Simpson&自适应Simpson

Simpson公式：

[int^{r}_{l}f(x)dxapproxfrac{(r-l)(f(l)+f(r)+4f(frac{l+r}{2}))}{6} ]

inline double simpson(double l, double r) {
	double mid = (l + r) / 2.0;
	return (r - l) * (f(l) + f(r) + 4.0 * f(mid)) / 6.0;
}

自适应Simpson法：
二分枚举精度，就可以了鸭！
以一道洛谷模板题为例LuoGu4525_自适应辛普森法1

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
double a, b, c, d, l, r;
inline double f(double x) {
	return (c * x + d) / (a * x + b);
} 
inline double simpson(double l, double r) {
	double mid = (l + r) / 2.0;
	return (r - l) * (f(l) + f(r) + 4.0 * f(mid)) / 6.0;
}
double asr(double l, double r, double eps, double ans) {
	double mid = (l + r) / 2;
	double l_ans = simpson(l, mid), r_ans = simpson(mid, r);
	if (fabs(l_ans + r_ans - ans) <= eps * 15) return l_ans + r_ans + (l_ans + r_ans - ans) / 15;
	return asr(l, mid, eps, l_ans) + asr(mid, r, eps, r_ans);
}
inline double asr(double l, double r, double eps) {
	return asr(l, r, eps, simpson(l, r));
}
int main() {
	scanf("%lf%lf%lf%lf%lf%lf", &a, &b, &c, &d, &l, &r);
	printf("%.6lf", asr(l, r, 1e-6));
	return 0;
}

对于收敛的函数，从1e-8到一个大致收敛成功的数值为止吧
再一道洛谷模板题为例LuoGu4526_自适应辛普森法2

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<algorithm>
using namespace std;
double a;
inline int read() {
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}
inline double f(double x) {
	return pow(x, a / x - x);
}
inline double simpson(double l, double r) {
	double mid = (l + r) / 2;
	return (r - l) * (f(l) + f(r) + 4 * f(mid)) / 6;
}
double asr(double l, double r, double eps, double ans) {
	double mid = (l + r) / 2;
	double l_ans = simpson(l, mid), r_ans = simpson(mid, r);
	if (fabs(l_ans + r_ans - ans) <= eps * 15) return l_ans + r_ans + (l_ans + r_ans - ans) / 15;
	return asr(l, mid, eps, l_ans) + asr(mid, r, eps, r_ans);
}
int main() {
	scanf("%lf", &a);
	if (a < 0) printf("orz");
	else printf("%.5lf", asr(1e-8, 20, 1e-7, simpson(1e-8, 20)));
	return 0;
}