Zstuacm summer camp flag bearer

Title Description
Annual Jiangsu “ Information and the future ” The summer camp for primary school students started again . Same as every year , The organizer also designed and arranged many interesting activities , The first one is still to choose the flag bearer of this summer camp ！ Because this is a very honorary role , So there are still a lot of campers who sign up for the flag bearer competition in the summer camp , The operating Committee then stipulated ： take N People in a row , Number 1～N. From 1 People start 1～M Forward counting , To report for duty M The people of , Continue from the next person 1 To M Number off 、 List out .（ Be careful ： When you count off in a certain direction and report to the tail , Continue counting in the opposite direction ）. Go on like this , Until there is one left , This man is the flag bearer of this summer camp . Xiao Ming hopes very much to become a flag bearer , Can you compile a program to help him realize his wish ？ If you can , Your program should output Xiao Ming's number .
Input
Include two integers N,M （2≤N,M≤300,N≥ M ）, Separate... With a space
Output
Include an integer , Indicates Xiao Ming's number in the queue .
The sample input Copy
【 sample input 1】
9 3
【 sample input 2】
8 3
Sample output Copy
【 sample output 1】
8
【 sample output 2】
8
Tips
This problem can be achieved by using a two-way linked list

I prefer to use stl To complete the linked list , After all, it's convenient , The only thing to pay attention to is not to use cin and cout
Use faster scanf and printf

The point to pay attention to is the position to the beginning and end .
Be careful :!list.end(), What is returned is a garbled code （ Because the default value of linked list creation is an empty value ）
but list.begin() What is returned is the first value entered

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<list>
using namespace std;

list<int>stu;
int main()
 {
    
	int n_total = 0;
	scanf("%d", &n_total);
	int m = 0;
	scanf("%d", &m);
	
	for (int i = 0; i < n_total; i++)
	{
    
		stu.push_back(i + 1);
	}
	int j = 1;
	bool flag = true;
	
	
	list<int>::iterator it=stu.begin();
	list<int>::iterator it1 = it;
		
	while (stu.size() != 1)
		{
    
			if (flag == true)
			{
    
				if (j != m)
				{
    
					it++;
					if (it == stu.end())
					{
    
						it--;// Go back to the last element 
						flag = false;
						continue;// At this point, you need to traverse from back to front 
					}
					else
					{
    
						j++;
					}
				}

				if (j == m)
				{
    
					j = 1;
					it1 = it;
					it++;// Take advantage of the situation and walk back one 
					stu.erase(it1);// Delete the data in this location 
				}

				if (it == stu.end())
				{
    
					it--;// The last element 
					flag = false;
					continue;
				}
			}
			else
			{
    
				if (j!=m)
				{
    
					j = j + 1;
					it--;
				}
				if (j == m)
				{
    
					j = 1;
					it1 = it;
					
					if (it == stu.begin())// If it happens to be the beginning, count off 
					{
    
						it++;
						flag = true;
						stu.erase(it1);// Delete the beginning 
						continue;
					}
					
					it--;
					stu.erase(it1);// Delete it1

				}
				
				if (it == stu.begin())
				{
    
					flag = true;
					continue;
				}
			}
		}
		
		printf("%d\n", *it);
		return 0;
}```