Problem E. Matrix from Arrays

PS：由于打表姿势没对，错过了很好写的一道题。经过打表你发发现最上角的矩阵，如果是奇数，大小为  L * L，如果为偶数，大小为  2L * 2L，是重复出现。为了便于计算，将循环节的大小统一为 2L * 2L，然后预处理二维前缀和，瞎搞搞就出来了。

#pragma warning(disable:4996)
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<vector>
#include<queue>
#include<map>
#include<string>
#include<stack>
#define ll long long
#define P pair<int, int>
#define PP pair<int,pair<int, int>>
#define pb push_back
#define pp pop_back
#define lson root << 1
#define INF (int)2e9 + 7
#define rson root << 1 | 1
#define LINF (unsigned long long int)1e18
#define mem(arry, in) memset(arry, in, sizeof(arry))
using namespace std;

int n, T;
int mp[1005][1005], a[20], dp[100][100];

ll get(int s, int t) {
　　 if (s == -1 || t == -1) return 0;

    int x = s % n, cnt_x = s / n;
    int y = t % n, cnt_y = t / n;
    return 1ll * dp[x][n - 1] * cnt_y + 1ll * dp[n - 1][y] * cnt_x + 1ll * dp[n - 1][n - 1] * cnt_x * cnt_y + 1ll * dp[x][y];
}

int main()
{
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
        int cur = 0;
        for (int i = 0; i <= 100 ; ++i) {
            for (int j = 0; j <= i; ++j) {
                mp[j][i - j] = a[cur];
                cur = (cur + 1) % n;
            }
        }

        dp[0][0] = mp[0][0];
        for (int i = 1; i < 2 * n; ++i) dp[0][i] = dp[0][i - 1] + mp[0][i];
        for (int i = 1; i < 2 * n; ++i) dp[i][0] = dp[i - 1][0] + mp[i][0];
        for (int i = 1; i < 2 * n; ++i) for (int j = 1; j < 2 * n; ++j) dp[i][j] = mp[i][j] + dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1];

        n = 2 * n;
    
        int q, x0, y0, x1, y1;
        scanf("%d", &q);
        while (q--) {
            scanf("%d %d %d %d", &x0, &y0, &x1, &y1);
            //cout << get(x1, y1) << endl;
            printf("%lld
", get(x1, y1) - get(x1, y0 - 1) - get(x0 - 1, y1) + get(x0 - 1, y0 - 1));
        }
    }
    return 0;
}