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, ans, l, n, i, v;

  scanf("%d", &N);

  for(l = 0, n = N; n > 0; l++, n/=10);
 
  ans = 0;

  if(l % 2){
    for(i = l - 1, v = 1; i > 0; i--) v *= 10;
    ans += N - v + 1;
  }
  l--;

  for(; l > 0; l--){
    if(l % 2){
      for(i = l - 1, v = 1; i > 0; i--) v *= 10;
      ans += v*10 - v;
    }
  }

  printf("%d\n", ans);

  return 0;
}