给你nn个长度为mm的0101串

你可以在每个串前面加一个运算符∨\lor或∧\land

每次询问一个长度为mm的0101串，问有多少总操作序列能得到这个串

n≤1000,m≤5000,q≤1000n\le1000, m\le 5000, q\le 1000

Luogu P4424

Solution

按位考虑

通过观察发现，只有形如∣1|1或&0\&0这样的运算才会改变这一位的答案

即最终结果是11的充要条件就是最后一个∣1|1的位置要在最后一个&0\&0的后面

于是把操作序列转化成一个0101序列∣→0,&→1|\rightarrow0, \&\rightarrow1

容易发现，在确定这一位最终结果的过程，实际上就是比较操作序列和原串的字典序的过程

那么就很好做了。按字典序排序，对于每个询问找到其对应的上下界（最后一个00出现的位置和第一个11出现的位置），直接统计答案即可

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }

inline void proc_status()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) <<endl;
}

inline int read ()
{
	int sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

const int Maxn = 1000 + 100, Maxm = 5000 + 100, Mod = 1e9 + 7;

int N, M, Q;
int A[Maxm][Maxn], Sum[Maxm], Rank[Maxm];

inline int cmp (int a, int b)
{
	for (int i = 1; i <= N; ++i)
		if (A[a][i] > A[b][i]) return 0;
		else if (A[a][i] < A[b][i]) return 1;
	return 0;
}

inline void Add (int &a, int b) { a += b; if (a >= Mod) a -= Mod; }

inline void Solve ()
{
	static int Pow[Maxn];
	Pow[0] = 1; for (int i = 1; i <= N; ++i) Pow[i] = 1ll * Pow[i - 1] * 2 % Mod;

	for (int i = 1; i <= M; ++i) reverse (A[i] + 1, A[i] + N + 1);
	for (int i = 1; i <= M; ++i) Rank[i] = i;
	sort(Rank + 1, Rank + M + 1, cmp);

	for (int i = 1; i <= M; ++i)
		for (int j = 1; j <= N; ++j) 
			Add (Sum[i], A[Rank[i]][j] * Pow[N - j]);
	Sum[M + 1] = Pow[N];

	while (Q--)
	{
		static char S[Maxm];
		int pos1 = 0, pos2 = M + 1;
		scanf("%s", S + 1);

		for (int i = M; i >= 1; --i) if (S[Rank[i]] == '0') {pos1 = i; break; }
		for (int i = 1; i <= M; ++i) if (S[Rank[i]] == '1') {pos2 = i; break; }

		printf("%d\n", pos1 <= pos2 ? (Sum[pos2] - Sum[pos1] + Mod) % Mod : 0);
	}
}


inline void Input ()
{
	N = read(), M = read(), Q = read();
	for (int i = 1; i <= N; ++i)
	{
		static char S[Maxm];
		scanf("%s", S + 1);
		for (int j = 1; j <= M; ++j)
			A[j][i] = S[j] - '0';
	}
}

int main()
{
#ifdef hk_cnyali
	freopen("A.in", "r", stdin);
	freopen("A.out", "w", stdout);
#endif
	Input();
	Solve();
	return 0;
}