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: Accepted)

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<cstdio>
#include<cmath>
#include<deque>
#include<numeric>
#include<queue>
#include<stack>
#include<cstring>
#include<limits>
#include<functional>
#include<unordered_set>
#include<iomanip>
#include<cassert>
#include<regex>
#include<bitset>
#include<complex>
#include<chrono>
#define rep(i,a) for(int i=(int)0;i<(int)a;++i)
#define pb push_back
#define eb emplace_back
using ll=long long;
constexpr ll mod = 1e9 + 7;
constexpr ll INF = 1LL << 60;
 
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
 
using namespace std;
 
template <typename A, typename B>
string to_string(pair<A, B> p);
 
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
 
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
 
string to_string(const string& s) {
  return '"' + s + '"';
}
 
string to_string(const char* s) {
  return to_string((string) s);
}
 
string to_string(bool b) {
  return (b ? "true" : "false");
}
 
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
 
template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
 
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
 
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
 
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";
}
 
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
 
void debug_out() { cerr << "\n"; }
 
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " <<to_string(H);
  debug_out(T...);
}
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

int dp[105][2][5];
void solve(){
  string s;
  cin>>s;
  int k;
  cin>>k;
  //桁DP
  memset(dp,0,sizeof(dp));
  dp[0][0][0]=1;
  rep(i,s.size()) {//桁数が大きいのでstringで読み取り、各位ごとに数字に変換する
		int D = s[i] - '0';
		rep(j,2) {
			rep(h,4) {
				for (int d = 0; d <= (j ? 9 : D); ++d)//i桁目がNのi桁目より小さければ9まではいる。そうでなければNのi桁目の数字以下が入る
          if(d==0&&h==k)dp[i+1][j||d<(j?9:D)][k]+=dp[i][j][k];
					else if(d!=0)dp[i + 1][j || d < (j ? 9 : D)][h+1] += dp[i][j][h];
          else dp[i + 1][j || d < (j ? 9 : D)][h] += dp[i][j][h];
			}
		}
//N未満ならdp[s.size()][1][条件]だけでよい
	}
  cout<<dp[s.size()][1][k]+dp[s.size()][0][k]<<endl;
}

int main(){
	ios::sync_with_stdio(false);
  cin.tie(0);
	cout<<fixed<<setprecision(15);
	solve();
	return 0;
}