hdu多校第4场E. Matrix from Arrays HDU 二维前缀和

Posted 2014slx

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了hdu多校第4场E. Matrix from Arrays HDU 二维前缀和相关的知识,希望对你有一定的参考价值。

Problem E. Matrix from Arrays

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 907    Accepted Submission(s): 403


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 dont 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
 

Recommend
chendu   |   We have carefully selected several similar problems for you:  6343 6342 6341 6340 6339 
 

 

 

 

技术分享图片

 

 

技术分享图片

 

如图二 根据容斥原理S=S1-S2-S3+S4;;S1, S2, S3, S4都是以(x, y)为右下角,以(0, 0)为左上角的矩阵,问题就转化成了求这样的矩阵图一;

米黄色的面积表示有多少个完整的循环矩阵,下方白条及右方白条表示只有长或宽不完整的矩阵,橙黄色面积表示不完整的循环矩阵;

 

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
#define ll long long
ll m[100][100];
ll a[100000];
ll sum[25][25];
ll len;
ll jisuan(int x,int y)
{
    ll ans=(x/len)*(y/len)*sum[len][len];//多少个重复规律
    ans+=sum[x%len][len]*(y/len)+sum[len][y%len]*(x/len);//左边和下面
    ans+=sum[x%len][y%len];//左下角
    return ans;

}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int l;
        scanf("%d",&l);
        for(int i=0;i<l;i++)
            scanf("%lld",&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;
            }
        }
        len=2*l;
        memset(sum, 0, sizeof(sum));
        for(ll i=1; i<=len; i++){
            for(ll j=1; j<=len; j++){
                sum[i][j]=sum[i][j-1]+sum[i-1][j]-sum[i-1][j-1]+m[i][j];//容斥原理
            }
        }
        int q;
        scanf("%d",&q);
        while(q--)
        {
            int x0,y0,x1,y1;
            scanf("%d%d%d%d",&x0,&y0,&x1,&y1);
            x0++,y0++,x1++,y1++;
            ll ans=0; ans=jisuan(x1,y1)+jisuan(x0-1,y0-1)-jisuan(x0-1,y1)-jisuan(x1,y0-1);
            cout<<ans<<endl;
            
        }
        
    }
    
    return 0;
}

 

以上是关于hdu多校第4场E. Matrix from Arrays HDU 二维前缀和的主要内容,如果未能解决你的问题,请参考以下文章

Problem E. Matrix from Arrays(杭电2018年多校第四场+思维+打表找循环节)

hdu多校第4场 B Harvest of Apples(莫队)

2017多校第4场 HDU 6078 Wavel Sequence DP

2016多校第4场 HDU 6076 Security Check DP,思维

2017多校第6场 HDU 6105 Gameia 博弈

2017多校第10场 HDU 6181 Two Paths 次短路