World is Exploding (容斥 + 统计)

　　题意：满足题目中的式子，a < b && c < d && Va < Vb && Vc > Vd

　　思路：先求不讨论位置重合的情况，把对应的2种关系相乘，然后得到的答案减去重合的地方。不想解释，我特么改着改着就对了。都不知道哪里错了，叫对了数据还是找不到。因为只有一组数据出错。

#include<bits/stdc++.h>
using namespace std;
 
const int maxn = 5e4 + ;
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 = ;
    while(x){
        ret += tr[x];
        x -= lowbit(x);
    }
    return ret;
}
 
int main(){
    while(~scanf("%d", &n)){
        long long ans = , sum1 = , sum2 = ;
        for(int i = ; i <= n; i ++) scanf("%d", &in[i]), sar[i] = in[i];
        sort(sar + , sar + n + );
        m = unique(sar + , sar + n + ) - sar;
        for(int i = ; i <= n; i ++)
            in[i] = lower_bound(sar + , sar + n + , in[i]) - sar;
        memset(tr, , sizeof(tr));
 
        for(int i = ; i <= n; i ++){
            ls[i] = query(in[i] - );
            lb[i] = query(n) - query(in[i]);
            sum1 += ls[i];
            sum2 += lb[i];
            add(in[i], );
        }
        for(int i = ; i <= n; i ++){
            rs[i] = query(in[i] - );
            rb[i] = query(n) - query(in[i]);
            add(in[i], -);
        }
        ans = sum1 * sum2;
        for(int i = ; 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 ;
}