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){
    /*配列に正の整数を格納*/
    int i = 0,n;
    scanf("%d",&n);
    int array[n];
    while(i < n){
        array[i] = i + 1;
        i++;
    }

    /*一つずつカウント*/
    int count = 0;
    i = 0;
    while(i < n){
        if((array[i] > 9999 && array[i] <100000) || (array[i] > 99 && array[i] < 1000) || array[i] < 10){
            count++;
        }
        i++;
    }
    printf("%d\n",count);

    return 0;
}