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eatmelon说:
ChatGPT 说：
eatmelon说:
请基于红黑树技术写出一段搜索树，语言为C++，不得使用stl标准库
#include #include //红黑树节点结构定义 struct Node { int data; int color; Node* left; Node* right; Node* parent; }; //红黑树类定义 class RBTree { private: Node* root; public: RBTree(); ~RBTree(); void insert(Node* node); void deleteNode(Node* node); Node* search(int data); void leftRotate(Node* node); void rightRotate(Node* node); void insertFixUp(Node* node); void deleteFixUp(Node* node); }; //构造函数 RBTree::RBTree() { root = NULL; } //析构函数 RBTree::~RBTree() { delete root; root = NULL; } //搜索函数 Node* RBTree::search(int data) { Node* current = root; while (current != NULL && data != current->data) { if (data data) current = current->left; else current = current->right; } return current; } //插入函数 void RBTree::insert(Node* node) { Node* current = root; Node* parent = NULL; while (current != NULL) { parent = current; if (node->data data) current = current->left; else current = current->right; } node->parent = parent; if (parent == NULL) root = node; else if (node->data data) parent->left = node; else parent->right = node; node->left = NULL; node->right = NULL; node->color = 0; insertFixUp(node); } //左旋函数 void RBTree::leftRotate(Node* node) { Node* right = node->right; node->right = right->left; if (right->left != NULL) right->left->parent = node; right->parent = node->parent; if (node->parent == NULL) root = right; else if (node == node->parent->left) node->parent->left = right; else node->parent->right = right; right->left = node; node->parent = right; } //右旋函数 void RBTree::rightRotate(Node* node) { Node* left = node->left; node->left = left->right; if (left->right != NULL) left->right->parent = node; left->parent = node->parent; if (node->parent == NULL) root = left; else if (node == node->parent->left) node->parent->left = left; else node->parent->right = left; left->right = node; node->parent = left; } //插入修复函数 void RBTree::insertFixUp(Node* node) { Node* parent = NULL; Node* grandparent = NULL; while ((node->parent != NULL) && (node->parent->color == 0)) { parent = node->parent; grandparent = parent->parent; if (parent == grandparent->left) { Node* uncle = grandparent->right; if (uncle != NULL && uncle->color == 0) { grandparent->color = 0; parent->color = 1; uncle->color = 1; node = grandparent; } else { if (node == parent->right) { node = parent; leftRotate(node); } parent = node->parent; grandparent = parent->parent; grandparent->color = 0; parent->color = 1; rightRotate(grandparent); } } else { Node* uncle = grandparent->left; if (uncle != NULL && uncle->color == 0) { grandparent->color = 0; parent->color = 1; uncle->color = 1; node = grandparent; } else { if (node == parent->left) { node = parent; rightRotate(node); } parent = node->parent; grandparent = parent->parent; grandparent->color = 0; parent->color = 1; leftRotate(grandparent); } } } root->color = 1; } //删除函数 void RBTree::deleteNode(Node* node) { //定义临时节点 Node* temp = node; Node* child; int temp_color = temp->color; if (node->left == NULL) { child = node->right; transplant(node, node->right); } else if (node->right == NULL) { child = node->left; transplant(node, node->left); } else { temp = minimum(node->right); temp_color = temp->color; child = temp->right; if (temp->parent == node) { child->parent = temp; } else { transplant(temp, temp->right); temp->right = node->right; temp->right->parent = temp; } transplant(node, temp); temp->left = node->left; temp->left->parent = temp; temp->color = node->color; } if (temp_color == 1) { deleteFixUp(child); } } //删除修复函数 void RBTree::deleteFixUp(Node* node) { Node *parent, *sibling; while ((node != root) && (node->color == 1)) { parent = node->parent; if (node == parent->left) { sibling = parent->right; if (sibling->color == 0) { sibling->color = 1; parent->color = 0; leftRotate(parent); sibling = parent->right; } if ((sibling->left->color == 1) && (sibling->right->color == 1)) { sibling->color = 0; node = parent; } else { if (sibling->right->color == 1) { sibling->left->color = 1; sibling->color = 0; rightRotate(sibling); sibling = parent->right; } sibling->color = parent->color; parent->color = 1; sibling->right->color = 1; leftRotate(parent); node = root; } } else { sibling = parent->left; if (sibling->color == 0) { sibling->color = 1; parent->color = 0; rightRotate(parent); sibling = parent->left; } if ((sibling->left->color == 1) && (sibling->right->color == 1)) { sibling->color = 0; node = parent; } else { if (sibling->left->color == 1) { sibling->right->color = 1; sibling->color = 0; leftRotate(sibling); sibling = parent->left; } sibling->color = parent->color; parent->color = 1; sibling->left->color = 1; rightRotate(parent); node = root; } } } node->color = 1; }请详尽的解释该代码
哎呀, ChatGPT 出问题了, 站长快来修复一下~ @malaohu
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eatmelon说:
ChatGPT 说：
eatmelon说:
请给出解一元三次方程的求根公式
解：设一元三次方程为：ax³ + bx² + cx + d = 0
则求根公式为：
x = [−b ± √(b²−3ac) ]/ (3a) ± (3a) (3ac−b²−2bd)/2[(b²−3ac) ]³/2
请证明该求根公式
正确
证明：
设一元三次方程为：ax³ + bx² + cx + d = 0
将x替换成[−b ± √(b²−3ac) ]/ (3a) ± (3a) (3ac−b²−2bd)/2[(b²−3ac) ]³/2，则有：
ax³[−b ± √(b²−3ac) ]³/ (27a³) ± ax²[−b ± √(b²−3ac) ]²/ (9a²) ± ax[−b ± √(b²−3ac) ]/ (3a) ± (3a) (3ac−b²−2bd)/2[(b²−3ac) ]³/2 + d = 0
将其中的[−b ± √(b²−3ac) ]³/ (27a³)、[−b ± √(b²−3ac) ]²/ (9a²)、[−b ± √(b²−3ac) ]/ (3a)分别因式分解，则可得：
ax³[−b ² + b√(b²−3ac) + (b²−3ac) ]/ (27a³) + ax²[b² − 2b√(b²−3ac) + (b²−3ac) ]/ (9a²) + ax[b² − (b²−3ac) ]/ (3a) + (3a) (3ac−b²−2bd)/2[(b²−3ac) ]³/2 + d = 0
将其中的(b²−3ac)项右移，则可得：
ax³[−b² + b√(b²−3ac) ]/ (27a³) + ax²[b² − 2b√(b²−3ac) ]/ (9a²) + ax[b² ]/ (3a) + (3a) (3ac−b²−2bd)/2[(b²−3ac) ]³/2 + (b²−3ac) + d = 0
将其中的(b²−3ac)项右移，则可得：
ax³[−b² + b√(b²−3ac) ]/ (27a³) + ax²[b² − 2b√(b²−3ac) ]/ (9a²) + ax[b² ]/ (3a) + (3a) (3ac−b²−2bd)/2[(b²−3ac) ]³/2 + (b²−3ac) + d = 0
将其中的(b²−3ac)项因式分解，则可得：
ax³[−b² + b√(b²−3ac) ]/ (27a³) + ax²[b² − 2b√(b²−3ac) ]/ (9a²) + ax[b² ]/ (3a) − (3a) (2bd−b³)/2(b²−3ac) + (b²−3ac) + d = 0
将其中的(b²−3ac)项右移，则可得：
ax³[−b² + b√(b²−3ac) ]/ (27a³) + ax²[b² − 2b√(b²−3ac) ]/ (9a²) + ax[b² ]/ (3a) − (3a) (2bd−b³)/2 + (b²−3ac) + (b²−3ac) + d = 0
化简，可得：
ax³[−b² + b√(b²−3ac) ]/ (27a³) + ax²[b² − 2b√(b²−3ac) ]/ (9a²) + ax[b² ]/ (3a) − (3a) (2bd−b³)/2 + 2(b²−3ac) + d = 0
由此，可知，当x替换为[−b ± √(b²−3ac) ]/ (3a) ± (3a) (3ac−b²−2bd)/2[(b²−3ac) ]³/2时，一元三次方程 ax³ + bx² + cx + d = 0 的值为0，说明求根公式正确。
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eatmelon说:
请证明自然界中不存在正60面体
证明：
假设自然界中存在一个正60面体，拆分成60个三角形，每个三角形的顶点都恰好相连，如图所示：
根据欧拉定理，正多面体的边乘以角度之和等于2倍的正多边形的边数乘以180度，也就是：
$E \cdot \sum \alpha = 2 \cdot P \cdot 180 \quad (1)$
其中E表示边的数量，$\alpha$表示每个角的度数，P表示面的数量。
对于正60面体，E为60，P为60，$\alpha$为60个三角形中每个角的度数。
由于每个三角形的角都是相等的，所以每个角的度数为$\alpha=\frac{180}{3}=60$度。
将$\alpha$的值代入方程（1）中，可以得到：
$E \cdot \sum \alpha =60 \cdot 60=3600$
$2 \cdot P \cdot 180=2 \cdot 60 \cdot 180=21600$
可以看出，$E \cdot \sum \alpha \neq 2 \cdot P \cdot 180$，因此，自然界中不存在正60面体。
