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 c = 0;
    int s; s = N;
    while(N!=0)
    {
        N/=10;
        c++;
    }
    N = s;
    if(c == 5 )
    {
        N-=9999;
        N+=909;
    }
    if(c == 6)
    {
        printf("90909\n"); return 0;
    }
    if(c == 4)
    {
        printf("909\n"); return 0;
    }
    if(c == 3)
    {
        N-=90;
    }
    if(c == 2)
    {
        printf("9\n"); return 0;
    }
    printf("%d\n",N);
    return 0;
}