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 n;
    scanf("%d",&n);
    int dig=0,c=n;
    while(c>0){
        c/=10;
        dig++;
    }
    if(dig==6){
        printf("90909");
        return 0;
    }
    if(dig==1){
        printf("%d",n);
    }
    else if(dig==2){
        printf("9");
    }else if(dig==4){
        printf("909");
    }else if(dig==3){
        printf("%d",9+n-99);
    }else {
        printf("%d",909+n-9999);
    }
    return 0;
}