using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace 牛顿法计算根号下2的值 { class Program { static void Main(string[] args) { , b = , t = ; while (b - a > 0.0000001) { t = (a + b) / ; ==
Power of Cryptography Time limit: 3.000 seconds http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=99&page=show_problem&problem=49 http://poj.org/problem?id=2109 Time Limit: 1000MS Memory Limit: 30000K Total
牛顿法 # coding:utf-8 import matplotlib.pyplot as plt import numpy as np def dataN(length):#生成数据 x = np.ones(shape = (length,3)) y = np.zeros(length) for i in np.arange(0,length/100,0.02): x[100*i][0]=1 x[100*i][1]=i x[100*i][2]=i + 1 + np.random.unifor
高斯牛顿法: function [ x_ans ] = GaussNewton( xi, yi, ri) % input : x = the x vector of 3 points % y = the y vector of 3 points % r = the radius vector of 3 circles % output : x_ans = the best answer % set up r equations r1 = @(x, y) sqrt((x-xi(1))^2+(y-y
不动点迭代 function xc = fpi( g, x0, tol ) x(1) = x0; i = 1; while 1 x(i + 1) = g(x(i)); if(abs(x(i+1) - x(i)) < tol) break end i = i + 1; end xc = x(i+1); end 牛顿法: function xk = funNewton(f, x0, max_steps, tol) syms x symbol_f = f(x); dif_f = matlabFunct