nts sin -s res a* max this 技術 eight

Ellipse

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2502 Accepted Submission(s): 1126

Problem Description 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 ) 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

題意

給定橢圓的a，b，求橢圓在[L,R]範圍內的面積，多組數據

題解

自適應辛普森積分裸題

直接對某個區間進行辛普森積分的話公式為（r - l ）*（f（l ）+4 * f（（ l + r ）/ 2）+f（ r ））/ 6

然後如果直接拆分所求區間的話，如果遇到鬼畜的函數就會使誤差變大

所以就有了自適應辛普森積分

就是說我們求這個區間的辛普森積分和左右部分的辛普森積分

如果相差小於eps的話，就直接返回答案

否則遞歸計算左右區間

就醬

代碼

#include<cstdio>
#include<iostream>
#include<cmath>
#define db double
using namespace std;

db a,b,l,r;
int t;

db f(db x)
{
    return sqrt(b*b*(1.0-x*x/a/a));
}

db xin(db l,db r)
{
    db mid=(l+r)/2;
    return (r-l)*(f(l)+4*f(mid)+f(r))/6.0;
}

db getans(db x,db y,db eps,db val)
{
    db mid=(x+y)/2;
    db aa=xin(x,mid),bb=xin(mid,y);
    if(fabs(val-aa-bb)<=eps*15.0) return aa+bb+(aa+bb-val)/15.0;
    return getans(x,mid,eps/2.0,aa)+getans(mid,y,eps/2.0,bb);
}

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf",&a,&b,&l,&r);
        printf("%.3lf\n",2.0*getans(l,r,0.00005,xin(l,r)));
    }
    return 0;
}

【自適應辛普森積分】hdu1724 Ellipse