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()
{
  int a;
  scanf("%d",&a);
  int i;
  i = 0;
  int t;
  int sum;
  while (i < a)
  {
    t = i + 1;
    while (i < a && i < t * 10 - 1)
    {
      sum++;
      i++;
      //printf("i:%d sum:%d\n",i,sum);
    }
    //printf("%d\n",i);
    i  = (i + 1)*10;
    i -= 1;
  }
  printf("%d",sum);
}