[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 = , j = len/;i < len-; 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/;
         while( j >= k) {
             j -= k;
             k /= ;
         }
         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 = ; h <= len; h <<= ) {
         Complex wn(cos(-on**PI/h),sin(-on**PI/h));
         for(int j = ;j < len;j+=h) {
             Complex w(,);
             for(int k = j;k < j+h/;k++) {
                 Complex u = y[k];
                 Complex t = w*y[k+h/];
                 y[k] = u+t;
                 y[k+h/] = u-t;
                 w = w*wn;
             }
         }
     }
     if(on == -) {
         for(int i = ;i < len;i++) {
             y[i].r /= len;
         }
     }
 }

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