HDU-6336-构造

Problem E. Matrix from Arrays

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 419    Accepted Submission(s): 180

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

 

 　　比赛的时候其实知道做法了，打表后可以发现这个矩阵的行列都是由循环节的而且行列的循环长度一致，

只不过一开始默认为是L，后来发现奇数是L，偶数是2*L,我们都按照2*L来做就好了，预处理出来行和列的

前缀和然后容斥的处理询问。

　　

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
#define pii pair<int,int>
#define mp make_pair
#define LL long long 
int cursor = 0;
int M[115][115],A[15],L,len;
LL pre1[25][25],pre2[25][25];
LL g[25];
LL sum(int x,int y){
    if(x<=0||y<=0) return 0;
    LL ans=0;
    for(int i=1;i<=len;++i){
        g[i]=g[i-1]+pre1[i][len]*(y/len)+pre1[i][y%len];
    }
    return g[len]*(x/len)+g[x%len];
}
int main(){
    int t,n,m,i,j,k,Q;
    int x1,y1,x2,y2;
    cin>>t;
    while(t--){
        scanf("%d",&L);
        len=L*2;
        memset(pre1,0,sizeof(pre1));
        memset(pre2,0,sizeof(pre2));
        for(i=0;i<L;++i) scanf("%d",A+i);
        int cursor = 0;
        for (int i = 0;i<=100; ++i) {
            for (int j = 0; j <= i; ++j) { 
                M[j+1][i - j+1] = A[cursor];
                cursor = (cursor + 1) % L;
            }
        }
        for(i=1;i<=len;++i){
            for(j=1;j<=len;++j){
                pre1[i][j]=M[i][j]+pre1[i][j-1];
                pre2[i][j]=M[j][i]+pre2[i][j-1];
            }
        }
        scanf("%d",&Q);
        while(Q--){
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            x1++,y1++,x2++,y2++;
            printf("%lld
",sum(x1-1,y1-1)+sum(x2,y2)-sum(x1-1,y2)-sum(x2,y1-1));
        }
    }
    return 0;
}