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

Problem E. Matrix from Arrays

Time Limit: / MS (Java/Others)    Memory Limit: / K (Java/Others)

Total Submission(s):     Accepted Submission(s):

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 = ;

for (int i = ; ; ++i) {

    for (int j = ; j <= i; ++j) {

        M[j][i - j] = A[cursor];

        cursor = (cursor + ) % 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 (≤T≤) denoting the number of test cases.

Each test case starts with an integer L (≤L≤) denoting the length of A.

The second line contains L integers A0,A1,...,AL− (≤Ai≤).

The third line contains an integer Q (≤Q≤) denoting the number of queries.

Each of next Q lines consists of four integers x0,y0,x1,y1 (≤x0≤x1≤,≤y0≤y1≤) 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

Sample Output

Source

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#include<iostream>

#include<stdio.h>

#include<string.h>

using namespace std;

#define ll long long

ll m[][];

ll a[];

ll sum[][];

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=;i<l;i++)

            scanf("%lld",&a[i]);

        int cursor = ;

        for (int i = ; i<; ++i)

        {

            for (int j = ; j <= i; ++j)

            {

                m[j+][i - j+] = a[cursor];

                cursor = (cursor + ) %l;

            }

        }

        len=*l;

        memset(sum, , sizeof(sum));

        for(ll i=; i<=len; i++){

            for(ll j=; j<=len; j++){

                sum[i][j]=sum[i][j-]+sum[i-][j]-sum[i-][j-]+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=; ans=jisuan(x1,y1)+jisuan(x0-,y0-)-jisuan(x0-,y1)-jisuan(x1,y0-);

            cout<<ans<<endl;

        }

    }

    return ;

}