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The editor of this article introduces in detail "how to realize the C++ Joseph Ring problem". The content is detailed, the steps are clear, and the details are handled properly. I hope that this article "how to realize the C++ Joseph Ring problem" can help you solve your doubts. Let's follow the editor's train of thought slowly and deeply, together to learn new knowledge.

The topics are as follows:

There is a company, this company has a boss and 13 programmers, the boss will organize them to play a game before leaving work every day, the winners of the game can not work overtime, losers need to work overtime for 2 hours. The rules of the game are as follows: a round table has 13 seats, from 1 to 13. Before the start of the game, the boss will say the seat number start and elimination serial number k. Then 13 programmers began to scramble for seats, each containing only one programmer, and each programmer had to choose a seat. Programmers whose seat number is start start at 1 and report the number in turn in the direction shown in the figure. Each time the programmers who reported k were eliminated and left their seats to work overtime, while the others continued to play until the last one left unrestrained.

There is a very smart programmer, every time the boss said start and k moment, can immediately choose a seat and win, so he has never worked overtime, other programmers envy him very much, asked him to win the magic weapon, saw that he slowly opened a .c file named IAMGOD, everyone showed admiring eyes.

Today, you are the smart programmer, please improve the contents of the IAMGOD.c file.

According to the prompts, improve the contents of the IAMGOD.c file in the editor on the right and find the seat number that you can work without overtime.

Input: start k

Output: selected seat number I

Example 1-enter: 2 3

Output: 13

# include#include / / create structure typedef struct Node {int data; struct Node* next;} NODE; / / create new node and insert node void insert (NODE* head) {int I; NODE* tail = head; / / each node is numbered 1,2,3.13 for (I = 2; I data = I / / join the linked list by tail insertion newnode- > next = NULL; tail- > next = newnode; tail = newnode;} / * this statement is used to print the linked list and check whether the linked list is connected correctly NODE* pmove = head While (pmove! = NULL) {printf ("% d->", pmove- > data); pmove = pmove- > next;} * / tail- > next = head; / / connects the tail node to the header node to form a ring} void serch (NODE* head) {int start_data,i,k; NODE* start = head Scanf ("% d% d", & start_data, & k); / / move to the start _ data node and treat this node as node 1 for (I = 2; I next;} NODE* front; / / front represents the previous node while of the k th node (start- > next! = NULL) {int j For (j = 2; j next; / / Mobile node} front- > next = start- > next; / / connect the last node of the k th node to its next node free (start); / / delete the specified node start = front- > next / / update the location of the start, that is, 1 / when the k th is still itself, that is, there is only one node left, jump out of the loop if (start- > data = = (start- > next)-> data) break;} printf ("% d", start- > data) } int main () {/ / create the linked list NODE* head; head = (NODE*) malloc (sizeof (NODE)); head- > data = 1; head- > next = NULL; / / create a new node and connection node insert (head); / / find the k node and delete it. Serch (head); return 0;} read here, this article "how to achieve the C++ Joseph ring problem" article has been introduced, want to master the knowledge of this article also need to practice and use to understand, if you want to know more about the article, welcome to follow the industry information channel.

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