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: Accepted)

#include <stdio.h>
int main(void){
    int N, count=0;
    scanf("%d", &N);
    int a, b=10, c;
    int i=1;
    int k=1;
    while(1){
        a = i/b;
        if(a==1){
            k*=100;
            i=k;
            b*=100;
        }
        if(i>N){
            break;
        }
        count++;
        if(i>=N){
            break;
        }
        i++;
    }
    printf("%d\n", count);
    return 0;
}