题意：
给定 $a,b,l,r,$求
与$x = l,x = r$ 围成的封闭图形的面积。
思路：
大佬可以直接算一下原函数就出来了，当没法计算或者很难计算的时候就可以用 自适应$simpson$ 积分来逼近真实值。
1 /* 2 * @Author: windystreet 3 * @Date: 2018-08-04 16:24:01 4 * @Last Modified by: windystreet 5 * @Last Modified time: 2018-08-04 16:24:33 6 */ 7 8 #include<bits/stdc++.h> 9 10 using namespace std; 11 12 #define X first 13 #define Y second 14 #define eps 1e-10 // 精度根据题目改变 15 #define gcd __gcd 16 #define pb push_back 17 #define PI acos(-1.0) 18 #define lowbit(x) (x)&(-x) 19 #define bug printf("!!!!! "); 20 #define mem(x,y) memset(x,y,sizeof(x)) 21 22 typedef long long LL; 23 typedef long double LD; 24 typedef pair<int,int> pii; 25 typedef unsigned long long uLL; 26 27 const int maxn = 1e5+2; 28 const int INF = 1<<30; 29 const int mod = 1e9+7; 30 31 double a,b; 32 33 double f(double x){ 34 return b*(sqrt(1.0-(x*x)/(a*a))); // 函数根据题目改变 35 } 36 double simpson(double L ,double R){ 37 return (R-L)*(f(L)+4.0*f((L+R)/2.0)+f(R))/6.0; 38 } 39 double asr(double L,double R){ 40 double mid = (L+R)/2.0; 41 double res = simpson(L,R); 42 double left = simpson(L,mid),right = simpson(mid,R); 43 if(fabs(left+right-res)<eps)return left + right; 44 else return asr(L,mid)+asr(mid,R); 45 } 46 47 48 void solve(){ 49 double l,r; 50 scanf("%lf%lf%lf%lf",&a,&b,&l,&r); 51 double ans = asr(l,r); 52 printf("%.3lf ",2.0*ans); 53 54 return; 55 } 56 57 int main() 58 { 59 // freopen("in.txt","r",stdin); 60 // freopen("out.txt","w",stdout); 61 // ios::sync_with_stdio(false); 62 int t = 1; 63 scanf("%d",&t); 64 while(t--){ 65 // printf("Case %d: ",cas++); 66 solve(); 67 } 68 return 0; 69 }