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: Accepted)

#include<stdio.h>
#include<math.h>

int keta(int n1){
  int a=0;
  while(n1!=0){
    n1=n1/10;
    a++;
  }
  return a;
}
  
int main(){
  int N,ans=0;
  scanf("%d",&N);
  int n=keta(N);
  if(n%2==1){
    if (n==1){
      printf("%d",N);
    }
    else{
      for(int i=1;i<n;i++){
        if(i%2==1){
          ans+=9*(pow(10,i-1));
        }
      }
      ans+=N-pow(10,n-1)+1;
      printf("%d",ans);
    }
  }
  else{
    for(int i=0;i<n;i++){
      if (i%2==1){
        ans+=9*(pow(10,i-1));
      }
    }
    printf("%d",ans);
  }
  return 0;
}