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>
#include <math.h>

int main(void)
{
  int n;
  scanf("%d", &n);
  int i = 0;
  while (n >= pow(10, i)){
    i++;
  }
  int ans = 0;
  for (int j = 1; j < i; j += 2){
    ans += 9 * pow(10, j - 1);
  }
  if (i % 2 != 0){
    ans += (n % (int) pow(10, i - 1)) + 1;
  }
  printf("%d\n", ans);
  return 0;
}