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;

int main() {
  int n;
  int sum;
  cin >> n;
  if(n<10) {
    for(int i=0; i<=n; i++) {
      sum = i;
    }
  }      
  else if(n<100) {
    cout << 9 << endl;
    return 0;
  }
  else if(n<1000) {
    for(int i=100; i<=n; i++) {
      sum = i-99;
    }
    sum += 9;
  }
  else if(n<10000) {
    cout << 909 << endl;
    return 0;
  }
  else if(n<100000) {
    for(int i=10000; i<=n; i++) {
      sum =i-9999;
    }
    sum += 909;
  }
  else
    sum = 90909;

  cout << sum << endl;
  return 0;
}