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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
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410523016
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answered
Dec 18, 2016
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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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-122
points)
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Hidden content!#include <stdio.h>
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#include <ctype.h>
int main()
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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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-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
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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
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
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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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** ** * * * * * ** ** * * **
** * * ** * ** * * * * * * *** ** * * *** * * ** * * ** 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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* *
** *** *** *
* * * * ** * **** * *** ** * * ** * ++;
* * ** * * * **** * *** * *
}
* * ** * * * * * * * (i = 0;t[i] != '\0';i++)
*
*** **** * ** ** * ** **** *** * ** ** * *
* *** * *** * ++;
* * * * * *
}
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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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