From e9c16803c796131305be22721eaf1883d1316825 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Mon, 11 Mar 2024 15:32:53 +0800
Subject: [PATCH] 'commit'

---
 .../ZhongGaoJi/LanQiao15STEMA202401/6.in      | 19 +++++
 .../LanQiao15STEMA202401/6_Test.cpp           | 73 +++++++++----------
 2 files changed, 52 insertions(+), 40 deletions(-)
 create mode 100644 TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/6.in

diff --git a/TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/6.in b/TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/6.in
new file mode 100644
index 0000000..9680da8
--- /dev/null
+++ b/TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/6.in
@@ -0,0 +1,19 @@
+5
+1 2 3 3 2
+
+5 
+2 3 2 3 1
+
+5 
+2 3 4 5 2
+
+5
+3 2 4 2 5
+
+7
+3 2 4 2 5 3 1
+
+5
+3 2 3 2 5
+
+0
\ No newline at end of file
diff --git a/TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/6_Test.cpp b/TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/6_Test.cpp
index 8ac8152..887f098 100644
--- a/TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/6_Test.cpp
+++ b/TangDou/LanQiaoBei/ZhongGaoJi/LanQiao15STEMA202401/6_Test.cpp
@@ -1,54 +1,47 @@
 #include <bits/stdc++.h>
 using namespace std;
-
+typedef pair<int, int> PII;
 const int N = 110;
 int a[N];
 int st[10];
-struct Node {
-    int st, ed;
-};
-vector<Node> q;
+vector<PII> q;
 int res;
-/*
-5
-1 2 3 3 2
-
-5
-3 2 4 2 3
+int n;
 
-7
-3 2 4 2 5 3 1
-*/
 int main() {
-    int n;
-    cin >> n;
-    for (int i = 1; i <= n; i++) cin >> a[i]; // 密码数组
-
-    for (int i = 1; i <= n; i++) {
-        if (!st[a[i]]) {
-            st[a[i]] = 1;
-            for (int j = n; j; j--)
-                if (a[i] == a[j]) {
-                    q.push_back({i, j});
-                    break;
-                }
-        }
-    }
-
-    // 从左到右枚举，所以，不再需再进行按左端点排序,现在就是按左端点排序完的
-    // Q:求一组区间的相交区间个数?如果存在一个相交区间，就多加1个1
-    res = q.size();
-
-    for (int i = 0; i < q.size(); i++) cout << q[i].st << " " << q[i].ed << endl;
-
-    for (int i = 0; i < q.size(); i++)
-        for (int j = i + 1; j < q.size(); j++) {
-            if (q[i].ed > q[j].st && q[j].ed > q[i].ed) {
-                res++;
+#ifndef ONLINE_JUDGE
+    freopen("6.in", "r", stdin);
+#endif
+
+    while (cin >> n && n) {
+        // 各种清空
+        memset(a, 0, sizeof a);
+        memset(st, 0, sizeof st);
+        q.clear();
+        res = 0;
+
+        for (int i = 1; i <= n; i++) cin >> a[i]; // 密码数组
+        for (int i = 1; i <= n; i++) {            // 遍历每个位置
+            if (!st[a[i]]) {                      // 如果此数字没有被处理过
+                st[a[i]] = 1;                     // 标识已处理
+                for (int j = n; j; j--)           // 从后向前找到它的终止位置
+                    if (a[i] == a[j]) {
+                        q.push_back({i, j}); // 记录有效区间
+                        break;
+                    }
             }
         }
 
-    cout << res << endl;
+        // 从左到右枚举，所以，不再需再进行按左端点排序,现在就是按左端点排序完的
+        // Q:求一组区间的相交区间个数?如果存在一个相交区间，就多加1个1
+        res = q.size();
 
+        // 如果存在区间相交，则res++
+        for (int i = 0; i < q.size(); i++)
+            for (int j = i + 1; j < q.size(); j++)
+                if (q[i].second > q[j].first && q[j].second > q[i].second) res++;
+
+        cout << res << endl;
+    }
     return 0;
 }
\ No newline at end of file