C :

#include<stdio.h>
#include<math.h>
int main()
{
	int n,i,j,a;
	double k;
	while(~scanf("%d",&n))
	{
		a=0;
	 for(i=2;i<=n;i++)
	 {
		k=sqrt(i);
		for(j=2;j<=k;j++)
		{
			if(i%j==0)
			{	
				break;
			}
			
		}
		if(j>k)
		{
			printf("%d",i);
				a++;
				if(a%5==0)
				printf("\n");
			else printf(" ");
		}
	}
	 printf("\n");
	}
	return 0;
}

C++ :

//
//  main.cpp
//  上机练习
//
//  Created by 朱世起 朱 on 2017/2/20.
//  Copyright © 2017年 朱世起 朱. All rights reserved.
//

#include <iostream>
#include <stdio.h>
#include <string>

using namespace std;

int main(){
    long n,temp;
    cin>>n;
    bool *a=new bool[n+1];
    for (long i=2; i<=n; ++i) {
        a[i]=true;
    }
    
    
    for (long i=2; i<=n+1; ++i) {
        for (long k=2; k<=i; ++k) {
            temp=i*k;
            if(temp<=n)a[temp]=false;
            else break;
        }
    }
    
    //输出
    long set=1;

        for (long j=2 ;j<=n; ++j)
        {
            if (set==6){cout<<endl;set=1;}
            if (set==1 && a[j]==true){cout<<j;++set;}
            else if(a[j]==true){cout<<" "<<j;++set;}
        }
   
    
    
    
    return 0;
}

Pascal :

var a:array[1..100]of boolean;
    n:integer;
    i,j,t:integer;
begin
  readln(n);
  fillchar(a,sizeof(a),true);
  a[1]:=false;

  for i:=2 to n do
    begin
      if a[i]=true then
      for j:=2 to (100 div i) do
        a[i*j]:=false;
    end;

  for i:=2 to n do
    if a[i]=true then
      begin
        inc(t);
        write(i,' ');
        if t mod 5=0 then writeln;
  end;
end.