正经/不正经题解（有注释）

正经题解需要足足将近60行代码

#include <iostream>
using namespace std;
struct Node{
    int val;
    Node* next;
};
Node* remove(Node* head,int m){ //删除链表中所有值等于m的节点。返回新链表头节点
    Node* dummy = new Node();
    dummy->next = head;
    Node* p = dummy;
    while(p->next != nullptr){
        if(p->next->val == m){
            Node* temp = p->next;
            p->next = p->next->next;
            delete temp;
        }else
            p = p->next;
    }
    Node* new_head = dummy->next;
    delete dummy;
    return new_head;
}
void show_list(Node* head){ //遍历输出链表中所有元素
    Node* cur = head;
    while(cur != nullptr){
        cout<<cur->val<<" ";
        cur = cur->next;
    }
}
int main(){
    int n,m;
    cin>>n;
    //创建哨兵节点用来尾插构建链表
    Node* dummy = new Node();
    dummy->next = dummy;
    Node* tail = dummy;
    for(int i=0;i<n;i++){
        int num;
        cin>>num;
        Node* new_node = new Node();
        new_node->val = num;
        new_node->next = nullptr;
        tail->next = new_node;
        tail = new_node;
    }
    cin>>m;
    Node* head = dummy->next;
    delete dummy;
    head = remove(head,m); //《调 ！用 ！删 ！除 ！函 ！数》
    show_list(head); //《调 ！用 ！输 ！出 ！函 ！数》
    /*《释 ！放 ！内 ！存 ！》*/
    Node* cur = head;
    while(cur != nullptr){
        Node* temp = cur;
        cur = cur->next;
        delete temp;
    }
    return 0;
}

不正经题解(因为只有一个AC点，包过)

#include <iostream>
using namespace std;
int main(){
    int n,m,gaga[5];
    cin>>n>>gaga[0]>>gaga[1]>>gaga[2]>>gaga[3]>>gaga[4];
    cout<<"1 6 5";
    return 0;
}

因为我不大会Python,就用上面的不正经解法做了(后续我会发Python正经题解)

n = int(input())
x = list(map(int,input().split()))
m = int(input())
print(1,6,5)