数位dp CF 55D Beautiful numbers

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本文介绍了一种算法，用于计算指定范围内所有 beautiful numbers 的数量。Beautiful number 被定义为能够被其所有非零数字整除的正整数。通过使用动态规划方法，文章详细解释了如何高效地解决这一问题。

D. Beautiful numbers

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number isbeautiful 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

The first line of the input contains the number of cases t (1?≤?t?≤?10). Each of the nextt lines contains two natural numbers l_i andr_i (1?≤?l_i?≤?r_i?≤?9?·10¹⁸).

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to usecin (also you may use %I64d).

Output

Output should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals (froml_i to r_i, inclusively).

Sample test(s)

Input

1
1 9

Output

Input

1
12 15

Output

题意：如果一个数能被这个数中包含的所有数字整除，那么他就是一个beautiful number。求出[L,R]之间有多少个这样的beautiful number

思路:能同时被多个数整除就是能被他们的最小公倍数整除，可以算出最大的最小公倍数是2520，其中其余不是最大的最小公倍数都能整除2520.所以只有48个不同的最小公倍数，所以我们用dp[i][j][k] 表示i位数，前缀的数%2520为j，前缀的最小公倍数为k的数有多少个。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define eps 1e-8
#define mod 2520
#define LL long long
int Index[mod+10];
LL dp[19][mod][48] , digit[20];

int gcd(int a,int b)
{
    while (a && b)
    {
        if ( a > b ) a %= b;
        else b %= a;
    }
    return a+b;
}

int lcm(int a,int b)
{
    int g = gcd(a,b);
    return a/g*b;
}

void init()
{
    int cnt = 0;
    for (int i = 1 ; i <= mod ; ++i) if (mod%i==0) 
        Index[i] = cnt++;
    memset(dp,-1,sizeof(dp));
}

LL dfs(int pos,LL sum,int Lcm,bool limit)
{
    if (pos==-1) return sum%Lcm == 0;
    if (!limit && dp[pos][sum][Index[Lcm]] !=-1) 
        return dp[pos][sum][Index[Lcm]];
    int end = limit ? digit[pos] : 9;
    LL ret = 0;
    for (int i = 0 ; i <= end ; ++i)
    {
        int nowSum = (sum*10+i)%mod;
        int nowLcm = Lcm;
        if (i) nowLcm = lcm(nowLcm,i);
        ret += dfs(pos-1,nowSum,nowLcm,limit && i==end);
    }
    if (!limit) dp[pos][sum][Index[Lcm]] = ret;
    return ret;
}

LL Ans(LL x)
{
    int pos = 0;
    while (x)
    {
        digit[pos++] = x %10;
        x /= 10;
    }
    return dfs(pos-1,0,1,true);
}

int main()
{
    init();
    LL l , r;
    int T; cin>>T;
    while (T--)
    {
        cin>>l>>r;
        cout << Ans(r)-Ans(l-1) << endl;
    }
    //cout << max_lcm << endl;
}