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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

using namespace std;
#define ll long long
#define rint register int
#define gc() getchar()
inline ll read(ll ans = 0, rint sgn = ' ', rint ch = gc())
{
	for(; ch < '0' || ch > '9'; sgn = ch, ch = gc());
	for(; ch >='0' && ch <='9';(ans*=10)+=ch-'0', ch = gc());
	return sgn-'-'?ans:-ans;
}
ll N; int a[20], L;
int DP(int L, int limit){if(!L) return 0; if(!limit) return 9*L; return max(DP(L-1,1)+a[L],a[L]>0?DP(L-1,0)+a[L]-1:0);}
int Calc(ll N){for(; N; a[++L] = N%10, N /= 10); return DP(L,1);}
int main(){N = read(), printf("%d\n",Calc(N)); return 0;}