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 <string.h>
int main(void)
{
  int n,i,x,A[6];
  char s[7];
  
  scanf("%d",&n);
  
  A[0]=n/100000;
  A[1]=(n%100000)/10000;
  A[2]=(n%100000%10000)/1000;
  A[3]=(n%100000%10000%1000)/100;
  A[4]=(n%100000%10000%1000%100)/10;
  A[5]=((((n%100000)%10000)%1000)%100)%10;
  x=0;
  for(i=0;i<6;i++)
    s[i]=A[i]+'0';
  s[6]='\0';
  for(i=1;i<=n;i++)
    if(strlen(s)%2==1) x++;
  printf("%d",x);
  return 0;
}