题意

题目链接

Sol

二维矩阵hash，就是对行和列分配一个不同的base，然后分别做一遍hash，这样会减少冲突的概率。

预处理出所有大小为\(A \times B\)的矩阵的hash值，判断一下即可

mdzz居然卡常数

#include<bits/stdc++.h>
#define ull unsigned int
using namespace std;
const int MAXN = 1010, mod = 100000007;
const ull base1 = 19260817, base2 = 998244353;
inline int read() {
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, A, B;
ull po1[MAXN], po2[MAXN], m[MAXN][MAXN], a[MAXN][MAXN];
bool ha[mod + 1];
int readch() {
    char c = '.';
    while(c != '0' && c != '1') c = getchar();
    return c;
}
main() {
    //freopen("1.in", "r", stdin);
    N = read(); M = read(); A = read(); B = read();
    po1[0] = 1; for(int i = 1; i <= N; i++) po1[i] = po1[i - 1] * base1;
    po2[0] = 1; for(int i = 1; i <= M; i++) po2[i] = po2[i - 1] * base2;
    for(int i = 1; i <= N; i++) for(int j = 1; j <= M; j++) m[i][j] = readch();
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= M; j++)
            m[i][j] += m[i - 1][j] * base1;
    for(int i = 1; i <= N; i++) 
        for(int j = 1; j <= M; j++)
            m[i][j] += m[i][j - 1] * base2;
    for(int i = A; i <= N; i++) {
        for(int j = B; j <= M; j++) {
            ull tmp = m[i][j] - m[i - A][j] * po1[A] - m[i][j - B] * po2[B] + m[i - A][j - B] * po1[A] * po2[B];
            ha[tmp % mod] = 1;
        }
    }
    int Q = read();
    while(Q--) {
        for(int i = 1; i <= A; i++) for(int j = 1; j <= B; j++) a[i][j] = readch();
        for(int i = 1; i <= A; i++)
            for(int j = 1; j <= B; j++)
                a[i][j] += a[i - 1][j] * base1;
        for(int i = 1; i <= A; i++) 
            for(int j = 1; j <= B; j++)
                a[i][j] += a[i][j - 1] * base2;
        putchar(ha[a[A][B] % mod] ? '1' : '0'); putchar('\n');
    }
    return 0;
}