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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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Hidden content!#include <stdio.h>
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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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-102
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Hidden content!#include<stdio.h>
* * ** * ** ** *
#include<ctype.h>
int main ( void )
{
* ** * * **** ** * *** ** * i,j;
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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 )
{
** *** * * * ** * * 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
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
**** * * ** * * * * ** *** i,j;
** * * * * *** * * * ** * * * ** * ** ** * **
** *** ** * * **** * *** **** * ** * * ** * *
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** ** ** * * * ** ** ** * **
* * ** * * ** * * * * ******
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* * *** * ** ** * ** *
* * * * * ** * ** ** * * = 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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* ** **
** ** * **
* * *** * * * * * *** * ++;
* * ** * * * * * ** ** *
}
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*
** * * * * *** * * ** * * * *
* * ** *** ** ** * ++;
* * *
}
** ** * * * ** * * * ** * ** ** * * ** ** *** ** *
* * * * * * ** * *
* ** ** * ** (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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