2 like 0 dislike
1.5k 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
| 1.5k 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:104.23.197.94
©2016-2026

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
| 3.8k 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
| 2.6k 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
| 1.9k views
12,783 questions
183,442 answers
172,219 comments
4,824 users