codeforces--55D--Beautiful numbers(数位dp，dfs+记忆化)

Beautiful numbers

Time Limit:4000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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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

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 l_i and r_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 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 l_i to r_i, inclusively).

Sample Input

Input

11 9

Output

9

Input

112 15

Output

2

题意：给出l，r 求在范围内的数x能乘除lcm(x)的数的个数

先求出1到n内的数的个数。

首先，如果两个数对1到9取余，得到的余数对应相同，那么在这两个数后面加任何相同的数，这两个数对lcm的余数相同。1到9的最小公倍数2520，所以当几个数能对2520取余的结果相同，同时这些数是由相同的数组成的，那么这些数可以分为一类，它们的后面加任何相同的数后，对自身lcm取余，仍然相同。

dp[i][256][2520],代表当前的数有i位时，其中256为状态压缩后的2到9，然后是对2520的取余结果。

由此，进行数位dp

 

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;#define LL __int64LL dp[20][256][2520] , digit[20] , cnt ;int f(int k,int mods){    int i ;    for(i = 0 ; i <= 7 ; i++)    {        if( k&(1<<i) )        {            if(mods%(i+2) != 0)                break ;        }    }    if( i <= 7 )        return 0 ;    return 1 ;}LL dfs(int cnt,int pre,int mods,int maxd,int zero){    if( cnt == 0 ) return !mods || ( f(pre,mods) ) ;    if( maxd && zero && dp[cnt][pre][mods] != -1 )        return dp[cnt][pre][mods] ;    LL i , r = maxd ? 9 : digit[cnt] , ans = 0 , k ;    for(i = 0 ; i <= r ; i++)    {        if( i >= 2 && (pre&( 1<<(i-2) ))==0 )        {            ans += dfs(cnt-1,pre|(1<<(i-2)) ,(mods*10+i)%2520,maxd||i<r,zero||i ) ;        }        else            ans += dfs(cnt-1,pre,(mods*10+i)%2520,maxd||i<r,zero||i ) ;    }    if( maxd && zero ) dp[cnt][pre][mods] = ans ;    return ans ;}LL solve(LL temp){    memset(digit,0,sizeof(digit)) ;    cnt = 0 ;    while( temp )    {        digit[++cnt] = temp%10 ;        temp /= 10 ;    }    return dfs(cnt,0,0,0,0) ;}int main(){    int t ;    LL n , m ;    memset(dp,-1,sizeof(dp)) ;    scanf("%d", &t) ;    while(t--)    {        scanf("%I64d %I64d", &n, &m) ;        printf("%I64d\n", solve(m)-solve(n-1)) ;    }    return 0;}