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 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using lint = long long;
using P = pair<int, int>;
using vec = vector<int>;
using mat = vector<vector<int>>;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define endl "\n"

constexpr int MOD = 1000000007;
const int INF = 1 << 30;

int main() {
    string s;
    cin >> s;
    int n = s.size();

    // dp[digit][smaller]
    int dp[20][2] = {};
    dp[0][1] = -INF;

    rep(i, n) {
        int ni = i+1;
        int nd = s[i] - '0';
        rep(j, 2) {
            rep(d, 10) {
                int nj = j;

                if (j == 0) {
                    if (d < nd) nj = 1;
                    if (d > nd) continue;
                }

                dp[ni][nj] = max(dp[ni][nj], dp[i][j] + d);
            }
        }
    }
    cout << max(dp[n][0], dp[n][1]) << endl;
    return 0;
}