1-11 Exercise 3: Prefix of a string

0 like 0 dislike
2.4k views

A string p is a prefix of a string s if the leading characters of s are identical to p. For example, "abc" is a prefix of "abcde". Develop a function prefix that checks if a C string p is a prefix of another C string s.

p是s的前綴若s開頭的一段字串等於p。例如，"abc"是"abcde"的前綴。寫一個程式 判斷字串p是否字串s的前綴。

Example input 1:

abc
abcde

Example output 1:

abc is a prefix of abcde

 

Example input 2:

bc
abcde

Example output 2:

bc is not a prefix of abcde

 

[Exercise] Coding (C) - asked in Chapter 13: Strings by (12.1k points)
ID: 41439 - Available when: Unlimited - Due to: Unlimited
edited by | 2.4k views
0 0
12345
commented by (-329 points)

27 Answers

0 like 0 dislike 
Hidden content!
#include<stdio.h>
#include<string.h>
char prefix(char string1[],char string2[])
{
    int i,s;
 *  * **** * **     *** *** * *
*  *   ***** *  * ***  *  * *  ***  *  *   * *   * **    
    {
 *      **** *  * * * * * *  *** * *   *** **     *    ** **   (i=0; i<=strlen(string2)-1;i++)
 ***     * *       **     *  *** *** *  ***  **  *  ****   
*****  **   **           * * * ***  *********  *  *  **   ** ** ** * ***    ****  * * (string2[i]==string1[i])
**  *  *  ** *    ***** ****   **   * * * * *** * ** * **  **   *  ** * *    * 
 ***  ***  **  * **  **    * ** ****  **   *****  **     * * *  *  **** **   *** ** ******   * * *** **  *
*  * ***  *    ****  **    *   *   ****    *  * * **  *  *  *    ** * *  **   ** *
*  ***  ** * * *** * *   *     *    **** ** *    *  **** 
***    *  ** **** ****  *  * *    * ***  * *  * * * *
 * * ** *** ***** * **  ***** **  **   *******  *  * 
  **  **   * *  * * * *     **   * * *   ** ** *  * ** ** ** *** * **** *   *  *   **   **** * is a prefix of %s", string2, string1);
* **        * *** *   **   *   *    ******* *  *   ** ** 
 **** *      *  **  *  * * ** *  ** * *** *   * * * ***  * 
**    * * **   * ** * ** ****   * ** *** *  *   * * **  * **
   * *** *  **  *  **  *    * *  *  ** ** **     *  * *  ** *    ** * ***  ** ***  * ***  ***     is not a prefix of %s", string2, string1);
   ***   *  *  *  *** *****   * ** * * * * ***** *
    }
 ****  *    ****  *   ********* * ***    *** ***  * *   
    {
  *** ***   **     ****     **   ****    * * ****    * **    (i=0; i<=strlen(string1)-1;i++)
  * * * **   * **  ** *   *** * ***  **   *** *   
*    ******* * *** * *        ** ** *  ***     *** * *   *** ****  **    **   **   *  * (string2[i]==string1[i])
**  * * *   *   **** * * * *  * * **  *   *****  *   * ** *  **  * * *** 
**    **   *** ***    * *  ***   ** * ****  *    ****       ** ** *    ** *   *    * *  *        ****  *    *****   ** * *
**    **   * *** ***        **  ***   ** *  *****  ** * ** *   *  *** *  * *
** ******   **   *** *** * ***  ** *   **  **  * ** 
* ***   **      * **   **  **   * **  ** **  *  *       * **  *  
*  *  ***** *****    **     *    ****   *  * *     * * *      
 ***   *  *  **  ***  * ***   *** *  * ** *** * ** **      *    *      *   **** ** **  *  *  * ** is a prefix of %s", string1, string2);
      **  ***** ** *  *** ***  **    ****      * ****  
**  * * *  * *****  ***  * ** ** * *  **  **  * *      ***
**  **   **    *  ** *     ** **** **    * *       
 **             *         *** **         ***  * *  *   ***      * *   *** *  *  * * * * * is not a prefix of %s", string1, string2);
 * *  *  *  **** ** *** **  ** **   *  * * *      
    }
}
int main (void)
{
    char chaine1[1001], chaine2[1001];
  * *****  ** ***   *****   *   *    *   %s", chaine1,chaine2);
       *  *     * * *** **  * *   ** * 
 * *  **            *** 0;
}
answered by (-168 points)
edited by
0 0
Case 0: Correct output
Case 1: Wrong output
Case 2: Correct output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Wrong output
Case 2: Correct output
commented by (4.3k points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>

#include <string.h>

int main()
{
   ***   * * *  **  *  ** *** * lenght1,lenght2,c;
    *****   *       * * *   *****  b[100];
*   **    * ** * ****  * *  a[100];
**  ** *  * ***    *** *  *     **  d[100];
  **  * ** *     ****  ****** **  **  **    *** *      
 * * * ** * * *** ***   * ** **  ** *** **  *   * 
*  ** *  **   *   *   ****    * *
   *   ****    * * *  * ***    *** * 
 *  **  **  * **   **    ******* * *   **  
 ** *  *****     *    ***** *
*  ***  ** *     ********** **  *  * **   **    **** is a prefix of %s",a,b);
  * *  *   *  *  *  *  *** 
 *  *    *   ** ** ** * ** ****
   ****  * ********* * *      
 ***   **    *** ***** **   *  * * ** *  *** *** ** * * ** * *   * is not a prefix of %s",a,b);
* **  * ***  * **   * **** ** **
  **   * *  * ****** *   0;
}
answered by (-140 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>
#include *    *   *  *   ***
#include       * * *  
//#include <ctype.h>
#define n 100
int aa(char A[],char B[])
{
   **    *  **    i;
 *  ****   **  *  *  * * *** * 
  {
 *** *   **   * * *  **  ***** * **   * *   A[i]!='\0' && A[i]!=B[i]   )
*  ****  *  ****** ** * * ****   **     *   *  *  ***** * * ** ****  *       0;
  ****    *  **    * * ** *  *** *    ***  * * **** *
  **   *  ***  *** **  ****  *** **  ***  ** *   ** *** **   ***   **
  }
* * **** * ** *  1;
}


int main()
{
   *** **** *   * *  ** **  a[n],b[n],c[n];
** ** * ****  * **    ***  * * i;
*** ** ** * *  *   * * * * *  * ******   *  * ***
* * **  *  *  ***    * ** *     **         * *    ** 
 **** ** *  *  * ** **  *  (aa(a,b)){
  *     * ** *   * ****   **  **  ****     *****    *     ****** * *** 
   ******  ** ** * *********** * *  *     *    * **  ** * ** **     *  * **   *****  *
* ** **      *****   * * *****  * *     **    ***   *  *       *  ****  * * *** **** *      * ** ****  *  
 * ** ***** *  ** *  *   **  ***** *   *  ***** * * ** *  * * *******        *  *  *** *

  **** **      *   *  ********   ***  ******  *****   
 * * *** *   *** *** * ****   *  *  ****  * *  *  *** *** * * * is a prefix of %s",c,b);
* * ***   *       **  *** * 
*   * * *      ** ** **  *** * 
* **** ** *   *    *** *      * * * *  **  *** **   ** **** *  is not a prefix of %s",a,b);
  * *  * ** **  *** *   **      0;
}
answered by (-32 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include  * *  * *   **
** *     * ** ***    *   * 
int main (){
char * *** **
char *  * * * *
* * *   **  *    string1: * *** **
* ** *** **  **** *     **  ** ** *
* *   *   *  *  string2:    * ***
*  **   ** * **     ** * 
 *  **  ** ** ****  * 
  * *****  ******* * *  **** *     *    is *** * * *    * of   *  *     * * * * ** *
else
 * ** * ** *    *     ****  * ****** *** is not prefix of **      * ***    ** *  *

return 0;
}
answered by (16 points)
0 0
prog.c: In function 'main':
prog.c:7:9: warning: format '%s' expects argument of type 'char *', but argument 2 has type 'char (*)[100]' [-Wformat=]
 scanf("%s",&string1);
         ^
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include  *   **  * **    
#include *        *******
#include    *   **** * 
#include *** *     * *** 
int prefix(char p[],char s[])
{
  ***   ****   *  *   * **  *** **     * *
 **  **** *   *   ***  *** * *  0;
*    **    * ***  ** *** if ** ** ******   
***   ** * * *  ******  ** ** * 1;
*    * * ***  *** ***** ***
 ***  **  *  * * ****** ** **** 0;
}

int main()
{
*   ** ** **  * * * p[999],s[999];
 ****  ** ***  ** *  *    **  *    ***   *  **
 ******* * * * ** * **   *   *  * 

 * ** ****  **   * ***  **   
 *   *   * *** *  **  *** *  * * * **** **   is a prefix of %s",p,s);
   ****  *****    **** 
* * * *    ** **** *   * *    ** *  *   **** is not a prefix of *  **  * *   

*** ** ** *     **  * *  0;
}
answered by (-16 points)
edited by
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>
#include   * *     **
int main()
{
 ***    *  *  **   * ***   * *  length1,length2,c;
 *  **   **  *   **  ** *  a[100];
* **  *  **  ** *  *   * **    * * b[100];
    * *   * *   *  ****   **  d[100];
**** *  ***     *** **  *****   * ** ** ***  *  
 *** **  * ** **  * * *   ** **    * *** *   *   
   * ******  ****    ** ****   * **
**** *   * * * * * ** **   ***   *** *   ** 
if (strcmp(a,d)==0)
  **  * *  ***     * ***  *  
* *** *  **  *    *** * *  ***  ** * **     ***   *         ** *  * * is a prefix of %s",a,b);
    *       *  ***   *  * *
 **  *  ****  *  *   * * * * * 
****  * *  *     ** **  *
* **** * *  *   * *   ** **  *   *** * **    * *    * *         * ****** is not a prefix of %s",a,b);
*  ** * ****** ******      *
   * **   **  * 0;
}
answered by (-229 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include ****   ** **** 
#include   **** * **  *   

int prefix (char *       str2[])
{
* **   **  *  ***** *  *  *   length=0,  *   **
  *   ** ***** * *   
**   *  * ****     **  **    **  =   *    **  * * *
      * * *   *** **** *** 
  ** *  ** * *  ** ** **** ****  * *   * * ** 
* *   *   ***   *  * ** * *     * *      *** * ***
*   *          ***      * if ** **   ** *  *
 ** ****       ** *  ** *        *  ** *** *   **   *           
      *  ****  *  * * ** **   *   }
 ***  ***             ** * *  **** **
**   **  *** **   *   * (count == length)
** * * **   *     *  ***  * *    *  ***  * 1;
  *  * **   *   * ** ***  else  *** ** 0;   **** *   *** **  *
* **      *  *    *  *    ***** 
}
int main(void)
{
*  *  *   **  ** **  * * *      *   *
 * *** * ** * * * *    * ***
     * *     ** * * ***  * * * *  ***    **  ****** ** str1, str2);

*   * ** ******  **** *  ** *  *  ***   ***
 ** *   *  *  {
 ****** *  *   *** *  **  * * **  ** ** * *** * * *      *** * * * *  is a prefix of ******* str1, str2);
*  ****** *  ****  *
 * ***   *  *** * *    ***  **
 * *  * ***     *    *   * ** * ** *    *  * * *  * * ** *  **   *  ***** is not a prefix of ******   str1, str2);

*  **   * * *  ** ***        ** * 0;
}
answered by (-304 points)
edited by
0 0
Case 0: Wrong output
Case 1: Correct output
Case 2: Wrong output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include    * *  *  **** 
#include  *     * *  ** *  
#include * * * *   *  *    

int main()
{
**        **     * **  * **   *  prefix[100];
**  ** * *** ** **** * * * word[100];
   **** **      *   * **  * ** *  ***** %s", prefix, word);
* * * *  *  *  ** *  **   i;
  *** * **   * * *     *    = 0; i< strlen(prefix); ++i)
  ** ***   *  **  ***   * ***
* ***  * ***   * ***        **  ****  * *     * * ****** * * ***
***********   *  * **    ***   * *****  * *** * *  ** ** *
   **  **** **   *  **  **** * **** * **  * * *  **** ****** * * **  *  *     *  ** *  * *** **  is not a prefix of %s", prefix, word);
 ** ***      *** *           **  ***  *****    * * **  * *    *** **** *** * *  * 0;
 **   * ** **     ****  *** **    *  * *** ***  *  ** 
  *** * ** **  *   *   *  *
* *******       **** *** ****    * is a prefix of %s", prefix, word);
*    *** *  *  * *     ***   * 0;
}
answered by (-108 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include *   ** ***   *  *
#include  **      *   ** ** *
#include *  *  *   ****

int main()
{
 *  *** *****  *** *  ** *****  prefix[100];
 **     *  **** ****** *  word[100];
  *     **   *  **  **** ** * ***** **  %s", prefix, word);
* **** * *    ** ***   ** * ****  i;
*  *  ** * *   **      * **  *      = 0; i< strlen(prefix); ++i)
*** * *** * ***** * * ****  * 
*****  **   ** ** ** ****  * * **   *****  **** *** *  **  **  
  * *   ** ****   *  ** **  ****  *   * * *** ****
  ***     ***    * *   ** ***** **  ***  **      **  *  ***    ** ** **** *** ***  * **  **** is not a prefix of %s", prefix, word);
* * **  *   *  ** ***** ** ** * *  ** ***  *** ** **     ** * *** ***  ** *** **  * *** 0;
**  **  *  *** **** **  *  * ****   *  ****  * *  ****  **
* ** *  ** *  **   * ** ** ***
   **   *     ***  **   * ** * ***  *  is a prefix of %s", prefix, word);
*** * **   *** ** * * **   *****  0;
}
answered by (-258 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
*   *** *** **     **  **
*** **  **   **   **    
 *****  *  * *  * ** 

int main()
{
  *      **    char   * * 
 *    ***  **** *  char  * ** ** 
 *  *  * * *    ***  **     **  ** *  *  *   word);
* * ***    ****   * int i;
* **  * **    ** for(i = 0; *****     **   *  ++i)
**  *    ***  *  {
   ** * *** *   *     * *  **  * * * **  * *
 **    *** *  **** * ****  *  {
*****       **   *  *  *     * *   **  *  * **  *** *** * ** *    is not a * * of  * *  ***** * ** word);
*       ***** * *  *  ** ** ** **   *****  * **   * *  0;
 **   * *  ***  *  *  ******* * }
** *     *  * * }
    *  ** ****  **  *** ***** is a prefix of * **   * *  *   word);
   *  *  *  **** *   * 0;
}
answered by (-249 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
Page:
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.71.254.173
©2016-2024

Related questions

0 like 0 dislike
57 answers
1-11 Exercise 1: UPPER and lower letters
[Exercise] Coding (C) - asked Jan 11, 2018 in Chapter 13: Strings
ID: 41431 - Available when: Unlimited - Due to: Unlimited | 3.1k views
0 like 0 dislike
29 answers
1-11 Exercise 2: Reverse a sentence
[Exercise] Coding (C) - asked Jan 11, 2018 in Chapter 13: Strings by thopd (12.1k points)
ID: 41436 - Available when: Unlimited - Due to: Unlimited | 2.1k views
0 like 0 dislike
53 answers
1-4 Exercise 3: String details
[Exercise] Coding (C) - asked Jan 4, 2018 in Chapter 13: Strings by semicolon (5.2k points)
ID: 40715 - Available when: 2018-01-04 18:00 - Due to: Unlimited | 2.9k views
12,783 questions
183,443 answers
172,219 comments
4,824 users