一图胜千言。

自适应辛普森模板：

const double eps=1e-10;double f(double x){    //你的函数。 } double simpson(double l,double r){    return (f(l)+4.0*f((l+r)/2.0)+f(r))/6.0*(r-l);}double integral(double l,double r){    double mid=(l+r)/2.0;    double res=simpson(l,r);    if(fabs(res-simpson(l,mid)-simpson(mid,r))

复合辛普森模板：

double f(double x) {    double l = max(s1, x-w);    double r = min(s2, x+w);    if(r-l < eps) return eps;    else return r-l;}double simpson(double l, double r) {    return (f(l) + 4.0*f((l+r)/2.0) + f(r))/6.0*(r-l);}double integral(double l, double r) {    double m = (l+r)/2.0;    double res = simpson(l, r);    if(fabs(res-simpson(l, m)-simpson(m, r)) < eps) return res;    else return integral(l, m) + integral(m, r);}double fuhesimpson(double l, double r, int n) {    double h = (r-l)/n, ans = 0;    for (int i = 0; i < n; ++i) {        ans += integral(l+i*h, l+(i+1)*h);    }    return ans;}

参考:

例题1：

代码：

#include
      
       using namespace std;#define ll long long#define pb push_back#define pi acos(-1.0)#define mem(a,b) memset(a,b,sizeof(a))const double eps=1e-10;double a,b;double f(double x){    return b*sqrt(1.0-(x*x)/(a*a)); } double simpson(double l,double r){    return (f(l)+4.0*f((l+r)/2.0)+f(r))/6.0*(r-l);}double integral(double l,double r){    double mid=(l+r)/2.0;    double res=simpson(l,r);    if(fabs(res-simpson(l,mid)-simpson(mid,r))
       
        >T;    double l,r;    while(T--)    {        cin>>a>>b>>l>>r;        cout<
        
         <
         
          <<2*integral(l,r)<