Codeforces 55d 数位dp

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

1
1 9

output

9

input

1
12 15

output

2

题意：给出区间(l,r)，找出区间所有符合以下条件的数的个数。

它能被自己所有非零位的数整除。

题解：数位dp，1-9的最小公倍数是2520，所以我们mod2520来保存即可。再用map来保存一下dp到每个位数的当前乘积，防止开太大空间mle。

#include<iostream>
#include<algorithm>
#include<string.h>
#include<map>
using namespace std;
typedef __int64 ll;
map<ll,ll>sp;
ll digit[20],dp[19][2520][50],cnt=0;
ll dfs(ll pos,ll sum,ll gs,ll lim){
	if(pos<=0){
		if(gs==0)return 0;
		else return sum%gs==0;
	}
	if(!lim&&dp[pos][sum][sp[gs]]!=-1){
		return dp[pos][sum][sp[gs]];
	}
	ll es=lim?digit[pos]:9;
	ll ret=0;
	for(ll i=0;i<=es;i++){
		ll sg=0;
		if(i==0)sg=gs;
		else if(gs==0)sg=i;
		else{
			ll sda=__gcd(i,gs);
			sg=gs*i/sda;
		}
		ret+=dfs(pos-1,(sum*10+i)%2520,sg,lim&&(i==es));
	}
	if(!lim){
		if(sp[gs]==0){
			sp[gs]=++cnt;
		}
		dp[pos][sum][sp[gs]]=ret;
	}
	return ret;
}
int main(){
	int t;
	cin>>t;
	memset(dp,-1,sizeof(dp));
	while(t--){
		ll l,r;
		cin>>l>>r;
		ll len;
		len=0;
		l--;
		while(l){
			digit[++len]=l%10;
			l/=10;
		}
		ll sol1=dfs(len,0,0,1);
		len=0;
		while(r){
			digit[++len]=r%10;
			r/=10;
		}
		ll sol2=dfs(len,0,0,1);
		cout<<sol2-sol1<<endl;
	}
	return 0;
}