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 KETA(int n){
    int i=0;
    while(n!=0){
        n = n/10;
        i++;
    }
    return i;
}

int main(void){
    int N;
    scanf("%d",&N);
    int count = 0;
    //printf("%d",KETA(N));
    for(int i=1;i<=N;i++){
        if(KETA(i)%2==1) count++;
    }
    printf("%d",count);
    return 0;
}