0 like 0 dislike
4.1k 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 | 4.1k 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:172.71.254.225
©2016-2025

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
| 629 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
| 2.2k 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
| 1.1k 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
| 2.8k 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
| 2.2k views
12,783 questions
183,442 answers
172,219 comments
4,824 users