Algorithm to calculate the number of 1s for a range of numbers in binary

Question

So I just got back for the ACM Programing competition and did pretty well but there was one problem that not one team got.

The Problem.

Answer 1

You can solve this efficiently as follows:

ret = 0;
for (i = 1; i <= 64; i++) {
  if (computeK(i) != desiredK) continue;
  ret += numBelow(HIGH, i) - numBelow(LO - 1, i);
}
return ret;

The function numBelow(high, numSet) computes the number of integers less than or equal to high and greater than zero that have numSet bits set. To implement numBelow(high, numSet) efficiently, you can use something like the following:

numBelow(high, numSet) {
  t = floor(lg(high));
  ret = 0;
  if (numBitsSet(high) == numSet) ret++;
  while (numSet > 0 && t > 0) {
    ret += nchoosek(t - 1, numSet);
    numSet--;
    while (--t > 0 && (((1 << t) & high) == 0));
  }
  return ret;
}