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>
#include<stdlib.h>
#include<math.h>
int main()
{
  int N,S,P;
  int i=1;
  int Z=0;
  int O=0;
 int k;
  scanf("%d",&N);
  S=N/10;
  while (S!=0) {
    i++;
  }
 if(i%2)
 {
   P=N-pow(10,i-1);
   for( k=2;k<i;k++)
   {
     Z+=pow(10,i-k-1)-1;
   }
 }
 printf ("%d\n",Z+P);
  if(i%2==0)
  {
 for( k=2;k<i;k++)
   {
      O+=pow(10,i-k-1)-1;
   }
 printf("%d\n",O);
}
}