用二分法求在(a,b)上单调的函数近似值 第八行的表达式可更改,第三行的kexi决定的精度,小数值计算可将第五行的extended更为real或double PROGRAM EQUANTION (input,output); CONST kexi=0.0000001; VAR a,b,c:extended; FUNCTION fx(x:extended):extended; BEGIN fx:=ln(x)/ln(2); END; BEGIN writeln('Please input a a
+二分法求平方根 x = float(raw_input('Enter the number')) low = 0 high = x guess = (low + high ) / 2 if x < 0: print 'Number Error' while abs(guess**2 - x) > 1e-5: if guess**2 < x: low = guess else: high = guess guess = (low + high) / 2 print 'The root o
进行单元测试数据____________________________________ public class Test { public static void main(String[] args) { // TODO Auto-generated method stub System.out.println(bin(-1000000000,9999,0.00000001)); } //提供的接口 //区间[a,b] 和 区间精度d static double bin(double a,
[抄题]: Implement int sqrt(int x). Compute and return the square root of x, where x is guaranteed to be a non-negative integer. Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned. E
exp=input().split() ls=list() def Cal(a,b,i): if i=="+": return a+b elif i=="-": return a-b elif i=="*": return a*b else: return a/b for i in exp: if i in "+-*/": a=ls.pop() b=ls.pop() ls.append(Cal(b,a,i)) else: ls