HDU-4609 3-idiots FFT

HDU-4609 3-idiots FFT

题意

给定\(n\) 根木棍，每根木棍有一长度\(a[i]\) ,问任意选出三根木棍，可以组成三角形的概率

\[T \leq 100,3\leq n \leq 10^5 ,1\leq a_i \leq 10^5 \]

分析

此题属于\(FFT\)入门模板题

这题刚开始看有点像是生成函数（好像就是

如果暴力枚举两根可能组成的长度，显然\(n^2\) 的复杂度不够

我们可以把问题抽象成两个多项式相乘，最后乘出来的指数或者说第几项就可以表示出能够组成的长度，系数就表示个数。

事实上就是在求卷积，也就是\(FFT\) 的工作。

注意实现上要删掉两个相同的组成，以及对剩下的除以2

特别注意\(FFT\) 传入的\(len\) 必须是\(2^k\)

代码

struct Complex {
    double x, y;
    Complex(double _x = 0.0, double _y = 0.0) {
        x = _x;
        y = _y;
    }
    Complex operator - (const Complex& b) const {
        return Complex(x - b.x, y - b.y);
    }
    Complex operator + (const Complex& b) const {
        return Complex(x + b.x, y + b.y);
    }
    Complex operator * (const Complex& b) const {
        return Complex(x * b.x - b.y * y, x * b.y + y * b.x);
    }
};

void change(Complex y[], int len) {
    int k;
    for (int i = 1, j = len / 2; i < len - 1; i++) {
        if (i < j) swap(y[i], y[j]);
        k = len / 2;
        while (j >= k) {
            j -= k;
            k /= 2;
        }
        if (j < k) j += k;
    }
}

void fft(Complex y[], int len, int on) {
    change(y, len);
    for (int h = 2; h <= len; h <<= 1) {
        Complex wn(cos(-on * 2 * PI / h), sin(-on * 2 * PI / h));
        for (int j = 0; j < len; j += h) {
            Complex w(1, 0);
            for (int k = j; k < j + h / 2; k++) {
                Complex u = y[k];
                Complex t = w * y[k + h / 2];
                y[k] = u + t;
                y[k + h / 2] = u - t;
                w = w * wn;
            }
        }
    }
    if (on == -1) {
        for (int i = 0; i < len; i++)
            y[i].x /= len;
    }
}

Complex x[maxn << 2];
ll num[maxn << 2];
ll sum[maxn << 2];
int a[maxn];

int main() {
    int T = readint();
    int n;
    while (T--) {
        n = readint();
        memset(num, 0, sizeof num);
        for (int i = 0; i < n; i++) a[i] = readint(), num[a[i]]++;
        sort(a, a + n);
        int len1 = a[n - 1] + 1;
        int len = 1;
        while (len < 2 * len1) len <<= 1;
        for (int i = 0; i < len1; i++) x[i] = Complex(num[i], 0);
        for (int i = len1; i < len; i++) x[i] = Complex(0, 0);
        fft(x, len, 1);
        for (int i = 0; i < len; i++) x[i] = x[i] * x[i];
        fft(x, len, -1);
        for (int i = 0; i < len; i++) num[i] = (ll)(x[i].x + 0.5);
        len = 2 * a[n - 1];
        for (int i = 0; i < n; i++) num[a[i] + a[i]]--;
        for (int i = 1; i <= len; i++) num[i] /= 2;
        sum[0] = 0;
        for (int i = 1; i <= len; i++) sum[i] = sum[i - 1] + num[i];
        ll cnt = 0;
        for (int i = 0; i < n; i++) cnt += sum[a[i]];
        ll tot = (ll)n * (n - 1) * (n - 2) / 6;
        printf("%.7f\n", 1.0 - 1.0 * cnt / tot);
    }
}