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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;

int main(){
  int n;
  cin >> n;
  string s = to_string(n);
  int ct = s.size();
  int res = 0;
  if (ct & 1){
    string tmp = s.substr(1, ct - 1);
    ct -= 2;
    res += stoi(tmp) + 1;
  }
  else {
  	ct--;
  }
  for (int i = 1; i <= ct; i += 2){
    res += pow(10, i - 1) * 9;
  }
  cout << res << endl;
}