2 like 0 dislike
1.2k 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
| 1.2k 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:172.71.254.192
©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
| 3k 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
| 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
| 1.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users