Counting 1s in a binary number with one loop

konesinarow2.cpp

// konesinarow2.cpp
// juanfc 2012-06-24
// Design a program that reads from keyboard two natural numbers $k$ and
//  $n$ and sends to the screen a message that indicates whether $n$ has
//  got at least $k$ 1s (ones) straight without a break, or not.  For example,
//  if $k=3$ and $n=135$ the program should display ‘Yes’ since 135 as binary
//  is ‘1000 0111’ that does have at least 3 ones in a row.

// One loop only (boolean helped)

#include <iostream>
using namespace std;

int main()
{
    int N, k;
    cout << "Enter n: ";
    cin >> N;
    cout << "Enter k: ";
    cin >> k;

    int n = N;
    int cnt = 0;
    bool lastisone = false;
    while (n > 0 and cnt < k) {
        if (n % 2 == 1) {
            lastisone = true;
            ++cnt;
        } else {
            lastisone = false;
            cnt = 0;
        }
        n /= 2;
    }

    if (cnt >= k)
        cout << "Yes, " << N << " has " << k << " 1s in a row" << endl;
    else
        cout << "No, " << N << " hasn't " << k << " 1s in a row" << endl;

    return 0;
}