本文已参与「新人创作礼」活动，一起开启掘金创作之路。

 题目链接：

登录—专业IT笔试面试备考平台_牛客网

解题思路：

​编辑

​编辑

AC代码：

#include<bits/stdc++.h>
#define up(i, x, y) for(ll i = x; i <= y; i++)
#define down(i, x, y) for(ll i = x; i >= y; i--)
#define bug prllf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define IO ios::sync_with_stdio(false),cin.tie(0)
typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-8;
const ll mod = 1e9 + 7;
const ll maxn = 1e5 + 7;
const double pi = acos(-1);
const ll inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fLL;
using namespace std;

ll n, a[maxn];

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

int main()
{
    while(~scanf("%lld", &n))
    {
        ll ans = 0;
        for(ll i = 1; i <= n; i++)
        {
            scanf("%lld", &a[i]);
        }
        for(ll i = 1; i <= n; i++)
        {
            ll t = 2 * a[i] % mod;
            for(ll j = 1; j <= n; j++)
            {
                if(i == j) continue;
                t = (t * ((a[j] * a[j] % mod) - (a[i] * a[i] % mod) + mod) % mod) % mod;
            }
            ans = (ans + 1LL * power(t, mod - 2) ) % mod;
        }
        cout<<ans<<'\n';
    }
}

​