Problem C: realize Joseph Ring with linked list

Problem C: Using linked list to realize Joseph Ring
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 1948 Solved: 1009
Description
Have you heard of the Joseph problem ？ The question is as follows ： First n A circle of individuals , Marked as 1 To n Number . next , from 1 We'll start counting on the （ from 1 Start ）, then 2 Number one , then 3 Number ... When someone reports in m when , This man is going to play , Then start with the next person to be kicked out , Recount （ from 1 Start ）. Go through like this n-1 Next time , There's only one person left , Ask the number of the last person left ？

Input
The first 1 Behavior T, Express T Group data ;

The first 2 Go to the first place T+1 Start , Enter... On each line n and m,n How many people are there ,m For each number reported above m One person will be kicked out at a time

1=<n<=100, 1=<m<=100

Output
a number , Which number is left in the end

Sample Input
2
5 3
6 4
Sample Output
4
5

#include <stdio.h>
#include <malloc.h>
typedef struct Y{
    
	int num;
	Y *next;
}Y;
int main()
{
    
	int T;
	scanf("%d",&T);
	while (T--)
	{
    
		int n,m;
		scanf("%d%d",&n,&m);
		int i,j;
		Y* head = (Y*)malloc(sizeof (Y));// Head node  
		head->num = 1;
		Y* tail = head;// The head pointer and tail pointer defined at the beginning point to the head node  
		for (i = 2;i <= n;i++)
		{
    
			Y* p = (Y*)malloc(sizeof (Y));// Use tail insertion , So you need an excessive pointer  
			tail->next = p;
			p->next = head;// The necessary steps to form a ring  
			p->num = i;// Remember to initialize the number  
			tail = p;// The tail pointer is assigned to the newly added node  
		}
		int sum ;//sum Used to count  
		if (n>1)// Particular attention , Want to consider n by 1 The case when  
		{
    
				while (head != head->next)
			{
    
				for (sum = 1;sum != m;sum++)// Cycle count  
				{
    
					tail = head;
					head = head->next;
				}
				// When the conditions are met, delete a node  
				Y* q = head;
				head = head->next;
				tail->next = head;// Steps necessary to delete a node  
				free(q);
			}
		}
	
		printf("%d\n",head->num);
	}
	return 0;
}

Sort out the problem making process ：

1. Creating a ring is difficult , Finally, the condition for the success of the ring is ：
   tail->next = head;
   Keep the head and tail pointers
2. Delete node ：tail To point to the previous node of the deleted node ,head Before pointing to the next node after the deleted node, define a pointer to save the deleted node
   Key steps ：
   p = head;
   head = p->next;
   tail->next = head;
   free§;
3. I should pay attention to n by 1 The case when