2 like 0 dislike
854 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
| 854 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:172.71.1.191
©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
| 2.2k 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
| 1.4k 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
| 1.1k views
12,783 questions
183,442 answers
172,219 comments
4,824 users