1 like 0 dislike
2.7k 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 | 2.7k 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:172.71.254.169
©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
| 1.5k 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
| 5.4k 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
| 2.7k 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
| 11.9k views
12,783 questions
183,442 answers
172,219 comments
4,824 users