python算法（一）

python算法（一）

一、求数x的因子

　　

 x=100
 divisors=()#初始化空的元组
 for i in range(1,x):
     if x%i==0:
         divisors=divisors+(i,)
         print divisors

二、求数x各个数位之和

 sumdigits=0
 for c in str(1952):
     sumdigits +=int(c)
     print sumdigits
 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？

 def slove(num_heads,num_legs):
     for chicken_num in range(0,num_heads+1):
         pig_num =num_heads-chicken_num
         top_legs=pig_num*4+chicken_num*2
         if top_legs==num_legs:
             return [chicken_num,pig_num]
     return [None,None]

 def barnYard():
     heads=int(raw_input("Enter the number of heads: "))
     legs=int (raw_input("Enter the number of legs: "))
     chicken,pig=slove(heads,legs)
     if chicken==None:
         print "不可解"
     else:
         print "the number of chicken is:",chicken
         print "the number of pig is:",pig

 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？

 def slove1(num_heads,num_legs):
     for spider_num in range(0,num_heads+1):
         for chicken_num in range(0,num_heads-spider_num):
             pig_num =num_heads-chicken_num-spider_num
             top_legs=pig_num*4+chicken_num*2+spider_num*8
             if top_legs==num_legs:
                 return [chicken_num,pig_num,spider_num]
     return [None,None,None]

 def barnYard1():
     heads=int(raw_input("Enter the number of heads: "))
     legs=int (raw_input("Enter the number of legs: "))
     chicken,pig,spider=slove1(heads,legs)
     if chicken==None:
         print "不可解"
     else:
         print "the number of chicken is:",chicken
         print "the number of pig is:",pig
         print "the number of spider is",spider

 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

 def slove2(num_heads,num_legs):
     solutionFound=False
     for spider_num in range(0,num_heads+1):
         for chicken_num in range(0,num_heads-spider_num):
             pig_num =num_heads-chicken_num-spider_num
             top_legs=pig_num*4+chicken_num*2+spider_num*8
             if top_legs==num_legs:
                 print "the number of chicken is:", chicken_num
                 print "the number of pig is:", pig_num
                 print "the number of spider is", spider_num
                 solutionFound=True
     if not solutionFound:
         print "不可解"

 def barnYard2():
     heads=int(raw_input("Enter the number of heads: "))
     legs=int (raw_input("Enter the number of legs: "))
     slove2(heads,legs)

 barnYard2()

四、递归判断字符串是否为回文

解法一：

 def isPlalindrome(s):
     if len(s)<=1:
         return True
     else :
         return s[0]==s[-1] and isPlalindrome(s[1:-1])

解法二：

 def isPlalindrome1(s,indent):
     print indent, 'hisPalindromel called with', s
     if 1 >= len(s):
          print indent, 'About to return True from base case',s
          return True
     else:
         ans= s[0] == s[-1] and isPlalindrome1(s[1:-1], indent + indent)
         print indent, 'About to return ',ans
     return ans

 isPlalindrome1("abccba",1)

五、斐波那契数列

 def fib(x):
     sum=1;
     if x==1 or x==0:
         return 1;
     else:
         return fib(x-1)+fib(x-2)

 print fib(4)

六、求数x平方根

1.二分法求解

 def squrtRootBi(x,epsilon):
     assert x>=0,"x must be positive"+str(x)
     assert epsilon>0,"epsilon must be positive"+str(epsilon)
     low=0
     #high=x
     high=max(x,1.0)
     guess=(low+high)/2.0
     ctr=1
     while abs(guess**2-x)>epsilon and ctr<=100:
         #print "low",low,"high",high,"guess",guess
         if guess**2<x:
             low=guess
         else:
             high=guess
         guess=(low+high)/2.0
         ctr+=1
     assert ctr<=100,"not perfect square number!"
     print "times of Iteration:",ctr," guess",guess
     return guess

 def testBi():
     squrtRootBi(4,0.0001)
     squrtRootBi(2, 0.0001)
     squrtRootBi(0.25, 0.0001)

 testBi()

2.牛顿迭代法求解

 def squrtRootNR(x,epsilon):
     assert x >= 0, "x must be positive" + str(x)
     assert epsilon > 0, "epsilon must be positive" + str(epsilon)
     x=float(x)
     guess=x/2.0
     #guess=0.001
     diff=guess**2-x
     ctr=1
     while abs(diff)>epsilon and ctr<=100:
        # print "error",diff,"guess",guess
         guess=guess-diff/(2.0*guess)
         diff=guess**2-x
         ctr+=1
     assert ctr <= 100, "not perfect square number!"
     print "times of Iteration:", ctr, " guess", guess
     return guess

 def testBi1():
     squrtRootNR(4,0.0001)
     squrtRootNR(2, 0.0001)
     squrtRootNR(0.25, 0.0001)

 testBi1()