Integration

题目链接

题意

已知

$$\int_{0}^{ \infty} \frac{1}{1+{x}^{2}} = \frac{ \pi}{2}.$$

计算

$$\frac{1}{ \pi} \int_{0}^{ \infty} \frac{1}{ \prod_{i} ({a_{i} }^{2} + {x}^{2} )}.$$

思路

这我哪想得出来。。。

$$\frac{1}{ \prod_{i} ({a_{i} }^{2} + {x}^{2})} = \sum_{i} \frac{c_{i} }{({a_{i} }^{2}+{x}^{2} )}.$$ $$1 = \sum_{i} \frac{c_{i} \prod_{j} ({a_{j} }^{2} + {x}^{2} )}{({a_{i} }^{2} + {x}^{2})}.$$ $$1 = c_{i} \prod_{i \neq j} ({a_{j} }^{2} + {x}^{2}) + \sum_{j \neq i} \frac{c_{j} \prod_{k \neq j} ({a_{k} }^{2} + {x}^{2})}{ {a_{j}}^{2} + {x}^{2} }.$$ $$x = a_{i} i.$$

$i$为单位虚根.

$$1 = c_{i} \prod_{j \neq i} ({a_{j} }^{2} - {a_{i} }^{2}) + 0.$$ $$c_{i} = \frac{1}{ \prod_{j \neq i} ({a_{j} }^{2} - {a_{i} }^{2})}.$$

Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#ifndef ONLINE_JUDGE
#define dbg(x...) do{cout << "\033[33;1m" << #x << "->" ; err(x);} while (0)
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
#else
#define dbg(...)
#endif
#define inf 1ll << 50

const ll mod = 1e9 + 7;
ll Pow(ll a, ll b)
{
	ll ans = 1;
	while (b)
	{
		if (b & 1)
			ans = ans * a % mod;
		a = a * a % mod;
		b >>= 1;
	}
	return ans;
}
const int N = 1e3 + 5;
ll a[N], c[N];
template<class T>
void read(T& ret)
{
	ret = 0;
	char c;
	while ((c = getchar()) > '9' || c < '0');
	while (c >= '0' && c <= '9')
	{
		ret = ret * 10 + c - '0';
		c = getchar();
	}
}
int main()
{
	int n;
	ll inv2 = (mod + 1) / 2;
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; i++)
			read(a[i]);
		ll ans = 0;
		for (int i = 1; i <= n; i++)
		{
			ll t = 1;
			for (int j = 1; j <= n; j++)
			{
				if (i == j)
					continue;
				ll x = a[j] * a[j] % mod - a[i] * a[i] % mod;
				x = (x + 2 * mod) % mod;
				t = t * Pow(x, mod - 2) % mod;
			}
			ans = (ans + t * Pow(a[i], mod - 2) % mod) % mod;
		}
		printf("%lld\n", ans * inv2 % mod);
	}
	return 0;
}