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