$$\sum_{(x,y)} C_{x,y}\\ =\sum_{a\le x\le c} \sum_{b\le y\le d} \sum_{k=1}^n A_{x,k}\times B_{k,y}\\ =\sum_{k=1}^n \sum_{a\le x\le c} \sum_{b\le y\le d} A_{x,k}\times B_{k,y}\\ =\sum_{k=1}^n \sum_{a\le x\le c} A_{x,k}\times \left(\sum_{b\le y\le d} B_{k,y}\right)\\ =\sum_{k=1}^n \bigg(\left(\sum A\right)\left(\sum B\right)\bigg) $$

维护两个前缀和即可。极其卡常。

卡常前代码：

#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#define int long long
using namespace std;
int n,m,a[2012][2012],b[2012][2012];
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>n>>m;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			cin>>a[i][j],a[i][j]+=a[i-1][j];
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			cin>>b[i][j],b[i][j]+=b[i][j-1];
	while(m--){
		int x1,y1,x2,y2;
		cin>>x1>>y1>>x2>>y2;
		if(x1>x2) swap(x1,x2);
		if(y1>y2) swap(y1,y2);
		int res=0;
		for(int k=1;k<=n;k++)
			res+=(a[x2][k]-a[x1-1][k])*(b[k][y2]-b[k][y1-1]);
		cout<<res<<'\n';
	} 
	return 0;
}

卡常工作由 DeepSeek 完成（？）

#pragma GCC optimize("O3,unroll-loops")
#include<bits/stdc++.h>
#define int long long
using namespace std;
using ll=long long;
const int MAXN=2012;
int a[MAXN][MAXN],b[MAXN][MAXN];
namespace io{
    const int S=1<<20;
    char buf[S],*p1,*p2;
    inline char gc(){
        if(p1==p2){
            p2=(p1=buf)+fread(buf,1,S,stdin);
            if(p1==p2) return EOF;
        }
        return *p1++;
    }
    template<class T>inline void read(T&x){
        x=0;char c=gc();bool f=0;
        while(c<'0'||c>'9'){
            if(c=='-')f=1;
            c=gc();
        }
        while(c>='0'&&c<='9'){
            x=x*10+(c-'0');
            c=gc();
        }
        if(f)x=-x;
    }
    char obuf[S],*op=obuf;
    inline void pc(char c){
        if(op-obuf==S){
            fwrite(obuf,1,S,stdout);
            op=obuf;
        }
        *op++=c;
    }
    inline void write(ll x){
        if(x==0){
            pc('0');
            pc('\n');
            return;
        }
        if(x<0){
            pc('-');
            x=-x;
        }
        char s[30];
        int n=0;
        while(x){
            s[n++]=char('0'+x%10);
            x/=10;
        }
        while(n--) pc(s[n]);
        pc('\n');
    }
    struct Flusher{
        ~Flusher(){
            if(op!=obuf) fwrite(obuf,1,op-obuf,stdout);
        }
    }flusher;
}
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n,m;
    io::read(n);
    io::read(m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            io::read(a[i][j]);
            a[i][j]+=a[i-1][j];
        }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            io::read(b[i][j]);
            b[i][j]+=b[i][j-1];
        }
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
            swap(b[i][j],b[j][i]);
    while(m--){
        int x1,y1,x2,y2;
        io::read(x1);
        io::read(y1);
        io::read(x2);
        io::read(y2);
        if(x1>x2) swap(x1,x2);
        if(y1>y2) swap(y1,y2);
        ll*A2=a[x2],*A1=a[x1-1],*B2=b[y2],*B1=b[y1-1];
        ll r0=0,r1=0,r2=0,r3=0,r4=0,r5=0,r6=0,r7=0;
        int k=1;
        for(;k+7<=n;k+=8){
            r0+=(A2[k]-A1[k])*(B2[k]-B1[k]);
            r1+=(A2[k+1]-A1[k+1])*(B2[k+1]-B1[k+1]);
            r2+=(A2[k+2]-A1[k+2])*(B2[k+2]-B1[k+2]);
            r3+=(A2[k+3]-A1[k+3])*(B2[k+3]-B1[k+3]);
            r4+=(A2[k+4]-A1[k+4])*(B2[k+4]-B1[k+4]);
            r5+=(A2[k+5]-A1[k+5])*(B2[k+5]-B1[k+5]);
            r6+=(A2[k+6]-A1[k+6])*(B2[k+6]-B1[k+6]);
            r7+=(A2[k+7]-A1[k+7])*(B2[k+7]-B1[k+7]);
        }
        ll res=r0+r1+r2+r3+r4+r5+r6+r7;
        for(;k<=n;k++)
            res+=(A2[k]-A1[k])*(B2[k]-B1[k]);
        io::write(res);
    }
    return 0;
}