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 n,i,c=0;

    scanf("%d",&n);
    while(n)
    {
       c+=cheak(n);
       n--;
    }

    printf("%d",c);



}
int cheak(int n)
{
    int count=0;
     while(n/10!=0)
  {
      n=n/10;
      count++;
  }
  ++count;
  if(count%2==0)
    return 0;
  else
    return 1;
}