树状数组。考虑ai（从0开始，则i左边共i个，右边n-i-1个），左边有x个比他大的，i-x个比他小的，右边有y个比他大的，n-i-1-y个比他大的。交叉乘一下就得到了以ai为裁判的比赛总数。把所有人都枚举一遍，加在一起就是答案，会超int。

如何才能知道ai左边有多少比他小的呢？假如aj<ai且j<i，用另一个数组b来标记这个数有没有出现过，那么b[aj] = 1;这样，比ai小的数的个数，就是b的前缀和。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<string>
#include<queue>
#include<cmath>
///LOOP
#define REP(i, n) for(int i = 0; i < n; i++)
#define FF(i, a, b) for(int i = a; i < b; i++)
#define FFF(i, a, b) for(int i = a; i <= b; i++)
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)
#define FDD(i, a, b) for(int i = a; i >= b; i--)
///INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RFI(n) scanf("%lf", &n)
#define RFII(n, m) scanf("%lf%lf", &n, &m)
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
#define RS(s) scanf("%s", s)
///OUTPUT
#define PN printf("\n")
#define PI(n) printf("%d\n", n)
#define PIS(n) printf("%d ", n)
#define PS(s) printf("%s\n", s)
#define PSS(s) printf("%s ", n)
///OTHER
#define PB(x) push_back(x)
#define CLR(a, b) memset(a, b, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(b))
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = 100000000;
const int INFI = 1e9 * 2;
const LL LINFI = 1e17;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int N = 111111;
const int M = 22;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};int a[N], b[N], c[N], n;void update(int x)
{while(x <= N){b[x]++;x += x & (-x);}
}int sum(int x)
{LL s = 0;while(x){s += b[x];x -= x & (-x);}return s;
}int main()
{//freopen("input.txt", "r", stdin);//freopen("output.txt", "w", stdout);int t, k;LL ans;RI(t);while(t--){RI(n);CLR(b, 0);REP(i, n){RI(a[i]);c[i] = sum(a[i]);update(a[i]);}CLR(b, 0);ans = 0;FD(i, n, 0){k = sum(a[i]);ans += LL(c[i]) * LL(n - i - 1 - k) + LL(i - c[i]) * LL(k);update(a[i]);}printf("%lld\n", ans);}return 0;
}