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<bits/stdc++.h>
using namespace std;
//#include<boost/multiprecision/cpp_int.hpp>
//using boost::multiprecision::cpp_int;

#define int long long
#define cint cpp_int
#define endl "\n"
#define REP(i,a,n) for(int i=a;i<n;++i)
#define REPR(i,a,n) for(int i=a;i>n;--i)
#define RUP(a,b) (((a)+(b)-1)/(b))
#define ALL(v) (v).begin(),(v).end()
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define MOD 1000000007
#define INF LLONG_MAX/2
#define PI acos(-1.0)
#define leng(n) (int)(log10(n)+1)


typedef long long ll;
typedef pair<int,int> Pii;
typedef tuple<int,int,int> Tiii;
typedef vector<int> Vi;
typedef vector<Vi> VVi;
typedef vector<Pii> VPii;
typedef vector<string> Vs;

template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T> void YesNo(T a) {cout<<(a?"Yes":"No")<<endl;}
template<class T> void YESNO(T a) {cout<<(a?"YES":"NO")<<endl;}
void vout(Vi &v){for(int i=0;i<(v).size();i++) cout<<v[i]<<" ";cout<<endl;}
void vvout(VVi &v){for(int i=0;i<(v).size();i++) vout(v[i]);}
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a,b)*b;}
int mypow(int x, int n, int m){if(n==0)return 1;if(n%2==0)return mypow(x*x%m,n/2,m);else return x*mypow(x,n-1,m)%m;}
int digit(int k,int i){string s = to_string(k);return s[s.size()-i]-'0';}
void uniq(Vi &v){sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end());}
int ctoi(char c){if(c>='0'&&c<='9'){return c-'0';}return 0;}
Vi accum(Vi v){Vi ret((v).size());ret[0]=v[0];REP(i,1,(v).size()) ret[i]=ret[i-1]+v[i];return ret;}
bool comp(Pii a,Pii b){
    if(a.second != b.second) return a.second<b.second;
    else return a.first<b.first;
}

signed main(){cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(12);
    string s;
    int k;
    cin>>s>>k;
    int n=s.size();

    vector<VVi> dp(n+1,VVi(2,Vi(n+1)));
    dp[0][0][0]=1;

    REP(i,0,n){
        REP(j,0,2){
            REP(l,0,n){
                int D = s[i]-'0';
                REP(d,0,(j?9:D)+1){
                    if(d==0){
                        dp[i+1][j || (d<D)][l] += dp[i][j][l];
                    }else{
                        dp[ i + 1 ][ j || ( d < D ) ][l+1] += dp[ i ][ j ][l];
                    }
                }
            }
        }
    }
    cout<<dp[n][0][k]+dp[n][1][k]<<endl;
    return 0;
}