方差

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题目大意：对于一个大小为 𝑛 的数集 {𝑎𝑖} ，设其平均值为 𝑥 =∑ 𝑎𝑖/𝑛 ，则它的方差为\(∑(𝑎𝑖−𝑥)^2/𝑛\) 。现在有一个数集，你需要求出这个数集的每个非空子集的方差之和对1e9 + 7 取模的结果

推式子可以得到：

\[\dfrac { \sum{a_i}^2} {n} - \dfrac {(\sum a_i)^2}{n^2}\]

代码如下：

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#include <bits/stdc++.h>
#define int long long
using namespace std;

const int N = 1e6 + 666, mod = 1e9 + 7;

int n, sum1, sum2, res;
int a[N], sum[N], fac[N];

inline int ksm(int a, int b)
{
    int ret(1);
    for (; b; b >>= 1)
    {
        if (b & 1) ret = ret * a % mod;
        a = a * a % mod;
    }
    return ret;
}

inline void pres_dou()
{
    fac[0] = 1;
    for (int i = 1; i <= n; ++ i) fac[i] = fac[i - 1] * i % mod;
    return;
}

inline int C(int x, int y)
{
    if (x < y) return 0;
    if (x == y || y == 0) return 1;
    return fac[x] * ksm(fac[y] * fac[x - y] % mod, mod - 2) % mod;
}

inline int ifac(int x)
{
    if (x == 1) return 1;
    return ksm(x, mod - 2) % mod;
}

signed main()
{
    int i, x, y;
    n = read();
    for (i = 1; i <= n; ++ i)
    {
        a[i] = read();
        sum[i] = ( sum[i - 1] + a[i]) % mod;
    }

    for (i = 1; i <= n; ++ i)
    {
	(sum1 += a[i] * a[i] % mod) %= mod, (sum2 += 2 * a[i] * ((sum[n] - sum[i]) % mod + mod) % mod) %= mod;
    }

    pres_dou();

    for (i = 2; i <= n; ++ i)
    {

	x = ifac(i), y = ifac(i * i % mod);

	res += ((sum1 * C(n - 1, i - 1) % mod * x
        - (sum1 * C(n - 1, i - 1) % mod + sum2 * C(n - 2, i - 2) % mod) * y + mod) % mod + mod) % mod;

	res = res % mod;
    }
    put(res % mod);
    return 0;
}