2007 Programming Contest

Recommend

lcy

1.古典方法，也就是直接求，需要用到椭圆扇形面积公式：S = a*b*arccos(x/a)/2

#include <iostream>
#include <cmath>
using namespace std;

#define PI 3.14159265
int main()
{
    int n, a, b, l, r;
    double Sl, Sr, S;
    scanf("%d", &n);
    while((n--) > 0)
    {
        scanf("%d %d %d %d", &a, &b, &l, &r);
        if(l == r)    S = 0;
        else{
            if(r <= 0)
            {
                Sl = 0.5*a*b*(PI-acos(l*1.0/a))
                    -0.5*abs(l)*b*sqrt(a*a-l*l)/a;
                Sr = 0.5*a*b*(PI-acos(r*1.0/a))
                    -0.5*abs(r)*b*sqrt(a*a-r*r)/a;
                S = 2*(Sr-Sl);
            }
            else if(l >= 0)
            {
                Sl = 0.5*b*(a*acos(l*1.0/a)-l*sqrt(a*a-l*l)/a);
                Sr = 0.5*b*(a*acos(r*1.0/a)-r*sqrt(a*a-r*r)/a);
                S = 2*(Sl-Sr);
            }
            else
            {
                Sl = Sl = 0.5*a*b*(PI-acos(l*1.0/a))
                    -0.5*abs(l)*b*sqrt(a*a-l*l)/a;
                Sr = 0.5*b*(a*acos(r*1.0/a)-r*sqrt(a*a-r*r)/a);
                S = PI*a*b - 2*(Sl+Sr);
            }
        }
        printf("%.3f\n", S);
    }
    return 0;
}

2.采用微积分计算

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    double t1, t2, S;
    int n, a, b, l, r;
    scanf("%d", &n);
    while(n--)
    {
        scanf("%d %d %d %d", &a, &b, &l,&r);
        t1 = asin(l*1.0/a);
        t2 = asin(r*1.0/a);
        S = a*b*(t2-t1+(sin(2*t2)-sin(2*t1))/2);
        printf("%.3f\n", S);
    }
    return 0;
}

第二种方法比较简单。