Problem:
Score : 300 points

Problem Statement

Find the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Constraints

1\leq N \leq 10^{16}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Sample Input 1

100

Sample Output 1

18

For example, the sum of the digits in 99 is 18, which turns out to be the maximum value.

Sample Input 2

9995

Sample Output 2

35

For example, the sum of the digits in 9989 is 35, which turns out to be the maximum value.

Sample Input 3

3141592653589793

Sample Output 3

137

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;

namespace agc021a
{
    class Program
    {
        static void Main(string[] args)
        {
            string N = Console.ReadLine();
            long num = long.Parse(N);
            long ans = 0;
            //10未満だったらその値
            if(num<10){
                ans = num;
            }else{
                double digit = Digit(num);
                Console.WriteLine(digit);
                int tekito = (int)(num/(Math.Pow(10,digit-1)));
                long sum = (long)((tekito-1)*Math.Pow(10,digit-1));
                for(int i = (int)digit-2; i >= 0;i--){
                    sum += (long)(9*Math.Pow(10,i));
                    //Console.WriteLine((long)(9*Math.Pow(10,i)));
                }
                ans = sum;
                //Console.WriteLine(ans);
            }
            Console.WriteLine(digit_sum(ans));
        }
        public static long digit_sum(long n)
        {
            if(n<10) return n;
            return digit_sum(n/10)+n%10;
        }
        public static double Digit(long num){
            return (num == 0) ? 1: ((int)Math.Log10(num)+1);
        }
    }
}