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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack> 
#include <set>
#include <map>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <bits/stdc++.h>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
using edge = struct {ll to; ll cost;};
using point = struct {ll x; ll y;};
typedef string str;
typedef std::pair<ll,ll> pl;
typedef std::pair<ll,pl> pl3;
typedef std::map<string, ll> msl;
typedef std::map<char, ll> mcl;
typedef std::map<ll, ll> mll;
typedef std::vector<ll> vl;
typedef std::vector<pl> vpl;
typedef std::vector<point> vxy;
typedef std::vector<pl3> vpl3;
typedef std::priority_queue<ll> pq;
typedef std::priority_queue<ll, vl, greater<ll>> pql; // priority queue taking from the lower value.
typedef std::vector<edge> gr;
const ll MOD = 1e9 + 7;
const ll INF = MOD * MOD;
const long double EPS = 1e-9;
const long double PI = 3.14159265358979323846;

#define rep(i, n) for(ll (i) = 0; (i) < (n) ; (i)++ )
#define revrep(i, n) for(ll (i) = n - 1; (i) >= 0 ; (i)--)
#define For(i, a, b) for(ll (i) = (a); (i) < (b) ; (i)++ )
#define revFor(i, b, a) for(ll (i) = (b) - 1; (i) >= (a) ; (i)--)
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define isUpper(c) ('a' - c > 0)
#define toLower(c) char((c)+0x20)
#define toUpper(c) char((c)-0x20)
#define pb push_back
#define mp make_pair
#define pr(a) cout << (a)
#define prl(a) cout << (a) << endl
#define prl2(a,b) cout << (a) << " " << (b) << endl
#define prl3(a,b,c) cout << (a) << " " << (b) << " " << (c) << endl
#define prl4(a,b,c,d) cout << (a) << " " << (b) << " " << (c) << " " << (d) << endl
#define prs(a) cout << (a) << " "
#define yn(condition) if ((condition)) prl("Yes"); else prl("No");
#define YN(condition) if ((condition)) prl("YES"); else prl("NO");
#define in1(a) cin >> (a)
#define in2(a,b) cin >> (a) >> (b)
#define in3(a,b,c) cin >> (a) >> (b) >> (c)
#define in4(a,b,c,d) cin >> (a) >> (b) >> (c) >> (d)
#define e1 first
#define e2 second
#define ctol(c) ll((c)) - ll('0')
#define ltos(n) to_string((n))
#define items(kv, v) for(auto &(kv): (v))
#define ndig(N, n) ctol(ll(ltos((N))[ll(ltos((N)).length()) - (n)]))
#define rsort(a,n) sort(a,a+n,greater<>())
#define Forchar(c,a,z) for(char (c) = (a); (c) <= (z); (c)++)
#define cntchar(s,c) count(all((s)), c)
#define substring(s, start, end) s.substr((start), (end) - (start) + 1)
#define prl_nd(num, digits) cout << fixed << setprecision(digits) << (num) << endl;
#define XOR(a, b) (a)^(b)
#define prl_time(s) prl3("Elapsed Time:", 1000.0 * (clock() - s) / CLOCKS_PER_SEC, "[ms]");
#define char_to_str(c) string(1, (c))

class UnionFind {
    vl parents, set_size;
public:
    UnionFind(){}
    UnionFind(ll n){
        parents = set_size = vl(n);
        rep(i, n) {parents[i] = i; set_size[i] = 1LL;}
    }

    ll root_find(ll x){
        if (parents[x] == x) return x;
        return parents[x] = root_find(parents[x]);
    }

    void unite(ll x, ll y){
        x = root_find(x); y = root_find(y);
        if (x == y) return;
        if (set_size[x] < set_size[y]){
            parents[y] = x; set_size[x] += set_size[y];
        }else{
            parents[x] = y; set_size[y] += set_size[x];
        }
    }

    bool is_same(ll x, ll y) {// connected or not
        return root_find(x) == root_find(y);
    }

    ll size(ll x) {
        return set_size[root_find(x)];
    }
};

/*
class LCA{
public:
    ll N, logN;
    vl depth, len;
    gr tree[200005]; // global declaration later.
    vector<vl> parents;

    LCA(ll n){
        N = n;
        logN = 0;
        while (N > (1LL << logN)) logN++;
        depth = vl(N); len = vl(N);
        parents = vector<vl>(logN, vl(N));
        init(0, -1, 0, 0);
        build();
    }

    void init(ll source, ll parent, ll d, ll l){
        depth[source] = d;
        parents[0][source] = parent;
        len[source] = l;
        rep(i, tree[source].size()){
            ll target = tree[source][i].to;
            ll cost = tree[source][i].cost;
            if (target == parent) continue;
            init(target, source, d + 1, cost + l);
        }
    }

    void build(){
        rep(k, logN - 1) rep(n, N){
            // if there is no parent, -1.
            // otherwise, the parent of the parent is the parent.
            if (parents[k][n] < 0) parents[k + 1][n] = -1;
            else parents[k + 1][n] = parents[k][parents[k][n]];
        }
    }

    ll query(ll u, ll v){
        if (depth[u] > depth[v]) swap(u, v);
        rep(k, logN) if ((depth[v] - depth[u]) >> k & 1) v = parents[k][v];
        if (u == v) return u;

        revrep(k, logN){
            if (parents[k][u] != parents[k][v]){
                u = parents[k][u]; v = parents[k][v];
            }
        }
        return parents[0][u];
    }

    ll distance(ll u, ll v){
        ll w = query(u, v);
        return len[u] + len[v] - 2 * len[w];
    }

};
*/

struct BIT{
    ll n; vl dat;
    BIT(ll n): dat(n + 1), n(n){};

    //       x: 1001 1010  1100 1011 1101  1111
    // x & - x: 0001 0010  0100 0001 0001  0001
    //      ->: 1010 1100 10000 1100 1100 10000

    ll sum(ll i){
        ll s = 0;
        while(i > 0){s += dat[i]; i -= i & -i;}
        return s;
    }

    ll sum(ll i, ll j){
        return sum(j - 1) - sum(i - 1);
    }

    void add(ll i, ll x){
        while(i <= n){dat[i] += x; i += i & -i;}
    }
};

ll gcd(ll m, ll n){
    ll a = max(m, n);
    ll b = min(m, n);
    while(b != 1 && b != 0){ a %= b; swap(a, b);}
    return b == 1 ? 1 : a;
}

ll lcm(ll m, ll n){ return m / gcd(m, n) * n;}

ll power_mod(ll a, ll power, ll mod){
    ll value = 1;
    while(power != 0){
        if(power & 1) value = (value * a) % mod;
        a = (a * a) % mod;
        power = power >> 1;
    }
    return value % mod;
}

ll modinv(ll a, ll mod){
    return power_mod(a, mod - 2, mod);
}

ll power_normal(ll a, ll power){
    ll value = 1;
    while(power != 0){
        if(power & 1) value = value * a;
        a = a * a;
        power = power >> 1;
    }
    return value;
}

ll comb_memo[55][55];

ll pascal_triangle(ll n){
    comb_memo[0][0] = 1;
    For(i, 1, n + 1) rep(j, i + 1){
        comb_memo[i][j] += comb_memo[i - 1][j];
        if (j > 0) comb_memo[i][j] += comb_memo[i - 1][j - 1];
    }
}

ll combination(ll n, ll r, ll mod){
    if (n == r && n == 0) return 1;
    else if (n <= 0 || r < 0 || r > n) return 0;
    ll numerator = 1;
    ll denomenator = 1;
    for(ll i = 0; i < r ; i++){
        (numerator *= ll(n - i)) %= mod;
        (denomenator *= modinv(i + 1, mod)) %= mod;
    }

    return (numerator * denomenator) % mod;
}

ll combination_memo(ll n, ll r, ll pre, ll mod){
    if (n == r && n == 0) return 1;
    else if (n <= 0 || r < 0 || r > n) return 0;
    ll numerator = ll(n - r + 1) % mod;
    ll denomenator = modinv(r, mod) % mod;
    ll val = (numerator * denomenator) % mod;
    val *= pre;
    return val % mod;
}

ll bin_search(ll key, ll A[], ll left, ll right){
    // return the index idx where A[idx] = key.
    // A[left] is start and A[right] is end..
    // In other words, A[right], not A[right - 1], must be defined.
    while(right >= left){
        ll mid = left + (right - left) / 2;
        if (A[mid] == key) return mid;
        else if (A[mid] > key) right = mid - 1;
        else if (A[mid] < key) left = mid + 1;
    }
    return -1;
}

/*
ll bin_search_temp(ll left, ll right, callable judge){
    while(right > left){
        // when seeking lower bound
        ll mid = (right + left) / 2;
        if (judge(mid)) right = mid;
        else left = mid + 1;
        // when seeking upper bound
        ll mid = (right + left + 1) / 2;
        if (judge(mid)) left = mid;
        else right = mid - 1;
    }

    return right;   
}
*/

ll lower_bound_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value right instead of -1.
    if (A[left] >= key) return left;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == left) {
            if (A[left] < key && key <= A[left + 1]) return left + 1;
            else return -1;
        }
        if (A[mid - 1] < key && key <= A[mid]) return mid;
        else if (A[mid - 1] >= key) right = mid - 1;
        else if (A[mid] < key) left = mid;
    }

    return -1; // all the elements < key
}

ll inf_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value right instead of -1.
    if (A[left] > key) return left;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == left) {
            if (A[left] <= key && key < A[left + 1]) return left + 1;
            else return -1;
        }
        if (A[mid - 1] <= key && key < A[mid]) return mid;
        else if (A[mid - 1] > key) right = mid - 1;
        else if (A[mid] <= key) left = mid;
    }

    return -1; // all the elements <= key
}

ll upper_bound_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value left instead of -1.
    if (A[right] <= key) return right;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == right) {
            if (A[right - 1] <= key && key < A[right]) return right - 1;
            else return -1;
        }
        if (A[mid] <= key && key < A[mid + 1]) return mid;
        else if (A[mid] > key) right = mid;
        else if (A[mid + 1] <= key) left = mid + 1;
    }

    return -1; // all the elements > key
}

ll sup_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value left instead of -1.
    if (A[right] < key) return right;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == right) {
            if (A[right - 1] < key && key <= A[right]) return right - 1;
            else return -1;
        }
        if (A[mid] < key && key <= A[mid + 1]) return mid;
        else if (A[mid] >= key) right = mid;
        else if (A[mid + 1] < key) left = mid + 1;
    }

    return -1; // all the elements >= key
}

ll bin_search_vector(ll key, vl v, ll left, ll right){
    // return the index idx where v[idx] = key.
    // v[left] is start and v[right] is end..
    // In other words, v[right], not v[right - 1], must be defined.
    while(right >= left){
        ll mid = left + (right - left) / 2;
        if (v[mid] == key) return mid;
        else if (v[mid] > key) right = mid - 1;
        else if (v[mid] < key) left = mid + 1;
    }
    return -1;
}

ll lower_bound_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N - 1] < key <= v[N]
    // N == -1 if all the elements < key
    return lower_bound(all(v), key) - v.begin();
}

ll inf_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N - 1] <= key < v[N] <= key + 1
    // N == -1 if all the elements <= key
    return lower_bound(all(v), key + 1) - v.begin();
}

ll upper_bound_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N] <= key < v[N + 1]
    // N == -1 if all the elements > key
    return upper_bound(all(v), key) - v.begin() - 1;
}

ll sup_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N] <= key - 1 < key <= v[N + 1]
    // N == -1 if all the elements >= key
    return upper_bound(all(v), key - 1) - v.begin() - 1;
}

ll fact(ll n){
    if (n == 0) return 1;
    return n * fact(n - 1);
}

bool is_prime(ll n){
    if (n <= 1) return false;
    for(ll i = 2; i * i <= n; i++){
        if (n % i == 0) return false;
    }
    return true;
}

ll bool_sum(ll a1, ll a2){
    if (a1 == 1 || a2 == 1) return 1;
    return 0;
}

mll prime_decomposition(ll n){
    ll i = 2;
    mll table;
    while(i * i <= n){
        while(n % i == 0){
            table[i] ++;
            n /= i;
        }
        i++;
    }
    if (n > 1) table[n] = 1;
    return table;
}

ll char_to_idx(char c){
    ll idx = 0;
    Forchar(cc, 'a', 'z'){
        if (c == cc) return idx;
        else idx ++;
    }
}

bool is_cross(ll x1, ll y1, ll x2, ll y2, ll x3, ll y3, ll x4, ll y4){
    ll val1 = (x1 - x2) * (y3 - y1) + (y1 - y2) * (x1 - x3);
    ll val2 = (x1 - x2) * (y4 - y1) + (y1 - y2) * (x1 - x4);
    ll val3 = (x3 - x4) * (y1 - y3) + (y3 - y4) * (x3 - x1);
    ll val4 = (x3 - x4) * (y2 - y3) + (y3 - y4) * (x3 - x2);

    return val1 * val2 < 0 && val3 * val4 < 0;
}

/*
const ll N_VERTEX = 310;
ll a, b, t;
ll dist[N_VERTEX][N_VERTEX];

void warshall_floyd(ll n){
    // rep(i, n) rep(j, n) dist[i][j] = INF * (i != j);
    rep(k, n) rep(i, n) rep(j, n) dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
}

int main(void){
    in2(n, m);
    rep(i, n) rep(j, n) dist[i][j] = INF * (i != j);
    rep(i, m){
        in3(a, b, t);
        a--; b--;
        dist[a][b] = t;
        dist[b][a] = t;
    }

    warshall_floyd(n);
}

const ll N_VERTEX = 310;
gr graph[N_VERTEX];
ll dist[N_VERTEX], a, b, t, n, m;

void dijkstra(ll start, ll n) {
    priority_queue<pl, vector<pl>, greater<pl>> edge_costs;
    fill(dist, dist + n, INF);
    dist[start] = 0;
    edge_costs.push(pl(0, start));

    while (!edge_costs.empty()) {
        pl edge_cost = edge_costs.top();
        edge_costs.pop();
        ll idx = edge_cost.second;
        ll cost = edge_cost.first;
        if (dist[idx] < cost) continue;

        rep(i, graph[idx].size()){
            edge e = graph[idx][i];
            if (dist[e.to] > dist[idx] + e.cost){
                dist[e.to] = dist[idx] + e.cost;
                edge_costs.push(pl(dist[e.to], e.to));
            }
        }
    }
}

int main(void){
    in2(n, m);
    rep(i, m){
        in3(a, b, t);
        a--; b--;
        G[a].pb({b, t});
        G[b].pb({a, t});
    }

    dijkstra(0, n);
}


# ABC061D

bool bellman_ford(ll start, ll n, ll goal){
    // if there is a closed circuit, it returns false. (when goal == -1)
    // if the distance to goal cannot be obtained, it returns false (when goal != -1)
    fill(dist, dist + n, INF);
    dist[start] = 0;
    rep(i, n) rep(v, n) rep(k, graph[v].size()){
        edge e = graph[v][k];
        if (dist[e.to] > dist[v] + e.cost){
            dist[e.to] = dist[v] + e.cost;
            if (i == n - 1 && (e.to == goal || goal == -1)) return false;
        }
    }
    return true;
}

*/

/*
    # 1. The usage of map pair
    map<pl, ll> cnt;
    cnt[{i, j}] = 0;
    items(kv, cnt){
        prl2(kv.first, kv.second);
    }

    # 2. The usage of next_permutation and combination (factorial search)
    ll a[8];
    rep(i, 8) a[i] = i;
    sort(a, a + 8);
    do{
    }while(next_permutation(a, a+n));

    // here, combination
    ll n, r;
    ll i1, i2, ..., ir;
    For(i1, n - r, n) For(i2, n - r - 1, i1) ... For(ir, n - 2 * r + 1, irr){ process;}

    # 3. bit search
    ll n;
    in1(n);
    const ll base = 3;
    ll upper = power_normal(base, n);
    
    rep(i, upper){
        ll tmp = i;
        rep(j, n){
            rep(k, base) if (tmp % base == k) prl(k);
            tmp /= base;
        }
    }

    # 4. imos method
    // used when we would like to count the number which
    // shows how many times the numbers between l and r belongs to smt.
    // This method is composed of three process.
    ll n, m, s[MAX_M], l, r;
    in2(n, m);
    rep(i, m) s[i] = 0;

    // 1st step
    rep(i, n){
        in3(l, r, c);
        l--; r--; // if l starts from 1.
        s[l] += c; s[r + 1] -= c;
    }

    // 2nd step
    rep(i, m - 1) s[i + 1] += s[i];
    // 3rd step: judgement...

    #5. shakutori method (syakutori)
    // 1. strech right side while the condition is met.
    // 2. renew the answer
    // 3. increments left side
    // 4. Back to 1. (l <= r must be satisfied all the time.)

    ll l = 0; ll r = 0;
    while (l < n){
        r = max(r, l);
        if (l == r) r++;
        while(r < n && cond) r++;
        answer += r - l; l++;
    }
    prl(answer);

    #6. priority queue

    pq q;
    ll answer = 0;
    ll v;

    rep(i, n) q.push(a[i]);
    rep(i, m){
        v = q.top(); q.pop(); // get the top value and dump the value from queue
        v /= 2; q.push(v); // add the new value
    }

    while(!q.empty()){
        answer += q.top();
        q.pop();
    }

    #7. The shortest path between the k-th edge and another edge (Tree)
    ll depth[MAX_N];
    gr tree[MAX_N];

    void path(ll source, ll parent, ll all_cost){
        depth[source] = all_cost;
        items(e, tree[source]){
            if (e.to == parent) continue;
            dfs(e.to, source, all_cost + e.cost);
        }
    }

    ll n, k, a, b, c;

    in2(n, k);
    rep(i, n - 1){
        in3(a, b, c);
        a--; b--;
        tree[a].pb({b, c});
        tree[b].pb({a, c});
    }

    k--;
    dfs(k, -1, 0);

    #10. Visiting Subtree using recurrent function (ABC138D)

    gr tree[MAX_N];
    ll c[MAX_N];
    bool visited[MAX_N];

    void dfs(ll source, ll parent, ll val){
        visited[source] = true;
        c[source] += val;
        rep(i, tree[source].size()){
            rep(i, m){ll res = n % match[i].e1;}
            ll vertex = tree[source][i].to;
            if (vertex == parent) continue;
            dfs(vertex, source, c[source]);
        }
    }

    #11. bfs ABC146D, ABC007C
    1. first create a tree.
    2. start searching from a node.
    3. do some processes and push nodes connected with a given target node in BFS.
    4. repeat a series of procedure until queue is empty.

    queue<pl> q;

    void bfs(ll source, ll parents){
        ll n_edge = G[source].size();
        if (parents != -1) dist[source] = min(dist[source], dist[parents] + 1);
        if (visited[source]) return;
        visited[source] = true;

        rep(idx, n_edge){
            ll target = G[source][idx].to;
            if (target == parents) continue;
            q.push(mp(target, source)); 
        }
    }

    q.push(mp(sg.e1, -1));
    while(!q.empty()){
        pl source = q.front(); q.pop();
        bfs(source.e1, source.e2);
    }

    #12. grid to distance matrix (dx, dy)
    ll w, h;
    ll pos_to_idx(ll x, ll y){
        return y * w + x;
    }

    pl idx_to_pos(ll idx){
        return mp(idx % w, idx / w); 
    }

    int main(void){
        in2(h, w);
        rep(y, h){
            in1(s);
            rep(x, w){
                if (s[x] == '#') wall[x][y] = true;
                else wall[x][y] = false;
            }
        }

        rep(i1, h * w)rep(i2, h * w) dist[i1][i2] = INF * (i1 != i2);

        rep(x, w)rep(y, h){
            ll idx1 = pos_to_idx(x, y); ll idx2;
            if (wall[x][y]) continue;
            if (x != 0 && !wall[x - 1][y]){
                idx2 = pos_to_idx(x - 1, y);
                // if warshall floyd
                dist[idx1][idx2] = 1;
                // if dijkstra
                // graph[idx1].pb({idx2, 1});
            }
            if (x != w - 1 && !wall[x + 1][y]){
                idx2 = pos_to_idx(x + 1, y);
                dist[idx1][idx2] = 1;
                // graph[idx1].pb({idx2, 1});
            }
            if (y != 0 && !wall[x][y - 1]){
                idx2 = pos_to_idx(x, y - 1);
                dist[idx1][idx2] = 1;
                // graph[idx1].pb({idx2, 1});
            }
            if (y != h - 1 && !wall[x][y + 1]){
                idx2 = pos_to_idx(x, y + 1);
                dist[idx1][idx2] = 1;
                // graph[idx1].pb({idx2, 1});
            }
        }
    }


*/

/*
# the operators regarding bit
& (AND), | (OR), ^ (XOR)
- (REVERSE), >> (SMALLER SHIFT)
<< (BIGGER SHIFT)


     x1: 0000 0001 0010 0101 0110 0111
     x2: xxxx 0001 0011 0100 0101 1000
x1 & x2: 0000 0001 0010 0100 0100 0000

      x: 1001 1010  1100 1011 1101  1111
x & - x: 0001 0010  0100 0001 0001  0001
    sum: 1010 1100 10000 1100 1100 10000

x << y is x * 2 ** y
x >> y is rep(i, y) x = x // 2 

#Conditional Operator
condition ? true : false;

#iterator
type declaration: auto
value reference: *itr
increment: itr++
decrement: itr--
substitution of value: *itr = smt
*/

const ll MAX_N = 100005;
bool okay = false;
ll answer = 0;
ll n, k;
str s;

void solve(){
    n = s.length();

    ll nine = power_normal(9, k);
    rep(i, n - k){
        answer += combination(n - 2 - i, k - 1, MOD) * nine;
    }

    ll m = ctol(s[0]);
    if (k == 1) answer += m;
    else if (k == 2){
        answer += (m - 1) * (n - 1) * 9; bool did_zero = true;
        For(i, 1, n){
            if (did_zero) answer += ctol(s[i]);
            else answer += 9;
            if (s[i] != '0') did_zero = false;
        }
    }else if (k == 3){
        answer += (m - 1) * (n - 1) * (n - 2) / 2 * 81;
        bool did_zero0 = true; bool did_zero1 = true; bool did_zero2 = true;
        For(i1, 1, n - 1){
            if (!did_zero0) did_zero1 = false;
            if (did_zero0 && s[i1] == '0') continue;
            else did_zero0 = false;
            if (did_zero1) answer += (ctol(s[i1]) - 1) * (n - 1 - i1) * 9;
            else {answer += 9 * (n - 1 - i1) * 9; continue;}

            did_zero2 = true;
            For(i2, i1 + 1, n){
                if (did_zero2) answer += ctol(s[i2]);
                else answer += 9;
                if (s[i2] != '0') did_zero2 = false;
            }
        }
    }

    prl(answer);
    
    // check negative MOD
    // check index flow
    // check overwrite of the input variables
}
 
int main(void){
    in2(s, k);
    // assert(s.length() <= 50);
    solve();

    return 0;
}