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>
int main(){
  int i,j=1,n,num,count=0,len=1;
  scanf("%d",&n);
  num=n;
  while(num/10>0){
    num/=10;
    len++;
    j*=10;
  }
  int k=1;
  for(i=1;i<len;i=i+2){
    count+=9*k;
    k*=100;
  }
  if(len%2==1)
    count+=n+1-j;
  printf("%d\n",count);
  return 0;
}