• HDU1724 Ellipse


    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 ) 

    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 }
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  • 原文地址:https://www.cnblogs.com/SilverNebula/p/6351794.html
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