dp[i]=dp[i/10]+i%10;

int n=arr.length;
for(int i=0; i<(1<<n); i++){
    for(int j=0; j<n; j++){
        if((i&(1<<j))>0){
            System.out.println(arr[j]+"  ");
        }
    }
}

algorithm Sieve of Eratosthenes is
    input: an integer n > 1.
    output: all prime numbers from 2 through n.

    let A be an array of Boolean values, indexed by integers 2 to n,
    initially all set to true.
    
    for i = 2, 3, 4, ..., not exceeding √n do
        if A[i] is true
            for j = i2, i2+i, i2+2i, i2+3i, ..., not exceeding n do
                set A[j] := false

    return all i such that A[i] is true.

Time - O(nloglogn)

Example problem: https://leetcode.com/problems/closest-prime-numbers-in-range/description