用单循环链表实现约瑟夫问题。
约瑟夫问题是个有名的问题：N个人围成一圈，从第一个开始报数，第M个将被杀掉，最后剩下一个，其余人都将被杀掉。例如N=6，M=5，被杀掉的人的序号为5，4，6，2，3。最后剩下1号。
#include <stdio.h>#include <stdlib.h>#include "declear.h"typedef struct Node{ElemType data;struct Node *next;}Node, *CycLinkList;void JOSEPHUS(int totalNum, int startNum, int interval){int index;CycLinkList node;CycLinkList temPtr;CycLinkList temPriorPtr;/*建立一个无头结点的循环链表*/CycLinkList L = (CycLinkList)malloc(sizeof(Node));L->data = 1;L->next = L;temPtr = L;for (index = 2; index <= totalNum; index++){node = (CycLinkList)malloc(sizeof(Node));node->data = index;temPtr->next = node;node->next = L;temPtr = node;}temPtr = L;/*找到起始位置*/for (index = 1; index < startNum; index++){temPriorPtr = temPtr;temPtr = temPtr->next;}while (temPtr != temPtr->next){/*从起始位置开始找到报数为interval的位置*/for (index = 1; index < interval; index++){temPriorPtr = temPtr;temPtr = temPtr->next;}printf("%d\n",temPtr->data);/*从循环链表中删除该点*/temPriorPtr->next = temPtr->next;free(temPtr);/*更新起始位置*/temPtr = temPriorPtr->next;}}int main(){JOSEPHUS(9,2,4);return 0;}