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)

// B - Uneven Numbers
#include <stdio.h>

int main(void){
    int n, i;
    int sum=0;
    scanf("%d", &n);
    for(i=1; i<=n; i++){
        if(0<i && i<10){
            sum++;
        }else if(99<i && i<1000){
            sum++;
        }else if(9999<i && i<100000){
            sum++;
        }
    }
    printf("%d\n", sum);
    return 0;
}