Fifth chapter conditions and recursion
"//" represents the down-divide method
"%" modulo operator
The code that determines the intertextuality:
#-*-Coding: UTF-8-*-
def First (word):
return word[0]
def Last (word):
return Word[-1]
def Middle(word):
return Word[1:-1]
def Huwen(word):
If Len (word) <=1:
Return True
If first (word)!=last (word):
Return False
Else
Return Huwen (middle (word))
Print (Huwen (' a '))
Print (Huwen ('afbhj'))
Print (Huwen ('adhsuia'))
Print (Huwen ('heeh'))
def Huwen(WORD1,WORD2):
If Len (word1)! = Len (word2):
Return False
I=0
J=len (WORD2)-1
While j>0:
If WORD1[I]!=WORD2[J]:
Return False
I+=1
J-=1
Return True
Print (Huwen ('hkl', 'lkh'))
def Is_huwen(word):
I=0
J=len (Word)-1
While I<=j:
If WORD[I]!=WORD[J]:
Return False
I+=1
J-=1
Return True
Print (Is_huwen ('guug'))
epsilon=0.0000001
Write the following code in Python
#-*-Coding:utf-8-*-
Import Math
def Jiecheng(n):
If n==0:
Return 1
Else
b=jiecheng (n-1)
C=n*jiecheng (n-1)
Return C
def Estimatepi():
D= (2*MATH.SQRT (2))/9801
K=0
Total=0
While True:
E= (Jiecheng (4*k)) * (1103+26390*k)
F= (Jiecheng (k)) **4
g=f* (396** (4*k))
Total1=d*e/g
Total+=total1
If ABS (TOTAL1) <1e-15:
Break
K+=1
Return 1/total
Print (Estimatepi ())
Two methods for traversing strings
Find usage
#-*-Coding: UTF-8-*-
word=' banana '
Print (Word.find ('na')) #2 look for the subscript when a string is encountered for the first time
Print (Word.find ('na', 3)) #4 indicates which subscript to start looking for
Print (Word.find (' A ', 0, 1)) #-1 indicates from which small label to start finding which of the following table ends
Find a word with three repeating letters
#-*-Coding:utf-8-*-
def Liayiyang(word):
Count=0
I=0
While I<len (word)-1:
If WORD[I]==WORD[I+1]:
Count+=1
If count==3:
Return True
I=i+2
Else
Count=0
I=i+1
Return False
def Wenjian():
Fin = open (' c:\\users\\chenxi\\Desktop\\words.txt ')
For line in Fin:
Word=line.strip ()
If Liayiyang (word):
Print (word)
Wenjian ()
Think like computer scientists Python (3)