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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
 
int main()
{
	  
   int N;
   int Ans;
   
    scanf("%d", &N);
    if(N> 100000) N=100000; 
    if(N<1) N=1;

   
    if(N==100000){
        Ans = 9 + (100*9) + (10000*9); // 90909
    }
    else if(N >= 10000){
        Ans = 9 + 900 + N%10000;
    }
    else if(N>=1000){
        Ans = 9 + 900;
    }
    else if(N>=100){
        Ans = 9 + N%100 + 1;
    }
    else if(N>=10){
        Ans = 9; 
    }
    else{
        Ans = N;
    }
    
    
    printf("%d",Ans);
    
  return 0;  
}