Beautiful numbers_beautiful numbers上海-优快云博客

本文链接：https://blog.youkuaiyun.com/DancingZ/article/details/83384299

本文介绍了一种复杂的数位动态规划方法，用于计算特定范围内满足特定条件（即每个非零数字都能整除该数）的美丽数字的数量。通过深入解析代码，读者将了解到如何使用动态规划解决这一问题，包括状态转移方程、边界条件和记忆化搜索的实现。

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Description

Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful numbers in given ranges.

Input

Multiple test cases.
Each line contains two natural numbers li and ri (1 ≤ li ≤ ri ≤ 9 · 10^18).

Output

One line specifies the answer.

Sample Input

1 9

12 15

Sample Output

一道很强的数位dp，感觉要去练下记搜了= =

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int Maxn=2520;
ll l,r,c[20],idx[Maxn+5],f[20][Maxn+5][50];
ll gcd(ll a,ll b){
	return !b?a:gcd(b,a%b);
}
#define lcm(a,b) a/gcd(a,b)*b
ll dp(int x,ll lef,ll lcm,bool flag){
	if(!x)return lef%lcm==0;
	if(!flag&&~f[x][lef][idx[lcm]])return f[x][lef][idx[lcm]];
	int limit=flag?c[x]:9;
	ll ret=0;
	for(int i=0;i<=limit;++i){
		ll Lcm=i?lcm(lcm,i):lcm;
		ret+=dp(x-1,(lef*10+i)%Maxn,Lcm,flag&&i==limit);
	}
	if(!flag)f[x][lef][idx[lcm]]=ret;
	return ret;
}
ll Solve(ll x){
	for(c[0]=0;x;x/=10)c[++c[0]]=x%10;
	return dp(c[0],0,1,1);
}
int main(){
	for(int i=1;i<=Maxn;++i)
		if(Maxn%i==0)idx[i]=++idx[0];
	memset(f,-1,sizeof(f));
	while(~scanf("%lld%lld",&l,&r)){
		cout<<Solve(r)-Solve(l-1)<<"\n";
	}
	return 0;
}