Simpson&自适应Simpson

Simpson公式：
\[\int^{r}_{l}f(x)dx\approx\frac{(r-l)(f(l)+f(r)+4f(\frac{l+r}{2}))}{6}\]

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;
}

标签：double,mid,适应,ans,inline,include,Simpson
来源： https://www.cnblogs.com/Agakiss/p/11568853.html