hdu 4609 3-idiots（FFT）

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题目链接：hdu 4609 3-idiots
代码

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn = 400040;typedef long long ll;/***************** FFT ************************/const double pi = acos(-1.0);struct Complex {    double r,i;    Complex(double r = 0,double i = 0): r(r), i(i) { }    Complex operator + (const Complex &u) { return Complex(r+u.r,i+u.i); }    Complex operator - (const Complex &u) { return Complex(r-u.r,i-u.i); }    Complex operator * (const Complex &u) { return Complex(r*u.r-i*u.i,r*u.i+i*u.r); }};Complex X[maxn];void change(Complex* y, int n) {    int k;    for(int i = 1, j = n/2; i < n-1; i++) {        if(i < j)swap(y[i],y[j]);        k = n/2;        while(j >= k) {            j -= k;            k /= 2;        }        if(j < k) j += k;    }}void dft(Complex* y, int n, int sign) {    change(y,n);    for (int h = 2; h <= n; h <<= 1) {        Complex wn(cos(-sign*2*pi/h),sin(-sign*2*pi/h));        for(int j = 0; j < n; 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 (sign == -1)        for(int i = 0;i < n;i++)            y[i].r /= n;}void fft (ll* cnt, int n) {    int m = 1;    while (m < 2 * n) m <<= 1;    for (int i = 0; i < n; i++)        X[i] = Complex(cnt[i], 0);    for (int i = n; i < m; i++)        X[i] = Complex(0, 0);    dft(X, m, 1);    for (int i = 0; i < m; i++)        X[i] = X[i] * X[i];    dft(X, m, -1);    for (int i = 0; i < m; i++)        cnt[i] = (ll)(X[i].r + 0.5);}/***********************************************/int N, M, A[maxn>>2];ll cnt[maxn], sum[maxn];void init () {    scanf("%d", &N);    memset(cnt, 0, sizeof(cnt));    for (int i = 0; i < N; i++) {        scanf("%d", &A[i]);        cnt[A[i]]++;    }    sort(A, A + N);}double solve () {    fft(cnt, A[N-1]+1);    int n = A[N-1] + 1;    for (int i = 0; i < N; i++)        cnt[A[i]+A[i]]--;    for (int i = 0; i < 2 * n; i++)        cnt[i] /= 2;    sum[0] = 0;    for (int i = 1; i < 2 * n; i++)        sum[i] = sum[i-1] + cnt[i];    double tot = 1.0 * N * (N-1) * (N-2) / 6, ret = 0;    for (int i = 0; i < N; i++)        ret += sum[A[i]] / tot;    return ret;}int main() {    int cas;    scanf("%d", &cas);    while(cas--) {        init();        printf("%.7lf\n", 1 - solve());    }    return 0;}
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