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("%i", &N);
    int n = N;
    int a = 1;
    while (N > 10){
        N /= 10;
        a++;
    }
    if (a == 1){
        printf("%i\n", n);
    } else if (a == 2){
        printf("%i\n", 9);
    } else if (a == 3){
        printf("%i\n", n - 99 + 9);
    } else if (a == 4){
        printf("%i\n", 999 - 99 + 9);
    } else if (a == 5){
        printf("%i\n", n - 9999 + 999 - 99 + 9);
    } else if (a == 6){
        printf("%i\n", 99999 - 9999 + 999 - 99 + 9);
    }
}