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最新推荐文章于 2025-05-18 15:51:46 发布

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本文介绍了一种算法，用于计算由平行于坐标轴的线段所形成的矩形的总面积。通过固定每一对平行于y轴的线段，枚举所有平行于x轴的线段组合，最终求得所有矩形面积之和。

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D - 井井井 / ###

Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

On a two-dimensional plane, there are m lines drawn parallel to the x axis, and n lines drawn parallel to the y axis. Among the lines parallel to the x axis, the i-th from the bottom is represented by y=yi. Similarly, among the lines parallel to the y axis, the i-th from the left is represented by x=xi.

For every rectangle that is formed by these lines, find its area, and print the total area modulo 109+7.

That is, for every quadruple (i,j,k,l) satisfying 1≤i<j≤n and 1≤k<l≤m, find the area of the rectangle formed by the lines x=xi, x=xj, y=yk and y=yl, and print the sum of these areas modulo 109+7.

Constraints

2≤n,m≤105
−109≤x1<…<xn≤109
−109≤y1<…<ym≤109
xi and yi are integers.

Input

Input is given from Standard Input in the following format:

n m
x1 x2 … xn
y1 y2 … ym

Output

Print the total area of the rectangles, modulo 109+7.

Sample Input 1

Copy

3 3
1 3 4
1 3 6

Sample Output 1

Copy

The following figure illustrates this input:

sample1-1

The total area of the nine rectangles A, B, ..., I shown in the following figure, is 60.

sample1-2

Sample Input 2

Copy

6 5
-790013317 -192321079 95834122 418379342 586260100 802780784
-253230108 193944314 363756450 712662868 735867677

Sample Output 2

Copy

835067060

题意：给定n条平行于y轴的线x[1~n]，和m条平行于x轴的线y[1~n]，计算x[i],x[j]和y[k],y[l]组成的矩形面积之和，1<=x<y<=n，1<=k<l<=m。

思路：每次固定两个y线，枚举所有x线二元组，发现x线和y线可以分开计算，即

# include <bits/stdc++.h>
using namespace std;
 
typedef long long LL;
const LL mod = 1e9+7;
LL x[100003], y[100003];
int main()
{
    int n, m;
    scanf("%d%d",&n,&m);
    for(int i=0; i<n; ++i)
        scanf("%lld",&x[i]);
    for(int i=0; i<m; ++i)
        scanf("%lld",&y[i]);
    LL W = 0, H = 0;
    for(int l=0, r=n-1; l<r; ++l,--r)//考虑每个小区间对答案的贡献。
        W = (W + (x[r]-x[l])*(r-l))%mod;
    for(int l=0, r=m-1; l<r; ++l,--r)
        H = (H + (y[r]-y[l])*(r-l))%mod;
    printf("%lld\n",(W*H)%mod);
    return 0;
}