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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
points)
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Hidden content!#include <stdio.h>
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#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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-102
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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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-102
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Hidden content!#include<stdio.h>
** * * * * * ***
#include<ctype.h>
int main ( void )
{
***** ** * **** ** * * * i,j;
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* ** * ** * ** * * ** * *
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** * * * * ** * * ** * *
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** * * * ** * ** * **** * * 0;
}
answered
Dec 18, 2016
by
Jim870729
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-102
points)
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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;
** ** * * * * * ** ** *** *
* * * * *** * ** * *** ** *
* * * ** * * * ***** **
* * ** * *** * * ** * * * ** * * * * * * ** * * * *** = s2[j];
*** * * ** * * ** * * * * *** ** * * * ** = 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<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
* * * * * * ** ** * *** i,j;
* * * * * ** **** ** ** * *** * * * * * * * ***** * ** *
*** * ** * * ** * * * * * * ** * ** *
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* * * * * ** * ** * * * * * * * * * *** * * * * = tolower(c1);
** * * * * * ** * * * *** * * * * *** * = c1;
* ***** * **** ** * * **
** ** * * *** * * ** * * * ** ** ***
** * * * * * *** *
* * ** * ** * * * * * **** * ** * *** * * * * = s2[j];
* * ** * **** * * * * *** * * ** * * = 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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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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