HDU 4609

实质上是两个多项式相乘注意去重

#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
const double pi = acos(-1.0);
const int maxn = 271111;
const double eps = 1e-6;
struct Complex {
    double a, b;
    Complex() {
    }
    Complex(double a, double b) :
            a(a), b(b) {
    }
    Complex operator +(const Complex& t) const {
        return Complex(a + t.a, b + t.b);
    }
    Complex operator -(const Complex& t) const {
        return Complex(a - t.a, b - t.b);
    }
    Complex operator *(const Complex& t) const {
        return Complex(a * t.a - b * t.b, a * t.b + b * t.a);
    }
    Complex operator =(const double &n) const {
        return Complex(1.0, 0);
    }
} a[maxn], b[maxn];
char s1[maxn], s2[maxn];
void brc(Complex *y, int l) {
    int i, j, k;
    for (i = 1, j = l >> 1; i < l - 1; i++) {
        if (i < j)
            swap(y[i], y[j]);

        k = l >> 1;
        while (j >= k) {
            j -= k;
            k >>= 1;
        }
        if (j < k)
            j += k;
    }
}
int num[maxn];
int cnt[maxn];
LL res[maxn];
LL sum[maxn];
void FFT(Complex *y, int l, int on)

{
    int h, i, j, k, p;
    double r;
    Complex u, t;
    brc(y, l);
    for (h = 2; h <= l; h <<= 1) {
        r = on * 2.0 * pi / h;
        Complex wn(cos(r), sin(r));
        p = h >> 1;
        for (j = 0; j < l; j += h) {
            Complex w(1, 0);
            for (k = j; k < j + p; k++) {
                u = y[k];
                t = w * y[k + p];
                y[k] = u + t;
                y[k + p] = u - t;
                w = w * wn;
            }
        }
    }
    if (on == -1)
        for (i = 0; i < l; i++)
            y[i].a = y[i].a / l;
}
int main() {

    ios::sync_with_stdio(false);
    int n, i, L = 1, ma, T;
    scanf("%d", &T);
    while (T--) {
        ma = 0;
        memset(cnt, 0, sizeof(cnt));
        scanf("%d", &n);
        for (i = 0; i < n; ++i) {
            scanf("%d", num + i);
            cnt[num[i]]++;
            ma = max(ma, num[i]);
        }
        if (n < 3) {
            printf("%.7f\n", 0.0);
            continue;
        }
        ++ma;
        L = 1;
        while (L < ma * 2) 
            L <<= 1;
        
        for (i = 0; i < ma; ++i) 
            a[i] = Complex(cnt[i], 0);
        
        for (i = ma; i < L; ++i) 
            a[i] = Complex(0, 0);
        
        FFT(a, L, 1);
        for (i = 0; i < L; ++i)
            a[i] = a[i] * a[i];
        FFT(a, L, -1);

        for (i = 0; i < L; ++i) 
            res[i] = (a[i].a + 0.5);
        

        for (i = 0; i <= ma; ++i) 
            res[i << 1] -= cnt[i];
        
        for (i = 0; i < L; ++i)
            res[i] >>= 1;
        
        for (i = 1; i < L; ++i) 
            sum[i] = sum[i - 1] + res[i];
            
        
        double tot = 0, den = 1.0 * n * (n - 1) * (n - 2);
        
        for (i = 0; i < n; ++i) 
            tot += sum[num[i]] / den;
        
    
        double ans = 1 - tot * 6.0;
        printf("%.7f\n", ans);
    }

    return 0;
}