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 ", ans); return 0; }