算法

(前缀和) $O(n)$

注意到$$\sum {{a_i}{a_j}} = {a_1}\left( {\sum\limits_{i = 2}^n {{a_i}} } \right) + {a_2}\left( {\sum\limits_{i = 3}^n {{a_i}} } \right) + \cdots + {a_{n - 1}}{a_n}$$

然后用前缀和处理一下即可。

C++ 代码

#include <iostream>
#define int long long

using namespace std;

const int N = 2e5 + 10;
const int MOD = 1e9 + 7;

int a[N];

signed main() {
    int n;
    cin >> n;

    int sum = 0;
    for (int i = 1; i <= n; ++i) cin >> a[i], sum += a[i];

    int ans = 0;
    for (int i = 1; i <= n; ++i) {
        sum -= a[i];
        ans += a[i] * (sum % MOD);
        ans %= MOD;
    }

    cout << ans << '\n';

    return 0;
}