codehouse

 1 1 # 整数部分十进制转二进制
 2  2
 3  3 num = int(raw_input(">>>"))
 4  4
 5  5 if num < 0:
 6  6     isNeg = True
 7  7     num = abs(num)
 8  8 else:
 9  9     isNeg = False
10 10 result = ‘‘
11 11 if num == 0:
12 12     result = ‘0‘
13 13 while num > 0:
14 14     result = str(num%2) + result
15 15     num = num/2
16 16 if isNeg:
17 17     result = ‘-‘ + result
18 18 # 小数部分十进制转二进制
19 19
20 20 x = float(raw_input(‘Enter a decimal number between 0 and 1: ‘))
21 21
22 22 p = 0
23 23 while ((2**p)*x)%1 != 0:
24 24     print(‘Remainder = ‘ + str((2**p)*x - int((2**p)*x)))
25 25     p += 1
26 26
27 27 num = int(x*(2**p))
28 28
29 29 result = ‘‘
30 30 if num == 0:
31 31     result = ‘0‘
32 32 while num > 0:
33 33     result = str(num%2) + result
34 34     num = num/2
35 35
36 36 for i in range(p - len(result)):
37 37     result = ‘0‘ + result
38 38
39 39 result = result[0:-p] + ‘.‘ + result[-p:]
40 40 print(‘The binary representation of the decimal ‘ + str(x) + ‘ is ‘ + str(result))
41 # 穷举法猜测检验平方根
42 x = 25
43 epsilon = 0.01
44 step = epsilon**2
45 numGuesses = 0
46 ans = 0.0
47 while (abs(ans**2 - x)) >= epsilon and ans <= x:
48     ans += step
49     numGuesses += 1
50 print(‘numGuesses = ‘ + str(numGuesses))
51 if abs(ans**2-x) >= epsilon:
52     print(‘Failed on square root of ‘ + str(x))
53 else:
54     print(str(ans) + ‘ is close to the square root of ‘ + str(x))
55 # 二分法猜测检验平方根
56 # bisection search for square root
57
58 x = 12345
59 epsilon = 0.01
60 numGuesses = 0
61 low = 0.0
62 high = x
63 ans = (high + low)/2.0
64 while abs(ans**2 - x) >= epsilon:
65     print(‘low = ‘ + str(low) + ‘ high = ‘ + str(high) + ‘ ans = ‘ + str(ans))
66     numGuesses += 1
67     if ans**2 < x:
68         low = ans
69     else:
70         high = ans
71     ans = (high + low)/2.0
72 print(‘numGuesses = ‘ + str(numGuesses))
73 print(str(ans) + ‘ is close to square root of ‘ + str(x))