题意：求一个序列中，有多少三元组（其中元素不重复）在任意的排列下能构成等差数列。
分析：等差数列：(A_j-A_i=A_k-A_j)，即(2A_j=A_i+A_k),枚举(A_i+A_j)的所有情况对应的个数，再扫一遍求解。
先统计出每个数对应的出现次数，FFT计算出和的组合情况。但是要减去(A_i+A_i)得到的结果以及(A_i+A_j)以及(A_j+A_i)重复的计算。
现在对于数(A_j)，假设(cnt=2*A_j)的系数，当然cnt中要减去(A_j)本身和一个值与(A_j)相等的数组合而成的情况。枚举完这个数以后，把这个数从序列中抹去，因为这个数对结果做出的贡献已经计算，之后的统计中该数以及该数对结果的贡献不能重复计算。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 10;
const double PI = acos(-1.0);
struct Complex{
double x, y;
inline Complex operator+(const Complex b) const {
return (Complex){x +b.x,y + b.y};
}
inline Complex operator-(const Complex b) const {
return (Complex){x -b.x,y - b.y};
}
inline Complex operator*(const Complex b) const {
return (Complex){x *b.x -y * b.y,x * b.y + y * b.x};
}
} va[MAXN * 2 + MAXN / 2], vb[MAXN * 2 + MAXN / 2];
int lenth = 1, rev[MAXN * 2 + MAXN / 2];
int N, M; // f 和 g 的数量
//f g和 的系数
// 卷积结果
// 大数乘积
int f[MAXN],g[MAXN];
vector<LL> conv;
vector<LL> multi;
void debug(){for(int i=0;i<conv.size();++i) cout<<conv[i]<<" ";cout<<endl;}
//f g
void init()
{
int tim = 0;
lenth = 1;
conv.clear(), multi.clear();
memset(va, 0, sizeof va);
memset(vb, 0, sizeof vb);
while (lenth <= N + M - 2)
lenth <<= 1, tim++;
for (int i = 0; i < lenth; i++)
rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (tim - 1));
}
void FFT(Complex *A, const int fla)
{
for (int i = 0; i < lenth; i++){
if (i < rev[i]){
swap(A[i], A[rev[i]]);
}
}
for (int i = 1; i < lenth; i <<= 1){
const Complex w = (Complex){cos(PI / i), fla * sin(PI / i)};
for (int j = 0; j < lenth; j += (i << 1)){
Complex K = (Complex){1, 0};
for (int k = 0; k < i; k++, K = K * w){
const Complex x = A[j + k], y = K * A[j + k + i];
A[j + k] = x + y;
A[j + k + i] = x - y;
}
}
}
}
void getConv(){ //求多项式
init();
for (int i = 0; i < N; i++)
va[i].x = f[i];
for (int i = 0; i < M; i++)
vb[i].x = g[i];
FFT(va, 1), FFT(vb, 1);
for (int i = 0; i < lenth; i++)
va[i] = va[i] * vb[i];
FFT(va, -1);
for (int i = 0; i <= N + M - 2; i++)
conv.push_back((LL)(va[i].x / lenth + 0.5));
}
void getMulti() //求A*B
{
getConv();
multi = conv;
reverse(multi.begin(), multi.end());
multi.push_back(0);
int sz = multi.size();
for (int i = 0; i < sz - 1; i++){
multi[i + 1] += multi[i] / 10;
multi[i] %= 10;
}
while (!multi.back() && multi.size() > 1)
multi.pop_back();
reverse(multi.begin(), multi.end());
}
int a[MAXN];
int cnt[MAXN];
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
int T; scanf("%d",&T);
while(T--){
int n; scanf("%d",&n);
int mx = -1;
memset(cnt,0,sizeof(cnt));
for(int i =1;i<=n;++i){
scanf("%d",&a[i]);
mx = max(mx,a[i]);
cnt[a[i]]++;
}
N = M = mx+1;
for(int i=0;i<N;++i){
f[i] = g[i] = cnt[i];
}
getConv();
int sz = conv.size();
for(int i=1;i<=n;++i){
conv[a[i]*2]--;
}
for(int i=0;i<sz;++i){
conv[i]>>=1;
}
LL res=0;
//debug();
//sort(a+1,a+n+1);
for(int i=1;i<=n;++i){
if(2*a[i]>=sz) continue;
LL tmp = conv[2*a[i]];
tmp -= cnt[a[i]]-1; //减去由自己构成的
conv[2*a[i]] -= cnt[a[i]]-1; //将Ai对结果的贡献抹去
cnt[a[i]]--; //将Ai从原序列中抹去
res += tmp;
}
printf("%lld
",res);
}
return 0;
}