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>
#include <math.h>

int main(void)
{
    int i,tmp;
    int N,keta=0;
    int count=0;

    scanf("%d",&N);

    for(i=0; i<=N; i++){
        keta=0;
        tmp = i;
        //桁を求める
        while(tmp > 0){
            tmp/=10;
            keta++;
        }
        //桁が奇数ならカウント
        if(keta%2 == 1) {
            //printf("i=%d\n",i);
            count++;
        }
    }

    printf("%d\n",count);

    return 0;
}