C :

#include<stdio.h>

int a[101],b[1001];
int main()
{
	int i,n,t,c=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%d",&a[i]);
		t=a[i];
		if(b[t]==0){
			c++;
		}
		b[t]++;
		
	}
	printf("%d\n",c);
	for(i=0;i<=1000;i++){
		if(b[i]>0){
			printf("%d\n",i);
		}
		
	}
	return 0;
}

C++ :

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int main(){
	int n,k,j,t,i,a[1000],s[1000]={0},c=0;
	bool f=false;
	cin>>n;
	for(i=0;i<n;i++){
		cin>>a[i];
	}
	for(i=0;i<n;i++){
	 for(j=0;j<n-i-1;j++){
	 	if(a[j]>a[j+1]){
	 		t=a[j];
	 		a[j]=a[j+1];
	 		a[j+1]=t;
		 }
	 }
}
    for(i=0;i<n;i++){
    	for(j=0;j<i;j++){
    		if(a[i]==s[j]){
    			f=true;
			}
		}
		if(f!=true){
			s[i]=a[i];
			c++;
		}
		f=false;
	}
	cout<<c<<endl;
	for(i=0;i<n;i++){
		if(s[i]!=0){
			cout<<s[i]<<endl;
		}
	}
} 

Pascal :

var a,b:array[1..1000000] of longint;
    i,c,s:longint;
begin
read(c);
for i:=1 to c do
read(a[i]);
for i:=1 to c do
b[a[i]]:=b[a[i]]+1;
for i:=1 to 1000000 do
if b[i]<>0 then inc(s);
writeln(s);
for i:=1 to 1000000 do
if b[i]<>0 then
writeln(i);
end.

Java :

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		List<Integer> sortList=new ArrayList<Integer>();
		HashSet<Integer> hs=new HashSet<Integer>();
		Scanner scanner = new Scanner(System.in);
		String n = scanner.nextLine();
		if(scanner.hasNext()){
			String  numberString=scanner.nextLine();
			String[] strArray=numberString.split(" ");
			for(int i=0;i<strArray.length;i++){
				if(!"".equals(strArray[i])){
					int a=Integer.valueOf(strArray[i]);
					hs.add(a);
				}
			}
			sortList.addAll(hs);
	        Collections.sort(sortList);
	        System.out.println(sortList.size());
	        for(Integer num:sortList){
	            System.out.println(num);
	        }	
		}	
	}
}

Python :

n=int(input())
l1=list(map(int,input().split()))
l2=[]
l3=[]
for i in range(0,1001):
    l2.append(0)
for i in l1:
    l2[i]+=1  #得出l1中所有数的重复次数
for i in l1:
    if l2[i]>1:
        l1.remove(i)
print(len(l1))
l1.sort()
for j in l1:
    print(j)