Solution for https://www.hackerrank.com/challenges/2s-complement/copy-from/32992023

copyfrom.cpp

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
      
        
long long onesFromTo(long long a,long long b){// counts number of ones for a>=0 and b>=0 and a<=b
    long long right = 0,count = 0,countedb = 0,counteda = 0,lastbitb = 0,lastbita = 0;
    // I go through the bits of a and b from LSB to MSB by shifting the bits in every round to right
    // counteda,countedb is what was already shifted away
    // EXAMPLE: a=1011 couteda=0000 => a = 0101 counteda = 0001 => a = 0010 counteda = 0011 
    // => a = 0001 counteda = 0011 => a = 0000 counteda = 1011
    // right is a counter counting how many digits we shifted away
    while(b!=0){
        lastbitb = b&1;
        lastbita = a&1;
        //this computes the number of ones to be added to the sum when the next digit is added thus not being zero anymore.
        //in the first loop iteration I compute the sum only for the last digit
        //then I increase it step by step one digit at a time
        count += 
            // this part computes only if the new bit added is 1, else this is not needed since our number does not change
            // the first term after multiplication is to produce the following sum: 
            // (number of digits already added)*2^(number of digits already added)/2
            // which gives us the number of 1's right of the current to get to 2^(right)-1 from 0
            // so for right=3 it gives us number of 1s from 000 to 011
            // now a 1 is added to get to 100
            // and now I add what I already counted, since I want to 1AB, for AB as in the number was before, not to 100
            lastbitb * ((right << max(0LL,right-1LL)) + 1LL + countedb)
            // same here, but to substract the part from 0 to target number that we didn't need to walk, so this only is important if
            // the current digit is not 0
            - lastbita * ((right << max(0LL,right-1LL)) + counteda);
        countedb += lastbitb<<right;
        counteda += lastbita<<right;
        right++;
        a = a>>1;
        b = b>>1;
    }
    return count;
}
long long onesFromToNeg(long long a,long long b){// counts number of ones for a,b<0 and a<=b
    // I just move the problem to the positives
    long long count = b-a+1LL;
    a = a&~(1<<31); //remove sign
    b = b&~(1<<31); //remove sign
    count += onesFromTo(a,b);
    return count;
}

long long ones(long long a, long long b){
    if (a<0 && b<0)
        return(onesFromToNeg(a,b));
    else if (a<0 && b>=0)
        return onesFromToNeg(a,-1)+onesFromTo(0,b);
    else if (a>=0 && b>=0)
        return onesFromTo(a,b);
    else       
        return 0;
}

int main() {
    long long n;
    cin >> n;
    for(int i = 0; i< n;i++){
        long long a,b;
        cin >> a;
        cin >> b;
        cout<<ones(a,b)<<endl;
    }
    return 0;
}

Please explain a little