12-28 Exercise 2: Prints a table showing how many time each digit appears

Question

Write a program that ask the user to input a serial character (max 1000). Prints a table showing how many time each digit appears in the number. You should see 'A' and 'a' as the same.

寫一個程式 求使用者輸入一字串(最大1000) 輸入字母出現的次數 (A跟a算一樣)

Example input

Write a program that ask the user to input a serial character. Prints a table showing how many time each digit appears in the number. You should see 'A' and 'a' as the same.

Example output

A 19
B 2
C 3
D 3
E 15
F 0
G 3
H 9
I 9
J 0
K 1
L 3
M 5
N 7
O 6
P 5
Q 0
R 10
S 10
T 13
U 5
V 0
W 3
X 0
Y 2
Z 0

 

Answer 1

Hidden content!
#include * *  **    *  ** 
#include * ******* ** *** 

#define MAX_SIZE 1000 //Maximum size of the string

int main()
{
** *   ***     * **   ****   *   string[MAX_SIZE];
* **  *  * *  *   * **     * * i, len;
   ** ** *  *   ** * * * ** frequency[26] = {0};

**    ****** **    *  **  *** * ** ***    *   * * * * string);

  *  *   ** ********* *** = strlen(string);

 *  **    **** * ***** **    *   i<len; i++){
** **   * **  ** *     ******   * ***  * *** ** * *  * * **** * **   *** * * && string[i]<='z'){
*****  * * * ** * * *** * *   ***    *   **** ** *    **  * *** * ** *    ***   *  * * * * - 97]++;
* *  **   **  *****   **** *  ***   ** **** ** *
 **  *   * ***** * **       ***     *   *****   *******      *  *** **  *  ***   &&   *         
  ***   *   * *   *  *  *     **    *   *  ***  ** *     *  ***  *****     *** * ***** *  *** *     *    - 65]++;
***    * *** ** *** * ***    *** **      ** *   ***
*  * ***** *  ****  * * *   * **

*   **    ***   ** ****** ** * i<26; i++)
* ** * * **   * *       ***    **
*    *  *** **    * *** * **  ** ****** **** *   **  <25)
* * *  *** *****      * * * * * ** *  *     *** * **** *  *  * **  * *  ***    **      * *   ***  %d\n", (i+65), frequency[i]);
  **  *  ***  ***     * *   ***  *** * ***  *  ***   *** * *
 **** * * * *** *     * *****   ** *    *   **** * * *  * *** ** **   ** **   ** *****    * %d", (i+65), frequency[i]);
* * *   *  *  *  *  *****   **

 * *  *         ** **** **  0;
}

Comments

Case 0: Wrong output
Case 1: Wrong output

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Case 0: Correct output
Case 1: Correct output