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, ans = 0;
    scanf("%d", &N);
    if (N / 10 == 0) {
        ans += N;
        printf("%d", ans);
        return 0;
    }
    else {
        ans += 9;
    }
    if (N < 100) {
        printf("%d", ans);
        return 0;
    }
    if (N / 1000 == 0) {
        ans += N - 99;
        printf("%d", ans);
        return 0;
    }
    else {
        ans += 900;
    }
    if (N < 10000) {
        printf("%d", ans);
        return 0;
    }
    if (N / 100000 == 0) {
        ans += N - 9999;
        printf("%d", ans);
        return 0;
    }
    printf("90909");
    return 0;
}