2 like 0 dislike
236 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
| 236 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:172.70.131.3
©2016-2024

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
| 522 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
| 323 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
| 171 views
12,783 questions
183,443 answers
172,219 comments
4,824 users