Hexadecimal's Numbers CodeForces - 9C

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本文介绍了一个有趣的算法问题：如何计算从1到给定整数n之间，有多少个整数仅由0和1构成。通过深度优先搜索（DFS）的方法，递归地生成所有符合条件的数，并统计其数量。

One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.

But his plan failed. The reason for this was very simple: Hexadecimal didn’t perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.

Input
Input data contains the only number n (1 ≤ n ≤ 10^9).

Output
Output the only number — answer to the problem.

Example

Input
10

Output
2

Note
For n = 10 the answer includes numbers 1 and 10.

大致题意：给你一个数n，问从1到n的范围内有多少数，只由0，1组成。

思路：假设一个数x只由0，1组成，那么下一个就只有x*10或者x*10+1这两种可能，所以我们从1开始dfs深搜即可

代码如下

#include <cstdio>  
#include <cstring>  
#include <iostream> 
#include <algorithm>   
#define ll long long

using namespace std; 
int sum=1;
int n;

void dfs(int x)
{
    if(x*10<=n)
    {
        sum++;
        dfs(x*10);
    }
    if(x*10+1<=n)
    {
        sum++;
        dfs(x*10+1);
    }
}
int main()
{
    cin>>n;
    dfs(1);
    cout<<sum<<endl;
    return 0;
}