python算法（一）

python算法（一）

一、求数x的因子

　　

1 x=1002 divisors=()#初始化空的元组3 for i in range(1,x):4     if x%i==0:5         divisors=divisors+(i,)6         print divisors

 

二、求数x各个数位之和

 

1 sumdigits=02 for c in str(1952):3     sumdigits +=int(c)4     print sumdigits5 print sumdigits

 

三、鸡兔同笼以及变形

　　1.有鸡兔两种，共有x个头，y只脚，求解鸡兔各有几只？

 1 def slove(num_heads,num_legs): 2     for chicken_num in range(0,num_heads+1): 3         pig_num =num_heads-chicken_num 4         top_legs=pig_num*4+chicken_num*2 5         if top_legs==num_legs: 6             return [chicken_num,pig_num] 7     return [None,None] 8  9 def barnYard():10     heads=int(raw_input("Enter the number of heads: "))11     legs=int (raw_input("Enter the number of legs: "))12     chicken,pig=slove(heads,legs)13     if chicken==None:14         print "不可解"15     else:16         print "the number of chicken is:",chicken17         print "the number of pig is:",pig18 19 barnYard()

2.有鸡兔，蜘蛛一共三种，共有x个头，y只脚，求解鸡兔，蜘蛛各有几只？

 1 def slove1(num_heads,num_legs): 2     for spider_num in range(0,num_heads+1): 3         for chicken_num in range(0,num_heads-spider_num): 4             pig_num =num_heads-chicken_num-spider_num 5             top_legs=pig_num*4+chicken_num*2+spider_num*8 6             if top_legs==num_legs: 7                 return [chicken_num,pig_num,spider_num] 8     return [None,None,None] 9 10 def barnYard1():11     heads=int(raw_input("Enter the number of heads: "))12     legs=int (raw_input("Enter the number of legs: "))13     chicken,pig,spider=slove1(heads,legs)14     if chicken==None:15         print "不可解"16     else:17         print "the number of chicken is:",chicken18         print "the number of pig is:",pig19         print "the number of spider is",spider20 21 barnYard1()

3.2中的问题或许不只有一个解答，依次输出符合要求的解答

 1 def slove2(num_heads,num_legs): 2     solutionFound=False 3     for spider_num in range(0,num_heads+1): 4         for chicken_num in range(0,num_heads-spider_num): 5             pig_num =num_heads-chicken_num-spider_num 6             top_legs=pig_num*4+chicken_num*2+spider_num*8 7             if top_legs==num_legs: 8                 print "the number of chicken is:", chicken_num 9                 print "the number of pig is:", pig_num10                 print "the number of spider is", spider_num11                 solutionFound=True12     if not solutionFound:13         print "不可解"14 15 16 def barnYard2():17     heads=int(raw_input("Enter the number of heads: "))18     legs=int (raw_input("Enter the number of legs: "))19     slove2(heads,legs)20 21 barnYard2()

四、递归判断字符串是否为回文

解法一：

1 def isPlalindrome(s):2     if len(s)<=1:3         return True4     else :5         return s[0]==s[-1] and isPlalindrome(s[1:-1])

解法二：

 1 def isPlalindrome1(s,indent): 2     print indent, 'hisPalindromel called with', s 3     if 1 >= len(s): 4          print indent, 'About to return True from base case',s 5          return True 6     else: 7         ans= s[0] == s[-1] and isPlalindrome1(s[1:-1], indent + indent) 8         print indent, 'About to return ',ans 9     return ans10 11 isPlalindrome1("abccba",1)

五、斐波那契数列

 

1 def fib(x):2     sum=1;3     if x==1 or x==0:4         return 1;5     else:6         return fib(x-1)+fib(x-2)7 8 print fib(4)

六、求数x平方根

1.二分法求解

 1 def squrtRootBi(x,epsilon): 2     assert x>=0,"x must be positive"+str(x) 3     assert epsilon>0,"epsilon must be positive"+str(epsilon) 4     low=0 5     #high=x 6     high=max(x,1.0) 7     guess=(low+high)/2.0 8     ctr=1 9     while abs(guess**2-x)>epsilon and ctr<=100:10         #print "low",low,"high",high,"guess",guess11         if guess**2<x:12             low=guess13         else:14             high=guess15         guess=(low+high)/2.016         ctr+=117     assert ctr<=100,"not perfect square number!"18     print "times of Iteration:",ctr," guess",guess19     return guess20 21 def testBi():22     squrtRootBi(4,0.0001)23     squrtRootBi(2, 0.0001)24     squrtRootBi(0.25, 0.0001)25 26 testBi()

2.牛顿迭代法求解

 1 def squrtRootNR(x,epsilon): 2     assert x >= 0, "x must be positive" + str(x) 3     assert epsilon > 0, "epsilon must be positive" + str(epsilon) 4     x=float(x) 5     guess=x/2.0 6     #guess=0.001 7     diff=guess**2-x 8     ctr=1 9     while abs(diff)>epsilon and ctr<=100:10        # print "error",diff,"guess",guess11         guess=guess-diff/(2.0*guess)12         diff=guess**2-x13         ctr+=114     assert ctr <= 100, "not perfect square number!"15     print "times of Iteration:", ctr, " guess", guess16     return guess17 18 def testBi1():19     squrtRootNR(4,0.0001)20     squrtRootNR(2, 0.0001)21     squrtRootNR(0.25, 0.0001)22 23 testBi1()