1 like 0 dislike
1.6k views
請設計一程式,將使用者輸入的數排序並剔除重複的數後輸出。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,程式會將接下來的N個數由小到大排好,剔除重複的數後輸出。

sample input:
10
5 3 7 6 4 5 3 7 6 4

sample output

3 4 5 6 7
[Exercise] Coding (C) - asked in 2016-1 程式設計(一)AC by (18k points)
ID: 15442 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00

edited by | 1.6k views

17 Answers

0 like 0 dislike
Hidden content!
#include<stdio.h>
** **** * ** ***** * *

int main()

{

int a = 0,b[100] = {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!
#include<stdio.h>





int main(){

    int x=0,i,j,tamp;
* * * * * * * ** * **** ** * *** * * * *

    int a[x];
** **** * ** * *** * ** ** **** ** * * *
* ***** * **** *** *** * *** * * ** * ** ******* *** *** ** * *

    }
**** **** * * * ** * *** ** **
* **** *** * ** **** **** *** * ***** ** * ** * **** *
* *** ** * ***** * * *** * *** * ******* * * *** * * * * **** * *
* * ** * * ** ** * ** * *** * * * **** ** ** ** ** * ** ** **** *** * ** ***** ** *
* ** ** * *** * ** ****** **** * **** ** * ** ***** ** * ** ** * ** *** * * *** ** * * *
* * ****** *** * ** *** * *** * * * *** ****** ** * ** ** ** **** *** ***** * **** **
* * * * **** **** * * ** ** * **** **** * * **** * * *** * **


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

    }
*** * * ******* *** ** *
** ** **** **** ***** * ** ** * *** * * * * != a[i-1]){
*** * * **** * * ** ** ** **** ****** **** * * * * * ** ** *** * * * *** ",a[i]);
***** * *** ** * * ****** ** ** ***



    }

    return 0;

}
answered by (-126 points)
0 like 0 dislike
Hidden content!
* * ** * *** *****

int main ()

{

int n, i, j, num[i],t;
***** * * ** * * * * *
** ****
***** * * * * * *
* * * * * *** * ***

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

 }
** * * ***** *****
** ***** ** *** ** ** ** *
** * * * 0;







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

int main ()

{

int n, i, j, num[i],t;
** ** * * * ** *** **
* **** ****
***** * ** * *
**** * * ** *

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

 }
** *** ** * ** *
** * * * ** * **** * * ****
* * * 0;







}
answered by (-124 points)
0 like 0 dislike
Hidden content!
#include<stdio.h>
***** * **** * * **

int main()

{

int a = 0,b[100] = {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!
#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!
#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)
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.71.254.74
©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
| 919 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
| 3k views
1 like 0 dislike
18 answers
[Exercise] Coding (C) - asked Oct 27, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 15440 - Available when: 2016-10-27 18:30:00 - Due to: 2016-10-27 21:00:00
| 1.6k 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
| 7.2k views
12,783 questions
183,442 answers
172,219 comments
4,824 users