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>

static int keta (int n)
{
  if (n<0)
    return keta(-n);
  
  int digits = 1;
  while (n /= 10)
    digits++;
  return digits;
}

int main(void)
{
  int n;
  int digits;
  int ans;
  
  scanf("%d",&n);
  digits = keta(n);
  
  switch(digits) {
    case 1: ans = n; break;
    case 2: ans = 9; break;
    case 3: ans = 9 + (n - 99); break;
    case 4: ans = 909; break;
    case 5: ans = 909 + (n - 9999); break;
    case 6: ans = 90909; break;
  }
  
  printf("%d",ans);
  
  return 0;
}