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AC 20161207 exercise 1
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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
Dec 7, 2016
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2016-1 程式設計(一)AC
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Shun-Po
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answered
Dec 18, 2016
by
410523016
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answered
Dec 18, 2016
by
410523016
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Hidden content!#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
int main()
{
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answered
Dec 18, 2016
by
Gulang
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-122
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Hidden content!#include <stdio.h>
#include * * ** * *
#include <ctype.h>
int main()
{
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answered
Dec 18, 2016
by
Gulang
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-122
points)
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
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* * * *** ** ** * 0;
}
answered
Dec 18, 2016
by
Jim870729
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
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** **** * * * ** * ** ** * 0;
}
answered
Dec 18, 2016
by
Jim870729
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Hidden content!#include<stdio.h>
* ** * ** * * ** *****
#include<ctype.h>
int main ( void )
{
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* * * *** **** * * ** 0;
}
answered
Dec 18, 2016
by
Jim870729
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
* * * * * * ***** * * i,j;
** * * * ** * ** * ** * * ** *** * ** * * * *
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* ** * * * * ** ** * ** * * ** ** * * * * = c2;
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** * ** * *** * * * *** * *
* * * ** * * * * * * *** * * * * ** * *** * *** ** are no %s in %s",b,a);
* * * ** * * * *** * *** ***
* * * *** * * * ** ** ** ** *** * * * * ** ** ** is %s in %s",b,a);
** * * * ** * * * * ** * 0;
}
answered
Dec 18, 2016
by
Jim870729
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-102
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
*** * * * ** * **** * * i,j;
**** ** ** * * * ****** * * * ** ** ****
* * **** * *** ** * * * * ** * ** * * **
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*** * * ** * * *** * * * * ** * * ** *** = c1;
* * * * * ** * *** * ***
* * **** ** * ** * ** * * * * * ** ***
* *** **** * * *** * * *
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* *** * * ** * ** * ** * *** * * * * ** **** * = tolower(c2);
*** * * ** * *** * * * * * * ** * * ** * *** * * * * = c2;
* ** * * * ** * **** *
* * * ** ** * *** * * * = strstr(a,b);
****** * ** ** * * * ** * * * * **
* * ** * *** * * * ** ** * **** * * ** ** * ** ***** * * * * are no %s in %s",b,a);
* *** * ** * ** * * * ** * ** *
* * * ** * ** ** ** * * * ** * ** * *** ** * *** * * * * ** is %s in %s",b,a);
*** *** * ** * *** * ** 0;
}
answered
Dec 18, 2016
by
Jim870729
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-102
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Hidden content!#include * * ** * * * *****
#include * * ** *
#include * ** ** *
#include * * * * * * * *
#include ** * * *** * ** **
int main(void){
*** *** * ** * ** * * * ** ** * * * ** * ** * *
** * * * * **
** * *** * * ** * * * *** * * ** *** * *** *
* ** * * * * ** * *** *
* * * * * * * *** *** * * * * *
* * * * *** * ** ** * **
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* * *
* * *
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* ** * *** * **** * * * ** *
}
**** * *** * ** * * * (i = 0;t[i] != '\0';i++)
* * **
** ** *** * * ** * * ** * * * * ***
** * * * * *** ** * ++;
* ** * *** **** **
}
* ** * ** ** ** * ** * * * * * ** * **** * * ** *
* * **** ** * * * ** *
** (i = 0;i <= sizes;i++)
* * * * *
** **** ***
* ** *** ** *** * = s[i];
* * * * * ** *
* * * * *
** *** *** * (i = 0;i <= sizet;i++)
* ** *
** *** *** ***
* ** * * * ** = t[i];
** ** ** ** ** * * *
* *** * *
*** * * * ** ** * *** *** ** * *** * *
* * * (i = 0;i <= sizes;i++)
{
*** ** ** *
* * * ** * * **
* * * * * ** * = p[i] - 32;
* ** * * * *
*** ** *
* * * * **** * ** = p[i];
}
* ** * * * ** * * * **** ** * * * *
for (i = 0;i <= sizet;i++)
{
* * * *
** * **** *
* * * * ** = f[i] - 32;
** * **** *
* * **
* * = f[i];
* ***** * * **
}
** * * ** ** * * *
//asplpSLP slp
* * * * (i = 0;p2[i] != '\0';i++)
** **
* * * * (p2[i] == f2[0])
* *** * * * * * ** *
*** ** * * * ** *
* * *** * * ** * *** * = i;
** ***** * * ** *
* ** * * * * (u = 0;u < sizet;u++)
* * * *
* * * * ** * *** * ** ***
** * * ** *
* * * * * **** * * ******* * (p2[j+u] == f2[u])
** * *** * * * * * * * * * ***
* *** * ** ** * * * += 1;
* ** ** * *** * * * * * *
**** ** ** ** * * * * * * *** ***
* * * * * * * *** **** ****
*** ** * ******* * * * *** ** * *
* ** * * * * * * * * * *** * * ** * * *
** * * * **** * * ** * ** = 0;
** * * *** * * *** ** * * * * * * *
** ** * **** * ** * * * **
* ** ** * * * * * * * ***
* * * * ** *** * * *
* * * ** ** *** * ** ** *
* * * * ** * * ** * * **
*** *** * * * *
** * * * ** ** * ** * * *
* * ** * *
* * * *
* * * * ** ***** (d > 0)
* *** ** * * ** * * * * **
** ** * * * ** * * ** *** is ** * ** * * * * in * * * * * * * * *** *** *
* * *** * * * *** * * *** **** *** **
* ** * * * *** * * * * *** ** ** *
* * * *** * * * * ** ** *
** * ** * * * ** ** * * ** ** * * are no * * * * * * ** in ** *** * *** * ** *** *
*** * ** * * * * * **** * 0;
** * * * * * * * * *
}
answered
Dec 10, 2016
by
蘇瑋朝
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