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, sum=0;
scanf("%d",&N);
/*while(N){
    dig= N%10;
    sum+=dig;
    N/=10;
}*/
for(int i=0;i<=N;i++){
if(i>=1 && i<=9){
     sum++;
}else if(i>=100 && i<=999) {
    sum++;
}else if  (i>=10000 && i<=99999) {
    sum++;
}
}
printf("%d\n",sum);
return 0;
}