[Codefoces 401d] Roman and numbers digital DP

Http://codeforces.com/problemset/problem/401/D

Given a number N, each digit of N is rearranged to find the number of methods in which N can be divisible.

Solution:

Brute force timeout ......

Most people write bit compression, but that would require 2 ^ 18*100 space, low efficiency, a large number of duplicate states, and inconvenient to process, this question provides a space of MB. However, if the space is doubled, the former method will be powerless.

It is found that there is a better state compression method for digital DP, which is directly compressed according to the number of times that digital X appears. For example, 333444 requires 2 ^ 6 = 64 space for compression using the former method, and only 3*3 space for Compression Based on the number of occurrences. For extreme data, by using the mean inequality, we only need (Ceil (18/10 + 1) ^ 10) = 59049 space, which improves the space utilization (Originally 2 ^ 18 = 262144) and the efficiency of heavy judgment.

The DP equation also easily establishes a part of the following State for all the numbers that require n to take the remainder of J for encoding I plus 10 plus the next digit. The result is that the remainder of all numbers is 0 (DP [s] [N]).

The key to this compression is codeit and decode functions.

Code

#include <cstdio>#include <cstring>#include <cmath>#include <map>#define SIZES 40000using namespace std;char n[20];int c[10],d[10],tim[10],m;long long dp[60000][100];int len;int codeit(int *te){    int res=0;    for (int i=0;i<10;i++)        res=res*(tim[i]+1)+te[i];    return res;}long long decode(int l,int *t){    for (int i=9;i>=0;i--){        t[i]=(l%(tim[i]+1));        l/=(tim[i]+1);        }}int main(){    int s;    while(~scanf("%s%I64d",n,&m)){    memset(dp,0,sizeof dp);    len=strlen(n);    memset(tim,0,sizeof(tim));    for (int i=0;i<len;i++)        ++tim[n[i]-'0'];    int s=codeit(tim);//    printf("%d\n",s);    dp[0][0]=1;    for (int i=0;i<s;i++){        int t[10];        decode(i,t);        for (int j=0;j<10;j++)            if (i+j)            if (tim[j]-t[j]){            t[j]++;            int pt=codeit(t);            for (int k=0;k<m;k++)            dp[pt][(k*10+j)%m]+=dp[i][k];            t[j]--;            }        }    printf("%I64d\n",dp[s][0]);    }}