2 like 0 dislike
733 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
| 733 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:172.70.130.162
©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
| 1.9k 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
| 1.2k 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
| 947 views
12,783 questions
183,442 answers
172,219 comments
4,824 users