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 ;

}