0 like 0 dislike
7.4k 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 | 7.4k 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:108.162.241.32
©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