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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>

main()
{
    int N, z, y;
    scanf("%d",&N);

    if(N>=1 && N<=9)
    {
        z = N-0;
        printf("%d\n",z);
    }

    else if(N>=10 && N<=99)
    {
        z = 9;
        printf("%d\n",z);
    }

    else if(N>=100 && N<=999)
    {
        z = (N - 100) + 1;
        y = z + 9;
        printf("%d\n",y);
    }

    else if(N>=1000 && N<=9999)
    {
        z = 900 + 9;
        printf("%d\n",z);
    }

    else if(N>=10000 && N<=99999)
    {
        z = (N - 10000) + 1;
        y = z + 900 + 9;
        printf("%d\n",y);
    }

    else if(N>=100000 && N<=999999)
    {
        z = 90000 + 900 + 9;
        printf("%d\n",z);
    }

    return 0;
}