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: Wrong Answer)

#include <stdio.h>

int main (void) {
  int N, ans;

  scanf ("%d", &N);

  if (N < 10) {
    ans = N;
  } else if (N > 9 && N < 100) {
    ans = 9;
  } else if (N > 99 && N < 1000) {
    ans = N - 90;
  } else if (N > 999 && N < 10000) {
    ans = 909;
  } else if (N > 9999 && N < 10000) {
    ans = N - 9090;
  } else if (N == 100000) {
    ans = 90909;
  }

  printf ("%d", ans);
}