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: Wrong Answer)

//#include <bits/stdc++.h>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <limits>
#include <map>
#include <vector>
#include <queue> 
#define LL long long
#define ft first
#define sd second
#define mp(x,y) make_pair(x,y)
//#define int LL
using namespace std;
const int N   = 233;
//const int mod = ;
const int INF =numeric_limits<int >::max();
#define For(n) for (int i=1;i<=n;++i)
#define rep(i,x,y) for (int i=x;i<=y;++i)
void read(int &x)
{
	x=0;
	char ch=getchar();
	int f=1;
	while (!isdigit(ch)) (ch=='-'?f=-1:0),ch=getchar();
	while ( isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	x*=f;
}

int n,ans;
string s;
int f[233];
int solve(int dep)

{
	if (dep>n) return 0;
	//if (dep==n) return k;
	if (f[dep]!=-1) return f[dep];
	int k=s[dep]-'0';
	if (dep==n) return k;
	int tmp=solve(dep+1)+k;
	if (dep!=n&&k>0) tmp=max(k-1+9*(n-dep-1),tmp);
	return f[dep]=tmp;
}
signed main()
{
	cin>>s;
	n=s.length();
	memset(f,-1,sizeof(f));
	return printf("%d",solve(0)),0;
}