CodeForces 9C Hexadecimal's Numbers

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.

题意是给定一个n，求n以内多少正整数是仅由0，1构成的。

第一个方案是直接枚举，复杂度约为9n，毫无疑问超时了。

很快思考一下，我们可以想到以下两种策略：

1. 用2进制进行枚举，再将二进制数直接变成与其“看起来” 一样的数字，这样顶多枚举到2的10次方便可以。优点：二进制计算快，缺点：略考验代码能力。

2.用字符串从1开始，每次dfs添加0或1，复杂度由于是递归和字符串操作稍微慢一些，但代码相对好实现。

下附AC代码。

#include
#include 
#include 
using namespace std;
string n;
string start="1";
int cnt=0;
int len;
void dfs(string s)
{
	if(s.size()>len) return;
	if(s.size()==len && s>n) return; 
	cnt++;
	dfs(s+'0');
	dfs(s+'1');
}
int main()
{
	cin>>n;
	len=n.size();
	dfs(start);
	cout<