<p>Given a positive integer <code>n</code>, return the number of the integers in the range <code>[0, n]</code> whose binary representations <strong>do not</strong> contain consecutive ones.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 5
<strong>Explanation:</strong>
Here are the non-negative integers &lt;= 5 with their corresponding binary representations:
0 : 0
1 : 1
2 : 10
3 : 11
4 : 100
5 : 101
Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule. 
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    int solve(int n) {
        if (n == 0) return 0;
        if (n == 1) return 1;
        int r = 1;
        int ans = 1;
        int prev = 1;
        while (n >= r) {
            prev = ((prev*(n+1-r+1)*(n-r+1)) / r) / (n+1);
            ans += prev;
            r++;
            n--;
        }
        return ans-1;
    }
    int findIntegers(int n) {
        if (n == 1) return 2;
        int x = 1 << ((int)log2(n)+1);
        int ans = 0;
        while (x > 0) {
            if (x & n) {
                ans += solve(log2(x));
                if (!(x<<1 & n)) ans++;
                else break;
            }
            x >>= 1;
            cout << ans << endl;
        }
        return ans+1;
    }
};
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