bzoj 1799: [Ahoi2009]self 同类分布 数位dp

1799: [Ahoi2009]self 同类分布

Time Limit: 50 Sec  Memory Limit: 64 MB
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Description

给出a,b，求出[a,b]中各位数字之和能整除原数的数的个数。

Input

 

Output

 

Sample Input

10 19

Sample Output

3

HINT

【约束条件】1 ≤ a ≤ b ≤ 10^18

Source

Day1

类似hdu 4389

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+,M=1e6+,inf=;
const ll INF=1e18+,mod=;
ll f[][][],bit[];
ll dp(int pos,int sum,int m,int p,int flag)
{
    if(pos==)return (sum==p&&m==);
    if(flag&&f[pos][sum][m]!=-)return f[pos][sum][m];
    int x=flag?:bit[pos];
    ll ans=;
    for(int i=;i<=x;i++)
    {
        ans+=dp(pos-,sum+i,(m*+i)%p,p,flag||i<x);
    }
    if(flag)f[pos][sum][m]=ans;
    return ans;
}
ll getans(ll x,int p)
{
    int len=;
    while(x)
    {
        bit[++len]=x%;
        x/=;
    }
    return dp(len,,,p,);
}
int main()
{
    ll l,r;
    scanf("%lld%lld",&l,&r);
    ll ans=;
    for(int i=;i<=;i++)
    {
        memset(f,-,sizeof(f));
        ans+=getans(r,i)-getans(l-,i);
    }
    printf("%lld\n",ans);
    return ;
}