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>
typedef long long ll;

int main(){
    int n,i,ans=0;
    scanf("%d",&n);
    int x=n;
    for(i=1;i<=6;i++){
        if(x<10) break;
        x/=10;
    }
    switch(i){
        case 1: ans+=n; break;
        case 3: ans+=n-99;
        case 2: ans+=9; break;
        case 5: ans+=n-9999;
        case 4: ans+=909; break;
        case 6: ans+=90909; break;
    }
    printf("%d",ans);
}