Implemented in C ++: FJ string Print

Problem Description

　　FJ in the sand table to write some string like this:
　　A1 = "A"
　　A2 = "ABA"
　　A3 = "ABACABA"
　　A4 = "ABACABADABACABA"
　　... ...
　　you can find out the rules and write them all columns AN it?

Input Format

　　Only a few: N ≤ 26.

Output Format

　　Please output corresponding string AN, to a newline character. Output must not contain extra space or line feed, carriage return.

Sample input

Sample Output

Spearheading

Thinking: Examples of the title is observed, are found each having a string symmetry, n = 1, A is the center of symmetry to it, n = 2, B is a center of symmetry on, n = 3, it is the center of symmetry with C ...... symmetrical and each time, are made on both sides of a string consisting of symmetry, so use a recursive function to solve this problem.

 1 #include<iostream>
 2 using namespace std;
 3 
 4 class print_string
 5 {
 6 public:
 7     int get_n()
 8     {
 9         cin>>n;
10         return n;
11     }
12 
13     void recursion(int n)    //递归函数
14     {
15         if(n==1)
16         {
17             cout<<"A " ;
 18 is          }
 . 19          the else 
20 is          {
 21 is              recursion (N- . 1 );
 22 is              T = ' A ' + N- . 1 ;
 23 is              COUT << T;
 24              recursion (N- . 1 );
 25          }
 26 is      }
 27  Private :
 28      int n-;
 29      char T;         // for each recursive control output 
30  };
 31 is  
32  int main(void)
33 {
34     print_string x;
35     int a;
36     a=x.get_n();
37     x.recursion(a);
38     return 0;
39 }

Looking at a code written in C language:

 1 # include <stdio.h>
 2 
 3 int main()
 4 
 5 {
 6 
 7     int i,j;
 8 
 9     char c[50][1000];
10 
11     int n;
12 
13     char cc='A';
14 
15     int count=1;
16 
17     int temp;
18 
19     scanf("%d", &n);
20 
21     c[1][1] = 'A';
22 
23     c[1][2] = '\0';
24 
25     for (i=2; i<=n; i++)
26 
27     {
28 
29         temp = count;
30 
31         count = count*2+1;
32 
33         for (j=1; c[i-1][j]!='\0'; j++)
34 
35         {
36 
37             c[i][j] = c[i-1][j];
38 
39             c[i][j+temp+1] = c[i-1][j];
40 
41             
42 
43         }
44 
45         c[i][temp+1] = ++cc;
46 
47         c[i][j+temp+1] = '\0';
48 
49     }   
50 
51     
52 
53     for (i=1; c[n][i]!='\0'; i++)
54 
55         printf("%c", c[n][i]);
56 
57     printf("\n");
58 
59             
60 
61     return 0;
62 
63 }

Original link: https: //blog.csdn.net/a237653639/article/details/21323641

Implemented in C ++: FJ string Print

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