class Lnput_lnitialization(): def __init__(self, top, Number_of_rows): self.top = top int(data_demo.window_width // 3.9875) int(data_demo.window_height // 1.138) self.frame = tk.Frame(self.top, bg='Black', highlightthickness=2, highlightbackground='Black') self.frame.place(relx=0.739, rely=0.083, width=int(data_demo.window_width // 3.87), height=int(data_demo.window_height // 1.15)) self.canvas = tk.Canvas(self.frame) self.canvas.place(relx=0, rely=0, width=int(data_demo.window_width // 3.9875), height=int(data_demo.window_height // 1.15)) scrollbar = tk.Scrollbar(self.frame, command=self.canvas.yview) scrollbar.pack(side="right", fill="y") self.canvas.configure(yscrollcommand=scrollbar.set) self.button_frame = tk.Frame(self.canvas, bg='Yellow') # self.button_frame.place(relwidth=0.98, relheight=1) self.canvas.create_window((0, 0), window=self.button_frame, anchor="nw") self.Content(Number_of_rows) # 在步骤栏创建填写框 def Content(self, Number_of_rows): x = 0.01 y = 0.01 for a in range(Number_of_rows): var = tk.IntVar() checkbutton1 = Checkbutton(self.button_frame, variable=var, text="{}".format(data_demo.Serial_Number), compound='right') checkbutton1.grid(row=data_demo.Serial_Number, column=1, padx=3, pady=5) # checkbutton1.place(relx=x, rely=y) data_demo.checkbutton1_boxs.append(checkbutton1) data_demo.checkbutton_bool.append(var)为什么只能用grid不能用plice

#include<stdio.h> #include<stdlib.h> //定义链表节点结构体 typedef struct Node{ int data; struct Node* next; }Node; //函数声明 Node* merge(Node* headA, Node* headB); void printList(Node* head); int main(){ //输入线性表A和B的元素个数 int m, n; scanf("%d%d", &m, &n); //创建链表A和B Node* headA = NULL; Node* headB = NULL; Node* pA = NULL; Node* pB = NULL; for(int i=0; i<m; i++){ int data; scanf("%d", &data); Node* node = (Node*)malloc(sizeof(Node)); node->data = data; node->next = NULL; if(headA == NULL){ headA = node; pA = node; } else{ pA->next = node; pA = node; } } for(int i=0; i<n; i++){ int data; scanf("%d", &data); Node* node = (Node*)malloc(sizeof(Node)); node->data = data; node->next = NULL; if(headB == NULL){ headB = node; pB = node; } else{ pB->next = node; pB = node; } } //合并链表A和B Node* headC = merge(headA, headB); //输出链表C printList(headC); return 0; } //合并链表函数 Node* merge(Node* headA, Node* headB){ if(headA == NULL){ return headB; } if(headB == NULL){ return headA; } //定义链表C的头节点 Node* headC = NULL; if(headA->data >= headB->data){ headC = headA; headA = headA->next; } else{ headC = headB; headB = headB->next; } //指向链表C的当前节点 Node* pC = headC; while(headA && headB){ if(headA->data >= headB->data){ pC->next = headA; headA = headA->next; } else{ pC->next = headB; headB = headB->next; } pC = pC->next; } if(headA){ pC->next = headA; } else{ pC->next = headB; } //反转链表C Node* pre = NULL; Node* cur = headC; Node* next = NULL; while(cur){ next = cur->next; cur->next = pre; pre = cur; cur = next; } headC = pre; return headC; } //输出链表函数 void printList(Node* head){ while(head){ printf("%d ", head->data); head = head->next; } }

。ChitGPT: 好的，我明白了。您要计算两个稀疏矩阵的乘法。首先，请输入第一个矩阵的行数和列数。 用户: 3 3 ChitGPT: 然后，请输入该矩阵的三元组形式，以0 0 0结束。 用户: 3 3 3 1 2 3 2 3 4 3 4 5 0 0 0 ChitGPT: 接下来，请输入第二个矩阵的行数和列数。 用户: 3 3 ChitGPT: 然后，请输入该矩阵的三元组形式，以0 0 0结束。 用户: 3 3 3 1 0 0 0 1 0 0 0 1 0 0 0 ChitGPT: 好的，现在我们有了两个稀疏矩阵，它们的乘积是： 1 2 3 1 0 0 1 2 3 2 3 4 x 0 1 0 = 2 3 4 3 4 5 0 0 1 3 4 5 请注意，这里假设矩阵的元素类型是实数。如果矩阵包含复数元素，则需要使用复数乘法规则计算乘积。