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, ketasu = 0, count = 0;
    scanf("%d", &n);
    int n_2 = n;
    while(n_2 != 0) {
        n_2 /= 10;
        ++ketasu;
    }
    if(ketasu % 2 == 0) {
        for(int i = 1; i <= ketasu/2; i++) {
            int j = i, ju = 1;
            while(j <= ketasu-i-1) {
                ju *= 10;
                j++;
            }
            count += 9*ju;
        }
    } else {
        for(int i = 1; i <= ketasu/2; i++) {
            int j = i, ju = 1;
            while(j <= ketasu-i-2) {
                ju *= 10;
                j++;
            }
            count += 9*ju;
        }
        if(ketasu == 5) {
            count += n-10000+1;
        } else if(ketasu == 3) {
            count += n-100+1;
        } else if(ketasu == 1) {
            count = n;
        }
    }
    printf("%d\n", count);
    return 0;
}