CodeForces55D Beautiful numbers

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 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).

Examples

input

11 9

output

9

input

112 15

output

2

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题目的意思是求一个区间有多少个数，数要求能被他的每一位整除

思路：数位dp，开3位数组记录长度位len的，到len为止和为sum的，sum容易很大，但可以知道1到9的最小公倍数为2520，所以可以对2520取模，LCM为lcm的书有多少个，然后dfs递归求解

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>#include <map>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 10000005#define Mod 10001using namespace std;#define LL long longLL dp[20][3000][50];int a[20],index[3000];int mod;int gcd(int x,int y){    return x%y==0?y:gcd(y,x%y);}int _lcm(int x,int y){    return x*y/gcd(x,y);}void init(){    mod=1;    for(int i=1; i<10; i++)    {        mod=_lcm(mod,i);    }    int ct=0;    for(int i=1; i<=mod; i++)    {        if(mod%i==0) index[i]=ct++;    }    memset(dp,-1,sizeof dp);}LL dfs(int len,int sum,int lcm,bool limit){    if(len<0)        return sum%lcm==0;    if(dp[len][sum][index[lcm]]!=-1&&!limit)        return dp[len][sum][index[lcm]];    int up=limit?a[len]:9;    LL ans=0;    for(int i=0; i<=up; i++)    {        ans+=dfs(len-1,(sum*10+i)%mod,i?_lcm(lcm,i):lcm,limit&&i==up);    }    return limit?ans:dp[len][sum][index[lcm]]=ans;}LL solve(LL x){    int cnt=0;    while(x>0)    {        a[cnt++]=x%10;        x/=10;    }    return dfs(cnt-1,0,1,1);}int main(){    int T;    LL n,m;    init();    for(scanf("%d",&T); T--;)    {        scanf("%lld%lld",&n,&m);        printf("%lld\n",solve(m)-solve(n-1));    }    return 0;}