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: Wrong Answer)

#include<stdio.h>
int main(){
  int a,copy;
  scanf("%d",&a);
  int keta=1;
  copy = a;
  while(a>9){
    keta += 1;
    a = a/10;
  }
  int ans;
  if(keta==6){
    ans = 90909;
  }
  else if(keta==5){
    ans = 109;
    ans += (copy-10000+1);
  }
  else if(keta==4){
    ans = 109;
  }
  else if(keta==3){
    ans = 9;
    ans += (copy-100+1);
  }
  else if(keta==2){
    ans = 9;
  }
  else if(keta==1){
    ans = copy;
  }
  printf("%d",ans);
  return 0;
}