From 573c77674e97857c8689e24496577aac624bd3bf Mon Sep 17 00:00:00 2001
From: HuangHai <10402852@qq.com>
Date: Sun, 5 Oct 2025 18:33:03 +0800
Subject: [PATCH] 'commit'

---
 TangDou/CSP-J/AcWingTraining/T1/442.md    | 115 ++++++++++++++++++++++
 TangDou/CSP-J/AcWingTraining/T1/442_2.cpp |  28 ++++++
 TangDou/Test/TD.cpp                       |  42 ++++----
 3 files changed, 160 insertions(+), 25 deletions(-)
 create mode 100644 TangDou/CSP-J/AcWingTraining/T1/442.md
 create mode 100644 TangDou/CSP-J/AcWingTraining/T1/442_2.cpp

diff --git a/TangDou/CSP-J/AcWingTraining/T1/442.md b/TangDou/CSP-J/AcWingTraining/T1/442.md
new file mode 100644
index 0000000..6da6cf1
--- /dev/null
+++ b/TangDou/CSP-J/AcWingTraining/T1/442.md
@@ -0,0 +1,115 @@
+
+## 排队接水
+
+### 一、数学证明
+首先我们要排的是所有的元素，但是为什么是从小到大呢？？？
+
+排队总是象征要排序，每个元素在此序列下都要满足条件，也就是从中拿出两个相邻元素同样满足条件：
+
+于是设：
+
+$a_i$ 和 $b_i$且$a_i<b_i$
+
+那么针对这两个元素：就有两种排列情况：
+
+1.$a_i$排在$b_i$前面那么有总时间：$t_1=a_i+a_i+b_i$.
+
+2.$b_i$排在$a_i$前面那么有总时间：$t_2=b_i+b_i+a_i$.
+
+##于是由$a_i<b_i$得出 $t_1<t_2$—》变一下式子—》$a_i+a_i+b_i<b_i+b_i+a_i$;
+
+再化简不等式得出$a_i<b_i$
+
+于是得出结论：**当$a_i$在$b_i$前面时，时间为最小值**。
+
+于是反推回总体，两两相较，那么越小的应该越排在前面，以至于总时间越小。
+
+### 二、代码
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 10010, M = 200;
+int n, m;
+int w[N];
+int q[M];
+
+int main() {
+  cin >> n >> m;
+  for (int i = 0; i < n; i++)
+    cin >> w[i];
+
+  for (int i = 0; i < n; i++) {
+    // 找出最小值
+    int t = 0;
+    for (int j = 0; j < m; j++) {
+      if (q[j] < q[t])
+        t = j;
+    }
+    // 将i号同学安排到第t个位置上
+    q[t] += w[i];
+  }
+  int res = 0;
+  for (int i = 0; i < m; i++)
+    res = max(res, q[i]); // 找出最大值就是答案
+  cout << res << endl;
+  return 0;
+}
+```
+
+### 三、变形
+
+**【题目描述】**
+有$n$个人在一个水龙头前排队接水，假如每个人接水的时间为$T_i$，请编程找出这$n$个人排队的一种顺序，使得$n$个人的平均等待时间最小。
+
+**【输入】**
+共两行，第一行为$n(1≤n≤1000)$；第二行分别表示第$1$个人到第$n$个人每人的接水时间$T_1，T_2，…，T_n$，每个数据之间有$1$个空格。
+
+**【输出】**
+有两行，第一行为一种排队顺序，即$1$到$n$的一种排列；第二行为这种排列方案下的平均等待时间(输出结果精确到小数点后两位)。
+
+**【输入样例】**
+```
+10							
+56 12 1 99 1000 234 33 55 99 812
+```
+**【输出样例】**
+```
+3 2 7 8 1 4 9 6 10 5
+291.90
+```
+
+```
+http://ybt.ssoier.cn:8088/problem_show.php?pid=1319
+```
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 1005;
+
+struct Node {
+    int num;
+    int sum;
+} a[N];
+bool cmp(Node x, Node y) {
+    return x.sum < y.sum;
+}
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 0; i < n; i++) {
+        cin >> a[i].sum;
+        a[i].num = i + 1;
+    }
+    sort(a, a + n, cmp);
+    double temp = 0.00;
+    for (int i = 0; i < n; i++) {
+        cout << a[i].num << " ";
+        temp += (n - (i + 1)) * a[i].sum;
+    }
+    cout << endl;
+    printf("%.2f", temp / n * 1.00);
+    return 0;
+}
+```
\ No newline at end of file
diff --git a/TangDou/CSP-J/AcWingTraining/T1/442_2.cpp b/TangDou/CSP-J/AcWingTraining/T1/442_2.cpp
new file mode 100644
index 0000000..208163c
--- /dev/null
+++ b/TangDou/CSP-J/AcWingTraining/T1/442_2.cpp
@@ -0,0 +1,28 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 1005;
+
+struct Node {
+    int num;
+    int sum;
+} a[N];
+bool cmp(Node x, Node y) {
+    return x.sum < y.sum;
+}
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 0; i < n; i++) {
+        cin >> a[i].sum;
+        a[i].num = i + 1;
+    }
+    sort(a, a + n, cmp);
+    double temp = 0.00;
+    for (int i = 0; i < n; i++) {
+        cout << a[i].num << " ";
+        temp += (n - (i + 1)) * a[i].sum;
+    }
+    cout << endl;
+    printf("%.2f", temp / n * 1.00);
+    return 0;
+}
diff --git a/TangDou/Test/TD.cpp b/TangDou/Test/TD.cpp
index 8e1cccc..b4b107c 100644
--- a/TangDou/Test/TD.cpp
+++ b/TangDou/Test/TD.cpp
@@ -1,31 +1,23 @@
 #include <bits/stdc++.h>
 using namespace std;
-long long n;
+int w[1001];
+int s[1001];
 int main() {
-    cin >> n;
-    bool first = true;
-    for (long long p = 2;p*p<=n;p++){
-        if (n % p != 0) continue;
-        int cnt = 0;
-        while (n % p == 0){
-            n /= p;
-            cnt++;
-        }
-            if (first){
-                first = false;
-                cout << p;
-            }else{
-                cout << " * " << p;
-            }
-        if (cnt > 1) cout << " ^ " << cnt;
+    int n, m;
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++) {
+        cin >> w[i];
     }
-    if (n != 1){
-        if (first){
-            first = false;
-            cout << n;
-        }else{
-            cout << " * " << n;
-        }
+    int ans = 0;
+    sort(w + 1, w + 1 + n);
+    int j = 0;
+    for(int i = 1; i <= n; i++){
+        j++;
+        if (j == m+ 1) j = 1;
+        s[j] += w[i];
+        ans += s[j];
     }
-        return 0;
+    
+    cout << ans << endl;
+    return 0;
 }
\ No newline at end of file