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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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Hidden content!*** * * * * * *
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answered
Dec 18, 2016
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410523016
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answered
Dec 18, 2016
by
410523016
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-136
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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>
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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>
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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
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Hidden content!#include<stdio.h>
* ****** * * *
#include<ctype.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 )
{
* *** * * * * ****** ** * i,j;
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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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* * ** * * ** * *
* * ** * (i = 0;i <= sizes;i++)
* * *
* * ** *
* * * ** * ** * = s[i];
* * *** **
* ** **
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* * *
* ** * **
* ** * * * * * = t[i];
* * **
** * ** * *
** * * * ** * **** * * ** *** *** *
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{
* * * * * * * *
** * * ** * * ** *
* * * ** ** ** = 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;
* ** * * * * * ** *
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* * * ** ** ** * ***
** * * * * * * * ** *
* ** * * * * * *** ** * * ** * **
* ** ** *** ** * * * * * (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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