递归练习题|集3-yiteyi-C++库

解释下面递归函数的功能。

null

问题1

C++

                         void                                     fun1(                                     int                                     n)                       
                         {                       
                                                 int                                     i = 0;                       
                                                 if                                     (n > 1)                       
                                                 fun1(n - 1);                       
                                                 for                                     (i = 0; i < n; i++)                       
                                                 cout <<                                     " * "                                     ;                       
                         }                       
           
                         // This code is contributed by shubhamsingh10                       

C

                         void                                     fun1(                                     int                                     n)                       
                         {                       
                                                 int                                     i = 0;                       
                                                 if                                     (n > 1)                       
                                                 fun1(n-1);                       
                                                 for                                     (i = 0; i < n; i++)                       
                                                 printf                                     (                                     " * "                                     );                       
                         }                       

JAVA

                         static                                     void                                     fun1(                                     int                                     n)                       
                         {                       
                                                 int                                     i =                                     0                                     ;                       
                                                 if                                     (n >                                     1                                     )                       
                                                 fun1(n -                                     1                                     );                       
                                                 for                                     (i =                                     0                                     ; i < n; i++)                       
                                                 System.out.print(                                     " * "                                     );                       
                         }                       
                                   
                         // This code is contributed by shubhamsingh10                       

蟒蛇3

                         def                                     fun1(n):                       
                                                 i                                     =                                     0                       
                                                 if                                     (n >                                     1                                     ):                       
                                                 fun1(n                                     -                                     1                                     )                       
                                                 for                                     i                                     in                                     range                                     (n):                       
                                                 print                                     (                                     " * "                                     ,end                                     =                                     "")                       
           
                         # This code is contributed by shubhamsingh10                       

C#

                         static                                     void                                     fun1(                                     int                                     n)                       
                         {                       
                                                 int                                     i = 0;                       
                                                 if                                     (n > 1)                       
                                                 fun1(n-1);                       
                                                 for                                     (i = 0; i < n; i++)                       
                                                 Console.Write(                                     " * "                                     );                       
                         }                       
           
                         // This code is contributed by shubhamsingh10                       

Javascript

                         <script>                       
           
                         function                                     fun1(n)                       
                         {                       
                                                 let i = 0;                       
                                   
                                                 if                                     (n > 1)                       
                                                 fun1(n - 1);                       
                                   
                                                 for                                     (i = 0; i < n; i++)                       
                                                 document.write(                                     " * "                                     );                       
                         }                       
           
                         // This code is contributed by gottumukkalabobby                       
           
                         </script>                       

答：打印的星星总数等于1+2+…。（n-2）+（n-1）+n，也就是n（n+1）/2。

问题2

C++

                         #define LIMIT 1000                       
                         void                                     fun2(                                     int                                     n)                       
                         {                       
                                                 if                                     (n <= 0)                       
                                                 return                                     ;                       
                                                 if                                     (n > LIMIT)                       
                                                 return                                     ;                       
                                                 cout << n <<                                     " "                                     ;                       
                                                 fun2(2*n);                       
                                                 cout << n <<                                     " "                                     ;                       
                         }                       
           
                         // This code is contributed by shubhamsingh10                       

C

                         #define LIMIT 1000                       
                         void                                     fun2(                                     int                                     n)                       
                         {                       
                                                 if                                     (n <= 0)                       
                                                 return                                     ;                       
                                                 if                                     (n > LIMIT)                       
                                                 return                                     ;                       
                                                 printf                                     (                                     "%d "                                     , n);                       
                                                 fun2(2*n);                       
                                                 printf                                     (                                     "%d "                                     , n);                       
                         }                       

JAVA

                         int                                     LIMIT =                                     1000                                     ;                       
                         void                                     fun2(                                     int                                     n)                       
                         {                       
                                                 if                                     (n <=                                     0                                     )                                     return                                     ;                       
                                                 if                                     (n > LIMIT)                                     return                                     ;                       
           
                                                 System.out.print(String.format(                                     "%d "                                     , n));                       
                                                 fun2(                                     2                                     * n);                       
                                                 System.out.print(String.format(                                     "%d "                                     , n));                       
                         }                       

蟒蛇3

                         LIMIT                                     =                                     1000                       
                         def                                     fun2(n):                       
                                                 if                                     (n <                                     =                                     0                                     ):                       
                                                 return                       
                                                 if                                     (n > LIMIT):                       
                                                 return                       
                                                 print                                     (n, end                                     =                                     " "                                     )                       
                                                 fun2(                                     2                                     *                                     n)                       
                                                 print                                     (n, end                                     =                                     " "                                     )                       
           
                         # This code is contributed by shubhamsingh10                       

C#

                         int                                     LIMIT = 1000                       
                         void                                     fun2(                                     int                                     n)                       
                         {                       
                                                 if                                     (n <= 0)                       
                                                 return                                     ;                       
                                                 if                                     (n > LIMIT)                       
                                                 return                                     ;                       
                                                 Console.Write(n+                                     " "                                     );                       
                                                 fun2(2*n);                       
                                                 Console.Write(n+                                     " "                                     );                       
                         }                       
           
                         // This code is contributed by Shubhamsingh10                       

Javascript

                         <script>                       
           
                         let LIMIT = 1000;                       
                         function                                     fun2(n)                       
                         {                       
                                                 if                                     (n <= 0)                       
                                                 return                                     ;                       
                                                 if                                     (n > LIMIT)                       
                                                 return                                     ;                       
           
                                                 document.write(n +                                     " "                                     ));                       
                                                 fun2(2 * n);                       
                                                 document.write(n +                                     " "                                     ));                       
                         }                       
           
                         // This code is contributed by gottumukkalabobby                       
           
                         </script>                       

答：对于正n，fun2（n）打印n、2n、4n、8n……的值，而该值小于极限。按递增顺序打印值后，它会按相反顺序再次打印相同的数字。例如，fun2（100）打印100、200、400、800、800、400、200、100。如果n为负，则立即返回函数。如果您发现任何答案/代码不正确，或者您想分享有关上述主题的更多信息，请写下评论。

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