思路：

一道数位DP的板子题

$code$

#include<iostream>
#include<cstring>
#include<cstdio>

#define ll long long

using namespace std;

ll n, ans, len;
ll f[21][163][163];
ll lim[20], a[20];

inline ll dfs(register ll x, register ll sum, register ll k, register ll r, register ll limit) {
    
	//cout<<x<<' '<<sum<<' '<<k<<' '<<r<<' '<<limit<<endl;
	if(x > len && sum == 0 && r == 0) {
    
		f[x][sum][r] = 1;
		return 1;
	}
	if(x > len) {
    
		f[x][sum][r] = 0;
		return 0;
	}
	if(!limit && f[x][sum][r] != -1) return f[x][sum][r];
	register ll tmp = 0;
	if(limit) {
    
		for(register ll i = 0; i <= lim[x]; ++ i)
			if(sum - i >= 0)
				tmp += dfs(x + 1, sum - i, k, (r * 10 + i) % k, i == lim[x]);
	}
	else {
    
		for(register ll i = 0; i <= 9; ++ i)
			if(sum - i >= 0)
				tmp += dfs(x + 1, sum - i, k, (r * 10 + i) % k, 0);
	}
// if(!limit && f[x][k][sum][r] != tmp && f[x][k][sum][r] != -1)
// cout<<"dsfdsafasdf"<<endl;
	if(!limit) f[x][sum][r] = tmp;
// cout<<tmp<<endl;
	return tmp;
}

int main() {
    
	freopen("number.in", "r", stdin);
	freopen("number.out", "w", stdout);
	scanf("%lld", &n);
	register ll m = n;
	while(m != 0) {
    
		++ len;
		a[len] = m % 10;
		m /= 10;
	}
	for(register ll i = 1; i <= len; ++ i) lim[i] = a[len - i + 1];
	for(register ll i = 1; i <= len * 9; ++ i)
	{
    
		memset(f, -1, sizeof(f));
		ans += dfs(1, i, i, 0, 1);
	}
	printf("%lld", ans);
	return 0;
}