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()
{
  int n;
  int ans=0;
  scanf("%d", &n);
  if(n<10){
    ans=n;
  }else{
    ans+=9;
    if(n>=100 && n<1000){
      ans+=n-99;
    }else if(n>=1000 && n<10000){
      ans+=900;
    }else if(n>=10000 && n<100000){
      ans+=900;
      ans+=n-9999;
    }else if(n==100000){
      ans=90909;
    }
  }
  printf("%d\n",ans);
  return 0;
}