Introduction to programming language Classic 100 cases--1

1 # Title: There are 1, 2, 3, 4 digits, how many different and no repetition of the number of three digits? How much are they?

    2 num = [1,2,3,4]  3 n = []  4  for x in num:  5     for y in num:   6         for z in num:  7              if x != y and  x != z and y != z:  8                   m = x*100 + y*10 +  z  9                   n.append (m)  10                    11 print (' Can make up a non-repeating three-digit number with%d '% (len (n)))  12 print (' They are: ')  13 print (n)

Operation Result:

[[email protected] code_100]# Python code_1.py can consist of three digits without duplicates with 24 of them are: [123, 124, 132, 134, 142, 143, 213, 214, 231, 234, 241 , 243, 312, 314, 321, 324, 341, 342, 412, 413, 421, 423, 431, 432]

Script Explanation:

    2 num = [1,2,3,4] #  Define these four numbers as a list for the for traversal of these numbers, or you can use rang (1,5) To represent the   3 n = []  #定义一个空list, used to store the required number, the list is used in order to use the Len () function to take the length   4  for x in num:   #用三个for循环来遍历所有可能的组合   5     for  y in num:  6         for z in  num:  7             if x  != y and x != z and y != z:  #  exclude three digits with duplicate numbers   (I don't know if there are any simple expressions to express this condition, not much)   8                   m = x*100 + y*10 + z     #取得符合要求的数字   9                  &Nbsp;n.append (m)         #将符合要求的数字用append方法添加到列表n中  10                    11 print ( ' Can compose a non-repeating three-digit number with%d '% (len (n)) ')     #  format the output list length, which is the number of lists  12 print (' They are: ')   13 print (n)                      #  lists the numbers that meet the requirements, or you can use the For loop to iterate through the list and take out the numbers

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Introduction to programming language Classic 100 cases--1