中文题目：http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=563&pid=1003
首先贴一下bc的题解：
首先我们考虑m=1的情况。给定两个数组A={a1,a2,…,an}和B={b1,b2,…,bk}，问B在A中出现了几次。令ci=ai+1ai,1≤i<n，同样令di=bi+1bi,1≤i<k，那么上述问题可以转化为ci和di的模式匹配问题，这个正确性显然，也没有什么好证明的。于是对于m=1的情况只有用个kmp即可搞定。
现在考虑m>1的情况，我们考虑用ac自动机来做。考虑到字符集并不是普通的数字，而是一个分数，我们不放搞个分数类，然后用map存转移边。用m个模式串（Bob的序列）建好自动机之后，把文本串（Alice的序列）在自动机上跑一遍即可统计出答案。
事实上，这个题的精华就在于序列变换，将问题转化为多模板匹配问题。而且由于转化后的序列有二维状态，可以用map映射重新为状态分配一个值（类似离散）。另外由于字符集比较大，所以ac自动机里面需要一点小小的修改，原来的二维数组的第二维需用map，另外需要同时记录所有儿子的字符值。原来的模板只需要小改就可以适应这种情况了。详见代码：
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define mem_1(a) memset(a, -1, sizeof(a)) 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define define_m int m = (l + r) >> 1 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 32 #define rep_down1(a, b) for (int a = b; a > 0; a--) 33 #define all(a) (a).begin(), (a).end() 34 #define lowbit(x) ((x) & (-(x))) 35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 38 #define pchr(a) putchar(a) 39 #define pstr(a) printf("%s", a) 40 #define sstr(a) scanf("%s", a) 41 #define sint(a) scanf("%d", &a) 42 #define sint2(a, b) scanf("%d%d", &a, &b) 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 44 #define pint(a) printf("%d ", a) 45 #define test_print1(a) cout << "var1 = " << a << endl 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 48 49 typedef long long LL; 50 typedef pair<int, int> pii; 51 typedef vector<int> vi; 52 53 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 54 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 55 const int maxn = 3e4 + 7; 56 const int md = 10007; 57 const int inf = 1e9 + 7; 58 const LL inf_L = 1e18 + 7; 59 const double pi = acos(-1.0); 60 const double eps = 1e-6; 61 62 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 63 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 64 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 65 template<class T>T condition(bool f, T a, T b){return f?a:b;} 66 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 67 int make_id(int x, int y, int n) { return x * n + y; } 68 69 LL ans; 70 71 struct AhoCorasickAutoMata { 72 const static int maxn = 250007; 73 int cc; 74 map<int, int> ch[maxn]; 75 vector<int> son[maxn]; 76 int val[maxn], last[maxn], f[maxn]; 77 78 void clear() { cc = 0; mem0(val); mem0(ch); mem0(last); mem0(f); rep_up0(i, maxn) { son[i].clear(); ch[i].clear(); } } 79 80 void Insert(int s[], int n) { 81 int pos = 0; 82 rep_up0(i, n) { 83 int id = s[i]; 84 if(!ch[pos][id]) { 85 ch[pos][id] = ++ cc; 86 son[pos].push_back(id); 87 } 88 pos = ch[pos][id]; 89 } 90 val[pos] ++; 91 } 92 93 void print(int j) { 94 if (j) { 95 ans += val[j]; 96 print(last[j]); 97 } 98 } 99 100 void find(int T[], int n) { 101 int j = 0; 102 rep_up0(i, n) { 103 int c = T[i]; 104 while (j && !ch[j][c]) j = f[j]; 105 j = ch[j][c]; 106 if (val[j]) print(j); 107 else { 108 if (last[j]) print(last[j]); 109 } 110 } 111 } 112 113 int getFail() { 114 queue<int> q; 115 f[0] = 0; 116 int SZ = son[0].size(); 117 rep_up0(i, SZ) { 118 int c = son[0][i]; 119 int u = ch[0][c]; 120 q.push(u); 121 } 122 while (!q.empty()) { 123 int r = q.front(), SZ = son[r].size(); q.pop(); 124 rep_up0(i, SZ) { 125 int c = son[r][i]; 126 int u = ch[r][c]; 127 q.push(u); 128 int v = f[r]; 129 while (v && !ch[v][c]) v = f[v]; 130 f[u] = ch[v][c]; 131 last[u] = val[f[u]]? f[u] : last[f[u]]; 132 } 133 } 134 } 135 }; 136 137 AhoCorasickAutoMata ac; 138 map<pii, int> mp; 139 int pa[100007], pb[100007]; 140 int a[100007], b[300007]; 141 142 int main() { 143 //freopen("in.txt", "r", stdin); 144 int T, n, m, tot = 0; 145 cin >> T; 146 while (T--) { 147 ans = 0; 148 ac.clear(); 149 mp.clear(); 150 cin >> n >> m; 151 rep_up0(i, n) { 152 sint(a[i]); 153 } 154 rep_up0(i, n - 1) { 155 int g = gcd(a[i], a[i + 1]); 156 int ga = a[i] / g, gb = a[i + 1] / g; 157 int &x = mp[make_pair(ga, gb)]; 158 if (!x) x = ++ tot; 159 pa[i] = x; 160 } 161 162 rep_up0(i, m) { 163 int k; 164 sint(k); 165 rep_up0(j, k) { 166 sint(b[j]); 167 } 168 if (k > n) continue; 169 if (k == 1) { 170 ans += n; 171 continue; 172 } 173 rep_up0(j, k - 1) { 174 int g = gcd(b[j], b[j + 1]); 175 int ga = b[j] / g, gb = b[j + 1] / g; 176 int &x = mp[make_pair(ga, gb)]; 177 if (!x) x = ++ tot; 178 pb[j] = x; 179 } 180 ac.Insert(pb, k - 1); 181 } 182 ac.getFail(); 183 ac.find(pa, n - 1); 184 cout << ans << endl; 185 } 186 return 0; 187 }