Starter 188-Inversion Queries · znzryb
## 思路讲解
two number <=L || >=R delete
```cpp
FOR(i,1,Q){
// two number <=L || >=R delete
cin>>L[i]>>R[i];
if(L[i]==R[i]){
cout<<0<<"\n";
continue;
}
ll lans=ans;
lans-=ansp.query(R[i],N);
lans-=anss.query(1,L[i]);
if(lans<0) cout<<"0\n";
else cout<<lans<<"\n";
}
```
## AC代码
[https://www.codechef.com/viewsolution/1168087702](https://www.codechef.com/viewsolution/1168087702)
::: 源代码
```cpp
// Problem: Inversion Queries
// Contest: CodeChef - START188
// URL: https://www.codechef.com/problems/INVQUER
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
using namespace std;
using ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
using i128 = __int128;using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e17)+9; // 1e18+9也是素数
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;
ll N,M,Q,X,K,lT,A[MAXN],L[MAXN],R[MAXN];
/*
*/
inline void init(){
}
struct BIT { // 普通的单点修改区间查询树状数组，含有查找第K个空位的功能
vector<ll> tr;
int n;
inline int lowbit(int x) { return x & (-x); }
BIT(int ln) { // 构造+初始化函数
n = ln;
tr.resize(n+5,0);
}
inline void add(int p, int x) {
if(p==0) return;
while (p <= n) {
tr[p] += x;
p += lowbit(p);
}
}
inline ll query(int l, int r) {
ll lres = 0, rres = 0;
l -= 1;
while (l > 0) {
lres += tr[l];
l -= lowbit(l);
}
while (r > 0) {
rres += tr[r];
r -= lowbit(r);
}
return rres - lres;
}
inline int findKth(int k) { // 找到第k个空位，当然应当是要求只能出现0和1
int cx = 0;
for (int i = 1 << 20; i > 0; i >>= 1) { // 这个你可以理解为不断缩小步长的搜索
if (cx + i <= n && lowbit(cx + i) - tr[cx + i] < k) {
// lowbit(cx)-tr[cx+i] 表示cx+i这个树状数组区间有多少空位
k -= lowbit(cx + i) - tr[cx + i];
cx += i;
}
}
return cx + 1; // 我们始终让cx < k，所以返回的就是最大的比k小的，+1就是新空位出现的地方
}
};
inline void solve(){
cin>>N>>Q;
init();
FOR(i,1,N){
cin>>A[i];
}
BIT trp(N),ansp(N),trs(N),anss(N);
// 找编号比自己小但是更大的
ll ans=0;
FOR(i,1,N){
ll res=trp.query(A[i],N);
ansp.add(A[i],res);
ans+=res;
trp.add(A[i],1);
}
// index big but value small
ROF(i,N,1){
ll res=trs.query(1,A[i]);
anss.add(A[i],res);
trs.add(A[i],1);
}
FOR(i,1,Q){
// two number <=L || >=R delete
cin>>L[i]>>R[i];
if(L[i]==R[i]){
cout<<0<<"\n";
continue;
}
ll lans=ans;
lans-=ansp.query(R[i],N);
lans-=anss.query(1,L[i]);
if(lans<0) cout<<"0\n";
else cout<<lans<<"\n";
#ifdef LOCAL
debug(lans);
#endif
}
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>lT;
while(lT--){
solve();
}
// solve();
return 0;
}
/*
AC
https://www.codechef.com/viewsolution/1168087702
*/
```
:::
## 心路历程（WA，TLE，MLE……）
那么就是要注意这个特判
```cpp
if(L[i]==R[i]){
cout<<0<<"\n";
continue;
}
```