Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2677 Accepted Submission(s): 1208 Problem Descrip ...
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2677 Accepted Submission(s):
1208
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 | We have carefully selected several similar problems for you: 1722 1727 1721 1726 1725 直接強上simpso積分就行 稍微有點卡精度
#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))); } }
