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python算法（一）
小程序：扫一扫查出行【扫一扫了解最新限行尾号】
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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()
