update: 添加问题“1039.多边形三角剖分的最低得分”的代码和题解

.vscode/settings.json

@@ -104,7 +104,16 @@
         "__node_handle": "cpp",
         "__verbose_abort": "cpp",
         "execution": "cpp",
-        "print": "cpp"
+        "print": "cpp",
+        "filesystem": "cpp",
+        "format": "cpp",
+        "ccomplex": "cpp",
+        "cstdbool": "cpp",
+        "ctgmath": "cpp",
+        "__assert": "cpp",
+        "ranges": "cpp",
+        "version": "cpp",
+        "bit": "cpp"
     },
     "editor.mouseWheelZoom": true,
     "workbench.tree.enableStickyScroll": true,

Codes/1039-minimum-score-triangulation-of-polygon.cpp

@@ -0,0 +1,35 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-09-29 18:44:48
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-10-01 19:58:47
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+#if false  // this doesn't work
+// 相邻两点乘积的和 + (n-1)条边乘积的2倍
+// 但是不能相交
+// 诶，好像直接最小的那个去连其他所有剩下的就好了
+class Solution {
+public:
+    int minScoreTriangulation(vector<int>& values) {
+        int ans = values[0] * values.back();
+        int m = values[0], loc = 0;
+        for (int i = 1; i < values.size(); i++) {
+            ans += values[i - 1] * values[i];
+            if (values[i] < m) {
+                m = values[i];
+                loc = i;
+            }
+        }
+        for (int i = 0; i < values.size(); i++) {
+            if (abs(i - loc) > 1) {
+                ans += m * values[i] * 2;
+            }
+        }
+        return ans;
+    }
+};
+#endif

Codes/1039-minimum-score-triangulation-of-polygon_Ok.cpp

@@ -0,0 +1,77 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-09-29 18:44:48
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-10-01 20:38:27
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+private:
+    unordered_map<int, int> cache;
+    vector<int> values;
+    int n;
+
+    int dfs(int i, int j) {
+        if (j - i < 2) {
+            return 0;
+        }
+        int key = i * n + j;
+        if (cache.count(key)) {
+            return cache[key];
+        }
+        if (j - i == 2) {
+            return cache[key] = values[i] * values[i + 1] * values[i + 2];
+        }
+        int ans = 1000000000;
+        /*
+        0 1 2 3 -> 0 1 2 + 0 2 3
+
+        0   1
+
+        3   2
+
+
+        0 1 2 3 4
+
+
+            3
+
+        4        2
+
+         0      1
+
+
+        (i,j,k) + dfs(i,k)+dfs(k,j)
+        */
+        for (int k = i + 1; k < j; k++) {
+            ans = min(ans, dfs(i, k) + dfs(k, j) + values[i] * values[k] * values[j]);
+        }
+        return cache[key] = ans;
+    }
+public:
+    int minScoreTriangulation(vector<int>& values) {
+        this->values = move(values);
+        n = this->values.size();
+        return dfs(0, n - 1);
+    }
+};
+
+#if defined(_WIN32) || defined(__APPLE__)
+/*
+[3,7,4,5]
+
+144
+*/
+int main() {
+    string s;
+    while (cin >> s) {
+        vector<int> v = stringToVector(s);
+        Solution sol;
+        cout << sol.minScoreTriangulation(v) << endl;
+    }
+    return 0;
+}
+#endif

Codes/lib.rs

@@ -2,7 +2,7 @@
  * @Author: LetMeFly
  * @Date: 2025-08-07 14:11:36
  * @LastEditors: LetMeFly.xyz
- * @LastEditTime: 2025-09-07 23:24:39
+ * @LastEditTime: 2025-10-01 20:40:27
  */
 pub struct Solution;
 include!("0976-largest-perimeter-triangle.rs");  // 这个fileName是会被脚本替换掉的

Codes/pinduoduo-01.py

@@ -0,0 +1,25 @@
+'''
+Author: LetMeFly
+Date: 2025-09-28 19:01:34
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-09-28 19:06:41
+'''
+a = input()
+if len(a) != 9:
+    print('Invalid')
+    exit(0)
+for i in range(2):
+    if not 'A' <= a[i] <= 'Z':
+        print('Invalid')
+        exit(0)
+for i in range(2, 8):
+    if not a[i].isdigit():
+        print('Invalid')
+        exit(0)
+should = 0
+for i in range(8):
+    should += ord(a[i])
+should %= 26
+should = chr(should + ord('A'))
+a = a[:-1] + should
+print(a)

Codes/pinduoduo-02.cpp

@@ -0,0 +1,118 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-09-28 19:10:08
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-09-28 19:31:14
+ */
+/*
+
+   1
+ 2   3
+0 1
+
+   2
+ 3   1
+0 2
+
+
+
+root 1->2
+times: 1
+2->3: 1
+need: 1
+1 -> 2: 1
+need: 1
+3->1: 
+need: 3
+times: 2
+sum: 1+2=3
+*/
+
+
+/*
+
+    1
+  2   3
+1 2  3 1
+
+
+    3
+  1   2
+3 1  2 1
+
+1->3: 2
+times: 2
+2->1: 4
+times: 2
+3->2: 4
+times: 2
+1->3: (5->3) 3
+times: 3
+2->1: (1->1) 0
+3->2: (2->2) 0
+1->1: (5->1) 1
+times 1
+sum: 2 + 2 + 2 + 3 + 1 = 10
+*/
+#include <bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+
+struct Node {
+    int original, to;
+    Node *left, *right;
+    int already;  // 由于father导致的已变更次数
+};
+
+// 变化范围是1，2，3，4，5
+inline int change(int original, int times) {
+    return (original + times - 1 + 5) % 5 + 1;
+}
+
+inline int calc(int from, int to) {
+    return (to - from + 5) % 5;
+}
+
+/*
+5
+1 2 3 0 1
+2 3 1 0 2
+
+*/
+
+int main() {
+    int n;
+    cin >> n;
+    queue<Node*> q;
+    Node* head = new Node();
+    q.push(head);
+    for (int i = 0; i < n; i++) {
+        Node* thisNode = q.front();
+        q.pop();
+        cin >> thisNode->original;
+        thisNode->left = new Node();
+        thisNode->right = new Node();
+        q.push(thisNode->left);
+        q.push(thisNode->right);
+    }
+    q = queue<Node*>();
+    q.push(head);
+    ll ans = 0;
+    for (int i = 0; i < n; i++) {
+        Node* thisNode = q.front();
+        q.pop();
+        cin >> thisNode->to;
+        q.push(thisNode->left);
+        q.push(thisNode->right);
+        if (thisNode->original == 0) {
+            continue;
+        }
+        int from = change(thisNode->original, thisNode->already);
+        int need = calc(from, thisNode->to);
+        ans += need;
+        thisNode->left->already = (thisNode->already + need) % 5;
+        thisNode->right->already = (thisNode->already + need) % 5;
+    }
+    cout << ans << endl;
+    return 0;
+}