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 )
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 )
InputInput 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).OutputFor 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
几何
simpson积分强行搞
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 using namespace std; 7 const double eps=1e-9; 8 const int mxn=100010; 9 int read(){ 10 int x=0,f=1;char ch=getchar(); 11 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 12 while(ch>='0' && ch<='9'){x=x*10-'0'+ch;ch=getchar();} 13 return x*f; 14 } 15 int a,b; 16 inline double f(double x){ 17 return b*sqrt((double)1-(x*x/(double)a/a)); 18 } 19 inline double sim(double l,double r){ 20 return (r-l)/6*(f(l)+4*f((l+r)/2)+f(r)); 21 } 22 double solve(double l,double r){ 23 double mid=(l+r)/2; 24 double res=sim(l,r); 25 // printf("l:%.3f r:%.3f res:%.3f ",l,r,res); 26 if(fabs(sim(l,mid)+sim(mid,r)-res)<=eps)return res; 27 return solve(l,mid)+solve(mid,r); 28 } 29 int main(){ 30 int T=read(); 31 double l,r; 32 while(T--){ 33 a=read();b=read(); 34 l=read();r=read(); 35 double ans=solve(l,r); 36 printf("%.3f ",2*ans); 37 } 38 return 0; 39 }