【问题描述】
约瑟夫环问题(Josephus)
用户输入M,N值,从1至N开始顺序循环数数,每数到M输出该数值,直至全部输出。写出C程序。(约瑟夫环问题 Josephus)
【解题思路】
构建一个循环链表,每个结点的编号为1,2,......,n。每次从当前位置向前移动m-1步,然后删除这个结点。
【C程序代码】
#include <stdio.h> #include <stdlib.h> typedef struct node { int num; struct node *next; }node; node *create_list(int n) { int i; node *head; node *p_cur, *p_pre; head = (node *)malloc(sizeof(node)); head->num = 1; head->next = NULL; p_pre = head; for(i = 2; i <= n; i++) { p_cur = (node *)malloc(sizeof(node)); p_cur->num = i; p_pre->next = p_cur; p_pre = p_cur; } p_cur->next = head; return head; } void print_list(node* head) { node *p; printf("%d ", head->num); p = head->next; while(p != head) { printf("%d ", p->num); p = p->next; } } void josephus(node *head, int m) { node *p_cur, *p_pre; int count; count = 1; p_cur = p_pre = head; while(1) { if(count == m) { p_pre->next = p_cur->next; printf("%d ", p_cur->num); free(p_cur); p_cur = p_pre->next; count = 1; } p_pre = p_cur; p_cur = p_cur->next; if(p_pre == p_cur) { printf("%d ", p_pre->num); free(p_pre); break; } count++; } } int main(void) { node *head; head = create_list(5); print_list(head); printf("\n"); josephus(head, 2); printf("\n"); return 0; }
约瑟夫环问题(Josephus),布布扣,bubuko.com
原文地址:http://blog.csdn.net/jjjcainiao/article/details/25106953