利用循环链表模拟该过程，无任何优化，$O(n*m)$

#include <iostream>
using namespace std;

// 节点定义
struct Node {
    int data;
    Node* next;
    Node* prev;
};

int main() {
    int n,m;
    cin >> n >> m;
    Node* Head = new Node();
    Head->data = 1;
    Node* tmp = Head;
    for(int i = 2;i<=n;i++){
        tmp->next = new Node();
        tmp->next->data = i;
        tmp = tmp->next;
    }
    tmp->next = Head;
    for(int i = 1;i<=n;i++)
    {
        for(int j = 1;j<m;j++)tmp = tmp->next;
        cout << tmp->next->data << " ";
        Node *p = tmp->next;
        tmp->next = p->next;
        delete(p);
    }
}