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 = 0;
  int n = 0;
  int i;
  scanf("%d",&a);
    
   for(i = 1 ; i <= a ; i++){
     if(1 <= i && i <= 9){
       n++;
       continue;
     }
     else if (10 <= i && i <= 99){
       continue;
     }
     else if(100 <= i && i <= 999){
       n++;
       continue;
     }
     else if(1000 <= i && i <= 9999){
       continue;
     }
     else if(10000 <= i && i <= 99999){
       n++;
       continue;
     }
     else if(i = 100000){
       continue;
     }
   }
   printf("%d",n);
   return 0;
}