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:172.70.127.57
©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