将1到N的所有数字相加，并将设置的位数设置为2

Question

I have a question where I have to add numbers from 1 to N which have their set bits as 2. Like for N = 5 we should get value 8, as number 3 and 5 have 2 bits set to one. I am implementing the same in java. I am getting the o/p correct for int value but when it comes to the long values, either it's taking a lot of time or freezing, and when I submit the same on code judge sites, it's giving run time exceeded message. Please guide me how may I optimise my code to run it faster, thanks :)

public static void main(String[] args)
{
    long n = 1000000L;
    long sum = 0;
    long start = System.currentTimeMillis();
    for(long i = 1L ; i <= n ; i++)
    {
        if(Long.bitCount(i) == 2)
        {
            sum += i;
        }
    }
    long end = System.currentTimeMillis();
    System.out.println(sum);
    System.out.println("time="+(end-start));
}

Answer 1

As @hbejgel notes, there is no point in looping over all numbers and checking their bit count. You can simply construct numbers with 2 bits and add them up.

You can construct a number with 2 bits by picking two different bit positions in the long, the "higher" bit and the "lower" bit":

long i = (1 << higher) + (1 << lower);

So, you can simply loop over all such numbers, until the value you have constructed exceeds your limit:

long sum = 0;
outer: for (int higher = 1; higher < 63; ++higher) {
  for (int lower = 0; lower < higher; ++lower) {
    long i = (1 << higher) + (1 << lower);
    if (i <= n) {
      sum += i;
    }
    if (i >= n) break outer;
  }
}

Answer 2

Let's say we know the closest number, x , equal to or lower than N with 2 set bits, then we can use the formula for power series to quickly sum all positions of the two set bits, for example, if x = b11000 , we sum

  4*2^0 + S(4)
+ 3*2^1 + S(4) - S(1)
+ 2*2^2 + S(4) - S(2)
+ x

where S(n) = 2 * (1 - 2^n) / (1 - 2) 
           = 2 + 2^2 + 2^3 ... + 2^n

Answer 3

With numbers encoded 2 out of 5 , exactly two bits are set in every one-digit number. The sum is 45, with the exception of N×(N-1)/2 for 0≤ N <9.

Answer 4

I think the question is supposed to discover the pattern.

1. Fast forward. Given a number N, you can tell the largest number should count by bitmask from the first two bits are set. So you have a smaller number M

2. Skip to next counted number Given any number with two bit set, next largest number is the shift the second bit by one, until underflow.

3. Skip to next order When underflow happens on set two, shift the highest bit by one and also the bit on it's right.

You don't really need a loop on N, but the bits it have.

Next question: can you answer a large number? which N >100,000,000

Next Next question: can you answer the same question for X bits when X>2