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子字串

請設計一個程式,使用者會輸入兩個字串,兩個字串間用一個空格隔開,每個字串的內容只會包含大小寫的英文字母,請判斷第二個字串是否出現在第一個字串內。請將大小寫視為同一個字。

輸入範例1:

friEND end

輸出範例1:

There is "end" in "friEND"

輸入範例2:

friEND friends

輸出範例2:

There are no "friends" in "friEND"
asked in 2016-1 程式設計(一)AD by (18k points) | 2k views

23 Answers

0 like 0 dislike
Hidden content!
#include<stdlib.h>

#include<stdio.h>

#include<string.h>

#include<ctype.h>



int main()



{
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* *** ** ** *** * * ch,xh;
** *** * * ** ** *** a[100],c[100];
*** * * * * ***** *** **** * * * b[100],d[100];
* ** ** * ** * ***
* *** * * * * *** *** ((ch = getchar()) != 32) {

*** ** ****** * * ** * * ** ****** *** * * * * * = ch;
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* * * * *** ** ** *
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*** * *** *** * **
**** * ** * * * *
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** * * * ** ** ** *** *

    
**** * ** ** * * * ** **** * * ** *** * **
* * ***** * ** * * 0;

}
answered by (122 points)
0 like 0 dislike
Hidden content!
#include *** ** * **

#include * ** * ** **

#include *** * * ** *

#include * ** ***



int main(void){
* * * ** * ***** **** **** ****
*** * **** i = 0;

  
*** * * * * * * * ************* *

  
** ** * * * *** *** != '\0';i++)
* * *** ******** ** * * **** **** = tolower(c[i]);
* ********* * ** * * = '\0';
*** * * *** *** * * ****** * **
* * * * ** *** * * != '\0';i++)
* * * * *** ***** ** **** = tolower(cc[i]);
** *** * ****** ** = '\0';
* **** * ***
** ***** * ***** ** **** != NULL)
** *** *** * *** ** ** * * * ** is ** ** * * ** in ** ** * * ******* ***
* * * *** ** *
** ** * ** ** * *** * ***** ** ***** * are no * ***** **** in ** *** * ** ****** *
* * *** * **** * ***
** ** * *
**** ** **** * **** * **** **** ** * ****  
* ** ** * 0;

}
answered by (-216 points)
0 like 0 dislike
Hidden content!
#include * * *

#include * *** * *

#include *** * * **

#include ** * ** ***



int main(int argc, char *argv[])

{
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* * *** **** i = 0;
*** *** * * **
** * * ** * * * ** ** * *****
* *** *
** * * ** **** != '\0';i++)
* ** ******** * ** * ***** = tolower(c[i]);
*** ***** * * * ** * * != '\0';i++)
* ****** ** * * **** ** ** ** * *** * * = tolower(cc[i]);
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* * * * ** * * ** ***** *** ** ***** ** * * is * *** ** **** in * * * *** * * * **
******* *
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*** ******** * *** ** * **

  
* * ***** ** * * *** * *** **  
**** *** * * * **** 0;

}
answered by (-216 points)
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