Python版本：python 3.2.2

电脑系统：win7旗舰

实例来源：python菜鸟教程100例

 #!/usr/bin/python

 # -*- coding: UTF-8 -*-

 import string

 import math

 import time

 import sys

 import os

 #import pygame

 #eg1:There are 1, 2, 3, 4 numbers, can be composed of a number of different and no duplication of the three digit number? How much is it?

 for i in range(1,5):

     for j in range(1,5):

         for k in range(1,5):

             if (i != k) and (i != j) and (j != k):

                 print("The number is ",i,j,k)

 #eg2:Bonuses paid by enterprises in accordance with the profit commission. Total number of 1.5% between 7.5% of the profits (I) less than or equal to 10 million yuan, bonus provided 10%; profit of more than 10 million yuan, less than 20 million yuan, less than 10 million yuan in 10% deduct a percentage from a sum of money, more than 10 million yuan, the Commission; 20 million to 40 million, more than 20 million yuan, to the Commission of 5%; 40 million to 60 million more than 40 million yuan, to the Commission of 3%; 60 million to 100 million, more than 60 million yuan, deduct a percentage from a sum of money, more than 100 million yuan, more than 100 million yuan according to the 1% commission, from the keyboard input month profit I, seeking to be bonuses?

 #Solution 1

 percent_less10mili = 0.1

 percent_less20mili = 0.075

 percent_less40mili = 0.05

 percent_less60mili = 0.03

 percent_less100mili = 0.015

 percent_last = 0.01

 I=input("Please enter a the profit for this year :")

 i=int(I)

 if i <= 10:

     Sum = i*percent_less10mili

 elif 10 < i and i <= 20:

     Sum = 1+(i-10)*percent_less20mili

 elif 20 < i and i <= 40:

     Sum = 1+0.75+(i-20)*percent_less40mili

 elif 40 < i and i <= 60:

     Sum = 1+0.75+1+(i-40)*percent_less60mili

 elif 60 < i and i <= 100:

     Sum = 1+0.75+1+0.6+(i-60)*percent_less100mili

 elif 100 < i:

     Sum = 1+0.75+1+0.6+0.6+(i-100)*percent_last

 else :

     print("Your have enter a wrong number!")

 print("The profit of this year for MaMiao",Sum)

 #Solution 2

 i = int(input("The profit:"))

 arr = [1000000,600000,400000,200000,100000,0]

 rat = [0.01,0.015,0.03,0.05,0.075,0.1]

 r = 0

 for idx in range(0,6):

     if i>arr[idx]:

         r+=(i-arr[idx])*rat[idx]

         print((i-arr[idx])*rat[idx])

         i=arr[idx]

 print(r)

 #eg3:An integer, which adds 100 and plus 268 is a perfect square, what is the number?

 #Solution 1

 n=0

 m=0

 for K_f in range(1,12):#This algorithm has a simple mathematical pretreatment and analysis, to a certain extent, the time complexity of the algorithm is simplified.

     K_s = 168/K_f

     if (K_s == int(K_s)):

         m=(K_s+K_f)/2

         n=(K_s-K_f)/2

     if n == int(n) and m == int(m) and n!=0 and m!=0:

         print(m,n)

         print(m*m-268)

 #Solution 2

 for i in range(10000):

     x = int(math.sqrt(i + 100))

     y = int(math.sqrt(i + 268))

     if(x * x == i + 100) and (y * y == i + 268):

         print(i)

 #eg4:Enter a certain day, judgment day is the first few days this year?

 year = int(input("Please enter the year:"))

 month = int(input("Please enter the month:"))

 day = int(input("Please enter the day:"))

 sum = 0

 month_day=(31,28,31,30,31,30,31,31,30,31,30,31)

 if (year%4==0 and year%100!=0) or year%400==0 :

     day_plus = 1

     print("This year is a leap year!")

 else :

     day_plus = 0

     print("This year is not a leap year!")

 if 2 < month :

     sum += day_plus

 for i in range(0,month-1):

     sum += month_day[i]

 sum += day

 print("The sum of today is equal to:",sum)

 #eg5:Enter three integers x, y, z, please put the three number of small to large output.

 #Solution 1

 def compare(A,B):

     if A>B:

         result = A

     else:

         result = B

     return result

 NUM1=int(input("Please enter the first number:"))

 NUM2=int(input("Please enter the second number:"))

 NUM3=int(input("Please enter the third number:"))

 Big1=compare(NUM1,NUM2)

 Big2=compare(Big1,NUM3)#caculate the biggest one

 if NUM1 == Big2:

     NUM1=0

 if NUM2 == Big2:

     NUM2=0

 if NUM3 == Big2:

     NUM3=0

 middle1=compare(NUM1,NUM2)

 middle2=compare(middle1,NUM3)#caculate the second biggest one

 if NUM1 == middle2:

     NUM1=0

 if NUM2 == middle2:

     NUM2=0

 if NUM3 == middle2:

     NUM3=0

 small1=compare(NUM1,NUM2)

 small2=compare(small1,NUM3)#caculate the second biggest one

 print(small2,middle2,Big2)

 #Solution 2

 l = []

 for i in range(3):

     x = int(input("Please enter the first number:"))

     l.append(x)

 l.sort()

 print(l)

 #eg6:Fibonacci sequence

 #Solution 1

 F = [0,1,1]

 print(F[0])

 print(F[1])

 for i in range(1,10):

     F[0]=F[1]

     F[1]=F[2]

     F[2]=F[0]+F[1]

     print(F[2])

 #Solution 2

 def fib(n):

     if n==1 or n==2:

         return 1

     return fib(n-1)+fib(n-2)

 print(fib(10))

 #Copy a list of data into another list.

 #Solution 1

 a = [1, 2, 3]

 b = a[:]

 print(b)

 #Solution 2

 c=[0,0,0]

 for i in range(0,3):

     c[i]=a[i]

 print(c)

 #eg8:The output of 9*9 multiplication table

 for i in range(1,10):

     for j in range(1,10):

         print(i,"*",j,"=",i*j)

 #eg9:Pause one second output

 #It need to import the headfiles named time like this:import time

 myD = {1: 'a', 2: 'b'}

 for key,value in dict.items(myD):

     print(key, value)

     print(time.strftime('%Y-%m-%d %H:%M:%S',time.localtime(time.time())))

     time.sleep(1)

     print(time.strftime('%Y-%m-%d %H:%M:%S',time.localtime(time.time())))

 #To determine how many prime numbers between 101-200, and the output of all prime numbers

 h = 0

 leap = 1

 from math import sqrt

 from sys import stdout

 for m in range(101,201):

     k = int(sqrt(m + 1))

     for i in range(2,k + 1):

         if m % i == 0:

             leap = 0

             break

     if leap == 1:

         print('%-4d' % m)

         h += 1

         if h % 10 == 0:

             print('')

     leap = 1

 print('The total is %d' % h)

 #eg10:Print out all the "daffodils", the so-called "daffodils" is a three digit number. The all digital cube and equal to the number itself. For example: 153 is a "daffodils", because 153=1^3+5^3+3^3

 for i in range(100,1000):

     sum = math.pow(int(i/100),3)+ math.pow(int((i%100)/10),3)+ math.pow(i%10,3)

     if sum==i:

         print("This number is a daffodils number equal to:",i)

 #eg11:A positive integer factorization. For example: enter 90, print out 90=2*3*3*5

 n = int(input("input number:\n"))

 print("n = %d" % n)

 for i in range(2,n + 1):

     while n != i:

         if n % i == 0:#All the number,which is not a prime number can't appear at there,because that the number(not prime number) have ever been diverse to be the premi number ,for Example:10 have already been diverse to be 2 multiple 5

             print(str(i))

             print("*")

             n = n / i

         else:

             break

 print("%d"% n)

 #eg11:Use the conditional operator to complete this problem: the study results >=90 points of the students with A, 60-89 points between the use of B, said the following 60 points with C

 Score = int(input("Please enter the score:"))#Python haven't the loop function for three member like this : A = expression ? "Q" : "P"

 if Score < 60:

     print("Your Grade is equal to C")

 elif 60<= Score < 90:

     print("Your Grade is equal to B")

 else :

     print("Your Grade is equal to A")

 #eg12:Enter a line of characters, respectively, the statistics of the English letters, spaces, numbers and other characters of the number

 write = "Come on,Baby!"

 write = write.encode()

 fo = open("eg_test.txt",'wb',1000)

 print("The filename is :",fo.name)

 print("The Rquest_Model of files :",fo.mode)

 print("This is the sentence which I have already written into the file named ",fo.name,write)

 fo.write(write)

 fo.close()

 fo = open("eg_test.txt","rb",1)

 Str = fo.read(15)

 Str = Str.decode()#You have to decode the string to count how many times the letter "C" have appear in the str!

 print("The string i have already read from txt file is :",Str)

 C = Str.count('C')

 Where = Str.find("Ba")

 print("The latter C have appear for ",C,"times!")

 print("The string om have appear at the address:",Where)

 fo.close()

 letters = 0

 space = 0

 digit = 0

 others = 0

 for c in Str:

     if c.isalpha():#remember the function to find alpha

         letters += 1

     elif c.isspace():#rememeber the function to find the space

         space += 1

     elif c.isdigit():#remember the function to find the digit

         digit += 1

     else:

         others += 1

 print('char = %d,space = %d,digit = %d,others = %d' % (letters,space,digit,others))#输出

 #eg13:Detect keyboard to control the size of the key

 print("Please enter the keyboard values,this example is used to detect the values of top bottom right and left values of keyboard!")

 top = input("Please enter the top key in keyboard!")

 #bottom = input("Please enter the bottom key in keyboard!")

 #left = input("Please enter the left key in keyboard!")

 #right = input("Please enter the right key in keyboard!")

 print("The top values is equal to ",top,int(top))

 #print("The bottom values is equal to ",bottom,int(bottom))

 #print("The left values is equal to ",left,int(left))

 #print("The right values is equal to ",right,int(right))

 #eg14:Find the value of s=a+aa+aaa+aaaa+aa... A, where a is a number. For example, 2+22+222+2222+22222 (at this time a total of 5 numbers add), a few numbers add a keyboard control

 K = int(input("Please enter a Number to calculate:"))

 N = int(input("Please enter a Number to define how many time you want to calculate:"))

 #for i in range(1,10):#范围是1到10-1=9，记住了！！！

 #    print(i)

 #for i in range(10):#范围是0到9之间，记住了！！！

 #    print(i)

 sum = 0#出初始化一定要做到啊！！！！

 for i in range(1,N+1):

     temp = i*math.pow(10,N-i)

     sum += temp

     print("The precudure is :",temp,"+")

 print(sum)

 print("The result is calculate to be :",sum*K)

 #eg15：If a number is exactly equal to the sum of its factors, this number is called "end of a few". For example 6=123. programming to find all completed within 1000

 i=1

 while (i<=1000):

     Buffer = i

     for j in range(1,i):

         if (i%j==0):

             Buffer -= j

     if (Buffer==0):

         print("This number is a complete number equal to:",i)

     i += 1

 #eg16:A ball dropped from a height of 100 m free. Each fall to the ground after jump back to the original height of half; to fall for it on the 10th floor, the total number of meters after? How high is the tenth bounce?

 Heigth = int(input("Please enter a Number for Heigth:"))

 sum = Heigth

 Times = int(input("Please enter a Number for calculate times:"))

 print("The Heigth of 10th bounce is equal to :",math.pow(0.5,Times)*Heigth)

 for i in range(1,Times):

     sum +=  2*Heigth*math.pow(0.5,i)

 print("The total length is equal to:",sum)

 #eg17:Two table tennis team for the game, each out of the three. A team for the A, B, C three, B team for the X, y, Z three. Draw a draw. Someone asked the team about the competition. A said he did not x than, C said he does not and Z, X than, please make up the program to find out the list of three teams race

 for i in range(ord('x'),ord('z') + 1):

     for j in range(ord('x'),ord('z') + 1):

         if i != j:

             for k in range(ord('x'),ord('z') + 1):

                 if (i != k) and (j != k):

                     if (i != ord('x')) and (k != ord('x')) and (k != ord('z')):

                         print('order is a -- %s\t b -- %s\tc--%s' % (chr(i),chr(j),chr(k)))

 #这种题目有很强的抽象性，一定要注意求解的方法

 #eg8:Print out the following pattern (diamond)

 print("   *")

 print("  ***")

 print(" *****")

 print("*******")

 print(" *****")

 print("  ***")

 print("   *")

 #eg19:There is a sequence of points: 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, and the sum of the first 20 items of this series.

 #分析题目的基本规律：

 #分子：2+3=5 3+5=8 5+8=13 8+13=21

 #分母：1+2=3 2+3=5 3+5=8 5+8=13

 Top_num1=2

 Bot_num1=1

 Top_num2=3

 Bot_num2=2

 sum=7/2

 for i in range(18):

     #Num = (Top_num1+Top_num2)/(Bot_num1+Bot_num2)

     print("The number is :",Top_num1,Bot_num1)

     Temp1=Top_num1#中间变量的申请和数据更新及存放注意啊！！

     Temp2=Bot_num1

     Top_num1 = Top_num2

     Bot_num1 = Bot_num2

     Top_num2 = (Temp1+Top_num2)

     Bot_num2 = (Temp2+Bot_num2)

     sum += Top_num2/Bot_num2

 print(sum)

 #eg20:Use recursive method for 5!

 def recursive(N):

     result = N

     if (N==1):

         return 1

     else :

         result *= recursive(N-1)

     return result

 result = recursive(5)

 print(result)

 #eg21:By using the recursive function call, the input of the 5 characters, in order to print out the opposite order

 #Solution 1:

 FIFO=['A','B','C','D','E']

 print("Please enter five letter for FIFO!")

 for i in range(5):

     FIFO[i] = input("Please enter the letter:")

 print(FIFO[0],FIFO[1],FIFO[2],FIFO[3],FIFO[4])

 print(FIFO[4],FIFO[3],FIFO[2],FIFO[1],FIFO[0])

 print(FIFO[4]+FIFO[3]+FIFO[2]+FIFO[1]+FIFO[0])

 #Solution 2(采用递归算法):

 def output(s,l):

     if l==0:

        return

     print(s[l-1])

     output(s,l-1)

 s = FIFO[4]+FIFO[3]+FIFO[2]+FIFO[1]+FIFO[0]

 l = len(s)

 output(s,l)

 #eg22:A 5 digit number, it is not a palindrome judgment. That 12321 is a palindrome, a bit and bit the same ten million, with thousands of the same

 #del Num

 Num = 12321

 print("The number for test is :",Num)

 if (int(Num/10000)==Num%10) and (int(Num/1000)%10==int((Num%100)/10)):

     print("This is a palindrome number!")

 #eg23:Please enter the first letter of the week to determine what is a week, if the first letter, then continue to determine the second letters

 days = ['Monday','Thuesday','Wednesday','Thursday','Friday','Saturday','Sunday']

 print("Please enter the first letter to determine the day in week:")

 Ch1=input("Please enter the first letter:")

 if Ch1=='M':

     print("Today is Monday!")

 elif Ch1=='W':

     print("Today is Wednsday!")

 elif Ch1=='F':

     print("Today is Friday!")

 elif Ch1=='T':

     print("Please enter the second letter for determined!")

     Ch2=input()

     if Ch2=='h':

         print("Today is Thuesday!")

     else :

         print("Today is Tursday!")

 else:

     print("Please enter the second letter for detemined!")

     Ch2=input()

     if Ch2=='a':

         print("Today is Saturday!")

     else :

         print("Today is Sunday!")

 #eg24:Comma separated list按逗号分隔列表

 L = [1,2,3,4,5]

 s1 = ','.join(str(n) for n in L)

 print(s1)

 #eg25:文本颜色设置--示例35

 class bcolors:

     HEADER = '\033[95m'

     OKBLUE = '\033[94m'

     OKGREEN = '\033[92m'

     WARNING = '\033[93m'

     FAIL = '\033[91m'

     ENDC = '\033[0m'

     BOLD = '\033[1m'

     UNDERLINE = '\033[4m'

 print(bcolors.WARNING + "Warning mesg color?" + bcolors.ENDC)
 #!/usr/bin/python

 # -*- coding: UTF-8 -*-

 import string

 import math

 import time

 import sys

 import os

 #import pygame

 #eg1:There are 1, 2, 3, 4 numbers, can be composed of a number of different and no duplication of the three digit number? How much is it?

 for i in range(1,5):

     for j in range(1,5):

         for k in range(1,5):

             if (i != k) and (i != j) and (j != k):

                 print("The number is ",i,j,k)

 #eg2:Bonuses paid by enterprises in accordance with the profit commission. Total number of 1.5% between 7.5% of the profits (I) less than or equal to 10 million yuan, bonus provided 10%; profit of more than 10 million yuan, less than 20 million yuan, less than 10 million yuan in 10% deduct a percentage from a sum of money, more than 10 million yuan, the Commission; 20 million to 40 million, more than 20 million yuan, to the Commission of 5%; 40 million to 60 million more than 40 million yuan, to the Commission of 3%; 60 million to 100 million, more than 60 million yuan, deduct a percentage from a sum of money, more than 100 million yuan, more than 100 million yuan according to the 1% commission, from the keyboard input month profit I, seeking to be bonuses?

 #Solution 1

 percent_less10mili = 0.1

 percent_less20mili = 0.075

 percent_less40mili = 0.05

 percent_less60mili = 0.03

 percent_less100mili = 0.015

 percent_last = 0.01

 I=input("Please enter a the profit for this year :")

 i=int(I)

 if i <= 10:

     Sum = i*percent_less10mili

 elif 10 < i and i <= 20:

     Sum = 1+(i-10)*percent_less20mili

 elif 20 < i and i <= 40:

     Sum = 1+0.75+(i-20)*percent_less40mili

 elif 40 < i and i <= 60:

     Sum = 1+0.75+1+(i-40)*percent_less60mili

 elif 60 < i and i <= 100:

     Sum = 1+0.75+1+0.6+(i-60)*percent_less100mili

 elif 100 < i:

     Sum = 1+0.75+1+0.6+0.6+(i-100)*percent_last

 else :

     print("Your have enter a wrong number!")

 print("The profit of this year for MaMiao",Sum)

 #Solution 2

 i = int(input("The profit:"))

 arr = [1000000,600000,400000,200000,100000,0]

 rat = [0.01,0.015,0.03,0.05,0.075,0.1]

 r = 0

 for idx in range(0,6):

     if i>arr[idx]:

         r+=(i-arr[idx])*rat[idx]

         print((i-arr[idx])*rat[idx])

         i=arr[idx]

 print(r)

 #eg3:An integer, which adds 100 and plus 268 is a perfect square, what is the number?

 #Solution 1

 n=0

 m=0

 for K_f in range(1,12):#This algorithm has a simple mathematical pretreatment and analysis, to a certain extent, the time complexity of the algorithm is simplified.

     K_s = 168/K_f

     if (K_s == int(K_s)):

         m=(K_s+K_f)/2

         n=(K_s-K_f)/2

     if n == int(n) and m == int(m) and n!=0 and m!=0:

         print(m,n)

         print(m*m-268)

 #Solution 2

 for i in range(10000):

     x = int(math.sqrt(i + 100))

     y = int(math.sqrt(i + 268))

     if(x * x == i + 100) and (y * y == i + 268):

         print(i)

 #eg4:Enter a certain day, judgment day is the first few days this year?

 year = int(input("Please enter the year:"))

 month = int(input("Please enter the month:"))

 day = int(input("Please enter the day:"))

 sum = 0

 month_day=(31,28,31,30,31,30,31,31,30,31,30,31)

 if (year%4==0 and year%100!=0) or year%400==0 :

     day_plus = 1

     print("This year is a leap year!")

 else :

     day_plus = 0

     print("This year is not a leap year!")

 if 2 < month :

     sum += day_plus

 for i in range(0,month-1):

     sum += month_day[i]

 sum += day

 print("The sum of today is equal to:",sum)

 #eg5:Enter three integers x, y, z, please put the three number of small to large output.

 #Solution 1

 def compare(A,B):

     if A>B:

         result = A

     else:

         result = B

     return result

 NUM1=int(input("Please enter the first number:"))

 NUM2=int(input("Please enter the second number:"))

 NUM3=int(input("Please enter the third number:"))

 Big1=compare(NUM1,NUM2)

 Big2=compare(Big1,NUM3)#caculate the biggest one

 if NUM1 == Big2:

     NUM1=0

 if NUM2 == Big2:

     NUM2=0

 if NUM3 == Big2:

     NUM3=0

 middle1=compare(NUM1,NUM2)

 middle2=compare(middle1,NUM3)#caculate the second biggest one

 if NUM1 == middle2:

     NUM1=0

 if NUM2 == middle2:

     NUM2=0

 if NUM3 == middle2:

     NUM3=0

 small1=compare(NUM1,NUM2)

 small2=compare(small1,NUM3)#caculate the second biggest one

 print(small2,middle2,Big2)

 #Solution 2

 l = []

 for i in range(3):

     x = int(input("Please enter the first number:"))

     l.append(x)

 l.sort()

 print(l)

 #eg6:Fibonacci sequence

 #Solution 1

 F = [0,1,1]

 print(F[0])

 print(F[1])

 for i in range(1,10):

     F[0]=F[1]

     F[1]=F[2]

     F[2]=F[0]+F[1]

     print(F[2])

 #Solution 2

 def fib(n):

     if n==1 or n==2:

         return 1

     return fib(n-1)+fib(n-2)

 print(fib(10))

 #Copy a list of data into another list.

 #Solution 1

 a = [1, 2, 3]

 b = a[:]

 print(b)

 #Solution 2

 c=[0,0,0]

 for i in range(0,3):

     c[i]=a[i]

 print(c)

 #eg8:The output of 9*9 multiplication table

 for i in range(1,10):

     for j in range(1,10):

         print(i,"*",j,"=",i*j)

 #eg9:Pause one second output

 #It need to import the headfiles named time like this:import time

 myD = {1: 'a', 2: 'b'}

 for key,value in dict.items(myD):

     print(key, value)

     print(time.strftime('%Y-%m-%d %H:%M:%S',time.localtime(time.time())))

     time.sleep(1)

     print(time.strftime('%Y-%m-%d %H:%M:%S',time.localtime(time.time())))

 #To determine how many prime numbers between 101-200, and the output of all prime numbers

 h = 0

 leap = 1

 from math import sqrt

 from sys import stdout

 for m in range(101,201):

     k = int(sqrt(m + 1))

     for i in range(2,k + 1):

         if m % i == 0:

             leap = 0

             break

     if leap == 1:

         print('%-4d' % m)

         h += 1

         if h % 10 == 0:

             print('')

     leap = 1

 print('The total is %d' % h)

 #eg10:Print out all the "daffodils", the so-called "daffodils" is a three digit number. The all digital cube and equal to the number itself. For example: 153 is a "daffodils", because 153=1^3+5^3+3^3

 for i in range(100,1000):

     sum = math.pow(int(i/100),3)+ math.pow(int((i%100)/10),3)+ math.pow(i%10,3)

     if sum==i:

         print("This number is a daffodils number equal to:",i)

 #eg11:A positive integer factorization. For example: enter 90, print out 90=2*3*3*5

 n = int(input("input number:\n"))

 print("n = %d" % n)

 for i in range(2,n + 1):

     while n != i:

         if n % i == 0:#All the number,which is not a prime number can't appear at there,because that the number(not prime number) have ever been diverse to be the premi number ,for Example:10 have already been diverse to be 2 multiple 5

             print(str(i))

             print("*")

             n = n / i

         else:

             break

 print("%d"% n)

 #eg11:Use the conditional operator to complete this problem: the study results >=90 points of the students with A, 60-89 points between the use of B, said the following 60 points with C

 Score = int(input("Please enter the score:"))#Python haven't the loop function for three member like this : A = expression ? "Q" : "P"

 if Score < 60:

     print("Your Grade is equal to C")

 elif 60<= Score < 90:

     print("Your Grade is equal to B")

 else :

     print("Your Grade is equal to A")

 #eg12:Enter a line of characters, respectively, the statistics of the English letters, spaces, numbers and other characters of the number

 write = "Come on,Baby!"

 write = write.encode()

 fo = open("eg_test.txt",'wb',1000)

 print("The filename is :",fo.name)

 print("The Rquest_Model of files :",fo.mode)

 print("This is the sentence which I have already written into the file named ",fo.name,write)

 fo.write(write)

 fo.close()

 fo = open("eg_test.txt","rb",1)

 Str = fo.read(15)

 Str = Str.decode()#You have to decode the string to count how many times the letter "C" have appear in the str!

 print("The string i have already read from txt file is :",Str)

 C = Str.count('C')

 Where = Str.find("Ba")

 print("The latter C have appear for ",C,"times!")

 print("The string om have appear at the address:",Where)

 fo.close()

 letters = 0

 space = 0

 digit = 0

 others = 0

 for c in Str:

     if c.isalpha():#remember the function to find alpha

         letters += 1

     elif c.isspace():#rememeber the function to find the space

         space += 1

     elif c.isdigit():#remember the function to find the digit

         digit += 1

     else:

         others += 1

 print('char = %d,space = %d,digit = %d,others = %d' % (letters,space,digit,others))#输出

 #eg13:Detect keyboard to control the size of the key

 print("Please enter the keyboard values,this example is used to detect the values of top bottom right and left values of keyboard!")

 top = input("Please enter the top key in keyboard!")

 #bottom = input("Please enter the bottom key in keyboard!")

 #left = input("Please enter the left key in keyboard!")

 #right = input("Please enter the right key in keyboard!")

 print("The top values is equal to ",top,int(top))

 #print("The bottom values is equal to ",bottom,int(bottom))

 #print("The left values is equal to ",left,int(left))

 #print("The right values is equal to ",right,int(right))

 #eg14:Find the value of s=a+aa+aaa+aaaa+aa... A, where a is a number. For example, 2+22+222+2222+22222 (at this time a total of 5 numbers add), a few numbers add a keyboard control

 K = int(input("Please enter a Number to calculate:"))

 N = int(input("Please enter a Number to define how many time you want to calculate:"))

 #for i in range(1,10):#范围是1到10-1=9，记住了！！！

 #    print(i)

 #for i in range(10):#范围是0到9之间，记住了！！！

 #    print(i)

 sum = 0#出初始化一定要做到啊！！！！

 for i in range(1,N+1):

     temp = i*math.pow(10,N-i)

     sum += temp

     print("The precudure is :",temp,"+")

 print(sum)

 print("The result is calculate to be :",sum*K)

 #eg15：If a number is exactly equal to the sum of its factors, this number is called "end of a few". For example 6=123. programming to find all completed within 1000

 i=1

 while (i<=1000):

     Buffer = i

     for j in range(1,i):

         if (i%j==0):

             Buffer -= j

     if (Buffer==0):

         print("This number is a complete number equal to:",i)

     i += 1

 #eg16:A ball dropped from a height of 100 m free. Each fall to the ground after jump back to the original height of half; to fall for it on the 10th floor, the total number of meters after? How high is the tenth bounce?

 Heigth = int(input("Please enter a Number for Heigth:"))

 sum = Heigth

 Times = int(input("Please enter a Number for calculate times:"))

 print("The Heigth of 10th bounce is equal to :",math.pow(0.5,Times)*Heigth)

 for i in range(1,Times):

     sum +=  2*Heigth*math.pow(0.5,i)

 print("The total length is equal to:",sum)

 #eg17:Two table tennis team for the game, each out of the three. A team for the A, B, C three, B team for the X, y, Z three. Draw a draw. Someone asked the team about the competition. A said he did not x than, C said he does not and Z, X than, please make up the program to find out the list of three teams race

 for i in range(ord('x'),ord('z') + 1):

     for j in range(ord('x'),ord('z') + 1):

         if i != j:

             for k in range(ord('x'),ord('z') + 1):

                 if (i != k) and (j != k):

                     if (i != ord('x')) and (k != ord('x')) and (k != ord('z')):

                         print('order is a -- %s\t b -- %s\tc--%s' % (chr(i),chr(j),chr(k)))

 #这种题目有很强的抽象性，一定要注意求解的方法

 #eg8:Print out the following pattern (diamond)

 print("   *")

 print("  ***")

 print(" *****")

 print("*******")

 print(" *****")

 print("  ***")

 print("   *")

 #eg19:There is a sequence of points: 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, and the sum of the first 20 items of this series.

 #分析题目的基本规律：

 #分子：2+3=5 3+5=8 5+8=13 8+13=21

 #分母：1+2=3 2+3=5 3+5=8 5+8=13

 Top_num1=2

 Bot_num1=1

 Top_num2=3

 Bot_num2=2

 sum=7/2

 for i in range(18):

     #Num = (Top_num1+Top_num2)/(Bot_num1+Bot_num2)

     print("The number is :",Top_num1,Bot_num1)

     Temp1=Top_num1#中间变量的申请和数据更新及存放注意啊！！

     Temp2=Bot_num1

     Top_num1 = Top_num2

     Bot_num1 = Bot_num2

     Top_num2 = (Temp1+Top_num2)

     Bot_num2 = (Temp2+Bot_num2)

     sum += Top_num2/Bot_num2

 print(sum)

 #eg20:Use recursive method for 5!

 def recursive(N):

     result = N

     if (N==1):

         return 1

     else :

         result *= recursive(N-1)

     return result

 result = recursive(5)

 print(result)

 #eg21:By using the recursive function call, the input of the 5 characters, in order to print out the opposite order

 #Solution 1:

 FIFO=['A','B','C','D','E']

 print("Please enter five letter for FIFO!")

 for i in range(5):

     FIFO[i] = input("Please enter the letter:")

 print(FIFO[0],FIFO[1],FIFO[2],FIFO[3],FIFO[4])

 print(FIFO[4],FIFO[3],FIFO[2],FIFO[1],FIFO[0])

 print(FIFO[4]+FIFO[3]+FIFO[2]+FIFO[1]+FIFO[0])

 #Solution 2(采用递归算法):

 def output(s,l):

     if l==0:

        return

     print(s[l-1])

     output(s,l-1)

 s = FIFO[4]+FIFO[3]+FIFO[2]+FIFO[1]+FIFO[0]

 l = len(s)

 output(s,l)

 #eg22:A 5 digit number, it is not a palindrome judgment. That 12321 is a palindrome, a bit and bit the same ten million, with thousands of the same

 #del Num

 Num = 12321

 print("The number for test is :",Num)

 if (int(Num/10000)==Num%10) and (int(Num/1000)%10==int((Num%100)/10)):

     print("This is a palindrome number!")

 #eg23:Please enter the first letter of the week to determine what is a week, if the first letter, then continue to determine the second letters

 days = ['Monday','Thuesday','Wednesday','Thursday','Friday','Saturday','Sunday']

 print("Please enter the first letter to determine the day in week:")

 Ch1=input("Please enter the first letter:")

 if Ch1=='M':

     print("Today is Monday!")

 elif Ch1=='W':

     print("Today is Wednsday!")

 elif Ch1=='F':

     print("Today is Friday!")

 elif Ch1=='T':

     print("Please enter the second letter for determined!")

     Ch2=input()

     if Ch2=='h':

         print("Today is Thuesday!")

     else :

         print("Today is Tursday!")

 else:

     print("Please enter the second letter for detemined!")

     Ch2=input()

     if Ch2=='a':

         print("Today is Saturday!")

     else :

         print("Today is Sunday!")

 #eg24:Comma separated list按逗号分隔列表

 L = [1,2,3,4,5]

 s1 = ','.join(str(n) for n in L)

 print(s1)

 #eg25:文本颜色设置--示例35

 class bcolors:

     HEADER = '\033[95m'

     OKBLUE = '\033[94m'

     OKGREEN = '\033[92m'

     WARNING = '\033[93m'

     FAIL = '\033[91m'

     ENDC = '\033[0m'

     BOLD = '\033[1m'

     UNDERLINE = '\033[4m'

 print(bcolors.WARNING + "Warning mesg color?" + bcolors.ENDC)

Python语言100例的更多相关文章

  1. 动态语言的灵活性是把双刃剑 －－ 以Python语言为例

    本文有些零碎,总题来说,包括两个问题:(1)可变对象(最常见的是list dict)被意外修改的问题,(2)对参数(parameter)的检查问题.这两个问题,本质都是因为动态语言(动态类型语言)的特 ...

  2. 动态语言的灵活性是把双刃剑 －－ 以 Python 语言为例

    本文有些零碎,总题来说,包括两个问题:(1)可变对象(最常见的是list dict)被意外修改的问题,(2)对参数(parameter)的检查问题.这两个问题,本质都是因为动态语言(动态类型语言)的特 ...

  3. python 02/100例

    题目 输入某年某月某日,判断这一天是这一年的第几天? 分析 例如:2018年6月19日 天数 = 19天 + 一月天数 + 2月天数 + ... + 6月天数 注意 闰年的2月是29天,如果年份是闰年 ...

  4. Python菜鸟100例

    题目地址 #-*- codeing = utf-8 -*- #@Time : 2021/3/18 21:17 #@Author : HUGBOY #@File : 1.py #@Software: P ...

  5. C语言100例02 PHP版（练习）

    问题: 企业发放的奖金根据利润提成. 利润(I)低于或等于10万元时,奖金可提10%: 利润高于10万元,低于20万元时,低于10万元的部分按10%提成,高于10万元的部分,可提成7.5%: 20万到 ...

  6. python练习题100例

    链接地址:http://www.runoob.com/python/python-100-examples.html

  7. 一，python编程100例

    1.有1.2.3.4个数字,能组成多少个互不相同且无重复数字的三位数?都是多少? #有1.2.3.4个数字,能组成多少个互不相同且无重复数字的三位数?都是多少? number = (1 ,2,3,4) ...

  8. C语言100例01 PHP版（练习）

    题目:有1.2.3.4个数字,能组成多少个互不相同且无重复数字的三位数?都是多少? 程序分析:可填在百位.十位.个位的数字都是1.2.3.4.组成所有的排列后再去 掉不满足条件的排列. 代码: for ...

  9. 100例Python代码带你从入门到进阶！

    以下所有代码全都至少运行一遍,确保可复现.易于理解.逐步完成入门到进阶的学习. 此教程经过我 反复打磨多遍 ,经常为此熬夜,真心不易,文章比较长,看完有用,帮我点个在看或分享支持. 教程包括 62 个 ...

随机推荐

  1. 利用name或id属性设置页面跳转的锚点

    理论准备         网页中的链接按照链接路径的不同,可以分为3种类型,分别是内部类型.锚点链接和外部链接:         按照使用对象的不同,网页中的链接又分为文本超链接,图像超链接,E-ma ...

  2. c++容器（vector、list、deque）

    vector ,deque 和 list 顺序性容器: 向量 vector :   是一个线性顺序结构.相当于数组,但其大小可以不预先指定,并且自动扩展.它可以像数组一样被操作,由于它的特性我们完全可 ...

  3. windbg学习进阶之——dump分析常用命令收集

    #重要说明 (1) windbg命令分为标准命令,元命令和扩展命令. 标准命令提供最基本的调试功能,不区分大小写.如:bp  g  dt  dv  k等 元命令提供标准命令没有提供的功能,也内建在调试 ...

  4. 【C#进阶系列】09 关于参数的故事

    可选参数和命名参数 不多说,上代码,自然懂 class Program { static void Main(string[] args) { var troy = new Troy(); troy. ...

  5. 【jQuery基础学习】05 jQuery与Ajax以及序列化

    好吧,这章不像上章那么水了,总是炒剩饭也不好. 关于AJAX 所谓Ajax,全名Asynchronous JavaScript and XML.(也就异步的JS和XML) 简单点来讲就是不刷新页面来发 ...

  6. HTML5 Wijmo：控制 Wijmo Grid 插件的编辑模式

    Wijmo jQuery 插件经常应用于在财务类网站中创建平滑和良好用户体验的交互表格.WijGrid 插件用于显示.排序.分组和编辑数据.今天我们来分享下如何控件WijGrid插件的编辑模式. 在本 ...

  7. [PHP] 自动加载的实现

    基于psr的规范,使用命名空间和spl_autoload_register()来实现自动加载 文件结构: |--Api |--Account.php |--User.php|--Service |-- ...

  8. Cookies简介和使用

    Cookie public class javax.servlet.http.Cookie  1作用(1)Cookie能使站点跟踪特定访问者的访问次数.最后访问时间和访问者进入站点的路径(2)Cook ...

  9. 自定义XML动态配置程序

    概述 1 在做程序开发时,我们往往要用到如下两个基本模块 1> 设置程序的基础参数,如分页的参数.邮件参数等: 2> 在基于表驱动开发时,即把一些判断的逻辑放在表数据中: 2 在这两个基本 ...

  10. C#生成条形码 Code128算法

    条形有很多种,Code128是比较常用的一种,是一种高密度条码, CODE128 码可表示从 ASCII 0 到ASCII 127 共128个字符,故称128码.其中包含了数字.字母和符号字符. Co ...