2 like 0 dislike
933 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
| 933 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:172.69.59.22
©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
| 2.4k 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
| 1.6k 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
| 1.2k views
12,783 questions
183,442 answers
172,219 comments
4,824 users