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;
    scanf("%d",&n);

    int sum = 0;
    if(n == 100000){
        printf("90909");
        return 0;
    } 

    if(n > 9999){
        for (;n > 9999;n--){
            sum++;
        }
    }

    if (n > 999){
        n = 999;
    }

    if(n > 99) {
        for (;n > 99;n--){
            sum++;
        }
    }

    if (n > 9){
        n = 9;
    }

    if(n > 0){
        for(;n > 0;n--){
            sum++;
        }
    }

    printf("%d",sum);
}