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>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
int max(const void *a, const void *b){return *(int*)b - *(int*)a;}


int main(void)
{
    int n;
    scanf("%d",&n);

    if(n == 100000)
    {
        printf("%d\n", 9+ 900+ 90000);
    }else if( n >= 10000)
    {
        printf("%d\n", 9+ 900+ (n-10000+1));
    }else if( n >= 1000)
    {
        printf("%d\n", 9+ 900);
    }else if( n >= 100)
    {
        printf("%d\n", 9+ (n-100+1));
    }else if( n >= 10)
    {
        printf("%d\n",9);
    }else 
    {
        printf("%d\n", n);
    }

    
    return 0;
}