2018 Multi-University Training Contest 4 Problem E. Matrix from Arrays 【打表+二维前缀和】

任意门：http://acm.hdu.edu.cn/showproblem.php?pid=6336

Problem E. Matrix from Arrays

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1711    Accepted Submission(s): 794

Problem Description

Kazari has an array A length of L, she plans to generate an infinite matrix M using A.
The procedure is given below in C/C++:

int cursor = 0;

for (int i = 0; ; ++i) {
    for (int j = 0; j <= i; ++j) { 
        M[j][i - j] = A[cursor];
        cursor = (cursor + 1) % L;
    }
}

Her friends don't believe that she has the ability to generate such a huge matrix, so they come up with a lot of queries about M, each of which focus the sum over some sub matrix. Kazari hates to spend time on these boring queries. She asks you, an excellent coder, to help her solve these queries.

 

Input

The first line of the input contains an integer T (1≤T≤100) denoting the number of test cases.
Each test case starts with an integer L (1≤L≤10) denoting the length of A.
The second line contains L integers A0,A1,...,AL−1 (1≤Ai≤100).
The third line contains an integer Q (1≤Q≤100) denoting the number of queries.
Each of next Q lines consists of four integers x0,y0,x1,y1 (0≤x0≤x1≤108,0≤y0≤y1≤108) querying the sum over the sub matrix whose upper-leftmost cell is (x0,y0) and lower-rightest cell is (x1,y1).

 

Output

For each test case, print an integer representing the sum over the specific sub matrix for each query.

Sample Input

1

3

1 10 100

5

3 3 3 3

2 3 3 3

2 3 5 8

5 1 10 10

9 99 999 1000

 

Sample Output

1

101

1068

2238

33076541

 

Source

2018 Multi-University Training Contest 4

题意概括：

给一个长度为 L 的序列，要求按照代码的要求构建一个矩阵。

Q次查询，每次查询输入矩阵的左上角和右下角坐标，输出矩阵的值的总和。

解题思路：

并没有什么过人的天分，老老实实把构建的矩阵输出来找规律。

发现大矩阵是长宽 为 2*L 的小矩阵构造而成的。

问题就转换为了类似于求矩阵面积的问题，也就可以通过二维前缀和 sum(x, y) 来求解答案。

例如要求解 （x1, y1, x2, y2) 的值就等于 sum(x2, y2) - sum(x1-1, y2) - sum(x2, y1-1) + sum(x1-1, y1-1)；

如何计算 sum(x, y)呢？首先统计有多少个 2L * 2L 的子矩阵，然后再单独加上这些除掉这些子矩阵的其余部分。

AC code：

#include<set>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
#include<cmath>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int MAXN = 1e3+10;
LL sum;
int mmp[MAXN][MAXN];
int N, M, L, Q;
LL sum1[MAXN], sum2[MAXN];
int A[MAXN];

void init()
{
    memset(mmp, 0, sizeof(mmp));
    memset(sum1, 0, sizeof(sum1));
    memset(sum2, 0, sizeof(sum2));
    int cnt = 0;
    for(int i = 0; i < 4*L; i++){
        for(int j = 0; j <= i; j++){
            mmp[j][i-j] = A[cnt];
            cnt = (cnt+1)%L;
        }
    }
    sum = 0;
    for(int i = 0; i < 2*L; i++){
        for(int j = 0; j < 2*L; j++){
            sum+=mmp[i][j];
            sum1[i] += mmp[i][j];
            sum2[j] += mmp[i][j];
        }
    }
}

LL query(int x, int y)
{
    int cnt_x = (x+1)/(2*L);
    int cnt_y = (y+1)/(2*L);
    LL res = sum*cnt_x*cnt_y;
    int lenx = (x+1)%(2*L);
    int leny = (y+1)%(2*L);

    for(int i = 0; i < lenx; i++) res+= cnt_y*sum1[i];
    for(int j = 0; j < leny; j++) res+= cnt_x*sum2[j];
    

    for(int i = 0; i < lenx; i++)
        for(int j = 0; j < leny; j++){
            res+=mmp[i][j];
        }

    return res;
}

int main()
{
    int T_case;
    int x1, x2, y1, y2;
    scanf("%d", &T_case);
    while(T_case--){
        scanf("%d", &L);
        for(int i = 0; i < L; i++){
            scanf("%d", &A[i]);
        }
        init();
        
        scanf("%d", &Q);
        while(Q--){
            scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
            LL ans = query(x2, y2)-query(x1-1, y2) - query(x2, y1-1) + query(x1-1, y1-1);
            printf("%lld
", ans);
        }
    }
    return 0;
}