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;
  scanf("%d", &N);
  int ans = N - ((N>=100000)? ((N>=1000000)? 900000: N-99999):0)
              - ((N>=1000)?   ((N>=10000)?   9000:   N-999):  0)
              - ((N>=10)?     ((N>=100)?     90:     N-9):    0);
  printf("%d\n", ans);
  
  return 0;
}