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;
		}
	}
} 

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)