1 like 0 dislike
2.2k views
請設計一程式,從使用者輸入的正整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。輸出時請注意小數點尾巴的零不要輸出。

sample input:

4

1 2 3 4

sample output

2.5
[Exercise] Coding (C) - asked in 2016-1 程式設計(一)AC by (18k points)
ID: 15440 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00

reshown by | 2.2k views

18 Answers

0 like 0 dislike
Hidden content!
** *** * ** *
*** * * * ** * ** * *

#include<time.h>

int main(void)

{
* * ** * * ** * * i,j,k,u;
*** ** ** * * * ** stu[100000];
** * * * * * * * *** * ** **** ** ** ***** **
*** *** * * * * * * = i/2;
* * * ** *** * ** = i/2+1;
***** ** ** * ** = (k+u)/2;
** * ** ** * * * *** - k == 0)
**** * * * ** ****** * * ** * ** *** *** *
** * ** ** * *
*** * *** * * * ** * * **** *
* *** * * * * ** 0;

}
answered by (-8 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

#include **** ** ****** * *

int main()

{
* * ** *** ***** ** a;
* * ** ** * * ** * ** *** *** *


* * * **** ** * num[a],i,m=0;
* * ** *** (i=1; i <= a; i++)
* * * ** * * ** ******* * **
* * ** ***** ** * * **** ** * * * * *** ** * &num[i]);


*** ** * ** * *** (a % 2 == 0)
**** * ****** * * *** ** * * * **** ** *
* * ******* ** ** **** ***
* *** * **** *** * * ***** *** ** *
** ** *** * *** *** * ** *** * *
** ** * * * ** * ** * ****** ** ****** m);

}
answered by (-42 points)
0 like 0 dislike
Hidden content!
#include<stdio.h>



int main()

{
* * *** ** *** * * i,j,k,n,num[n],m;
* **** ** * * * * * * * ** * &n);
* * ** * * * ** *** ***


* ****** * * **** * **** *****
* *** * * * * * * ** * **** *** &num[n]);


** * *** * * * * ** *** ***


* * * * * * ***** *** ** **


***** ****** ** * **** ****

   {
** * ** ** * *** ** * * ** ***** * *
*** * ** ******** * *** * ** ** * **
* * * * *** ** * ** * * * * ** * * *
** * * **** *
* * ** ** * ******** * *


*** * *** *** * *** * ** * ",num[i]);


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





}
answered by (-103 points)
0 like 0 dislike
Hidden content!
* * * ** *
* * ***** ** * ***

#include<time.h>

int main()

{
** * ** ** * *** * * i,j,k,u;
* * **** * * * * stu[100000];
*** * *** ** ** *** * ** * * * ** *
* * * ** ** ***** = i/2;
* * * * * **** * = i/2+1;
** * * * * ** * * = (k+u)/2;




****** ** ** * * * * * - k == 0)
*** ** * * ** ***** * ** **** *** *
** * * * ** ******
* * **** * *** * * ******** ** *** *
**** ** *** ** ** **** 0;









}
answered by (-8 points)
0 like 0 dislike
Hidden content!
#include<stdio.h>

int main ()

{
* ** ******* i ,h, num[h] ;


** * * * ** *** ** * * * ** * *** *** &h);


** * * * * ****
* ** **** *** * ** ******* * * * * * * ** ** * " ,&num[i]);

   }
** *** * * *** ** == 0){
** * ** ** * * * ** **** * ** * **** * * **** * ** * ** * **
*** * * ** * *** * * ** * ** ***
*** * **** *** *
* * * *** ** ****** * ** * *** * * *** * ** *** **
* * * * * * ** * * * * ** ** ** * * *
* * * * 0 ;

}
answered by (-82 points)
0 like 0 dislike
Hidden content!
* * ** * ** ****
*** ** ** * **** * * *

int main()

{

int a = 0,b[100] = {0},c = 0,i = 0,j = 0,k = 0,l = 0;
*** * ****** * ** *

for(i=0;i<a;i++){
** ** * **** * ******* * *
** * *** * * *** *
*** ** ** * *** *** **
* ** * **** ********
* ** * ** ** ** = 0;
* * *** * *** **** * ** * **

{
* * * **** *** * ** ** *** ** ********
* * ** *** * ** **** ** * * ** **** *** ** * ** * *
* * * * ** ** * **** * ** ** * *** * * * ******** **** = b[l];
* * ** *** * * ** * ** * * ** ** ** * ** * * **** ** = b[i];
* * ** *** *** * * *** * * **** ** **** * * **** * ** *** ** = k;
** * ** ** ** * ** ** ** ** ** ** *


* **** *** ***** * * * * *

}



if(a%2==0){

 i = a/2;

 j= a/2+1;
* ** *** * ** * * ** ** * ***

}else{

i = (a+1)/2;
**** * * ** * *

}









return 0;



}
answered by (-140 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

int main()

{
** **** * ** * ***** a,i;
******* * * **** * ** ** * ** *** * ** **
*** * ** ** ** * * * ****** m,j,k;


** ****** * * math[100000];
*** * *********** * ** * *** **** * ** **
* * * ** *** *** ** * *********** * ** ***** ***


******* ** ** * ** * (a%2==0)

    {
* * ***** ******** * ** * * * * * * *
* * * ** * ** * * * ******* * * *** *** * *
** ** *** *** * *** * * ** * * * * *** * ** **
** * * * ** ** * * *** *** * * * **** * * **** * **
** * ** * *** * *** **
* * * * * * *** **** ** * if(a%2!=0)

    {    m=math[a/2];
** * ** * ** **** *** * * * * * ** *** * * * * * ** ** ** ** * * *

}

}
answered by (-124 points)
0 like 0 dislike
Hidden content!
** * ***** * * * **
**** * *** **
** ** * * * *
* ** *
**
* * * * * * *
** ** * * * ** **
* ** *


* (i = * * * * ** *** * * **** * *****
** * (j = * ** * z - **
* (i = j + * ** **** * *** * * * *** * = * * * = ** *** ** = * **

}



c = z % 2;

d = z / 2;
* = d - *
*** * * = * *** * + * / **** ***** *** ** ** * **
*** ** = * * * ****** ** ** **


* * 0;

}
answered by (-276 points)
0 like 0 dislike
Hidden content!
#include<stdio.h>
* * * * *** * *

int main()

{

int a = 0,b[20000] = {0},c = 0,i = 0,j = 0,k = 0,l = 0;
** ** * * ******* * *

for(i=1;i<a+1;i++){
** * *** ** * ** *


** * **** *********** * * * *
*** ** *** ***** **
* *** * * * * ***** = 1;
** * * * * **** *
*** * ** *** **** *


* **** ***** * ** * *
* * * ***** **
* * * * * *
* ** *** ** ** **


** * * ** * ** ** = 0;
** ** ** *** * ** * *

   i = 0;
**** **** ** * * * * **

{
* * * ** ***** *** * *** *** ** * *
**** * **** * * ******* ** ** ** * * ******* * * ** ** * **
* **** ** ** * * *** * * * ** ** * ** * * = b[l];
********* * ** ** **** * * ***** * * *** *** * ** = b[i];
** **** * * * **** * **** **** ** * ***** *** = k;
**** ** * *** * *** * ** *** * * *** * * ** ** = 0;
* *** ******** ** ** * * ** ** ** *** *** * **** **


** * * * * ** *

}



for(i=0;i<j;i++){
* * ** * * * * **** * *** **** ",b[i]);

}





return 0;



}
answered by (-140 points)
0 like 0 dislike
Hidden content!
* * * **
* * * ** *
** ** * ******* * **
* * *
*
* *** * *
* * ****** * **
** ***


*** (i = * ****** * * * * * * * * **** ** *
**** (j = * z - * ***
* * (i = j + * * * ** ** * * ** * = ** * = ***** * * = ***

}



c = z % 2;

d = z / **
* = d +
*** ** = * * * + * *** * / * * * * * * * * * ******
** * = * * * * **** *** * * *


*** 0;

}
answered by (-276 points)
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:108.162.216.250
©2016-2025

Related questions

0 like 0 dislike
0 answers
[Resource] asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15475 - Available when: Unlimited - Due to: Unlimited
| 13 views
0 like 0 dislike
5 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15447 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 1.3k views
1 like 0 dislike
37 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15446 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 4.3k views
1 like 0 dislike
17 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15442 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 2.3k views
0 like 0 dislike
112 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15426 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 10.2k views
12,783 questions
183,442 answers
172,219 comments
4,824 users