2019牛客暑期多校训练营（第一场）B.integration

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

题目描述

Given n distinct positive integers [a1, a2, a3 … an], find the value of
$${\int }_0^{\infty }\frac{1}{{\prod }_{i=1}^n\left(a_i^2+x^2\right)}dx$$
It can be proved that the value is a rational number $p/q$
Print the result as $p\cdot q^{-1}\mathrm{m}\mathrm{o}\mathrm{d}(1e9+7)$

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 [a1 to an]

输出描述:

For each test case, print an integer which denotes the result.

InputSample

1
1
1
2
2
1 2

OutputSample

500000004
250000002
83333334

Solution：

We know that:
$${\int }_0^{\infty }\frac{1}{{\prod }_{i=1}^n\left(a_i^2+x^2\right)}dx=\sum _{i=1}^n{\frac{\arctan \left(\frac{x}{a_i}\right)}{a_i{\prod }_{forallj!=i}\left({a_j}^2-a_i^2\right)}|}_0^{\infty }$$

So we have:
$$\frac{1}{\pi }{\int }_0^{\infty }\frac{1}{{\prod }_{i=1}^n\left(a_i^2+x^2\right)}dx=\frac{1}{\pi }\sum _{i=1}^n\frac{\frac{\pi }{2}}{a_i{\prod }_{forallj!=i}\left({a_j}^2-a_i^2\right)}-0=\sum _{i=1}^n\frac{1}{2a_i{\prod }_{forallj!=i}\left({a_j}^2-a_i^2\right)}$$

AC Code:

/*
 * Copyright (c) 2019 Ng Kimbing, HNU, All rights reserved. May not be used, modified, or copied without permission.
 * @Author: Ng Kimbing, HNU.
 * @LastModified:2019-07-18 T 16:09:56.104 +08:00
 */

package NowCoder;

import static ACMProblems.ACMIO.*;

public class Int {
    private static final int MOD = 1000000007;

    private static long powMod(long a, long n, long mod) {
        long ret = 1;
        long temp = a % mod;
        while (n != 0) {
            if ((n & 1) != 0)
                ret = ret * temp % mod;
            temp = temp * temp % mod;
            n >>= 1;
        }
        return ret;
    }

    private static long dist(long a, long b) {
        return (a * a % MOD - b * b % MOD + MOD) % MOD;
    }

    public static void main(String[] args) {
        setStream(System.in);
        int[] a = new int[100005];
        try {
            while (true) {
                int n = nextInt();
                for (int i = 0; i < n; ++i)
                    a[i] = nextInt();
                long ans = 0;
                long foo;
                for (int i = 0; i < n; ++i) {
                    foo = a[i];
                    for (int j = 0; j < n; ++j) {
                        if (i != j)
                            foo = (foo * dist(a[j], a[i])) % MOD;
                    }
                    ans = (ans + powMod(foo, MOD - 2, MOD)) % MOD;
                }
                out.println((ans * powMod(2, MOD - 2, MOD)) % MOD);
            }
        } catch (Exception ignored) {
        }
        out.flush();
    }
}