[HDOJ4609]3-idiots（FFT，计数）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4609

题意：n个数，问取三个数可以构成三角形的组合数。

FFT预处理出两个数的组合情况，然后枚举第三个数，计数去重。

#include <bits/stdc++.h>
using namespace std;

const double PI = acos(-1.0);
//复数结构体
typedef struct Complex {
    double r,i;
    Complex(double _r = 0.0,double _i = 0.0) {
        r = _r; i = _i;
    }
    Complex operator +(const Complex &b) {
        return Complex(r+b.r,i+b.i);
    }
    Complex operator -(const Complex &b) {
        return Complex(r-b.r,i-b.i);
    }
    Complex operator *(const Complex &b) {
        return Complex(r*b.r-i*b.i,r*b.i+i*b.r);
    }
}Complex;
/*
 * 进行FFT和IFFT前的反转变换。
 * 位置i和 （i二进制反转后位置）互换
 * len必须是2的幂
 */
void change(Complex y[],int len) {
    int i,j,k;
    for(i = 1, j = len/2;i < len-1; i++) {
        if(i < j)swap(y[i],y[j]);
        //交换互为小标反转的元素，i<j保证交换一次
        //i做正常的+1，j左反转类型的+1,始终保持i和j是反转的
        k = len/2;
        while( j >= k) {
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}
/*
 * 做FFT
 * len必须为2^k形式，
 * on==1时是DFT，on==-1时是IDFT
 */
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].r /= len;
        }
    }
}

typedef long long LL;
const int maxn = 400300;
Complex x1[maxn];
int a[maxn/4];
LL num[maxn], s[maxn];
int n, len, q;

int main() {
    // freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d",&n);
        memset(a, 0, sizeof(a));
        memset(s, 0, sizeof(s));
        memset(num, 0, sizeof(num));
        int maxx = 0;
        for(int i = 0; i < n; i++) {
            scanf("%I64d", &a[i]);
            num[a[i]]++;
            maxx = max(maxx, a[i]);
        }
        int len1 = maxx + 1;
        len = 1;
        while(len < len1 * 2) len <<= 1;
        for(int i = 0; i < len; i++) x1[i] = Complex(0, 0);
        for(int i = 0; i < len1; i++) x1[i] = Complex(num[i], 0);
        fft(x1, len, 1);
        for(int i = 0; i < len; i++) x1[i] = x1[i] * x1[i];
        fft(x1, len, -1);
        for(int i = 0; i < len; i++) num[i] = (LL)(x1[i].r + 0.5);
        len = 2 * maxx;
        for(int i = 0; i < n; i++) num[a[i]*2]--; 
        for(int i = 1; i <= len; i++) num[i] /= 2;
        for(int i = 1; i <= len; i++) s[i] = s[i-1] + num[i];
        LL ret = 0;
        for(int i = 0; i < n; i++) {
            ret += s[len] - s[a[i]];
            ret -= (LL)(n - i - 1) * i;
            ret -= (n - 1);
            ret -= (LL)(n - i - 1) * (n - i - 2) / 2;
        }
        LL sum = (LL)n * (n - 1) * (n - 2) / 6;
        // cout << (double)ret/(double)((n*(n-1)*(n-2))/6) << endl;
        printf("%.7lf
", (double)ret/sum);
    }
    return 0;
}