題目鏈接

序列上的Hash和前綴和差不多，二維Hash也和二維前綴和差不多了。
求大矩陣所有rc的小矩陣hash值，再對詢問的矩陣Hash就可以了。
類比於序列上s[r]-s[l-1]pow[r-l+1]，比如s[i-r][j-c]多算了rc次，乘個pw[r]pw[c]就行了。

//106864kb  1544ms
#include <cstdio>
#include <cctype>
#include <algorithm>
//#define gc() getchar()
#define MAXIN 500000
#define gc() (SS==TT&&(TT=(SS=IN)+fread(IN,1,MAXIN,stdin),SS==TT)?EOF:*SS++)
#define P 100000000
#define base1 12289
#define base2 786433
typedef unsigned int uint;//在這用unsigned long long好像沒太大用吧...吧...
const int N=1005;

uint A[N][N],sum[N][N],pw1[N],pw2[N];
bool Hash[P];
char IN[MAXIN],*SS=IN,*TT=IN;

inline int read()
{
    int now=0;register char c=gc();
    for(;!isdigit(c);c=gc());
    for(;isdigit(c);now=now*10+c-'0',c=gc());
    return now;
}
inline int read01()
{
    register char c=gc();
    for(;!isdigit(c);c=gc());
    return c-'0';
}

int main()
{
    int n=read(),m=read(),r=read(),c=read();
    for(int i=1; i<=n; ++i) 
        for(int j=1; j<=m; ++j) sum[i][j]=read01();
    pw1[0]=pw2[0]=1;
    for(int i=1; i<=n; ++i) pw1[i]=pw1[i-1]*base1;
    for(int i=1; i<=m; ++i) pw2[i]=pw2[i-1]*base2;
    for(int i=2; i<=n; ++i)//Row
    {
        uint *s=sum[i], *las=sum[i-1];
        for(int j=1; j<=m; ++j) s[j]+=las[j]*base1;
    }
    for(int i=1; i<=n; ++i)//Column
    {
        uint *s=sum[i];
        for(int j=2; j<=m; ++j) s[j]+=s[j-1]*base2;
    }
    for(int i=r; i<=n; ++i)
        for(int j=c; j<=m; ++j)
        {
            uint hash=sum[i][j]-sum[i-r][j]*pw1[r]-sum[i][j-c]*pw2[c]+sum[i-r][j-c]*pw1[r]*pw2[c];
            Hash[hash%P]=1;
        }
    for(int Q=read(); Q--; )
    {
        for(int i=1; i<=r; ++i)
            for(int j=1; j<=c; ++j) A[i][j]=read01();
        for(int i=2; i<=r; ++i)
        {
            uint *s=A[i], *las=A[i-1];
            for(int j=1; j<=c; ++j) s[j]+=las[j]*base1;
        }
        for(int i=1; i<=r; ++i)
        {
            uint *s=A[i];
            for(int j=2; j<=c; ++j) s[j]+=s[j-1]*base2;
        }
        puts(Hash[A[r][c]%P]?"1":"0");
    }

    return 0;
}
 

BZOJ.2462.[BeiJing2011]矩陣模板(二維Hash)