Codeforces 55D Beautiful Numbers (digital dp+ math)

Title Link: Http://codeforces.com/problemset/problem/55/D

Test instructions: A beautiful number is a number that can be divisible by the number of each of its digits.

Given an interval, the number of beautiful numbers is obtained.

This is obviously a digital DP, which satisfies a number that can be divisible by the LCM of all bits.

The dp[len][mod][lcm],mod is usually set to represent the remainder, and the LCM represents the LCM of the front len bit.

But if the direct naked mod will be very complex, then want to lcm{0,1,2,3,4,5,6,7,8,9}=2520;

and Lcm{a,b,c,d ...} {a,b,c,d ... Indicates the number of individual digits) to be lcm{0,1,2....9 after going to the weight}

Divisible. We are asking for SUM%LCM (A,b,c,d.} ==0, so just meet

SUM%LCM (0,1,2,... 9}%LCM (A,b,c,d ...} ==0 can be. The MoD can then be expressed as

SUM%LCM (0,1,2,... 9} is the number. But mod<=2520 && lcm<=2520 this

Must not survive, so to consider how to deal with the LCM, after all, it is obvious that the 0~9 maximum common multiple species will not

Of more than 48. So you can consider discretization of the LCM,

If 2520% num = = 0-lcm[num]=temp++;

So the three-dimensional DP can be set to dp[20][2520][48];

#include <iostream> #include <cstring> #include <cmath> #include <cstdio>using namespace std; typedef long LONG ll;const int mmax = 2520;ll N, m, Dp[20][mmax][50];int temp, dig[20], Lcm[mmax + 10];ll gcd (ll A, l L b) {return b > 0 gcd (b, a% B): A;} ll LCM (ll A, ll b) {return A/GCD (A, b) * b;}    void init () {temp = 0;        for (int i = 1; I <= Mmax; i++) {if (mmax% i = = 0) {Lcm[i] = temp++;        } else {Lcm[i] = 0;    }}}ll dfs (int len, int count, int mod, int flag) {if (!len) {return mod% count = = 0;    } if (!flag && dp[len][mod][lcm[count]]! =-1) {return dp[len][mod][lcm[count]]; } int t = flag?    Dig[len]: 9;    ll sum = 0;        for (int i = 0; I <= t; i++) {int nextmod = (mod * + i)% Mmax;        int nextcount;        if (i = = 0) {Nextcount = count;        } else {nextcount = (int) LCM (count, i); }       Sum + = DFS (len-1, Nextcount, NEXTMOD, flag && i = = t);    } if (!flag) dp[len][mod][lcm[count]] = sum; return sum;}    ll Gets (ll x) {memset (dig, 0, sizeof (DIG));    int len = 0;    if (x = = 0) {Dig[++len] = 0;        } while (x) {Dig[++len] = x 10;    x/= 10; } return Dfs (len, 1, 0, 1);}    int main () {int t;    scanf ("%d", &t);    Init ();    Memset (DP,-1, sizeof (DP));        while (t--) {scanf ("%i64d%i64d", &n, &m);    printf ("%i64d\n", gets (m)-gets (n-1)); } return 0;}

Codeforces 55D Beautiful Numbers (digital dp+ math)