From b2c80a8161907f627cb8d78119292190348f3718 Mon Sep 17 00:00:00 2001
From: Zengtudor <Zengtudor@outlook.com>
Date: Sat, 11 Oct 2025 12:10:13 +0800
Subject: [PATCH] =?UTF-8?q?feat:=20=E6=B7=BB=E5=8A=A0P6006.cpp=E5=AE=9E?=
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实现一个动态规划算法来计算满足条件的三元组数量，用于解决特定范围的查询问题。算法通过预处理和动态规划表优化查询效率，并处理了数组边界条件和内存使用问题。
---
 src/10/10/P6006.cpp | 58 +++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 58 insertions(+)
 create mode 100644 src/10/10/P6006.cpp

diff --git a/src/10/10/P6006.cpp b/src/10/10/P6006.cpp
new file mode 100644
index 0000000..6295990
--- /dev/null
+++ b/src/10/10/P6006.cpp
@@ -0,0 +1,58 @@
+#include <cstdint>
+#include <iostream>
+#include <istream>
+#include <vector>
+
+using ll = int64_t;
+
+const ll maxn = 5e3+5;
+ll n,q, dp[maxn][maxn], a[maxn];
+
+struct T{
+    const static ll of = 2e6+3;
+    ll v[4*ll(1e6+5)]; // 注意限制的大小要*2
+    inline ll&operator[](const ll n){
+        return v[n+of];
+    }
+}tot;
+
+std::vector<ll> used;
+
+int main(){
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+
+    std::cin>>n>>q;
+    for(ll i=1;i<=n;i++){
+        std::cin>>a[i];
+    }
+
+    for(ll i=n-2;i>=1;i--){ // 注意循环顺序！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！
+        for(ll j:used)tot[j]=0; // 注意清空耗时
+        used.clear();
+        tot[a[i+1]]++;
+        used.emplace_back(a[i+1]);
+        for(ll j=i+2;j<=n;j++){
+            dp[i][j] = tot[-(a[i]+a[j])];
+            // printf("before dp[%lld][%lld]=%lld\n",i,j,dp[i][j]);
+            dp[i][j] += dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1];
+            tot[a[j]]++;
+            used.emplace_back(a[j]);
+            // printf("after dp[%lld][%lld]=%lld\n",i,j,dp[i][j]);
+        }
+    }
+
+    while(q--){
+        ll a,b;
+        std::cin>>a>>b;
+        std::cout<<dp[a][b]<<"\n";
+    }
+}
+
+/*
+
+dp[i][j] = i~j的
+dp[i][j] = dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1]
+
+
+*/
\ No newline at end of file