Generate as many distinct primes P such that reverse (P) is also prime and is not equal to P.
Output: Print per line one integer( ≤ 10^15 ). Don't print more than 10^6 integers in all.
Scoring: Let N = correct outputs. M = incorrect outputs. Your score will be max(0,N-M).
Note: Only one of P and reverse(P) will be counted as correct. If both are in the file, one will be counted as incorrect.
Sample Output 107 13 31 17 2
Explanation
Score will be 1. Since 13,107,17 are correct. 31 is incorrect because 13 is already there. 2 is incorrect.
Here is the code I've written which is giving me output Out Of Memory
error in Eclipse.
Since memory requirement is 256 MB, I set -Xmx256M
, but since it's giving me an Out Of Memory
error, I must have misunderstood the question or my code is buggy in terms of memory utilization. What am I doing wrong here? I'm getting the desired output for smaller lONGMAX
like 10000
or 1000000
.
public class ReversePrime {
final static long lONGMAX=1000000000000000L;
final static int MAXLISTSIZE=1000000;
final static boolean[] isPrime=isPrime();
public static void main(String...strings ){
Set<Long> reversedCheckedPrime = new LinkedHashSet<Long>();
int isPrimeLength=isPrime.length;
for(int i = 0; i < isPrimeLength ; i++){
if( isPrime[i]){
long prime = 2 * i + 3;
long revrse= reversePrime(prime);
if ( (!(prime==revrse)) && (!reversedCheckedPrime.contains(revrse)) &&
(reversedCheckedPrime.size()<=MAXLISTSIZE)){
reversedCheckedPrime.add(prime);
}
if (reversedCheckedPrime.size()==MAXLISTSIZE){
break;
}
}
}
for (Long prime : reversedCheckedPrime){
System.out.println(prime);
}
}
private static long reversePrime(long prime) {
long result=0;
long rem;
while(prime!=0){
rem = prime % 10;
prime = prime / 10;
result = result * 10 + rem ;
}
return result;
}
private static boolean[] isPrime() {
int root=(int) Math.sqrt(lONGMAX)+1;
root = root/2-1;
int limit= (int) ((lONGMAX-1)/2);
boolean[] isPrime=new boolean[limit];
Arrays.fill(isPrime, true);
for(int i = 0 ; i < root ; i++){
if(isPrime[i]){
for( int j = 2 * i * (i + 3 ) + 3, p = 2 * i + 3; j < limit ; j = j + p){
isPrime[j] = false;
}
}
}
return isPrime;
}
}