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: Wrong Answer)

#include<stdio.h>
int main()
{
    int num ,i,j,k,l,ans ;
    scanf("%d",&num);
    if(num>100 && num<1000)
    {
        ans = num-91 ;
    }
    else if(num<10)
    {
        ans=num;
    }
    else if(num>=1000 && num<10000)
    {
        ans = 909;
    }
    else if(num>=10000 &&num<100000)
    {
        ans = num-10000+1+909;
    }
    printf("%d",ans);

    return 0;
}