1. Java, Evgeni Pavlidis

import java.io.*;  
import java.util.*;   

class Main {
        private static int MAX = (int)Math.sqrt(Integer.MAX_VALUE);
        private static int SQRT = (int)Math.sqrt(MAX);
        private static boolean[] isPrime = new boolean[1000002];
        private static long[] prime = new long[50000];
        private static int maxPrime = 0;

        private static void calcPrimes()      {
            isPrime[2] = true;
            prime[maxPrime++] = 2;
            for(int i = 3; i <= MAX; i += 2)
              isPrime[i] = true;
              for(int i = 3; i <= SQRT; i++)
                 if(isPrime[i])
                   for(int j = i+i; j <= MAX; j+= i)
                      isPrime[j] = false; 
                               for(int i = 3; i <= MAX; i+=2)
                                if(isPrime[i])
                                 prime[maxPrime++] = i;
      } 

        public static void main(String...args)      {
           Scanner sc = new Scanner(System.in);
                    calcPrimes();
                    long l,u, d1,d2, s1,s2;
                    long last, minDist, maxDist;
        	         while(sc.hasNextInt())                            
                    {              
                       d1 = d2 = s1 = s2 = last = -1;
                       minDist = Integer.MAX_VALUE; 
                       maxDist = 0;
                       l = sc.nextInt();
                       u = sc.nextInt(); 

              	       	    // don't take 1 as prime	  	   
                       if(l == 1)  		l++;    	    

                       // init whole range(l - u) as primes  	    
                       for(long i = l; i <= u; i++)  		
                           isPrime[(int)(i-l)] = true;    	    

						// don't sieve with primes bigger than sqrt(u)  	    
						int squareRoot = (int)Math.sqrt(u);    	    

						// sieve range  	    
                       for(int i = 0; i < maxPrime && prime[i] <= squareRoot; i++)  		
						for(long j = (int)Math.floor(l/prime[i]) * prime[i]; j <= u; j += prime[i])  		    
						if(j != prime[i] && j >= l)
				  			isPrime[(int)(j-l)] = false;    	    
						// find distances
		              for(long i = l; i <= u; i++)
		                  if(isPrime[(int)(i-l)])
                  {
                                            if(last != -1 && i - last > maxDist)
                      {
                          d1 = last;
                          d2 = i;
                          maxDist = d2 - d1;
                      }
                                            if(last != -1 && i - last < minDist)
                      {
                          s1 = last;
                          s2 = i;
                          minDist = s2 - s1;
                      }
                         last = i;
                  }
                  if(s1 == -1) 
                 System.out.println("There are no adjacent primes.");
              else 
                 System.out.printf("%d,%d are closest, %d,%d are most distant.\n", s1,s2, d1,d2);   

                     }         
    }  
}  

2. C++, Evgeni Pavlidis

   
   #include <cmath>  
	#include <cstdio>
  #include <iostream>
    using namespace std;
   #define MAX 50000 
 #define SQRT 216  
  bool isPrime[1000002]; 
 long prime[MAX];  long maxPrime = 0; 
    void calcPrimes()
  {     
 isPrime[2] = true;     
 prime[maxPrime++] = 2;
           for(int i = 3; i <= MAX; i += 2) 
 	    isPrime[i] = true;    
        for(int i = 3; i <= SQRT; i++)  
	    if(isPrime[i])  	    
    for(int j = i+i; j <= MAX; j+= i)  		
        isPrime[j] = false;  		 
     for(int i = 3; i <= MAX; i+=2)  
	    if(isPrime[i])  	    
    prime[maxPrime++] = i;  }          
    int main()  {    
  calcPrimes();  
         long l,u, d1,d2, s1,s2;
      long last, minDist, maxDist;
          while(cin >> l >> u)      { 
 	    d1 = d2 = s1 = s2 = last = -1;  
	    minDist = 2000000; 
 	    maxDist = 0;    	
    // don't take 1 as prime	              
if(l == 1)  	       
 l++;    	   
 // init whole range(l - u) as primes
  	    for(int i = l; i <= u; i++)
  	        isPrime[i-l] = true;
    	    // don't sieve with primes bigger than sqrt(u)
  	    int squareRoot = (int)sqrt(u);
    	    // sieve range
  	    for(int i = 0; i < maxPrime && prime[i] <= squareRoot; i++)
  	        for(long j = (int)(l/prime[i]) * prime[i]; j <= u; j += prime[i])
  		    if(j != prime[i] && j >= l)
  		        isPrime[j-l] = false;
      	    // find distances
  	    for(int i = l; i <= u; i++)
  	        if(isPrime[i-l])
  	        {  		      
  if(last != -1 && i - last > maxDist)  		        
{  		            d1 = last;  
		            d2 = i;
  		            maxDist = d2 - d1;
  		        }
  		  		        if(last != -1 && i - last < minDist) 
 		        {  		            s1 = last;  		
            s2 = i;  		            minDist = s2 - s1;  	
	        }  		  		        last = i;  	   
     }              if(s1 == -1)     
         printf("There are no adjacent primes.\n");      
    else              printf("%ld,%ld are closest, %ld,%ld are most distant.\n", s1,s2, d1,d2);   
          }            return 0;  
   }