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)

int main(){
  int n;
  int cnt;
 scanf("%d",&n);
 
  if(n>=100000){
    n=n-(n/100000*100000);
  }
  if(n>=10000){
    cnt=cnt+(n-10000);
    n=n-(n/10000*10000);
  }
  if(n>=1000){
    n=n-(n/1000*1000);
  }
  if(n>=100){
    cnt=cnt+(n-100);
    n=n-(n/100*100);
  }
  if(n>=10){
    n=n-(n/10*10);
  }
  if(n>=0){
    cnt=cnt+n;
  }
  printf("%d",cnt);
  
}