<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 findIntegers(int n) {
        vector <int> f;
        while (n){
            if (n%2) f.push_back(1);
            else f.push_back(0);
            n/=2;
        }
        int sz=f.size();
        vector <vector<int>> dp(sz,vector<int>(2));
        dp[0][0]=1,dp[0][1]=1;
        for (int i=1;i<sz;i++){
            dp[i][0]=dp[i-1][0]+dp[i-1][1];
            dp[i][1]=dp[i-1][0];
        }
        int ans=0;
        bool ok=true;
        for (int i=sz-1;i>=0;i--){
            if (f[i]){
                if (i<sz-1 && f[i+1]==1){
                    ok=false;
                }
                ans+=dp[i][0];
            }
            if (!ok) break;
        }
        if (ok) ans++;
        return ans;
    }
};
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