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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int dp[105][4][2];

int main() {
  string s;
  cin >> s;
  int n = s.size();
  int K;
  cin >> K;
  dp[0][0][0] = 1;
  rep(i,n)rep(j,4)rep(k,2) {
    int nd = s[i]-'0';
    rep(d,10) {
      int ni = i+1, nj = j, nk = k;
      if (d != 0) nj++;
      if (nj > K) continue;
      if (k == 0) {
        if (d > nd) continue;
        if (d < nd) nk = 1;
      }
      dp[ni][nj][nk] += dp[i][j][k];
    }
  }
  int ans = dp[n][K][0] + dp[n][K][1];
  cout << ans << endl;
  return 0;
}