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 ;

}