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>

int main(void)
{
    int n;

    scanf("%d",&n);

    int keta=0,i,limit;
    if(n>=0)
    {
        if(n>=10)
            limit=10;
        else
            limit=n;
        for(i=0;i<limit-1;i++)
        keta++;
    }
    if(n>=100)
    {
        if(n>=1000)
            limit=1000;
        else
            limit=n;
        for(i=100;i<limit+1;i++)
        keta++;
    }
    if(n>=10000)
    {
        if(n>=100000)
            limit=100000;
        else
            limit=n;
        for(i=10000;i<limit-1;i++)
        keta++;
    }

    printf("%d",keta);
    return 0;
}