题意:
问A到B之间的所有整数，转换成BCD Code后，
有多少个不包含属于给定病毒串集合的子串,A,B <=10^200,病毒串总长度<= 2000.
BCD码这个在数字电路课上讲了，题干也讲的很详细。
数位DP的实现是通过0~9 ,并不是通过BCD码
所有我们需要先把字串放入AC自动机，建立一个BCD数组
因为BCD码是一个4位二进制数，但是tire图上全是0，1，
所以对于一个数字，我们的要在转移4次，
如果中间出现了病毒串就return -1 表示不能转移，
BCD【i】【j】表示在AC自动机 i 这个节点转移到数字 j 对应的在AC自动机上的节点标号。
然后就是简单的数位DP了，然而我写搓了，由于没有前导0所以前导0要处理掉。
但是你不转移的时候，不能 bcd[idx][i] != -1 就直接continue ，
因为有0的情况，i==0 但是(bcd[idx][i] == -1) 但是这个0是前导0所以不影响。
1 #include <set>
2 #include <map>
3 #include <stack>
4 #include <queue>
5 #include <cmath>
6 #include <ctime>
7 #include <cstdio>
8 #include <string>
9 #include <vector>
10 #include <cstring>
11 #include <iostream>
12 #include <algorithm>
13
14 #define pi acos(-1.0)
15 #define eps 1e-9
16 #define fi first
17 #define se second
18 #define rtl rt<<1
19 #define rtr rt<<1|1
20 #define bug printf("******\n")
21 #define mem(a, b) memset(a,b,sizeof(a))
22 #define name2str(x) #x
23 #define fuck(x) cout<<#x" = "<<x<<endl
24 #define sfi(a) scanf("%d", &a)
25 #define sffi(a, b) scanf("%d %d", &a, &b)
26 #define sfffi(a, b, c) scanf("%d %d %d", &a, &b, &c)
27 #define sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
28 #define sfL(a) scanf("%lld", &a)
29 #define sffL(a, b) scanf("%lld %lld", &a, &b)
30 #define sfffL(a, b, c) scanf("%lld %lld %lld", &a, &b, &c)
31 #define sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
32 #define sfs(a) scanf("%s", a)
33 #define sffs(a, b) scanf("%s %s", a, b)
34 #define sfffs(a, b, c) scanf("%s %s %s", a, b, c)
35 #define sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
36 #define FIN freopen("../in.txt","r",stdin)
37 #define gcd(a, b) __gcd(a,b)
38 #define lowbit(x) x&-x
39 #define IO iOS::sync_with_stdio(false)
40
41
42 using namespace std;
43 typedef long long LL;
44 typedef unsigned long long ULL;
45 const ULL seed = 13331;
46 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
47 const int maxn = 1e6 + 7;
48 const int maxm = 8e6 + 10;
49 const int INF = 0x3f3f3f3f;
50 const int mod = 1000000009;
51 int T, n;
52 char buf[maxn], num1[maxn], num2[maxn];
53 int bcd[2020][12], bit[maxn];
54 LL dp[205][2020];
55
56 struct Aho_Corasick {
57 int next[2010][2], fail[2010], End[2010];
58 int root, cnt;
59
60 int newnode() {
61 for (int i = 0; i < 2; i++) next[cnt][i] = -1;
62 End[cnt++] = 0;
63 return cnt - 1;
64 }
65
66 void init() {
67 cnt = 0;
68 root = newnode();
69 }
70
71 void insert(char buf[]) {
72 int len = strlen(buf);
73 int now = root;
74 for (int i = 0; i < len; i++) {
75 if (next[now][buf[i] - '0'] == -1) next[now][buf[i] - '0'] = newnode();
76 now = next[now][buf[i] - '0'];
77 }
78 End[now] = 1;
79 }
80
81 void build() {
82 queue<int> Q;
83 fail[root] = root;
84 for (int i = 0; i < 2; i++)
85 if (next[root][i] == -1) next[root][i] = root;
86 else {
87 fail[next[root][i]] = root;
88 Q.push(next[root][i]);
89 }
90 while (!Q.empty()) {
91 int now = Q.front();
92 Q.pop();
93 if (End[fail[now]]) End[now] = 1;
94 for (int i = 0; i < 2; i++)
95 if (next[now][i] == -1) next[now][i] = next[fail[now]][i];
96 else {
97 fail[next[now][i]] = next[fail[now]][i];
98 Q.push(next[now][i]);
99 }
100 }
101 }
102
103 int get_num(int cur, int num) {
104 if (End[cur]) return -1;
105 int now = cur;
106 for (int i = 3; i >= 0; --i) {
107 if (End[next[now][1 & (num >> i)]]) return -1;
108 now = next[now][1 & (num >> i)];
109 }
110 return now;
111 }
112
113 void get_bcd() {
114 for (int i = 0; i < cnt; ++i)
115 for (int j = 0; j < 10; ++j)
116 bcd[i][j] = get_num(i, j);
117 }
118
119 LL dfs(int pos, int idx, int flag, int limit) {
120 if (pos == -1) return 1;
121 if (!limit && dp[pos][idx] != -1) return dp[pos][idx];
122 int num = limit ? bit[pos] : 9;
123 LL ans = 0;
124 for (int i = 0; i <= num; ++i) {
125 if (flag && i == 0) ans = (ans + dfs(pos - 1, idx, 1, limit && i == num)) % mod;
126 else if (bcd[idx][i] != -1) ans = (ans + dfs(pos - 1, bcd[idx][i], 0, limit && i == num)) % mod;
127 }
128 if (!limit && !flag) dp[pos][idx] = ans;
129 return ans;
130 }
131
132 LL solve() {
133 get_bcd();
134 int len1 = strlen(num1), len2 = strlen(num2);
135 for (int i = len1-1; i>=0; --i) {
136 if (num1[i] > '0') {
137 num1[i]--;
138 break;
139 } else num1[i] = '9';
140 }
141 for (int i = 0; i < len1; ++i) bit[i] = num1[len1 - 1 - i] - '0';
142 LL ans1 = dfs(len1 - 1, 0, 1, 1);
143 for (int i = 0; i < len2; ++i) bit[i] = num2[len2 - 1 - i] - '0';
144 LL ans2 = dfs(len2 - 1, 0, 1, 1);
145 return (ans2 - ans1 + mod) % mod;
146 }
147
148 void debug() {
149 for (int i = 0; i < cnt; i++) {
150 printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
151 for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
152 printf("]\n");
153 }
154 }
155 } ac;
156
157 int main() {
158 // FIN;
159 sfi(T);
160 while (T--) {
161 sfi(n);
162 ac.init();
163 for (int i = 0; i < n; ++i) {
164 sfs(buf);
165 ac.insert(buf);
166 }
167 ac.build();
168 sffs(num1, num2);
169 mem(dp, -1);
170 printf("%lld\n", ac.solve());
171 }
172 return 0;
173 }