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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(){
	int N=0;
  	int tmp=1;
  	int i;
  	int ans=0;
    scanf("%d",&N);
  for(i=1;i<7;i++){
  	if(tmp>N){
    	break;
    }
    tmp=tmp*10;
  }
  i--;
  if(i==1){
    	ans=N;
  }else if(i==2){
  	ans=9;
  }else if(i==3){
  	ans=9+N-100+1;
  }else if(i==4){
  	ans=900+9;
  }else if(i==5){
  	ans=9+900+N-10000+1;
  }else if(i==6){
  	ans=9+900+90000;
  }
	printf("%d",ans);

	return 0;

}