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;
#define int long long
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rrep(i,n) for(int (i)=((n)-1);(i)>=0;(i)--)
#define itn int
#define all(x) (x).begin(),(x).end()
#define F first
#define S second
using P = pair <int,int>; //to, cost
using WeightedGraph = vector<vector <P>>; //to cost
const long long INF = 1LL << 60;
const int MOD = 1000000007;
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; }
//https://www.creativ.xyz/dump-cpp-652/
#define repi(itr, ds) for (auto itr = ds.begin(); itr != ds.end(); itr++)
 
// vector
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
    for (T &x : vec) is >> x;
    return is;
}
// pair
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
}
// vector
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
    os << "{";
    for (int i = 0; i < vec.size(); i++) {
        os << vec[i] << (i + 1 == vec.size() ? "" : ", ");
    }
    os << "}";
    return os;
}
// map
template <typename T, typename U>
ostream &operator<<(ostream &os, map<T, U> &map_var) {
    os << "{";
    repi(itr, map_var) {
        os << *itr;
        itr++;
        if (itr != map_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
// set
template <typename T>
ostream &operator<<(ostream &os, set<T> &set_var) {
    os << "{";
    repi(itr, set_var) {
        os << *itr;
        itr++;
        if (itr != set_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
 
#define DUMPOUT cerr
 
void dump_func() {
    DUMPOUT << endl;
}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
    DUMPOUT << head;
    if (sizeof...(Tail) > 0) {
        DUMPOUT << ", ";
    }
    dump_func(std::move(tail)...);
}

#ifdef DEBUG_
#define DEB
#define dump(...)                                                              \
    DUMPOUT << "  " << string(#__VA_ARGS__) << ": "                            \
            << "[" << to_string(__LINE__) << ":" << __FUNCTION__ << "]"        \
            << endl                                                            \
            << "    ",                                                         \
        dump_func(__VA_ARGS__)
#else
#define DEB if (false)
#define dump(...)
#endif

signed main(void) { cin.tie(0); ios::sync_with_stdio(false);
    string s; cin>>s;

    int n = s.size();
    int k; cin>>k;
    
    vector<vector<vector<int>>> dp (2, vector<vector<int>>(k+10, vector<int>(n+10)));
     //以下なら0 ピタリは1
    dp[1][0][0] = 1;
   
    rep(j,n){
        rep(i,k+1){ //iこ使った
            if(s[j] == '0'){
                //使う
                dp[0][i+1][j] += 9*dp[0][i][j];
                //使わない
                dp[1][i][j+1] += dp[1][i][j];
                dp[0][i][j+1] += dp[0][i][j];


            }else if(s[j] != '0'){

                //使う
                dp[1][i+1][j+1] += dp[1][i][j];
                dp[0][i+1][j+1] += (s[i] - '0'-1)*dp[1][i][j];
                dp[0][i+1][j+1] += 9*dp[0][i][j];

                //使わない
                dp[0][i][j+1] += dp[1][i][j];
                dp[0][i][j+1] += dp[0][i][j];

            }
        }
    }
    
    cout<<dp[0][k][n] + dp[1][k][n]<<endl;
}