1 like 0 dislike
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 | 237 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:3.235.179.79
©2016-2021

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
| 138 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
| 308 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
| 218 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
| 850 views
11,846 questions
152,046 answers
155,432 comments
4,159 users