hdu 1714 RedField

RedField

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 639    Accepted Submission(s): 213

Problem Description

As the graph you see below, we named the red part "RedField".The read part is the intersection of an ellipse and a triangle.Now, 8600's not good at math, so he wants you to help him calculate the area of the "RedField".

 

Input

The first line of input contains n, the number of test cases. n lines follow.Each test case contains four numbers a, b, x, y(0 < b <= a <= x),and a is the lenth of OA, b is the lenth of OB,and x, y representing the coordinate of the point P.

 

Output

For each test case, output the area of the "RedField",accurate up to 2 decimal places.

 

Sample Input

1 1.00 1.00 2.00 2.00

 

Sample Output

0.39

 

Source

HDU 2007-Spring Programming Contest

 

Recommend

lcy

非常简单的高数题！

1.先求出交点！

2.再对交点右边的含有抛物线的部分积分，再加上三角形就好了！

有一个要注意的地方：就是y可以为负数，即p可以在x轴的下方。（这里我WA了3次，哎真的没有仔细考虑啊！！）

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        double A,B,k,j_x,j_y,ans;
        double p_x,p_y;
        cin>>A>>B>>p_x>>p_y;
        p_y = p_y>0?p_y:-p_y;
        k=p_y/p_x;
        double tt=(A*A*B*B)/(B*B+A*A*k*k);
        j_x=sqrt(tt);
        j_y=k*j_x;
        double t1=asin(1.0);
        double t2=asin(j_x/A);
        ans=A*B*(  ( t1+sin(2*t1)/2.0) -(t2+sin(2*t2)/2.0) )/2+j_x*j_y/2;
        printf("%.2lf
",ans);
    }
    return 0;
}