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;
   scanf("%d", &n);
   if(n < 10){
       printf("%d", n);
   } else if(n < 100){
       printf("%d", 9);
   } else if(n < 1000){
       printf("%d", 9+(n-100+1));
   } else if(n < 10000){
       printf("%d", 9+900);
   } else {
       printf("%d", 9+900+90000);
   }
    return 0;
}