What is the term for numbers represented in base 16, known as hexadecimal?

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在Mainframe中，一个不幸的事件发生了：病毒Megabyte潜入系统。为了找到并消除这个病毒，我们需要通过递归深度优先搜索（DFS）来探索可能的感染路径。下面是解题思路的简化版：

开始于小于n+1位的二进制数，每个数表示一种可能的感染模式。例如，对于n=3，我们考虑从2^3-1=7位数的模式开始，即从111开始。

将这些数分解为2^n-1=7个部分，然后对每个部分应用DFS来检测TLE（时间限制错误）。如果找到TLE，那么就找到了病毒感染的一个路径。

这种方法的关键在于递归地应用DFS，直到找到病毒感染的确切路径。记住，不要超出100个字，并且不要使用复杂的数学术语。

• 刚开始发现小于 n 位的二级制形式数为 2^n-1 个，然后再在剩下的区间处理判断，发现TLE ,于是只好 DFS。

• 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.

  Examples input Copy 10 output Copy 2 Note

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

  
  

#include<iostream> #include<cstdio> #include<string> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<set> #include<map> using namespace std; typedef long long ll; #define maxn 2000005 #define inf 0x3f3f3f3f #define ii 0x3f const int mod = 1e9 + 7; // dp[i]=min(dp[j]+(i-j-1+sum[i]-sum[j]-l)^2) // dp[i] 表示前 i 个玩具装箱的费用 // f[i]=sum[i]+i; // c=1+l; // f[i] 单增 ll read() { ll x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') { f = -1; } ch = getchar(); } while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); } return x * f; } ll quickpow(ll a, ll b) { ll ans = 1; while (b > 0) { if (b % 2)ans = ans * a; b = b / 2; a = a * a; } return ans; } //char s[maxn]; int ans; ll n; void dfs(ll x) { if (x > n)return; ans++; dfs(x * 10); dfs(x * 10 + 1); } int main() { ios::sync_with_stdio(false); //ll n; cin >> n; ans = 0; dfs(1); cout << ans << endl; }