HackerRank "Xor and Sum"

Actually I think it should belong to category of 'Bits Manipulation'.. but still, a really good one.

My first reaction was, how to derive from a ^ (b << i) from a ^ (b << (i - 1)), but after some thoughts, I figured out that that should not be the way to go: there's no easy way to do that induction. 

This one resembles with another Bit one: we are to calculate sum, so we consider bit by bit: at bit[i], it contributes bit[i] << i to the final sum..

Lesson learnt: 1) power2 can be moduled too; 2) VISUALIZE the problem space\process in your mind - it is crucial.

#include <cmath>
#include <cstdio>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

#define MOD 1000000007

unsigned sum(vector<unsigned> &ones, int l, int r)
{
    l = std::max(0, l);
    r = std::min(int(ones.size() - 1), r);
    if (l > r) return 0;
    return ones[r] - (l == 0 ? 0 : ones[l - 1]);
}

int main() 
{
    const unsigned N    = 314159;
    const unsigned NMAX = 1e6;

    string a, b;
    std::cin >> a >> b;
    
    std::reverse(a.begin(), a.end());
    std::reverse(b.begin(), b.end());

    //    Power2 % MOD
    vector<unsigned> pw2(NMAX);
    pw2[0] = 1;
    for (unsigned i = 1; i < NMAX; i++)
        pw2[i] = (pw2[i - 1] * 2LL) % MOD;

    //    Count accumulative count of ones
    size_t lena = a.length();
    size_t lenb = b.length();
    vector<unsigned> ones(lenb, 0);
    for (int i = 0; i < lenb; i++)
    {
        ones[i] = b[i] - '0';
        if (i > 0) ones[i] += ones[i - 1];
    }

    //    Count accumulative count of zeros
    unsigned ans = 0;
    for (int i = 0; i < NMAX; i++)
    {
        int bita = (i >= lena) ? 0 : a[i] - '0';
        int cnt_ones = sum(ones, i - N, i);
        if (bita)
        {
            ans = (unsigned)((ans + (N + 1 - cnt_ones) * 1LL * pw2[i]) % MOD);
        }
        else
        {
            ans = (unsigned)((ans + cnt_ones * 1LL * pw2[i]) % MOD);
        }
    }    
    cout << ans << endl;
    return 0;
}