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 Go (1.6) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"math"
	"os"
	"strconv"
)

//Util
func nextStr() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func maxInt(a, b int) int {
	if a > b {
		return a
	} else {
		return b
	}
}

func minInt(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}

func absInt(a int) int {
	if a >= 0 {
		return a
	} else {
		return -a
	}
}

func absFloat64(a float64) float64 {
	if a >= 0 {
		return a
	} else {
		return -a
	}
}

func maxFloat64(a, b float64) float64 {
	if a > b {
		return a
	} else {
		return b
	}
}

func minFloat64(a, b float64) float64 {
	if a > b {
		return b
	} else {
		return a
	}
}

func str2Int(s string) int {
	n, _ := strconv.Atoi(s)
	return n
}

func reverse(s string) string {
	runes := []rune(s)
	for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
		runes[i], runes[j] = runes[j], runes[i]
	}
	return string(runes)
}

func powInt(p, q int) int {
	return int(math.Pow(float64(p), float64(q)))
}

func isPrime(x int) bool {
	if x == 1 {
		return false
	}
	if x == 2 {
		return true
	}
	if x%2 == 0 {
		return false
	}

	b := true
	rootx := int(math.Sqrt(float64(x)))
	i := 3
	for i <= rootx {
		if x%i == 0 {
			b = false
			break
		}
		i += 2
	}
	return b
}

func PrimeFactors(n int) (pfs []int) {
	// Get the number of 2s that divide n
	for n%2 == 0 {
		pfs = append(pfs, 2)
		n = n / 2
	}

	// n must be odd at this point. so we can skip one element
	// (note i = i + 2)
	for i := 3; i*i <= n; i = i + 2 {
		// while i divides n, append i and divide n
		for n%i == 0 {
			pfs = append(pfs, i)
			n = n / i
		}
	}

	// This condition is to handle the case when n is a prime number
	// greater than 2
	if n > 2 {
		pfs = append(pfs, n)
	}

	return
}

func sumInts(x []int) int {

	total := 0
	for _, v := range x {
		total += v
	}

	return total
}

//Main
const (
	initialBufSize = 10000
	maxBufSize     = 1000000
	mod            = 1e9 + 7
)

var (
	sc *bufio.Scanner = func() *bufio.Scanner {
		sc := bufio.NewScanner(os.Stdin)
		buf := make([]byte, initialBufSize)
		sc.Buffer(buf, maxBufSize)
		return sc
	}()
)

var N string

func main() {
	sc.Split(bufio.ScanWords)

	N = nextStr()
	d := len(N)
	n, _ := strconv.Atoi(N)
	a, _ := strconv.Atoi(string(N[0]))

	if a*powInt(10, d-1)+powInt(10, d-1)-1 == n {
		fmt.Println(a + 9*(d-1))
	} else {
		fmt.Println((a - 1) + 9*(d-1))
	}

}