3-idiots

个人理解

第一道\(FFT\)练习题，尽管开始会用模板，但还是心存疑惑

• \(IDFT\)的过程还是不会证
• 自底向上的迭代，取值过程依然模糊
• 线性卷积和循环卷积为什么在\(L > (n + m - 1)\)时相等
• ········

当你学会了\(FFT\)，那么这道题剩下的就是一丢丢容斥（枚举不合法的情况）kuangbin聚聚的博客讲的很清楚，就是自己写的时候要仔细！

long long cnt = 0;
for (int i = 0; i < n; ++i) {
    cnt += (sum[len - 1] - sum[a[i]]);  // 0, 1, ···,len - 1
    cnt -= (long long)i * (n - i - 1); // 一个取大一个取小
    cnt -= (n - 1); // 一个取本身，一个取其它数
    cnt -= (long long)(n - i - 1) * (n - i - 2) / 2; 
}

代码

struct Complex {
    double real, image;
    Complex() {}
    Complex(double real, double image) : real(real), image(image) {}

    Complex operator + (const Complex &a) const {
        return Complex(real + a.real, image + a.image);
    }
    Complex operator - (const Complex &a) const {
        return Complex(real - a.real, image - a.image);
    }
    Complex operator * (const Complex &a) const {
        return Complex(real * a.real - image * a.image, image * a.real + real * a.image);
    }
};

int rev(int id, int len) {
    int pos = 0;
    for (int i = 0; (1 << i) < len; ++i) {
        pos <<= 1;
        if (id & (1 << i)) pos |= 1;
    }
    return pos;
}

Complex A[500005];

void FFT(Complex *a, int len, int DFT) {
    rep(i, 0, len) A[rev(i, len)] = a[i];
    for (int s = 1; (1 << s) <= len; ++s) {
        int m = (1 << s);
        Complex wm = Complex(cos(DFT * 2 * PI / m), sin(DFT * 2 * PI / m));
        for (int i = 0; i < len; i += m) {
            Complex w = Complex(1, 0);
            for (int j = 0; j < (m >> 1); ++j) {
                Complex t = A[i + j];
                Complex u = w * A[i + j + (m >> 1)];
                A[i + j] = t + u;
                A[i + j + (m >> 1)] = t - u;
                w = w * wm;
            }
        }
    }
    if (DFT == -1) rep(i, 0, len) A[i].real /= len, A[i].image /= len;
    rep(i, 0, len) a[i] = A[i];
}

const int N = 500005;

int n;
int cnt[N], b[N];
Complex a[N];

LL sum[N], num[N];

int main()
{
    BEGIN() {
        mem(cnt, 0);
        mem(num, 0);

        sc(n);
        int ma = 0;

        rep(i, 0, n) {
            sc(b[i]);
            int x = b[i];
            cnt[x]++;
            num[x + x]--;
            ma = max(ma, x);
        }

        int sa = 0;
        while((1 << sa) < (ma + 1)) sa++;

        int m = 1 << (sa + 1);
        rep(i, 0, m) a[i] = Complex(cnt[i], 0);

        FFT(a, m, 1);
        rep(i, 0, m) a[i] = a[i] * a[i];
        FFT(a, m, -1);

        rep(i, 0, m) num[i] += (LL)(a[i].real + 0.5), num[i] /= 2;
        rep(i, 1, m) sum[i] = sum[i - 1] + num[i];

        sort(b, b + n);

        LL cnt = 0;
        rep(i, 0, n) {
            cnt += (sum[m - 1] - sum[b[i]]);
            cnt -= (LL)(n - 1 - i) * i;
            cnt -= (n - 1);
            cnt -= (LL)(n - 1 - i) * (n - i - 2) / 2;
        }

        LL tot = (LL)n * (n - 1) * (n - 2) / 6;
        printf("%.7f\n", (double)cnt / tot);
    }
    return 0;
}