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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int main()
{
  int n,x;
  scanf("%d",&n);
  if(n>=1&&n<=9)
  {
   x=n-0;
   printf("%d\n",x);
  }
  else if(n>=10&&n<=99)
  {
    x=9;
    printf("%d\n",x);
  }
  else if(n>=100&&n<=999)
  {
    x=(n-100)+1+9;
    printf("%d\n",x);
  }
  else if(n>=1000&&n<=9999)
  {
    x=(n-100)+1+9;
    printf("%d\n",x);
  }
  else if(n>=10000&&n<99999)
  {
    x=(999-100)+1+9+(n-10000)+1;
    printf("%d\n",x);
  }
  else if(n>=100000&&n<=999999)
  {
    x=(999-100)+1+9+(n-10000)+1;
    printf("%d\n",x);
  }
  return 0;
}