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(){
    int N;
    while(~scanf("%d",&N)){
        int sum = 0;
        if(N < 10) sum = N;
        else if(N<100) sum = 9;
        else if(N<1000) sum = 9+(N-100+1);
        else if(N<10000) sum = 9+900;
        else if(N<100000) sum = 9 + 900 + (N-10000+1);
        else sum = 90909;

        printf("%d\n",sum);
    }
}