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 getDigit(int d) {
  int digit = 0;
  while (d != 0) {
    d = d / 10;
    ++digit;
  }

  return digit;
}

int main(int argc, char* argv[]) {
  int n;
  scanf("%d", &n);

  int i, cnt = 0;
  for (i = 1; i <= n; i++) {
    if (getDigit(i)%2 != 0) cnt++;
  }

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

  return 0;
}