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<iostream>
#include<cstdio>
#include<cstring>
using std::max;
inline void umx(int &a,int b){a=max(a,b);}

const int MAXN=20;

int n;
char s[MAXN];
int sum[MAXN];

int main()
{
	int i;scanf("%s",s);
	n=strlen(s);
	if(n==1) {printf("%d",s[0]-'0');return 0;}
	for(i=0;i<n;i++) sum[i]+=(i?sum[i-1]:0)+s[i]-'0';
	int ans=0;
	for(i=0;i<=n;i++)
		umx(ans,(i?sum[i-1]:0)+s[i]-1-'0'+(n-i-1)*9);
	printf("%d",ans);
	return 0;
}