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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>
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{
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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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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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* ** ** * * * *** * ** * * * * 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 )
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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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** * ** * * * * * 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;
** * * * * * * * * * **** *
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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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**** ** * * * * * * * *
* ** (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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