1 like 0 dislike
2.3k views
請設計一程式,從使用者輸入的正整數中找出中位數輸出。使用者輸入的第一個數字N代表接下來會有多少個數字會被輸入,程式會從接下來的N個數中找出中位數輸出。輸出時請注意小數點尾巴的零不要輸出。

sample input:

4

1 2 3 4

sample output

2.5
[Exercise] Coding (C) - asked in 2016-1 程式設計(一)AC by (18k points)
ID: 15440 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00

reshown by | 2.3k views

18 Answers

0 like 0 dislike
Hidden content!
*** *** * * *



int main()

{
** * * ** * ** * * * *** math[3], a, i;
**** * ** *** ***** ** * ***** sum, ava;


*** * * *** * * * * * *** *** ** ***** * ** &a);
* **** * * **** * **** = 0;i < 4;i ++)
*** * *** * ***** ** * ** ***** * ** &math[i]);


* ** ** * ** * * * ** = 0;i < 4;i ++)
* **** * * * * ** * ** * **** * * *** += math[i];
* * ** ** ** ** * ** * * * *** ** * ***** * * * = sum / 4;


** * * * * * * *** **** **** ** **** * **** ** *


** ** * * *** * ** * ** * * ***** ** * * * ******* 0;

}
answered by (-114 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

int main()

{
****** * * ****** * * a,i;
***** * * *** * * * ** * ** * * *****
* ** * * * * * ***** ** m,j,k;


****** **** *** *** **** * math[100000];
* ***** * **** ** * * *************** **
* *** * * **** *** ****** ** ** ** * * ****


** * * * ** * *** * (a%2==0)

    {
* * * * * * *** * * * * ** * * * *
* * ** * ** * * * **** *** * *** ***** * ** ** *
** * * * ** * * * **** ** ** **** ** * *** ** *
** * *** * ** ** * ** ** * ** * * * * ** * ***** ** * ****

    }
* * *** ** ** ** * * if(a%2!=0)
** * * ** ** ** * * ** * * * ** *** ** * ******

}
answered by (-124 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

int main()

{
*** * * * *** *** *** ** a,i;
** * ***** * * ** ** * * ** * * * ** **
* **** ** * ******* * *** m,j,k;


** * * * ** * ** * math[100000];
***** ** ** * * *** **** * * ** ***
* * * ** * *** * ** * * * * * * * * ** ** ** * ** * ***


** * * * *** * *** ** (a%2==0)

    {
***** ** *** * ** * **** ** * ** * **** * ** *
** ** * **** *** ****** ** * * * * ** ****** * **
*** * * *** *** * * **** * * ** ** * * * * * ** *********
* * * ** * ****** * **** * ** ** * * **** *** **** ** * *
* * * * * ***** **** ***
** **** * * * * * * * if(a%2!=0)

    {    m=math[a/2];
*** * ** ** ** * * * ****** * * * * * * *** ** * ********* * * **** ** * ***

}

}
answered by (-124 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

int main()

{
*** ** *** ** ** ** * * * a,i;
* ********* * * * * * * * **** * * ** * *** **
* **** * ** ** *** m,j,k;


** ******** *** * ** ** math[100000];
* **** * ** * *** * * ** *** *
* * * * *** * ** **** * * ** ** * *****


***** * * ** *** * * ***** (a%2==0)

    {
* * *** ***** *** *** ** ** ** *** * * **** ***
* ***** **** * ** *** * ** * * * ** * **** *** ***
* *** ** ** * **** ** * * * * * * * ** * **
* * ** * ** * ******* * * ** *** * ** * * * ***** ** * ** *** **

    }
*** * ****** *** **** * ** ** if(a%2!=0)
* *** *** ** *** * * * * *** * ****** * ** ** * ** *

}
answered by (-124 points)
0 like 0 dislike
Hidden content!
#include ** * * ** ** **

#include * *** ***

#include ** ** * * **

int main()

{
** * * ** * ** * * c;
*** ** * ** * * * * * a;
* **** * ** * ** *** * ****** ** ****
**** * * ** * ***** * b[a],i,j;




*** ***** * ** * * * * (i=0;i<a;i++)
* ** * ** * * * *** * ***** * * *** *
** ** * * ***** * * * (j=0;j<a;j++)
** ** ********** **** * *** = c+b[j];


* * *** * = c/a;


* ** ** * **** ** * * * * ***** * * *
* **** ** ** * *** **** ** 0;

}
answered by (-136 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>



int main()

{
*** * * * **** *** N;
* *** *** **** * *** ** * * **** * *** *
***** ********* * * * num[N],i;
* *** * * ** ** ***** * * * ave,sum=0;
* * * ** * *** * * (i=0; i<N; i++)
******** ** * * *** **
*** * ** * * ** * ** ***** ** ** *** * * *** * * * * ** **
*** ** * * * ** ***
* *** * * * ** ** * (i=0; i<N; i++)
*** *** ** ** * * * ****
*** * **** **** * * *** **** * * *** * ** = sum + num[i];
* ** ** ** ***
** ***** ** **** * ** = sum / N;
* * ** ** * ** * *** ** ** ******



}
answered by (-162 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>



int main()

{
* * ******** ** * * ** * N;
* * **** * * * * *** * * ** * * ******
* * * ** ** ** * * num[N],i;
* * * * * ** * * * * ave,sum=0;
** *** * * * ****** (i=0; i<N; i++)
* * **** * ** **
* **** *** * * * * ** * *** * *** **** ** ** * ** ** *** *** * *
* * ** * **
* ** * * *** * ** ** (i=0; i<N; i++)
**** *** * * * * * * *
**** ***** ** ******* * *** * *** ** = sum + num[i];
***** * * * *****
** * * * *** * = sum / N;
**** ** * * ** * * * * ** * ** *



}
answered by (-162 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

#include * * ** * *

int main()

{
* ** *** * ** * ** * n,anynum,p,sum,ans;
**** * *** **** * ******** * * ** *** ** *


* * ** *** ** * * (p = 0;p < n;p ++)
*** * *** * *** * ***** ** *
* ** * *** ** * ** *** * ** * ***** * **** * ** * **
* ********* * * ** * * * * * * * * * ** * += anynum;


***** **** ** * **
* *** **** ** * * = sum / n;
*** ** ***** ***** ***** *** ** ***


* ** * ** * 0;

}
answered by (-320 points)
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.69.59.98
©2016-2025

Related questions

0 like 0 dislike
0 answers
[Resource] asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15475 - Available when: Unlimited - Due to: Unlimited
| 13 views
0 like 0 dislike
5 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15447 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 1.3k views
1 like 0 dislike
37 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15446 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 4.4k views
1 like 0 dislike
17 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15442 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 2.3k views
0 like 0 dislike
112 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15426 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 10.3k views
12,783 questions
183,442 answers
172,219 comments
4,824 users