题目链接：点击打开链接
题意：
我们认为一个数 num 能被每一位上的数字整除(expect 0) 那么这个数num就是合法的。
给出区间[l,r] ，问这个区间内有多少个合法的数。
首先solve(long x) 返回 [0, x] 内的合法个数，答案就是 solve(r) – solve(l-1);
以1234567为例
flag表示当前这位是能任意填，还是只能填<=该位对应的数字
若当前搜索的是第三位，且第二位已经填了0或1，则此时第三位可以任意填。
若第二位填的是2，则第三位只能填 [0, 3] ，所以dfs时传一个标记，标记这位是否能随便填。
若当前搜索的是第i位，，显然 pre_lcm和pre_mod 就是[最高位, i+1] 的mod值和lcm
pre_mod的作用是在填充完所有数字(n位数字都填充完)判断能否被这n位的lcm整除，
若能整除则 num = lcm*x (而lcm最大是 lcm(1,2··9) = 2520，所以2520%lcm == 0)
设num = x*2520+y，则 num%lcm = (x*2520+y)%lcm = y%lcm
=>num %lcm = num%2520%lcm
而2520的所有因子都有可能是2~9的组合的公倍数，（当然只是部分的因子）Init()离散化一下2520的所有因子
dp[pos][pre_mod][pre_lcm]代表前pos位数对2520取余为pre_mod并且非零位的lcm位pre_lcm的个数。
而答案是最低位往上回溯的，所以可以记忆化
import java.io.BufferedReader;import java.io.InputStreamReader;import java.io.PrintWriter;import java.math.BigInteger;import jav<span style="background-color: rgb(255, 255, 255);"> xcode </span><span style="font-family: Arial, Helvetica, sans-serif;">a.text.DecimalFormat;</span>import java.util.ArrayDeque;import java.util.ArrayList;import java.util.Arrays;import java.util.Collections;import java.util.Comparator;import java.util.Deque;import java.util.HashMap;import java.util.Iterator;import java.util.LinkedList;import java.util.Map;import java.util.PriorityQueue;import java.util.Scanner;import java.util.Stack;import java.util.StringTokenizer;import java.util.TreeMap;import java.util.TreeSet;import java.util.Queue;import java.io.File;import java.io.FileInputStream;import java.io.FileNotFoundException;import java.io.FileOutputStream;public class Main {int[] bit = new int[30], hash = new int[2550];long[][][] dp = new long[30][2550][50];long dfs(int pos, int pre_mod, int pre_lcm, boolean flag){if(pos == 0)return pre_mod%pre_lcm==0?1:0;if(flag && dp[pos][pre_mod][hash[pre_lcm]]!=-1)return dp[pos][pre_mod][hash[pre_lcm]];int u = flag?9:bit[pos];long ans = 0L;for(int d = 0; d <= u; d++){int next_mod = (pre_mod*10+d)%mod;int next_lcm = pre_lcm;if(d>0)next_lcm = Lcm(pre_lcm, d);ans += dfs(pos-1, next_mod, next_lcm, flag||d<u);}if(flag) dp[pos][pre_mod][hash[pre_lcm]] = ans;return ans;}long solve(long x){int len = 0;while(x>0){bit[++len] = (int) (x%10L); x/=10L;}return dfs(len, 0, 1, false);}void Init(){int cnt = 0;for(int i = 1; i <= mod; i++)if(mod%i==0)hash[i]=++cnt;for(int i = 0; i < 30; i++)for(int j = 0; j < 2520; j++)for(int k = 0; k < 50; k++)dp[i][j][k] = -1;}void work() throws Exception{Init();int T = Int();while(T–>0){long l = Long(), r = Long();out.println(solve(r) – solve(l-1L));}}public static void main(String[] args) throws Exception{Main wo = new Main();in = new BufferedReader(new InputStreamReader(System.in));out = new PrintWriter(System.out); // in = new BufferedReader(new InputStreamReader(new FileInputStream(new File("input.txt")))); // out = new PrintWriter(new File("output.txt"));wo.work();out.close();}static int N = 3*100050;static int M = N*N * 10;DecimalFormat df=new DecimalFormat("0.0000");static long inf = 1000000000000L;static long inf64 = (long) 1e18*2;static double eps = 1e-8;static double Pi = Math.PI;static int mod = 2520 ;private String Next() throws Exception{while (str == null || !str.hasMoreElements())str = new StringTokenizer(in.readLine());return str.nextToken();}private int Int() throws Exception{return Integer.parseInt(Next());}private long Long() throws Exception{return Long.parseLong(Next());}StringTokenizer str;static BufferedReader in;static PrintWriter out;/*class Edge{int from, to, nex;Edge(){}Edge(int from, int to, int nex){this.from = from;this.to = to;this.nex = nex;}}Edge[] edge = new Edge[M<<1];int[] head = new int[N];int edgenum;void init_edge(){for(int i = 0; i < N; i++)head[i] = -1; edgenum = 0;}void add(int u, int v){edge[edgenum] = new Edge(u, v, head[u]);head[u] = edgenum++;}/**/int upper_bound(int[] A, int l, int r, int val) {// upper_bound(A+l,A+r,val)-A;int pos = r;r–;while (l <= r) {int mid = (l + r) >> 1;if (A[mid] <= val) {l = mid + 1;} else {pos = mid;r = mid – 1;}}return pos;}int Pow(int x, int y) {int ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;y >>= 1;x = x * x;}return ans;}double Pow(double x, int y) {double ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;y >>= 1;x = x * x;}return ans;}int Pow_Mod(int x, int y, int mod) {int ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;ans %= mod;y >>= 1;x = x * x;x %= mod;}return ans;}long Pow(long x, long y) {long ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;y >>= 1;x = x * x;}return ans;}long Pow_Mod(long x, long y, long mod) {long ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;ans %= mod;y >>= 1;x = x * x;x %= mod;}return ans;}int Gcd(int x, int y){if(x>y){int tmp = x; x = y; y = tmp;}while(x>0){y %= x;int tmp = x; x = y; y = tmp;}return y;}long Gcd(long x, long y){if(x>y){long tmp = x; x = y; y = tmp;}while(x>0){y %= x;long tmp = x; x = y; y = tmp;}return y;}int Lcm(int x, int y){return x/Gcd(x, y)*y;}long Lcm(long x, long y){return x/Gcd(x, y)*y;}int max(int x, int y) {return x > y ? x : y;}int min(int x, int y) {return x < y ? x : y;}double max(double x, double y) {return x > y ? x : y;}double min(double x, double y) {return x < y ? x : y;}long max(long x, long y) {return x > y ? x : y;}long min(long x, long y) {return x < y ? x : y;}int abs(int x) {return x > 0 ? x : -x;}double abs(double x) {return x > 0 ? x : -x;}long abs(long x) {return x > 0 ? x : -x;}boolean zero(double x) {return abs(x) < eps;}double sin(double x){return Math.sin(x);}double cos(double x){return Math.cos(x);}double tan(double x){return Math.tan(x);}double sqrt(double x){return Math.sqrt(x);}}
流转的时光，都成为命途中美丽的点缀，
