根据下面的文字，我如何编码一个只以大写字母写成的文本的程序？

Question

  1 2 3 4 5 6  
1 A B C D E F
2 G H I J K L
3 M N O P R S
4 T U V Y Z W
5 X 1 2 3 4 5
6 6 7 8 9 ? !
7 - + * 

我想这么做。

Hello1234等效-> 221526263352535455

我得到的是输出。

636465

我从一个简单的逻辑开始，但不能继续。

我的密码。

#include <stdio.h>
#include <string.h>

int main(){
    int i,j;
    
  
    char string[20];
  char b[7][6]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','R','S','T','U'
  ,'V','Y','Z','W','X','1','2','3','4','5','6','7','8','9','?','!','-','+','*'};
  printf("please enter an input");
  scanf("%s",&string);
  
  
  for(i=0;i<7;i++){
    for(j=0;j<6;j++){
        if(string[i]==b[i][j])
            printf("%d%d",i,j);
            
      }
  }
  
  return 0;
}

Answer 1

您可以迭代string字符，并检查当前字符是否在b中找到。如果找到了，你只需打印row+1和column+1

#include <stdio.h>
#include <string.h>

char b[7][6]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','R','S','T','U'
  ,'V','Y','Z','W','X','1','2','3','4','5','6','7','8','9','?','!','-','+','*'};

void findRowCol(char c,int *row,int *col){// change row,col to the founded index, if not found to -1
     *row=-1;
     *col=-1;
     for(int i=0;i<7;i++){
         for(int j=0;j<6;j++){
            if(c==b[i][j]){
              *row=i;
              *col=j;
          }
       }
    }
}

int main()
{
    char string[20] = "HELLO1234";

    for(int i=0;i<strlen(string);i++){
       int row,col;
       findRowCol(string[i],&row,&col);
       if(row!=-1) printf("%d%d",row+1,col+1);   
    }
}

Answer 2

我创建了一个由256个元素组成的char数组，以解释所有的ASCII字符。我将字符的对应值存储在数组中。例如:a->11，alpha_num['A'] = 11。类似地，我分配了所有剩下的字符。

#include <stdio.h>
#include <string.h>

int main()
{
    int i = 1, j = 1;
    int k = 0;
    char alpha_num[256];
    char string[20];
    printf("Please enter an input\n");
    scanf("%s", string);

    for (char c = 'A'; c <= 'Z'; c++)
    {
        if (j <= 6 && i <= 7)
        {
            alpha_num[c] = i * 10 + j;
            j++;
        }
        if (j > 6)
        {
            j = 1;
            i++;
        }

        if (i > 7)
            i = 1;
    }

    i = 5;
    j = 2;
    for (char c = '1'; c <= '9'; c++)
    {
        if (j <= 6 && i <= 7)
        {
            alpha_num[c] = i * 10 + j;
            j++;
        }
        if (j > 6)
        {
            j = 1;
            i++;
        }

        if (i > 7)
            i = 1;
    }

    alpha_num['?'] = 65;
    alpha_num['!'] = 66;
    alpha_num['-'] = 71;
    alpha_num['+'] = 72;
    alpha_num['*'] = 73;

    for (int i = 0; i < strlen(string); i++)
    {
        printf("%d", alpha_num[string[i]]);
    }

    return 0;
}

产出如下：

Please enter an input
HELLO1234
221526263352535455

Answer 3

我追求的是简洁和简单：

(IDEOne链路)

#include <stdio.h>
#include <string.h>

void encode(char* input, char* output, char* matrix)
{
    while(*input)
    {
        size_t idx = strchr(matrix, *input++) - matrix;
        *output++ = '1' + idx / 6;
        *output++ = '1' + idx % 6;
    }
    *output = 0;
}

int main(void) {
    char matrix[]="ABCDEFGHIJKLMNOPRSTUVYZWX123456789?!-+* ";
    char input[50];
    
    printf("please enter an input");
    fgets(input, 50, stdin);
    *strchr(input, '\n') = '\0';
    
    size_t len = strlen(input);
    char output[2*len +1];
    
    encode(input, output, matrix);
    printf("\nEncoded String: %s\n", output);
    return 0;
}