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 a, i;
  int b = 0;
  
  scanf("%d", &a);
  
  for(i = 0; i <= a; i++){
    if((0 < i) && (i < 10)){
      b++;
    }
    else if((99 < i) && (i < 1000)){
      b++;
    }
    if((9999 < i) && (i < 100000)){
      b++;
    }
  }
  
  printf("%d\n", b);
  
  return 0;
}