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: Runtime Error)

#include <stdio.h>
#include <math.h>

void main()
{
    long int n=0;
    scanf("%ld",&n);
    long c=0;
    for(long i=1;i<=n;i++)
    {
        for(int k=0;k<=4;k+=2)
        {
            if((i/(int)(pow(10,k)))!=0 && (i/(int)(pow(10,k+1))==0))
            {
                c++;
            }
        }

    }
    printf("%ld",c);
}