python算法（一）
一、求数x的因子
1 x=100 2 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=0 2 for c in str(1952): 3 sumdigits +=int(c) 4 print sumdigits 5 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:",chicken 17 print "the number of pig is:",pig 18 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:",chicken 18 print "the number of pig is:",pig 19 print "the number of spider is",spider 20 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_num 10 print "the number of spider is", spider_num 11 solutionFound=True 12 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 True 4 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 ans 10 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",guess 11 if guess**2<x: 12 low=guess 13 else: 14 high=guess 15 guess=(low+high)/2.0 16 ctr+=1 17 assert ctr<=100,"not perfect square number!" 18 print "times of Iteration:",ctr," guess",guess 19 return guess 20 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",guess 11 guess=guess-diff/(2.0*guess) 12 diff=guess**2-x 13 ctr+=1 14 assert ctr <= 100, "not perfect square number!" 15 print "times of Iteration:", ctr, " guess", guess 16 return guess 17 18 def testBi1(): 19 squrtRootNR(4,0.0001) 20 squrtRootNR(2, 0.0001) 21 squrtRootNR(0.25, 0.0001) 22 23 testBi1()