Problem:
Score : 200 points

Problem Statement

Given is an integer N. Find the number of positive integers less than or equal to N that have an odd number of digits (in base ten without leading zeros).

Constraints

1 \leq N \leq 10^5

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of positive integers less than or equal to N that have an odd number of digits.

Sample Input 1

11

Sample Output 1

9

Among the positive integers less than or equal to 11, nine integers have an odd number of digits: 1, 2, \ldots, 9.

Sample Input 2

136

Sample Output 2

46

In addition to 1, 2, \ldots, 9, another 37 integers also have an odd number of digits: 100, 101, \ldots, 136.

Sample Input 3

100000

Sample Output 3

90909

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define F first
#define S second
#define mp make_pair
#define pb push_back
#define f(i,x,n) for(int i=x;i<n;i++)
#define all(c) c.begin(),c.end()
#define int ll
#define print(x) cerr << (#x) << "is "<< x << "\n"
using ll = long long;
using pii = pair<int,int>;
const int MOD = 1e9+7, N = 1e5 + 10;
 
int32_t main(){
   ios::sync_with_stdio(false);
   cin.tie(NULL);
   int n;
   cin >> n;
   string ss;
   int ans = 0, l;
   f(i,1,n+1){
      ss = to_string(i);
      l = ss.size();
      if(l & 1) ans++;
   }
   cout << ans << '\n';
   return 0;
}