Problem:
Score : 500 points

Problem Statement

Find the number of integers between 1 and N (inclusive) that contains exactly K non-zero digits when written in base ten.

Constraints

1 \leq N < 10^{100}

1 \leq K \leq 3

Input

Input is given from Standard Input in the following format:

N
K

Output

Print the count.

Sample Input 1

100
1

Sample Output 1

19

The following 19 integers satisfy the condition:

1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100

Sample Input 2

25
2

Sample Output 2

14

The following 14 integers satisfy the condition:

11,12,13,14,15,16,17,18,19,21,22,23,24,25

Sample Input 3

314159
2

Sample Output 3

937

Sample Input 4

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
3

Sample Output 4

117879300

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;

string n;
int k,inf[2]= {-1,-1};
ull ans=0;

inline ull comb(const int &v,const int &l) {
	if(l<v) return 0;
	if(!v) return 1;
	if(v==1) return 9*l;
	return 81*l*(l-1)/2;
}

int main() {
	cin>>n>>k;
	for(int i=1; i<n.length(); ++i)
		if(n[i]!=48) {
			if(inf[0]==-1) inf[0]=i;
			else {
				inf[1]=i;
				break;
			}
		}
	for(int i=k,x,y; i<=n.length(); ++i) {
		if(i!=n.length()) ans+=9*comb(k-1,i-1);
		else {
			if(k==1) {
				ans+=n[0]-48;
				break;
			}
			ans+=(n[0]-49)*comb(k-1,i-1);
			if(inf[0]==-1) break;
			if(k==2) {
				ans+=n[inf[0]]-48+comb(1,n.length()-inf[0]-1);
				break;
			}
			if(inf[1]==-1) break;
			for(x=inf[0]; x<n.length()-1; ++x)
				for(y=x+1; y<n.length(); ++y) {
					if(x==inf[0]) ans+=(n[x]-49)*9+(y<=inf[1] ? n[y]-48:9);
					else ans+=81;
				}
		}
	}
	cout<<ans<<endl;
	return 0;
}