摘要: 自适应辛普森法 double a, b, c, d, l, r; const double eps = 1e-8; double F(double x) { //需要积分的公式 return (c * x + d) / (a * x + b); } double simpson(double l, 阅读全文
posted @ 2024-10-02 12:32 Ke_scholar 阅读(10) 评论(0) 推荐(0) 编辑