Non-negative Integers without Consecutive Ones
<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>
Dynamic Programming
class Solution {
public:
    int findIntegers(int n) {
        int m = 32 - __builtin_clz(n);
        int f[m][2];
        memset(f, -1, sizeof(f));
        auto dfs = [&](this auto&& dfs, int i, int pre, bool limit) -> int {
            if (i < 0) {
                return 1;
            }
            if (!limit && f[i][pre] != -1) {
                return f[i][pre];
            }
            int up = limit ? (n >> i & 1) : 1;
            int ans = 0;
            for (int j = 0; j <= up; ++j) {
                if (j && pre) {
                    continue;
                }
                ans += dfs(i - 1, j, limit && j == up);
            }
            if (!limit) {
                f[i][pre] = ans;
            }
            return ans;
        };
        return dfs(m - 1, 0, true);
    }
};
