传送门

题意：

思路：

比较套路的题，首先也有个明显的状态$f[pos][num][sum]$表示到了$pos$位，当前数为$num$，各位数字之和为$sum$。由于$a,b\le 1e18$，所以显然是不行的。看到整除就可以考虑是否可以通过将第二维取模来优化状态呢？我们发现第三维最多只有$9\ast 18$个数，所以我们枚举第三维的约数，让后将$num$模上约数即可，这样状态只有$20\ast 9\ast 18\ast 9\ast 18$个了，再乘上枚举约数$9\ast 18$，复杂度约为$85030560$，显然可以过掉。

// Problem: P4127 [AHOI2009]同类分布
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P4127
// Memory Limit: 125 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;LL a,b;
int A[20],tot;
int limit;
LL f[20][9*18+10][9*18+10];LL dp(int pos,int num,int sum,int flag) {if(sum>limit) return 0;if(pos==0) return num==0&&sum==limit;if(f[pos][num][sum]!=-1&&flag) return f[pos][num][sum];int x=flag? 9:A[pos];LL ans=0;for(int i=0;i<=x;i++) {ans+=dp(pos-1,(num*10+i)%limit,sum+i,flag||i<x);}if(flag) f[pos][num][sum]=ans;return ans;
}LL solve(LL x) {tot=0;while(x) A[++tot]=x%10,x/=10;return dp(tot,0,0,0);
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);//cout<<20*9*18*9*18*9*18<<endl;cin>>a>>b;LL ans=0;for(int i=1;i<=9*18;i++) {limit=i; memset(f,-1,sizeof(f));ans+=solve(b)-solve(a-1);}cout<<ans<<endl;return 0;
}
/**/