From cc3380434cd14727d4392e41c99b52db1b5de05e Mon Sep 17 00:00:00 2001
From: Zengtudor <Zengtudor@outlook.com>
Date: Wed, 3 Sep 2025 17:27:32 +0800
Subject: [PATCH] =?UTF-8?q?feat:=20=E6=B7=BB=E5=8A=A0P3660=E9=A2=98?=
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实现了一个使用树状数组的解法来计算数组中成对元素的交叉对数。通过维护一个访问标记数组和树状数组来高效统计区间和，优化了计算过程。
---
 src/9/3/P3660.cpp | 55 +++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 55 insertions(+)
 create mode 100644 src/9/3/P3660.cpp

diff --git a/src/9/3/P3660.cpp b/src/9/3/P3660.cpp
new file mode 100644
index 0000000..e752dd5
--- /dev/null
+++ b/src/9/3/P3660.cpp
@@ -0,0 +1,55 @@
+#include <cstdint>
+#include <iostream>
+#include <istream>
+#include <vector>
+using ll = int64_t;
+
+ll n,ans=0;
+std::vector<ll> vis,t,a;
+
+static inline ll lb(const ll n){
+    return n&(-n);
+}
+
+static inline void upd(ll idx,ll add){
+    while(idx<=2*n){ // 注意要*2
+        t[idx]+=add;
+        idx+=lb(idx);
+    }
+}
+
+static inline ll get(ll idx){
+    ll res=0;
+    while(idx){
+        res+=t[idx];
+        idx-=lb(idx);
+    }
+    return res;
+}
+
+int main(){
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+
+    std::cin>>n;
+    a.resize(n*2+1);
+    vis.resize(n+1);
+    t.resize(n*2+1);
+
+    for(ll i=1;i<=n*2;i++){
+        std::cin>>a[i];
+    }
+
+    for(ll i=1;i<=n*2;i++){
+        if(vis[a[i]]){
+            ll nans=get(i)-get(vis[a[i]]);
+            ans+=nans;
+            // printf("%lld += %lld\n",i,nans);
+            upd(vis[a[i]], -1);
+        }else{
+            upd(i, 1);
+            vis[a[i]]=i;
+        }
+    }
+    std::cout<<ans<<"\n";
+}
\ No newline at end of file