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;
#include<cstdlib>

#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
typedef long long ll;

int main(){
  int n;
  cin >> n;

  int ans=0;

  rep(i,n+1){
    int check=i;
    int keta = 0;
    
    while(check>0){
      check /= 10;
      keta++;
    }

    if(keta%2==1) ans++;
  }

  cout << ans;
}