hdu 6336 Problem E. Matrix from Arrays-优快云博客

本文介绍了一个使用固定长度数组生成无限矩阵的问题，并提供了一种高效解决矩阵中子矩阵求和查询的方法。通过预处理和数学推导，实现了快速响应大量查询。

Problem E. Matrix from Arrays

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1286 Accepted Submission(s): 583

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.

思路：
M[i][j]=A[((i+j)*(i+j+1))%L]。
进一步推得M[i][j]=M[i+2*L][j]+M[i][j+2*L]。

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=33;
ll a[maxn][maxn],A[maxn],sum[maxn],r[maxn];
int L;
ll query(int x0,int y0,int x1,int y1)
{
    ll ans=0;
    for(int i=0;i<L;i++) sum[i]=(y1-y0+1)/L*r[i];
    y1-=(y1-y0+1)/L*L;
    for(int i=0;i<L;i++)
    {
        for(int j=y0;j<=y1;j++) sum[i]+=a[i][j%L];
        ans+=sum[i];
    }
    ans=(x1-x0+1)/L*ans;
    x1-=(x1-x0+1)/L*L;
    for(int i=x0;i<=x1;i++) ans+=sum[i%L];
    return ans;
}
int main()
{
    int T;scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&L);
        for(int i=0;i<L;i++) scanf("%lld",&A[i]);
        for(int i=0;i<2*L;i++)
            for(int j=0;j<2*L;j++)
            a[i][j]=A[((i+j)*(i+j+1)/2+i)%L];
        L*=2;
        memset(r,0,sizeof(r));
        for(int i=0;i<L;i++)
        {
            for(int j=0;j<L;j++) r[i]+=a[i][j];
        }
        int Q;scanf("%d",&Q);
        while(Q--)
        {
            int x0,x1,y0,y1;
            scanf("%d%d%d%d",&x0,&y0,&x1,&y1);
            printf("%lld\n",query(x0,y0,x1,y1));
        }
    }
    return 0;
}