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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void)
{
  int N, i, con = 0;
  scanf("%d", &N);
  //100000
  for(i = 1; i <= N; i ++){
    if(i < 100000){
      if(i < 10000){
        if(i < 1000){
          if(i < 100){
            if(i < 10){
              con ++;
            }
            else{}
          }
          else{con ++;}
        }
        else{}
      }
      else{con ++;}
    }
    else{}
  }

  printf("%d\n", con);
  return 0;
}