題意：滿足題目中的式子，a < b && c < d && Va < Vb && Vc > Vd

　　思路：先求不討論位置重合的情況，把對應的2種關係相乘，然後得到的答案減去重合的地方。不想解釋，我特麼改著改著就對了。都不知道哪裡錯了，叫對了資料還是找不到。因為只有一組資料出錯。

#include<bits/stdc++.h>
using namespace std;

const int maxn = 5e4 + 7;
int tr[maxn], in[maxn], sar[maxn], ls[maxn], lb[maxn], rs[maxn], rb[maxn];
 
int n, m;

int lowbit(int x){
    return x & -x;
}

void add(int x, int d){
    while(x <= n){
        tr[x] += d;
        x += lowbit(x);
    }
}

int query(int x){
    int ret = 0;
    while(x){
        ret += tr[x];
        x -= lowbit(x);
    }
    return ret;
}

int main(){
    while(~scanf("
 
%d", &n)){
        long long ans = 0, sum1 = 0, sum2 = 0;
        for(int i = 1; i <= n; i ++) scanf("%d", &in[i]), sar[i] = in[i];
        sort(sar + 1, sar + n + 1);
        m = unique(sar + 1, sar + n + 1) - sar;
        for(int i = 1; i <= n; i ++)
            in[i] = lower_bound(sar + 1
 
, sar + n + 1, in[i]) - sar;
        memset(tr, 0, sizeof(tr));

        for(int i = 1; i <= n; i ++){
            ls[i] = query(in[i] - 1);
            lb[i] = query(n) - query(in[i]);
            sum1 += ls[i];
            sum2 += lb[i];
            add(in[i], 1);
        }
        for(int i = 1; i <= n; i ++){
            rs[i] = query(in[i] - 1);
            rb[i] = query(n) - query(in[i]);
            add(in[i], -1);
        }
        ans = sum1 * sum2;
        for(int i = 1; i <= n; i ++){
            ans -= 1LL * ls[i] * lb[i];
            ans -= 1LL * ls[i] * rs[i];
            ans -= 1LL * rb[i] * lb[i];
            ans -= 1LL * rs[i] * rb[i];
        }
        printf("%lld\n",ans);

    }
    return 0;
}