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 isbeautifulif 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 casest(1≤t≤10). Each of the nexttlines contains two natural numbersliandri(1≤li≤ri≤9·1018).
Please, do not use%lldspecificator to read or write 64-bit integers in C++. It is preffered to usecin(also you may use%I64d).
Output
Output should containtnumbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals (fromlitori, inclusively).
Sample test(s)
input
11 9
output
9
input
112 15
output
2
/**CF 55D 数位dp（一个数是组成它的所有位数对应数的倍数）题目大意：找出给定区间内有多少数，其本身是所有组成它的各个位上的数的倍数解题思路：（kuangbin）一个数能被它的所有非零数位整除，则能被它们的最小公倍数整除，而1到9的最小公倍数为2520，数位DP时我们只需保存前面那些位的最小公倍数就可进行状态转移，到边界时就把所有位的lcm求出了，为了判断这个数能否被它的所有数位整除，我们还需要这个数的值，显然要记录值是不可能的，其实我们只需记录它对2520的模即可，这样我们就可以设计出如下数位DP：dfs(pos,mod,lcm,f),pos为当前位，mod为前面那些位对2520的模，lcm为前面那些数位的最小公倍数，f标记前面那些位是否达到上限，，这样一来dp数组就要开到19*2520*2520，明显超内存了，考虑到最小公倍数是离散的，1-2520中可能是最小公倍数的其实只有48个，经过离散化处理后，dp数组的最后一维可以降到48，这样就不会超了。*/#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;const int mod=2520;typedef long long LL;LL dp[25][mod][50];int index[2530],bit[25];int gcd(int a,int b){if(b==0)return a;else return gcd(b,a%b);}int lcm(int a,int b){return a/gcd(a,b)*b;}void init(){int num=0;for(int i=1; i<=mod; i++){if(mod%i==0)index[i]=num++;}}LL dfs(int pos,int presum,int prelcm,bool flag){if(pos==-1)return presum%prelcm==0;if(!flag&&dp[pos][presum][index[prelcm]]!=-1)return dp[pos][presum][index[prelcm]];LL ans=0;int end=flag?bit[pos]:9;for(int i=0; i<=end; i++){int nowsum=(presum*10+i)%mod;int nowlcm=prelcm;if(i)nowlcm=lcm(nowlcm,i);ans+=dfs(pos-1,nowsum,nowlcm,flag&&i==end);}if(!flag)dp[pos][presum][index[prelcm]]=ans;return ans;}LL solve(LL x){int len=0;while(x){bit[len++]=x%10;x/=10;}return dfs(len-1,0,1,1);}int main(){int T;init();memset(dp,-1,sizeof(dp));scanf("%d",&T);while(T–){LL r,l;scanf("%I64d%I64d",&l,&r);printf("%I64d\n",solve(r)-solve(l-1));}return 0;}
只要功夫深，铁棒磨成绣花针。
