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 data contains the only number n (1?≤?n?≤?109).
Output the only number — answer to the problem.
10
2
For n = 10 the answer includes numbers 1 and 10.
题目分析 ： 给你一个 n ，表示 1 - n 共 n 个数字，问有多少个数字中只含有 0 和 1
思路分析 ： 对于一个只含有 01 数字的十进制数字，他一定可以通过这样的变换得来， a*10 和 a*10+1 ， 既然有这样的规律，那么写一个深搜就可以直接过了。
代码示例 ：
int ans = 0;
ll n;
void dfs(ll x){
if (x > n) return;
ans++;
dfs(x*10);
dfs(x*10+1);
}
int main() {
cin >> n;
dfs(1);
printf("%d\n", ans);
return 0;
}
