1 like 0 dislike
1.9k 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 | 1.9k 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:172.69.17.62
©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
| 1.1k 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
| 3.5k 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
| 1.8k 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
| 8.1k views
12,783 questions
183,442 answers
172,219 comments
4,824 users