0 like 0 dislike
7.3k 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 | 7.3k 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:172.69.59.58
©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
| 1.1k 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
| 4.1k 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
| 2.1k 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
| 5.1k 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
| 3.8k views
12,783 questions
183,442 answers
172,219 comments
4,824 users