解题思路：根据dp规则，我们可以把d[i]值相同的放到一起，然后算一个总的DP就行了。
解题代码
1 // BEGIN CUT HERE
2 /*
3
4 */
5 // END CUT HERE
6 /*#line 7 "Nine.cpp"
7 #include <cstdlib>
8 #include <cctype>
9 #include <cstring>
10 #include <cstdio>
11 #include <cmath>
12 #include <algorithm>
13 #include <vector>
14 #include <string>
15 #include <iostream>
16 #include <sstream>
17 #include <map>
18 #include <set>
19 #include <queue>
20 #include <stack>
21 #include <fstream>
22 #include <numeric>
23 #include <iomanip>
24 #include <bitset>
25 #include <list>
26 #include <stdexcept>
27 #include <functional>
28 #include <utility>
29 #include <ctime>
30 using namespace std;
31
32 #define PB push_back
33 #define MP make_pair
34
35 #define REP(i,n) for(i=0;i<(n);++i)
36 #define FOR(i,l,h) for(i=(l);i<=(h);++i)
37 #define FORD(i,h,l) for(i=(h);i>=(l);--i)
38 #define M 1000000007
39
40 typedef vector<int> VI;
41 typedef vector<string> VS;
42 typedef vector<double> VD;
43 typedef long long LL;
44 typedef pair<int,int> PII;
45
46 const int maxn = 9*9*9*9*9;
47 int dp2[5005][10];
48 int pow9[6];
49 int dp[maxn][32];
50 void get_nx()
51 {
52 dp2[0][0] = 1 ;
53 for(int i = 1;i<= 5000 ;i ++)
54 for(int j = 0 ;j <= 9; j ++){
55 for(int s = 0 ; s <= 9 ; s ++){
56 dp2[i][(j+s) % 9] = (dp2[i][(j+s) %9] + dp2[i][j]) % M;
57 }
58 }
59 }
60 class Nine
61 {
62 public:
63 int count(int m, vector <int> d){
64 memset(dp2,0,sizeof(dp2));
65 get_nx();
66 memset(dp,0,sizeof(dp));
67 for()
68 return dp[1][0];
69 }
70 public:
71 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(); }
72 private:
73 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(); }
74 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; } }
75 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)); }
76 void test_case_1() { int Arg0 = 2; int Arr1[] = {1,2,3}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 16; verify_case(1, Arg2, count(Arg0, Arg1)); }
77 void test_case_2() { int Arg0 = 1; int Arr1[] = {0,0,1}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 200; verify_case(2, Arg2, count(Arg0, Arg1)); }
78 void test_case_3() { int Arg0 = 5; int Arr1[] = {1,3,5,8,24,22,25,21,30,2,4,0,6,7,9,11,14,13,12,15,18,17,16,19,26,29,31,28,27,10,20,23}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 450877328; verify_case(3, Arg2, count(Arg0, Arg1)); }
79 void test_case_4() { int Arg0 = 5; int Arr1[] = {31,31,31,31,31}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 11112; verify_case(4, Arg2, count(Arg0, Arg1)); }
80
81 // END CUT HERE
82
83 };
84
85 // BEGIN CUT HERE
86 int main()
87 {
88 Nine ___test;
89 clock_t be,en;
90 ___test.run_test(-1);
91 return 0;
92 }
93
94 // END CUT HERE
95 */
96
97 #include <vector>
98 #include <list>
99 #include <map>
100 #include <set>
101 #include <deque>
102 #include <stack>
103 #include <bitset>
104 #include <algorithm>
105 #include <functional>
106 #include <numeric>
107 #include <utility>
108 #include <sstream>
109 #include <iostream>
110 #include <iomanip>
111 #include <cstdio>
112 #include <cmath>
113 #include <cstdlib>
114 #include <ctime>
115
116 using namespace std;
117
118 class Nine {
119 public:
120 int count(int, vector <int>);
121 };
122
123 const int md = 1000000007;
124
125 inline void add(int &a, int b) {
126 a += b;
127 if (a >= md) a -= md;
128 }
129
130 inline int mul(int a, int b) {
131 return (long long)a * b % md;
132 }
133
134 int f[33][100010];
135 int ways[5555][13];
136
137 int Nine::count(int que, vector <int> d) {
138 int ds = d.size();
139 for (int i = 0; i <= ds; i++) {
140 for (int j = 0; j < 9; j++) {
141 ways[i][j] = 0;
142 }
143 }
144 ways[0][0] = 1;
145 for (int i = 0; i < ds; i++) {
146 for (int j = 0; j <= 9; j++) {
147 for (int k = 0; k <= 9; k++) {
148 int new_j = (j + k) % 9;
149 add(ways[i + 1][new_j], ways[i][j]);
150 }
151 }
152 }//预处理 多少位 余数为多少的情况数
153 int n = (1 << que);
154 vector <int> cnt(n, 0);
155 for (int i = 0; i < ds; i++) {
156 cnt[d[i]]++;
157 }
158 int MAX = 1;
159 for (int i = 0; i < que; i++) {
160 MAX *= 9;
161 }
162 for (int t = 0; t <= n; t++) {
163 for (int z = 0; z < MAX; z++) {
164 f[t][z] = 0;
165 }
166 }
167 f[0][0] = 1;
168 int cur[42];
169 for (int t = 0; t < n; t++) {
170 for (int z = 0; z < MAX; z++) {//max代表5个问题的情况压缩
171 int zz = z;
172 for (int i = 0; i < que; i++) {
173 cur[i] = zz % 9;
174 zz /= 9;
175 }//cur[i] 代表的是每一种状态的余数
176
177 for (int now = 0; now <= 9; now++) { //枚举位数
178
179 int new_z = 0;
180 for (int i = que - 1; i >= 0; i--) {//que代表的是问题个数
181 new_z *= 9;
182 int digit = cur[i];
183 if (t & (1 << i)) {
184 digit += now;
185 if (digit >= 9) digit -= 9;
186 }
187 //
188 new_z += digit;
189 }
190 add(f[t + 1][new_z], mul(f[t][z], ways[cnt[t]][now]));//其实是一下加 cnt[t] 个值
191 }
192 }
193 }
194 return f[n][0];
195 }
196
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