树状数组求逆序对以及相关例题

//sum[i]为位置i的逆序对数量
for(int i=n;i>=1;i--){
	sum[i]=query(a[i]-1);
	//查询该数字后面是否存在比它小的数
	add(a[i],1)
	//从a[i]到N的个数加1
}

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <set>
#include <stack>
#include <iomanip>
#include <unordered_map>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define seteps(N) setprecision(N)
#define lson (ind<<1)
#define rson (ind<<1|1)
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef long long ll;
typedef pair<ll,ll> PLL;
typedef __int128 lll;
constexpr int N=1e6+10;
constexpr int M=1e6+10;
constexpr int INF=0x3f3f3f3f;
constexpr int mod=1e9+7;
constexpr lll one=1;
int a[N],c[N],sum[N];
struct BIT{
    void init(){
        rep(i,0,N+1) c[i]=0;
    }

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

    void add(int i,int v){
        while(i<=N){
            c[i]+=v;
            i+=lowbit(i);
        }
    }

    ll query(int i){
        ll res=0;
        while(i>0){
            res+=c[i];
            i-=lowbit(i);
        }
        return res;
    }
};
void solve(){
    int n;
    scanf("%d",&n);
    BIT bit;
    bit.init();
    rep(i,1,n+1) scanf("%d",&a[i]);
    rep(i,1,n+1){
        sum[i] = bit.query(N) - bit.query(a[i]);
        bit.add(a[i], 1);
    }
    bit.init();
    rrep(i,n,1){
        sum[i] += bit.query(a[i] - 1);
        bit.add(a[i], 1);
    }
    ll res=0;
    rep(i,1,n+1) res += sum[i]*(sum[i] + 1ll) / 2;
    printf("%lld\n", res);
}
signed main(){
    //IOS;
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    //int t;cin>>t;
    //while(t--)
        solve();
    return 0;
}