HackerRank - "2's complement"

This one is marked as "Advanced".. i don‘t tink so, not that hard if you can visualize all the bits from a to b. Two key points here:

1. Say both a and b are positive, bits form an interesting pattern vertically - all bit[i] of all numbers
2. negative numbers, count(a) = 32 - count(1 - a)

The above observation leads to this code:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

long long _calc(long long a, long long b)
{
    int bb = b;
    int pc = 1, sz = 2;
    long long ttl = 0;
    while (bb)
    {
        int sa = a / sz;
        int sb = b / sz;
        long long cnt = (long long)(sb - sa + 1) * (long long)pc;

        int ra = a % sz;
        if (ra > pc)
            cnt -= std::max(ra - pc, 0);
        int rb = b % sz;
        if (rb < sz - 1)
            cnt -= std::min(pc, sz - rb - 1);

        ttl += cnt;

        //    Move on
        bb >>= 1;
        pc *= 2;
        sz *= 2;
    }

    return ttl;
}

long long calc(long long a, long long b)
{
    long long cnt = 0;

    //    negative
    int na, nb;
    if (a < 0)
    {
        na = a; nb = std::min((long long)-1, b);
        long long aa = -(nb + 1);
        long long bb = -(na + 1);
        long long ret = _calc(aa, bb);
        cnt = (long long)(bb - aa + 1) * 32 - ret;
    }

    long long pa, pb;
    if (b >= 0)
    {
        pa = std::max(a, (long long)0);
        pb = b;
        cnt += _calc(pa, pb);
    }
    return cnt;
}

int main()
{
    int t; cin >> t;
    while (t--)
    {
        long long a, b; cin >> a >> b;
        auto r = calc(a, b);
        cout << r << endl;
    }
    return 0;
}

HackerRank - "2's complement"