codeforces 55D Beautiful numbers（数位DP）

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

The first line of the input contains the number of cases t (1 ≤ t ≤ 10). Each of the next t lines
contains two natural numbers li and ri (1 ≤ li ≤ ri ≤ 9 ·1018).

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (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 (from li to ri,
inclusively).

Sample Input

Input

1
1 9

Output

9

Input

1
12 15

Output

2

美丽数就是该数可以被他的每一位数字整除。
dp[i][j][k]表示处理到数位i,该数对2520取模为j,各个数位的LCM为k

分析：一个数能被它的所有非零数位整除，则能被它们的最小公倍数整除，而1到9的最小公倍数为2520，

数位DP时我们只需保存前面那些位的最小公倍数就可进行状态转移，到边界时就把所有位的lcm求出了，

为了判断这个数能否被它的所有数位整除，我们还需要这个数的值，显然要记录值是不可能的，其实我们只

需记录它对2520的模即可，这样我们就可以设计出如下数位DP：dfs(pos,mod,lcm,f),pos为当前

位，mod为前面那些位对2520的模，lcm为前面那些数位的最小公倍数，f标记前面那些位是否达到上限，

这样一来dp数组就要开到19*2520*2520，明显超内存了，考虑到最小公倍数是离散的，1-2520中可能

是最小公倍数的其实只有48个，经过离散化处理后，dp数组的最后一维可以降到48，这样就不会超了。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef __int64 ll;
const int mod = 2520;
ll dp[25][2520][48];
int index[2600], bit[25];
int gcd(int a, int b)
{
return b == 0 ? a : gcd(b, a%b);
}
int lcm(int a, int b)
{
return a*b / gcd(a, b);
}
void init()
{
int num = 0;
for (int i = 1;i <= mod;i++)
{
if (mod%i == 0)
index[i] = num++;
}
}
ll dfs(int pos, int presum, int prelcm, bool flag)
{
if (pos == -1) return presum%prelcm == 0;
if (!flag&&dp[pos][presum][index[prelcm] ]!=-1)
return dp[pos][presum][index[prelcm]];
ll ans = 0;
int end = flag ? bit[pos] : 9;
for (int i = 0;i <= end;i++)
{
int nowsum = (presum * 10 + i) % mod;
int nowlcm = prelcm;
if (i) nowlcm = lcm(prelcm, i);
ans += dfs(pos - 1, nowsum, nowlcm, flag&&i == end);
}
if (!flag) dp[pos][presum][index[prelcm]] = ans;
return ans;
}
ll calc(ll x)
{
int pos = 0;
while (x)
{
bit[pos++] = x % 10;
x = x / 10;
}
return dfs(pos - 1, 0, 1, 1);
}
int main()
{
ll t, l, r;
init();
memset(dp, -1, sizeof(dp));
scanf("%I64d", &t);
while (t--)
{
scanf("%I64d%I64d", &l, &r);
printf("%I64d\n", calc(r) - calc(l - 1));
}
return 0;
}