任意多个集合的笛卡尔积-yiteyi-C++库

给定N个集合。任务是编写一个程序，按给定顺序对所有集合进行笛卡尔积。实例 :

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Input:1st set: 1 22nd set: A 3rd set: x 4th set: 5 6Output:[['1', 'A', 'x', '5'], ['1', 'A', 'x', '6'], ['2', 'A', 'x', '5'], ['2', 'A', 'x', '6']]Input:1st set: 1 22nd set: A 3rd set: x y z Output:[['1', 'A', 'x'], ['1', 'A', 'y'], ['1', 'A', 'z'], ['2', 'A', 'x'], ['2', 'A', 'y'], ['2', 'A', 'z']]

推荐：请尝试你的方法 {IDE} 首先，在进入解决方案之前。

方法： 该方法是在开始时计算集合1和集合2的乘积，然后集合1和集合2的结果将与集合3有一个乘积，然后集合1、集合2、集合3的结果将与集合4有一个笛卡尔积，依此类推，直到集合n。以下是上述方法的实施情况。

Python3

                         # Python program for cartesian                       
                         # product of N-sets                       
           
                         # function to find cartesian product of two sets                       
                         def                                     cartesianProduct(set_a, set_b):                       
                                                 result                                     =                                     []                       
                                                 for                                     i                                     in                                     range                                     (                                     0                                     ,                                     len                                     (set_a)):                       
                                                 for                                     j                                     in                                     range                                     (                                     0                                     ,                                     len                                     (set_b)):                       
           
                                                 # for handling case having cartesian                       
                                                 # product first time of two sets                       
                                                 if                                     type                                     (set_a[i]) !                                     =                                     list                                     :                       
                                                 set_a[i]                                     =                                     [set_a[i]]                       
                                   
                                                 # coping all the members                       
                                                 # of set_a to temp                       
                                                 temp                                     =                                     [num                                     for                                     num                                     in                                     set_a[i]]                       
                                   
                                                 # add member of set_b to                       
                                                 # temp to have cartesian product                       
                                                 temp.append(set_b[j])                       
                                                 result.append(temp)                       
                                   
                                                 return                                     result                       
           
                         # Function to do a cartesian                       
                         # product of N sets                       
                         def                                     Cartesian(list_a, n):                       
                                   
                                                 # result of cartesian product                       
                                                 # of all the sets taken two at a time                       
                                                 temp                                     =                                     list_a[                                     0                                     ]                       
                                   
                                                 # do product of N sets                       
                                                 for                                     i                                     in                                     range                                     (                                     1                                     , n):                       
                                                 temp                                     =                                     cartesianProduct(temp, list_a[i])                       
                                   
                                                 print                                     (temp)                       
           
                         # Driver Code                       
                         list_a                                     =                                     [[                                     1                                     ,                                     2                                     ],                                     # set-1                       
                                                 [                                     'A'                                     ],                                     # set-2                       
                                                 [                                     'x'                                     ,                                     'y'                                     ,                                     'z'                                     ]]                                     # set-3                       
                                   
                         # number of sets                       
                         n                                     =                                     len                                     (list_a)                       
           
                         # Function is called to perform                       
                         # the cartesian product on list_a of size n                       
                         Cartesian(list_a, n)                       

输出：

[[1, 'A', 'x'], [1, 'A', 'y'], [1, 'A', 'z'],  [2, 'A', 'x'],  [2, 'A', 'y'],  [2, 'A', 'z']]

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Python