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>
     
    int main(){
      int n;
      int i;
      int count=0;
      
      scanf("%d",&n);
      
      for(i=1;i<=n;i++){
        if(i>=1 || i<10){
          count++;
        } else if(i>=100 || i<1000){
          count++;
        } else if(i>=10000 || i<100000){
          count++;
        }
      }
      printf("%d",count);
    }