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>
main()
{   int N,a,b,c,d,e;
scanf("%d",&N);
     if (N < 10)
        printf("%d",N);
    else if (N / 10 < 10)
        {   a=9;
            printf("%d",a);}
    else if (N / 100 < 10)
        {
            b=(9 + N - 99);
            printf("%d",b);
        }
    else if (N / 1000 < 10)
        {
            c=909;
            printf("%d",b);
        }
    else if (N / 10000 < 10)
       {

        d= (909 + N - 9999);
    printf("%d",d);
       }
        else {e=90909;
        printf("%d",e);
        }
    return 0;
}