AtCoder 2394

 

Problem Statement

 

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

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=x_i, x=x_j, y=y_k and y=y_l, and print the sum of these areas modulo 10⁹+7.

Constraints

 

• 2≤n,m≤10⁵
• −10⁹≤x₁<…<x_n≤10⁹
• −10⁹≤y₁<…<y_m≤10⁹
• x_i and y_i 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

 

3 3
1 3 4
1 3 6

Sample Output 1

 

60

The following figure illustrates this input:

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

Sample Input 2

 

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

Sample Output 2

 

835067060

 

题意：在一个二维坐标系上，有 n 条平行 y 轴的线和 m 条平行于 x 轴的线，求这些线组成的矩形的面积和

 

根据组合数学的原理，可以发现每个x/y 各出现 2*i-n次，所以遍历统计一下就好了。 

 

#include       <set>
#include       <map>
#include     <queue>
#include     <cmath>
#include    <cstdio>
#include    <cctype>
#include    <vector>
#include   <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const ll mod = 1e9+7;
const double eps = 1e-7;
const int maxn = 1e5+5;

ll fast_pow(ll a,ll b,ll mod){
    ll ans = 1;
    while(b){
        if(b&1)
            ans = ans*a%mod;
        a = a*a%mod;
        b>>=1;
    }
    return ans;
}

int main(){
    ll n, m;
    ll a, x=0, y=0;
    scanf("%lld%lld",&n,&m);

    for(int i=1;i<=n;i++){
        scanf("%lld",&a);
        x = (x+(2*i-n-1)*a)%mod;
    }
    for(int i=1;i<=m;i++){
        scanf("%lld",&a);
        y = (y+(2*i-m-1)*a)%mod;
    }
    printf("%lld\n",x*y%mod);
    return 0;
}