题目：点击打开链接
题意：给你一个长度为l的a数组，然后按题目所给代码构造m矩阵，q次询问，每次查询给出左上角(l,r)和右下角(ll,rr)，求m数组中这个矩形里所有数的和。
分析：
打表也可以发现规律，求2L*2L单位矩阵的二维前缀和。
代码：
#pragma comment(linker, "/STACK:102400000,102400000")
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
typedef pair<int,int> PII;
const ll mod = 1e9+7;
const int N = 1e6+10;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};int t,l,q,a[20],m[100][100],x2,y2,x3,y3;void init() {int cur=0;for(int i=0;i<4*l;i++)for(int j=0;j<=i;j++)m[j][i-j]=a[cur],cur=(cur+1)%l;for(int i=0;i<2*l;i++)for(int j=0;j<2*l;j++)if(i&&j) m[i][j] += m[i-1][j]+m[i][j-1]-m[i-1][j-1];else if(i==0&&j) m[i][j] += m[i][j-1];else if(j==0&&i) m[i][j] += m[i-1][j];
}ll gs(int x,int y) {if(x<0||y<0) return 0;ll s1 = 1LL*(x/(2*l))*(y/(2*l))*m[2*l-1][2*l-1];ll s2 = 1LL*y/(2*l)*m[x%(2*l)][2*l-1]+1LL*x/(2*l)*m[2*l-1][y%(2*l)];ll s3 = m[x%(2*l)][y%(2*l)];return s1+s2+s3;
}int main() {ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);cin>>t;while(t--) {cin>>l;for(int i=0;i<l;i++) cin>>a[i];init();cin>>q;while(q--) {cin>>x2>>y2>>x3>>y3;cout<<gs(x3,y3)-gs(x2-1,y3)-gs(x3,y2-1)+gs(x2-1,y2-1)<<endl;}}return 0;
}
