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 n,out;
        scanf("%d",&n);
        if(n>9999)
        {
            out=n-9999+900+9;
            if(n==100000)
                out--;
        }
        else if(n>999)
            out=999-99+9;
        else if(n>99)
            out=n-99+9;
        else if(n>=9)
            out=9;
        else
            out=n;
        printf("%d",out);
        return 0;
    }