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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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points)
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Hidden content!#include <stdio.h>
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#include <ctype.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
points)
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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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* * ** * * *** * * * **
* * *** ** ** * ******* ** * * *** * * * * * * * *** ** * ** 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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* ** ** ** *** * * * *
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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 *** * *** * *
#include * * * * * *
#include * * * *** * **
#include * * * * *** * *
#include * * * * * *
int main(void){
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* **** * ***
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* *** * * * *** ** * ** * **
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* * * *
** * * * *
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* * * * ** ** ** * * *
}
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** * *
* * * ** * * *** ** * **
* * * * ** * * ++;
* * * * ** *
}
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* * * **** ** * * * *
* * * ** * (i = 0;i <= sizes;i++)
* *
* * ** * * * * *
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* * * ** *
* * *
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* ** * **
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* *** **
* * *
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* * * *** (i = 0;i <= sizes;i++)
{
* ** * *** * *
*** ** * * ** * *
* ** * * ** * ** = p[i] - 32;
** ** * * *
****
* ** ** **** = p[i];
}
* * * * * * * * **
for (i = 0;i <= sizet;i++)
{
** ** * ** *
* * *
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* * * * *** ** * **
** * * *
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* ** * * *
}
* * * **** * * * ** * * **
//asplpSLP slp
* *** * (i = 0;p2[i] != '\0';i++)
*
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* * * *
**** ** ** ** * * *
* * ** * * *** * = 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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