hdu 4609 3-idiots (快速傅里叶求卷积)

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define pi acos(-1.0)
int len1,len2;
long long a[500100],num[500100],sum[500100];
struct Complex{
    double r,i;
    Complex(){};
    Complex(double x,double y){
        r=x,i=y;
    }
    Complex operator +(const Complex &t)const{
       return Complex(r+t.r,i+t.i);
    }
    Complex operator -(const Complex &t)const{
       return Complex(r-t.r,i-t.i);
    }
    Complex operator *(const Complex &t)const{
       return Complex(r*t.r-i*t.i,r*t.i+i*t.r);
    }
}x1[500100];
void fft(Complex y[],int n,int rev)
{
    for(int i=1,j,k,t;i<n;i++){
       for(j=0,k=n>>1,t=i;k;k>>=1,t>>=1) j=j<<1|t&1;
       if(i<j) swap(y[i],y[j]);
    }
    for(int s=2,ds=1;s<=n;ds=s,s<<=1){
       Complex wn=Complex(cos(rev*2*pi/s),sin(rev*2*pi/s)),w=Complex(1,0),t;
       for(int k=0;k<ds;k++,w=w*wn){
          for(int i=k;i<n;i+=s){
              t=w*y[i+ds];
              y[i+ds]=y[i]-t;
              y[i]=y[i]+t;
          }
       }
    }
    if(rev==-1) for(int i=0;i<n;i++) y[i].r/=n;
}
long long temp,ans;
int main()
{
    int i,j,k,t,n,m;
    while(scanf("%d",&t)!=EOF)
    {
        while(t--)
        {
            memset(num,0,sizeof(num));
            memset(sum,0,sizeof(sum));
            ans=0;
            scanf("%d",&n);
            for(i=0;i<n;i++){
               scanf("%d",&a[i]);
               num[a[i]]++;
            }
            sort(a,a+n);
            len1=a[n-1];
            int len=1;
            while(len<=len1*2) len<<=1;
            for(i=0;i<=len1;i++) x1[i]=Complex(num[i],0);
            for(;i<len;i++) x1[i]=Complex(0,0); //扩展
            fft(x1,len,1); //DFT
            for(i=0;i<len;i++) x1[i]=x1[i]*x1[i]; //点乘
            fft(x1,len,-1); //IDFT
            for(i=0;i<len;i++)
                num[i]=(long long)(x1[i].r+0.5); //结果
            len=2*a[n-1];
            for(i=0;i<n;i++)
                num[a[i]*2]--;  //两根棒相同
            for(i=1;i<=len;i++)
                num[i]/=2;      //不同顺序只取一种
            for(i=1;i<=len;i++){
               sum[i]=sum[i-1]+num[i];  
            }
            for(i=0;i<n;i++) //枚举三边中最长边
            {
                ans+=sum[len]-sum[a[i]]; //两边之和大于第三边
                ans-=(long long)(n-i-1)*i; //一大一小
                ans-=(n-1);  //一边与枚举边相同
                ans-=(long long)(n-i-1)*(n-i-2)/2; //两大
            }
            temp=(long long)n*(n-1)*(n-2)/6; //一定要long long
            printf("%.7f\n",(double)ans/temp);
        }
    }
    return 0;
}