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++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <sstream>
#include <iomanip> 
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <functional>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <bitset>

using namespace std;

long long digitsum(long long n) {
	long long ans = 0;
	while (n > 0) {
		ans += n % 10;
		n /= 10;
	}
	return ans;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);

	long long n, i, j, k, x, v, d, ans;
	string s;
	cin >> n;
	s = to_string(n);
	ans = digitsum(n);
	d = 1;
	for (i = 1; i < s.length(); i++) {
		n /= 10;
		d *= 10;
		x = (n - 1) * d + (d - 1);
		ans = max(ans, digitsum(x));
	}
	cout << ans << "\n";

	return 0;
}