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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
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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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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>
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{
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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
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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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* * ** * *** ** * ****** * **** * * * * ** * * * = c1;
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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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* * * * * ** * * * *
* * *** ** * * * * * * * * = strstr(a,b);
* ** ** * * ** * * * * ** * *
* * ** ** * * * ** * ** * ** ** * * * ** * * *** * * 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];
* *** ** *
* * * * *
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*** ** **
* * ** *** *
* ** ** ** ** = t[i];
* * ** * * * * *
** ** *
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** *** * (i = 0;i <= sizes;i++)
{
* * ** *** *
** * * * *
* * ** * * = p[i] - 32;
** * * *
*** * * *
* ** *** ** * = p[i];
}
*** * * * * * * * *** *
for (i = 0;i <= sizet;i++)
{
* *** **** ** *
** * * * * ** ****
* * ** * = f[i] - 32;
* * * * *
** * * *
** ** * * * * = f[i];
* ** * * *** *
}
*** * **** * * ** ** ** **
//asplpSLP slp
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** * * *
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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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