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(void)
{
  int i,j,k=1,l=1,m=0;
  
  for(i=0;i<=4;i++){
    
    for(j=1;j<=i;j++){
       k *= 10;
    }
    
      
    for(l=1;l<=i+1;l++){
       l *= 10;
    }
    
    if((i%2)==0){ 
      m = m+(l-k-1);
    }
    
  }
  
  printf("%d",m);
  
  
  return 0;
}