Topcoder SRM655 DIV2 950 NineEasy 状压 + 数位 dp

题意：要你构造一个n位的数子 ，给你m（1-5）个询问，每一次询问取一些位数的数组成一个新数且这个数 %9 ==  0 ， 问你要满足这m个询问的数字有多少个(允许前缀0)。

解题思路：把每一种情况状压，得到一个最多  9x9x9x9x9 的情况，然后根据 每个数的询问决定状态转移方程。

解题代码: 

// BEGIN CUT HERE
/*

*/
// END CUT HERE
#line 7 "NineEasy.cpp"
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,n) for(i=0;i<(n);++i)
#define FOR(i,l,h) for(i=(l);i<=(h);++i)
#define FORD(i,h,l) for(i=(h);i>=(l);--i)

typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;

#define M 1000000007
const int MaxS = 9*9*9*9*9*2;

long long dp[MaxS][25];
long long pow9[6];

class NineEasy
{
        public:
        int count(int m, vector <int> d)
        {
            memset(dp,0,sizeof(dp));
            pow9[0] = 1;  for(int i = 1;i <= 5;i ++)  pow9[i] = pow9[i-1] * 9 ; 
            int n = d.size();
            dp[0][0] = 1; 
            for(int i = 0 ;i < n; i ++){
                for(int j = 0 ;j < pow9[m]; j ++){                    
                  for(int p = 0 ;p <= 9 ; p ++){
                    int ns = 0; 
                    for(int s = 0 ; s < m ; s ++){
                        if((1 << s ) & d[i]){
                            ns += ((j/pow9[s] + p )%9)*pow9[s] ;    
                        }else{
                            ns += ((j/pow9[s])%9)*pow9[s];
                        }
                    }
                    dp[ns][i+1] = (dp[ns][i+1] + dp[j][i]) % M; 
                  }
                }
            }
            return dp[0][n];
        }
        
// BEGIN CUT HERE
    public:
    void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); }
    private:
    template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
    void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
    void test_case_0() { int Arg0 = 2; int Arr1[] = {1,2}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 4; verify_case(0, Arg2, count(Arg0, Arg1)); }
    void test_case_1() { int Arg0 = 2; int Arr1[] = {3,3}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 12; verify_case(1, Arg2, count(Arg0, Arg1)); }
    void test_case_2() { int Arg0 = 2; int Arr1[] = {1,3,2}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 16; verify_case(2, Arg2, count(Arg0, Arg1)); }
    void test_case_3() { int Arg0 = 5; int Arr1[] = {1,2,4,8,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 893703876; verify_case(3, Arg2, count(Arg0, Arg1)); }
    void test_case_4() { int Arg0 = 1; int Arr1[] = {0,0,1}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 200; verify_case(4, Arg2, count(Arg0, Arg1)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main()
{
        NineEasy ___test;
        ___test.run_test(-1);
        return 0;
}
// END CUT HERE