BZOJ 3513: [MUTC2013]idiots

忽略不能取同一根的要求。

构造多项式 \(A(x) = \sum x_{a_i}\)

那么 \(A^2(x)\) 就是取两个木棍，组成长度为 \(s\) 的方案。

容斥后得到

\[(\sum x)^2 = \sum x^2 + 2\sum xy \]

\[\sum xy = \dfrac{(\sum x)^2 - \sum x^2}{2} \]

这里的 \(x\) 和 \(y\) 就可以看成有序对了。

设多项式 \(B(s)\) 为上述多项式 。

枚举最长边，假设最长边为 \(s\)，在其前面的木棍（也就是长度小于它的木棍和长度等于它但不包括它的木棍的个数）有 \(i - 1\) 根，在其后面的木棍有 \(n-i\) 根。

那么符合情况的有序对对数为 \(\sum_{j > s} B(j) - ((i - 1)(n - i + 1) + n - i + C_{n - i}^{2})\)。

然后除一下 \(C_n ^3\) 就可以了。

#include <bits/stdc++.h>

struct Complex {
  double r, i;
  void clear() { r = i = 0.0; }
  Complex(double r = 0, double i = 0): r(r), i(i) {}
  Complex operator + (const Complex &p) const { return Complex(r + p.r, i + p.i); }
  Complex operator - (const Complex &p) const { return Complex(r - p.r, i - p.i); }
  Complex operator * (const Complex &p) const { return Complex(r * p.r - i * p.i, r * p.i + i * p.r); }
};

const double pi = acos(-1.0);
const int N = 5e5 + 7;
int r[N];

void FFT(Complex *a, int n, int pd) {
  for (int i = 0; i < n; i++)
    if (i < r[i])
      std::swap(a[i], a[r[i]]);
  for (int mid = 1; mid < n; mid <<= 1) {
    Complex wn(cos(pi / mid), pd * sin(pi / mid));
    for (int l = mid << 1, j = 0; j < n; j += l) {
      Complex w(1.0, 0.0);
      for (int k = 0; k < mid; k++, w = w * wn) {
        Complex u = a[k + j], v = w * a[k + j + mid];
        a[k + j] = u + v;
        a[k + j + mid] = u - v;
      }
    }
  }
  if (pd < 0)
    for (int i = 0; i < n; i++)
      a[i] = Complex(a[i].r / n, a[i].i / n);
}

#define ll long long
int len, n, l;
Complex A[N];
int a[N], cntA[N], cntB[N];
ll sum[N];

ll C(int n) {
  return 1LL * n * (n - 1) / 2;
}

ll C3(int n) {
  return 1LL * n * (n - 1) * (n - 2) / 6;
}

int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    int m = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
      m = std::max(m, a[i]);
      cntA[a[i]]++;
      cntB[a[i] * 2]++;
    }
    len = 1;
    l = 0;
    while (len <= 2 * m) len <<= 1, l++;
    for (int i = 0; i < len; i++)
      r[i] = r[i >> 1] >> 1 | ((i & 1) << (l - 1));
    for (int i = 0; i < len; i++)
      A[i] = Complex((double)cntA[i], 0.0);
    FFT(A, len, 1);
    for (int i = 0; i < len; i++)
      A[i] = A[i] * A[i];
    FFT(A, len, -1);
    for (int i = 0; i < len; i++)
      sum[i] = (ll)((A[i].r - cntB[i]) / 2.0 + 0.5);
    for (int i = 1; i < len; i++)
      sum[i] += sum[i - 1];
    std::sort(a + 1, a + 1 + n);
    ll ans = 0;
    for (int i = 1; i <= n; i++) {
      ans += sum[len - 1] - sum[a[i]];
      ll temp = 1LL * (i - 1) * (n - i + 1) + n - i + C(n - i);
      ans -= temp;
    }
    ll all = C3(n);
    double res = (double)(ans * 1.0 / all);
    printf("%.7f\n", res);
    for (int i = 0; i < len; i++) {
      A[i].clear();
      cntA[i] = cntB[i] = 0;
      sum[i] = 0;
    }
  }
}