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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>
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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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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* * * * * * *** ** 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
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
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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* * * * ** * * * * * ** ** 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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* * * * ** * ** * * * * ** * *** * * * ** * ** = 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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** ** *
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* ** ** *
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***** *** ** ***
* *** *
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* * *
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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
* *** (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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