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 a){
  int i;
  for(i=0;;i++){
    a=a/10;
    if(a==0)break;
  }
  return i+1;
}

int main(){
  int n,i,sum=0;
  scanf("%d",&n);
  for(i=1;i<n+1;i++){
    if(keta(i)%2==1)sum++;
  }
  printf("%d\n",sum);
}