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(void){
    int N,M,p=1,q=1;
    scanf("%d",&N);
    p=N;
    while(N>0){
        N=N/10;
        M++;
        q=q*10;
    }
    if(M%2==0){
        if(M==2){
            printf("9");
        }else if(M==4){
            printf("909");
        }else{
            printf("90909");
        }
    }else{
        if(M==1){
            printf("%d",p);
        }else if(M==3){
            printf("%d",p-90);
        }else{
            printf("%d",p-9090);
        }
    }
}