1 like 0 dislike
2.3k 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 | 2.3k 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:172.70.100.82
©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
| 4.4k 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
| 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