链接：https://ac.nowcoder.com/acm/contest/881/B
来源：牛客网

Integration
时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 524288K，其他语言1048576K
64bit IO Format: %lld
题目描述
Bobo knows that
∫
∞
0
1
1

  • x
    2

d
x
=
π
2
.
∫0∞11+x2 dx=π2.

Given n distinct positive integers
a
1
,
a
2
,
…
,
a
n
a1,a2,…,an, find the value of
1
π
∫
∞
0
1
∏
n
i
=
1
(
a
2
i

  • x
    2
    )

d
x
.
1π∫0∞1∏i=1n(ai2+x2) dx.

It can be proved that the value is a rational number
p
q
pq.
Print the result as
(
p
⋅
q
−
1
)
mod
(
10
9

  • 7
    )
    (p⋅q−1)mod(109+7).
    输入描述:
    The input consists of several test cases and is terminated by end-of-file.

The first line of each test case contains an integer n.
The second line contains n integers
a
1
,
a
2
,
…
,
a
n
a1,a2,…,an.

  • 1
    ≤
    n
    ≤
    10
    3
    1≤n≤103
  • 1
    ≤
    a
    i
    ≤
    10
    9
    1≤ai≤109
  • {
    a
    1
    ,
    a
    2
    ,
    …
    ,
    a
    n
    }
    {a1,a2,…,an} are distinct.
  • The sum of
    n
    2
    n2 does not exceed
    10
    7
  1. 输出描述:
    For each test case, print an integer which denotes the result.
    示例1
    输入
    复制
    1
    1
    1
    2
    2
    1 2
    输出
    复制
    500000004
    250000002
    83333334

题意：

思路：

事实上我并不会上面的处理，真正积分的话要用裂项相消来出来。

但是有强大的自动积分软件啊: https://www.wolframalpha.com/

输入一个n=5的 情况，就可以看出规律。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll a[maxn];
int n;
const ll mod = 1e9 + 7;
int main() {
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    gbtb;
    while (cin >> n) {
        repd(i, 1, n) {
            cin >> a[i];
        }
        ll ans = 0ll;
        repd(i, 1, n) {
            ll sum = 1ll;
            repd(j, 1, n) {
                if (i != j)
                    sum = sum * ((a[j] * a[j] - a[i] * a[i]) % mod) % mod;
            }
            // sum = (sum + mod) % mod;
            sum = (sum * a[i]) % mod;
            sum = (sum * 2ll) % mod;
            sum = powmod(sum, mod - 2ll, mod);
            ans = (ans + sum) % mod;

        }
        ans = (ans + mod) % mod;
        cout << ans << endl;
    }

    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    } else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}