0 like 0 dislike
9.8k views
請設計一程式,從使用者輸入的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。

sample input:

4

1 2 3 4

sample output

2.5
[Exercise] Coding (C) - asked in 2016-1 程式設計(一)AD by (18k points)
ID: 15439 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00

reshown by | 9.8k views

73 Answers

0 like 0 dislike
Hidden content!
#include <stdio.h>

#include <stdlib.h>



int main()

{
* ** * * *** ** ** n = 4, i, j, c;
** * ** **** * ** * k = 0, f;

   int num[n];
*** * ** * ** * **** ** * * &n);
** * ** * *** * * = 0;i <= n-1;i ++)
***** * * ** ** * ** *** * ** ** * **** ** * &num[i]);
* *** * * * * ** = 0;i <= n-1;i ++)
**** ** ** * **** **** * ** ** * * = i+1;j <= n-1;j ++)
* * * * * * *** * * ** * * * * * ***** * *** * > num[j])
* ** * ** *** * ** ** * * * ** ****** * *** *** ********* * * **
* *** * * ** ** *** * *** *** *** *** * * * **** * ** * **** * ** **** ** * * = num[i];
** * ** * * *** ** ** *** ** * * * * * * * ** ** *** ** ***** * * * * ** * * ** * = num[j];
* * * * * ** ** * * * *** * * ** ** ** ** * *** *** * *** * * ** * ** ** * * = c;
** **** * **** ** * ** ******** * *** *** ** *** ** *** * ****** * *
* ** * ******** * ** * != 0)
** **** ** **** ** ** *** * = n/2;
* * * * * * ** * *** f = (n+1)/2;
** ***** * ** ** ****** * = f-1;i <= f;i ++)
*** * * ** ** ** ***** * * ** **
*** *** * * **** * * * * ** * * *** ** * * *** = k + num[i];
***** *** * * ** ** * * * ** **
* **** * * * * *** = k/2;
* *** * * * * * * ** ****** *** ** *** k);


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

 }
answered by (-188 points)
0 like 0 dislike
Hidden content!
#include<stdio.h>
* * * *** * *



int main()

{
****** * *** * * * ** * i,j,k,l;
* * * *** ** * *
* **** **** * * * *** * *** * * ** ** * *


*** * * ** * *** ** * ** *
**** * * ** * * * * * n[i];
* ** * *** ****** ** *
* ** * **** * * ** * = 0;k < i;k++)
* ********* ** ******* * ** *** * ** * ** ** ** n[k]);
** ****** ****** **** *
* **** ** **** * * ** *
* * ** * ** * = 0;
* ** * * * **
** *** **** * * ** * * * ** ** * = 0;k < i;k++)
* **** * * * *** * ** = l + n[k];

   
**** * * * * ***** * **** ** m;
***** * **** ** ** *** = l / i;

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

}
answered by (-264 points)
0 like 0 dislike
Hidden content!
#include<stdio.h>
* ***** *** * *



int main()

{
**** * ** * * * **** ** * ** * i,j,k,l;
* * * * * *** * **
* ***** **** ** ** ** * *** * * ** ** ** ** **


**** * ** * ** ** * *
*** ** **** * * * n[i];
* * **** *** * * ** *
* ** * ** **** ** ****** = 0;k < i;k++)
* * * * ** *** ** *** * ** ** * * **** * **** * n[k]);
**** * * * ** *
*** ** ** * ***
* ********* *** * = 0;
*** ** **** * * ** * *
* ***** * *** * * ** ** * = 0;k < i;k++)
* * **** * * *** * * **** = l + n[k];

   
*** ** * ** * **** m;
* * ** * * *** ** * = l / i;

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

}
answered by (-264 points)
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:104.23.197.65
©2016-2026

Related questions

0 like 0 dislike
5 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 15448 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 1.5k views
0 like 0 dislike
33 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 15445 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 5.3k views
0 like 0 dislike
11 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 15441 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 2.9k views
0 like 0 dislike
41 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 15427 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 6.5k views
0 like 0 dislike
28 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 15421 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users