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>
#include <string.h>

int main() {

    int n;
    scanf("%d", &n);
    int mysum = 0;

    if (n >= 1 && n < 10) {
        mysum += n % 10;
    } else if (n >= 10 && n < 100) {
        mysum += 10 - 1;
    } else if (n >= 100 && n < 1000) {
        mysum += n % 100;
        mysum += 10 - 1;
    } else if (n >= 1000 && n < 10000) {
        mysum += 10 - 1;
        mysum += 1000 - 100;

    } else if (n >= 10000 && n < 100000) {
        mysum += n % 10000;
        mysum += 10 - 1;
        mysum += 1000 - 100;
    } else {
        mysum += 10 - 1;
        mysum += 1000 - 100;
        mysum += 100000 - 10000;

    }

    printf("%d", mysum);

    return 0;
}