FFT

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn=220005;
const double eps(1e-8);
typedef long long LL;
const double PI = acos(-1.0);

struct Complex
{
double real, image;
Complex(double _real, double _image)
{
real = _real;
image = _image;
}
Complex(){}
};

Complex operator + (const Complex &c1, const Complex &c2)
{
return Complex(c1.real + c2.real, c1.image + c2.image);
}

Complex operator - (const Complex &c1, const Complex &c2)
{
return Complex(c1.real - c2.real, c1.image - c2.image);
}

Complex operator * (const Complex &c1, const Complex &c2)
{
return Complex(c1.real*c2.real - c1.image*c2.image, c1.real*c2.image + c1.image*c2.real);
}

int rev(int id, int len)
{
int ret = 0;
for(int i = 0; (1 << i) < len; i++)
{
ret <<= 1;
if(id & (1 << i)) ret |= 1;
}
return ret;
}

Complex A[maxn<<1];
void FFT(Complex* a, int len, int DFT)//对a进行DFT或者逆DFT, 结果存在a当中,len必须是2的幂而且len大于多项式最高次数
{
for(int i = 0; i < len; i++)
A[rev(i, len)] = a[i];
for(int s = 1; (1 << s) <= len; s++)
{
int m = (1 << s);
Complex wm = Complex(cos(DFT*2*PI/m), sin(DFT*2*PI/m));
for(int k = 0; k < len; k += m)
{
Complex w = Complex(1, 0);
for(int j = 0; j < (m >> 1); j++)
{
Complex t = w*A[k + j + (m >> 1)];
Complex u = A[k + j];
A[k + j] = u + t;
A[k + j + (m >> 1)] = u - t;
w = w*wm;
}
}
}
if(DFT == -1) for(int i = 0; i < len; i++) A[i].real /= len, A[i].image /= len;
for(int i = 0; i < len; i++) a[i] = A[i];
return;
}
Complex a[maxn<<1];
int san[maxn];
int num[maxn<<1];
LL xishu[maxn<<1];
int main()
{
int t,n;
cin>>t;
while(t--){
scanf("%d",&n);
int len=1;
int s=0;
for(int i=1;i<=n;i++){
scanf("%d",&san[i]);
s=max(s,san[i]);
}
while(len<=s){
len<<=1;
}
len<<=1;
sort(san+1,san+1+n);
for(int i=0;i<len;i++){
a[i]=Complex(0,0);
num[i]=0;
}
for(int i=1;i<=n;i++){
a[san[i]].real+=1;
num[san[i]]++;
}
FFT(a,len,1);
for(int i=0;i<len;i++){
a[i]=a[i]*a[i];
}
FFT(a,len,-1);
}
}