BZOJ 2462 [BeiJing2011]矩阵模板 矩阵哈希

昨天卡了一天常数。。。然后发现吧$unsigned\space long\space long$改成$unsigned$就可以过了$qwq$

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先把每一行的前缀哈希求出，然后再竖着把每个前缀哈希值哈希起来（相当于二维前缀哈希）

注意横着和竖着哈希的$Base$不能相同

#include<cstdio>
#include<iostream>
#define ll long long
#define ull unsigned
#define R register int
using namespace std;
const int B=,B2=,N=;
namespace Fread {
    static char B[<<],*S=B,*D=B;
    #define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
    inline int g() {
        R ret=,fix=; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-:fix;
        do ret=ret*+(ch^); while(isdigit(ch=getchar())); return ret*fix;
    } inline bool isempty(const char& ch) {return ch<=||ch>=;}
    inline void gs(char* s) {register char ch; while(isempty(ch=getchar())); do *s++=ch; while(!isempty(ch=getchar()));}
}using Fread::g; using Fread::gs;
int n,m,h,l;
ull hsh[N][N],p[N],p2[N],hs[N/][N/];
char s[N][N],ss[N/][N/];
const int MOD=;
bool v[MOD];
signed main() {
    p[]=p2[]=; for(R i=;i<=;++i) p[i]=p[i-]*B; for(R i=;i<=;++i) p2[i]=p2[i-]*B2;
    n=g(),m=g(),h=g(),l=g(); for(R i=;i<=n;++i) gs(s[i]+);
    for(R i=;i<=n;++i) for(R j=;j<=m;++j) hsh[i][j]=hsh[i][j-]*B+s[i][j]-'';
    for(R i=;i<=n;++i) for(R j=;j<=m;++j) hsh[i][j]=hsh[i-][j]*B2+hsh[i][j];
    for(R i=h;i<=n;++i) for(R j=l;j<=m;++j) v[(hsh[i][j]-hsh[i-h][j]*p2[h]-hsh[i][j-l]*p[l]+hsh[i-h][j-l]*p2[h]*p[l])%MOD]=true;
    for(R q=g();q;--q) {
        for(R i=;i<=h;++i) gs(ss[i]+);
        for(R i=;i<=h;++i) for(R j=;j<=l;++j) hs[i][j]=hs[i][j-]*B+ss[i][j]-'';
        for(R i=;i<=h;++i) for(R j=;j<=l;++j) hs[i][j]=hs[i-][j]*B2+hs[i][j];
        printf("%d\n",v[hs[h][l]%MOD]);
    }
}

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2019.06.12