Codeforces--55d--beautiful numbers (digital dp,dfs+ memory)

Beautiful numbersTime limit:4000MS Memory Limit:262144KB 64bit IO for Mat:%i64d &%i64u Submit Status

Description

Volodya is a odd boy and his taste are strange as well. It seems to him, a positive integer number is beautiful if and only if it's divisible by E Ach of its nonzero digits. We'll not be 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-natural numbers   l _i   and   R _i   ( 1?≤? L _{i ? ≤? R i ? ≤?9? 10¹⁸} ).

%lld specificator to read or write 64-bit integers in C + +. It is the preffered to use cin (also -use %i64d).

Output

Output should contain t numbers-answers to the queries, one number per line-quantities of b Eautiful numbers in given intervals (from l_i to R _i, inclusively).

Sample Input

Input

11 9

Output

9

Input

112 15

Output

2

Test instructions: L,r the number of X-multiplication LCM (x) in the range to be obtained.

The number of numbers within 1 to n is calculated first.

First, if the two-digit number is 1 to 9, and the remainder corresponds to the same, then the two numbers are appended with any of the same numbers, and the two numbers are the same as the remainder of the LCM. 1 to 9 of least common multiple 2520, so when several numbers can be the same as the results of 2520, and these numbers are composed of the same number, then these numbers can be divided into a class, after they add any of the same number, to their own LCM residual, still the same.

DP[I][256][2520], which represents the current number with I bit, where 256 is the state after the compression of 2 to 9, and then the result of the remainder of 2520.

In this way, the digital DP

#include <cstdio> #include <cstring> #include <algorithm>using namespace std; #define LL __int64ll dp[    20][256][2520], digit[20], cnt; int f (int k,int mods) {int i; for (i = 0; I <= 7; i++) {if (k& (1<<i)) {if (mods% (i+2)! = 0) b        Reak;    }} if (I <= 7) return 0; return 1;} LL dfs (int cnt,int pre,int mods,int maxd,int zero) {if (cnt = = 0) return!mods | |    (f (pre,mods));    if (maxd && zero && dp[cnt][pre][mods]! =-1) return dp[cnt][pre][mods]; LL I, r = Maxd?    9:DIGIT[CNT], ans = 0, k; for (i = 0; I <= r; i++) {if (I >= 2 && (pre& (1<< (i-2))) ==0) {An s + = DFS (cnt-1,pre| ( 1<< (I-2)), (mods*10+i)%2520,maxd| | i<r,zero| |        i); } else ans + = DFS (Cnt-1,pre, (mods*10+i)%2520,maxd| | i<r,zero| |    i); } if (Maxd && zero) dp[cnt][pre][mods] = ans;    return ans;}    ll solve (ll temp) {memset (digit,0,sizeof (digit));    CNT = 0;        while (temp) {digit[++cnt] = temp%10;    Temp/= 10; } return Dfs (cnt,0,0,0,0);}    int main () {int t;    LL N, M;    Memset (Dp,-1,sizeof (DP));    scanf ("%d", &t);        while (t--) {scanf ("%i64d%i64d", &n, &m);    printf ("%i64d\n", Solve (m)-solve (n-1)); } return 0;}

Codeforces--55d--beautiful numbers (digital dp,dfs+ memory)