题目
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
分析
所有的数位是1到9，最小公倍数是5 7 8* 9，是2520，开数组为数位，和，最小公倍数，分别对应dp【20】【2520】【2520】，但是超内存，所以利用hash，把最后的最小公倍数压缩在50以内，因为可能的最小公倍数是48个。
代码
#include <iostream> #include <cstring> #include <cstdio> using namespace std; long long a[22]; int hash[2522]; long long dp[22][2522][50]; int gcd(int a,int b){ if(a<b)swap(a,b); while(b){ int w=a; a=b; b=w%b; } return a; } long long dpp(int pos,int sum,int b,int limit){ if(pos==-1)return sum%b==0; if(!limit&&dp[pos][sum][hash[b]]!=-1)return dp[pos][sum][hash[b]]; long long sun=0; int end=limit?a[pos]:9; for(int i=0;i<=end;i++){ int q=b; if(i>=2)q=q*i/gcd(q,i); sun+=dpp(pos-1,(sum*10+i)%2520,q,limit&&i==a[pos]); } if(!limit)dp[pos][sum][hash[b]]=sun; return sun; } long long go(long long x){ int pos=0; while(x){ a[pos++]=x%10; x/=10; } return dpp(pos-1,0,1,1); } int main(){ long long n,m,t; int j; memset(dp,-1,sizeof(dp)); for(int i=1,j=0;i<=2520;i++){ if(2520%i==0)hash[i]=j++; } scanf("%I64d",&t); while(t--){ scanf("%I64d%I64d",&n,&m); printf("%I64d ",go(m)-go(n-1)); } return 0; }