2 like 0 dislike
274 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
| 274 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:172.69.58.147
©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
| 663 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
| 412 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
| 217 views
12,783 questions
183,443 answers
172,219 comments
4,824 users