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, i;
  i = 0;
  scanf("%d", &n);
  if(n<10){
    do{
      i = i + 1;
    }while(i = n);
      printf("%d",i);
  }
  else if(n<100){
    do{
      i = i + 1;
    }while(i = n-100);
      printf("%d",i);
  }
  else if(n<10000){
    do{
      i = i + 1;
    }while(i = n-10000);
      printf("%d",i);
  }
  return 0;
}