2 like 0 dislike
794 views
請依下列的定義完成函數。

 

#include <stdio.h>

 

///

///函數名稱 count

///函數功能 計算字串當中出現特定子字串的次數

///參數說明

///str 要被搜尋的字串

///key 子字串

///

int count(const char str[], const char key[]);

 

int main()

{

   char str[1000], key[1000];

   gets(str);

   gets(key);

   printf(“子字串共出現 %d 次”, count(str, key));

   return 0;

}

 

int count(const char str[], const char key[])

{

 

}
[Normal] Essay (Open question) - asked in 2017-1 程式設計(一)AD by (30k points)
ID: 38147 - Available when: Unlimited - Due to: Unlimited
| 794 views

5 Answers

0 like 0 dislike
Hidden content!
#include ** ** ** ****
#include * ** ** ****

int **** *** char str[], const char key[]);

int main()
{
* ***** * * * *** str[1000], key[1000];
* * * ** *** * ***
** * ** ** ** *** ** ** **
***** * * * ** ** * **** %d 次", count(str, key));
***** * ** *** *** 0;
}

int ** * char str[], const char key[])
{
** * * *** * * **** * * ** ** * ***
** ** * * ** * ******** * * ** ** * key);
* * ** ** ** ** *
* *** *** * ** ** *
* * * * ** * * ** * ** * ** ** * *** *** *
* **** * * **** *** * * ** **** ** ***
* ** * ** *** ** *** * **** * *** * ** * * key);
** * **** ** * *****
** ** ** ** **** * * count;
}
answered
0 like 0 dislike
Hidden content!
#include ** ** * *

#include ** ** ** *



int ***** char str[], const char key[]);



int main()

{
**** * * ** **** str[1000], key[1000];
* * ** * * *****
** *** ** * * *
* * *********** ** * ** * *** ** %d * * count(str, key));
** **** * ** * ***** 0;

}



int count(const char str[], const char key[])

{
** **** * * * * * * **** *** ****
*** * ** *** * ** * *** ** key);
* * * * *** * * * * * ****
* * ** ***** * ***
* ******* ** ************ * * * ** *** *
* ** ***** * * **** * * *** *** *** ** **
* * * ***** ** * * * * * * *** *** ** * ** * key);
*** *** * * ** *** ****
** *** * ** *** ****** count;

}
answered by (190 points)
0 like 0 dislike
Hidden content!
#include *** ** * **
#include ** ** **

int * * * char str[], const char key[]);

int main()
{
* ** ** ** ** * ** str[1000], key[1000];
  * ** ** ***
  ** ****** **
****** * * * * *** * *** * *** * %d * * ** **** key));
* * * * * * * 0;
}

int ** * char str[], const char key[])
{
* * * ******* * * * ** ** * ** * *** * **
***** * ** * ** * ** ** key);
* * * * ** * ** * ** ** * ***
** * * * * * **
** ** ***** ** ** * *** * * ** ** * *** * * * *
* ** * * * **** *** ** ****** ** * ** * * ** *
* ** ** *** ** * *** * ** * * * * * *** * *** key);
* * **** * ** * ****
* * * ** * * * * count;
}
answered by (174 points)
edited by
0 like 0 dislike
Hidden content!
* * * ** * ***



int char ** ** * ***



int *



{


*** * * **** ** * * * *


*** ** * * ** ** *


*** *** * ** *


*** ** * *** * * * * * * *** **** * %d *** * *** ***


** * ****** * ** 0;



}



int ** * char * ** ** **



{



int x=0;


* * *p;



do



{


** * *** * ** * * * ** ** * *


* * * * * * * *** != NULL)



{


*** ******* ***** * *** *** * ****


** * * * ** * * * *** *** *** * ** *



}



}


* * *


** x;



}
answered by (160 points)
0 like 0 dislike
Hidden content!
#include * **** * **
#include ** ** ** * ***

int char str[], const char key[]);

int main()
{
* * ** ** ** *** * * str[1000], key[1000];
******* * *** * * *** *
* * *** ***** ** *
* ** * ***** * * * ** **** * %d 次", count(str, key));
** * *** * * *** * 0;
}

int **** char str[], const char key[])
{
*** **** ** ** *** * * ***** *
*** * ** * ** ** * *** ** ** **** key);
* ** ** * ** * **** * * * ** *
* * *** ** ****
* * *** * ** **** * ** * **** ** ** * **** *
** * * * *** * ** ** * ** ** *** *
* ** ** * ** *** * * ** * ** * ** * *** * *** key);
*** * * ******* ** *
* * *** * * ** ** ***** count;
}
answered by (30k points)
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.70.100.189
©2016-2025

Related questions

2 like 0 dislike
8 answers
[Normal] Coding (C) - asked Dec 20, 2017 in 2017-1 程式設計(一)AD by 楊修俊 (30k points)
ID: 38155 - Available when: Unlimited - Due to: Unlimited
| 2k views
2 like 0 dislike
6 answers
[Normal] Coding (C) - asked Dec 20, 2017 in 2017-1 程式設計(一)AD by 楊修俊 (30k points)
ID: 38154 - Available when: Unlimited - Due to: Unlimited
| 1.3k views
8 like 0 dislike
3 answers
[Exam] asked Dec 8, 2017 in 2017-1 程式設計(一)AD by Shun-Po (18k points)
ID: 36175 - Available when: 2017-12-08 18:30 - Due to: 2017-12-08 21:00
| 1k views
12,783 questions
183,442 answers
172,219 comments
4,824 users