建立一颗trie树，记录下来每个点以它为结尾的字符串的个数cnt，和它的子树内有多少字符串size
于是查询的时候就只需要把沿途的cnt加起来，再加上最后的size就好了
1 /************************************************************** 2 Problem: 1590 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:320 ms 7 Memory:3892 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 12 using namespace std; 13 14 inline int read(); 15 16 struct trie { 17 trie *son[2]; 18 int sz, tail; 19 20 #define Len (1 << 16) 21 void* operator new(size_t) { 22 static trie *mempool, *c; 23 if (c == mempool) 24 mempool = (c = new trie[Len]) + Len; 25 c -> son[0] = c -> son[1] = NULL; 26 c -> sz = 0, c -> tail = 0; 27 return c++; 28 } 29 #undef Len 30 31 void insert(int x) { 32 ++this -> sz; 33 if (x == 0) { 34 ++this -> tail; 35 return; 36 } 37 int t = read(); 38 if (!this -> son[t]) this -> son[t] = new()trie; 39 this -> son[t] -> insert(x - 1); 40 } 41 42 int query(int x) { 43 if (x == 0) return this -> sz; 44 int t = read(); 45 if (!this -> son[t]) { 46 while (--x) read(); 47 return this -> tail; 48 } 49 return this -> son[t] -> query(x - 1) + this -> tail; 50 } 51 } *t; 52 53 int n, m; 54 55 int main() { 56 int i; 57 n = read(), m = read(); 58 t = new()trie; 59 for (i = 1; i <= n; ++i) t -> insert(read()); 60 for (i = 1; i <= m; ++i) printf("%d ", t -> query(read())); 61 return 0; 62 } 63 64 inline int read() { 65 static int x; 66 static char ch; 67 x = 0, ch = getchar(); 68 while (ch < '0' || '9' < ch) 69 ch = getchar(); 70 while ('0' <= ch && ch <= '9') { 71 x = x * 10 + ch - '0'; 72 ch = getchar(); 73 } 74 return x; 75 }