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
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;
- }