自学教程：Python用二分法求平方根的案例

def sq2(x,e):  e = e ＃误差范围    low= 0   high = max(x,1.0) ＃处理大于0小于1的数  guess = (low + high) / 2.0  ctr = 1    while abs(guess**2 - x) > e and ctr<= 1000:    if guess**2 < x:      low = guess    else:      high = guess          guess = (low + high) / 2.0    ctr += 1  print(guess)

fun (input x) return x^2+x-6newton (input a, input b, input e)//a是区间下界，b是区间上界，e是精确度 x <- (a + b) / 2 if abs(b - 1) < e: return x else: if fun(a) * fun(b) < 0:  return newton(a, x, e) else:  return newton(x, b, e)

#include <iostream>#include <cmath>using namespace std; double fun (double x);double newton (double a, double b,double e); int main(){ cout << newton(-5,0,0.5e-5); return 0;} double fun(double x){ return pow(x,2)+x-6;} double newton (double a, double b, double e){ double x; x = (a + b)/2; cout << x << endl; if ( abs(b-a) < e) return x; else if (fun(a)*fun(x) < 0)  return newton(a,x,e); else  return newton(x,b,e);}

def fun(x):  return x ** 2 + x - 6def newton(a,b,e):  x = (a + b)/2.0  if abs(b-a) < e:    return x  else:    if fun(a) * fun(x) < 0:      return newton(a, x, e)    else:      return newton(x, b, e)print newton(-5, 0, 5e-5)