题目链接

题意：



题解：














AC代码

/*
    Author : zzugzx
    Lang : C++
    Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1e9 + 7;
const int MOD = 998244353;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 4e3 + 10;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};

ll a[maxn], w[maxn], s[maxn];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        s[i] = (s[i - 1] + a[i]) % mod;
    }
    ll sum = 0, ans = 0;
    for (int i = 1, w; i <= n; i++) {
        cin >> w;
        sum = (sum + s[n - i + 1] - s[i - 1] + mod) % mod;
        ans = (ans + w * sum) % mod;
    }
    cout << ans;
    return 0;
}