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 X=0,d=0,c=0,count=0;
    scanf("%d",&X);
    for(int i=1;i<=X;i++)
        {
    for(d=i;d!=0;++c)
    {
        d=d/10;
    }
    if(c%2!=0)
    {
        count++;
    }
    c=0;
}
   printf("%d ",count);
    return 0;
}