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