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>
int main()
{
    int n, i , x, y,sum = 0;

    scanf("%d", &n);

    if(n > 0 && n < 10)
    {
        for(i = 0; i < n ; i++)
        {
            sum++;
        }
        printf("%d\n",sum);
    }
    else if(n >= 10 && n <= 99)
        printf("9\n");
    else if(n > 99 && n < 1000)
    {
        x = n - 90;
        printf("%d\n",x);
    }
    else if(n > 999 && n < 10000)
    {
        printf("909\n");
    }
    else if(n > 9999 && n < 100000)
    {
        y = n - 9090 ;
        printf("%d\n", y);
    }
    else if(n > 99999)
        printf("90909\n");

    return 0;
}