codeforce 55 D. Beautiful numbers(数位dp,好题)

题目链接

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

题意:

一个美丽数就是可以被它的每一位的数字整除的数。给定一个区间，求美丽数的个数。

题解：

传送门

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>using namespace std;typedef long long ll;const int MOD=2520;ll d[20][MOD][50];int H[MOD],bit[20];int cnt=0;int gcd(int a,int b){return b==0? a:gcd(b,a%b);}int lcm(int a,int b){return a/gcd(a,b)*b;}void init(){cnt=0;for(int i=1;i<=MOD;i++) if(MOD%i==0) H[i]=++cnt;}ll dp(int pos,int mod,int l,int f){if(pos==-1) return (mod%l==0);if(!f&&d[pos][mod][H[l]]!=-1) return d[pos][mod][H[l]];ll ans=0;int end=f? bit[pos]:9;ans+=dp(pos-1,mod*10%MOD,l,f&&(bit[pos]==0));for(int i=1;i<=end;i++)ans+=dp(pos-1,(mod*10+i)%MOD,lcm(l,i),f&&(i==end));if(!f) d[pos][mod][H[l]]=ans;return ans;}ll cal(ll x){int num=0;while(x){bit[num++]=x%10;x/=10;}return dp(num-1,0,1,1);}int main(){int cas;scanf("%d",&cas);memset(d,-1,sizeof(d));init();while(cas--){ll a,b;scanf("%I64d%I64d",&a,&b);printf("%I64d\n",cal(b)-cal(a-1));}return 0;}