C. Hexadecimal's Numbers
题目连接：
http://www.codeforces.com/contest/9/problem/C
Description
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 ≤ 109).
Output
Output the only number — answer to the problem.
Sample Input
10
Sample Output
2
Hint
题意
问你小于等于n的数里面，有多少个只由0和1组成的
题解：
感觉上没多少个数嘛，那就直接dfs就好了啦
代码
#include<bits/stdc++.h>
using namespace std;
long long n;
map<int,int> vis;
long long ans = 0;
void dfs(long long x)
{
if(x>n)return;
if(vis[x])return;
vis[x]=1;
ans++;
dfs(x*10);
dfs(x*10+1);
}
int main()
{
cin>>n;
dfs(1);
cout<<ans<<endl;
}