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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>
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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
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
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}
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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}
answered
Dec 18, 2016
by
Jim870729
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-102
points)
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Hidden content!#include<stdio.h>
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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 )
{
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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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* * * * * ***
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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 * * * * *
#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++)
* ** *
* ** * *
* * ** ** * * * = s[i];
* * * * **
*
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* * * *
* * * ***
* * * * = 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++)
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** ** * * * * * ****
** ** * *** * * * ** *** ****
* ** * * * * * * * * * * * * ** (p2[j+u] == f2[u])
** ** * ** *** * ** * * * *
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* ** * * * ** ***** * * * * *
** * * * * ** *** * ** ** ** ** *
*** **** * * * * * * * ** * * *** *
*** ** * * ** ** *** * ** *
* * * ** * **** * * * ****** ** ** *
* ** ** *** ** * * * * *** ** = 0;
** * ** ** ** * * * * * ** * **
* * * * * *** ** * * ** * * *
* * * ** **** ** * * * * * *
* * * ** ***** * **** *
*** *** ** * * * * *** * * **
* *** ** *** ** *** *
** * ** * * *** * *
* * * ** *** * *
** ** * * * * * * *
******
** * **** * ** ** * ** ** * (d > 0)
* * * *** * * * * * * * ** *
* * * ** * ** * * * * * * *** * * is * ** * * * * in * ** * * *****
** **** * * ** * ** * *** * * * * *
* * * *** ** **** * ** * ***
* * ** * ** * ** **** *** *
* * * * ** * ** ** ** **** ** * ** * are no * * *** ** in *** * * * * * * * * ** * *
* * * * * ** ** ** ** 0;
** * * * * * ** ** *
}
answered
Dec 10, 2016
by
蘇瑋朝
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