Two implementations of Joseph's ring

Array Implementation and linked list implementation, I do not understand the formula, tragedy!

The linked list method is the simplest. Initialize a circular linked list and naturally form a Joseph ring. You do not need to move any elements. You only need to delete the M elements in sequence until the linked list is empty.

Code

# Include <iostream> # include <algorithm> using namespace STD; typedef struct node {int data; struct node * Next;} node, * List; List createlistwithouthead (int n) // create a circular linked list with no leading nodes {If (n = 0) return NULL; List head = (list) malloc (sizeof (node )); head-> DATA = 1; head-> next = NULL; list p = head; List q = NULL; For (INT I = 2; I <= N; I ++) {q = (list) malloc (sizeof (node); q-> DATA = I; q-> next = NULL; P-> next = Q; P = Q ;} p-> next = head; // the end A pointer points to the first node return head;} void juseph (list head, int N, int m) {int COUNT = 0; list p = head, q = NULL; for (INT I = 0; I <n; I ++) // There are n people in total {COUNT = m; while (-- count) // pick out the M-th individual {q = P; P = p-> next;} cout <p-> data <""; q-> next = p-> next; // Delete the M-th individual P = Q-> next; // count from the next person} void traversewithouthead (list head) // traverse the linked list with no leading node {If (Head = NULL) return; list p = head; do {cout <p-> data <""; P = p-> next;} while (P! = Head); cout <Endl;} int main () {int n = 3; List head = createlistwithouthead (n); traversewithouthead (head); juseph (Head, 3, 5 ); getchar (); Return 0 ;}

If array representation is followed by the previous question, the elements need to be moved every time. Here, we can trick you into applying for a number with a length of N. Each element is initialized as a non-zero element. If this element has been removed, set it to 0, when counting, only one element is counted. The ring is simulated by I = (I + n + 1) % N.

Code

# Include <stdio. h> # include <malloc. h> # include <stdlib. h> int main () {int * P; int M, N; printf ("input m and n \ n"); scanf ("% d", & M, & N); fflush (stdin); P = (int *) malloc (sizeof (INT) * m); For (INT I = 0; I <m; I ++) // create an array with the array element numbered. If a person leaves the queue, the number is 0 p [I] = I + 1; int COUNT = 0; // int sum = 0 when reporting data; // calculate the number of players in the team. int K = 0; // record the subscript of the current element while (sum <m) // loop until m members are both out of the queue {If (P [k] = 0) // if the person in this position has already been out of the queue, skip and do not count, that is, Count does not change {k = (k + m + 1) % m; // skip this position} else // if the person in this position is not in the queue {count ++; // Add 1 to count, if (COUNT = N) // if n is reported, this person will be outputted {printf ("% d ", P [k]); P [k] = 0; // after leaving, set the element in this position to zero sum ++; // Total number of people leaving the team + 1 COUNT = 0; // The report counter starts from scratch again} k = (k + m + 1) % m ;}} system ("pause ");}