学步园

约瑟夫环问题有多种解法，例如穷举法，或者采用链表法。

一、穷举法。设置一个N位的标志位数组，用以标志该位置是否已经出列。每出列一个位置，则将该标志位置位。当所有位置都置位时，则循环结束。

#include <iostream>
#include <bitset>
using namespace std;

#define N 7
#define M 3

int G_josephus_ring[N] = {8, 3, 6, 4, 1, 5, 2};
int G_josephus_output[N] = {0};
bitset<N> G_josephus_flag(0);

int main()
{
	int cur_pos = 0;
	int out_pos = 0;
	int count = 1;
	while(1){
		cur_pos = (cur_pos + 1) % N;
		if(!G_josephus_flag.test(cur_pos)){
			if(++count == M){
				count = 0;
				G_josephus_output[out_pos++] = G_josephus_ring[cur_pos];
				G_josephus_flag.set(cur_pos);
				if(G_josephus_flag.count() == N){
					break;
				}
			}
		}
	}

	for(int i = 0; i < N; ++i){
		cout << G_josephus_output[i] << " ";
	}
	cout << endl;

	return 0;
}

二、链表法。将序列构造成一个循环链表，并依次访问该链表。当某位置需要出列时，将该位置从循环链表里面删除，直到链表里面只有一个元素，则循环结束。

#include <stdio.h>

typedef int ElemType;
typedef struct _node{
	ElemType data;
	struct _node *next;
}Node;

int main()
{
	Node node1, node2, node3, node4, node5, node6, node7;
	Node *preNode = NULL;
	Node *curNode = NULL;
	int count = 1;
	
	node1.data = 8;
	node1.next = &node2;

	node2.data = 3;
	node2.next = &node3;

	node3.data = 6;
	node3.next = &node4;

	node4.data = 4;
	node4.next = &node5;

	node5.data = 1;
	node5.next = &node6;

	node6.data = 5;
	node6.next = &node7;

	node7.data = 2;
	node7.next = &node1;

	preNode = &node1;
	curNode = preNode->next;
	while(curNode->next != curNode){
		if(++count == 3){
			count = 0;
			printf("%d ", curNode->data);
			preNode->next = curNode->next;
		}else{
			preNode = curNode;
		}
		curNode = curNode->next;
	}
	printf("%d ", curNode->data);

	return 0;
}

如果只要求最后的胜利者，则可以采用动态规划法，状态转移方程：

f(1) = 0;
#只有一个人

f(n) = (f(n - 1) + M) % n;

如果采用递归法，则：

int Josephus(int n)
{
	if(1 == n){
		return 0;
	}

	return (Josephus(n - 1) + M) % n;
}

也可不使用递归：

#include <stdio.h>

#define N 7
#define M 3

int G_josephus_ring[N] = {8, 3, 6, 4, 1, 5, 2};

int main()
{
	int i;
	int s = 0;
	for (i = 2; i <= N; ++i)
	{
		s = (s + M) % i;
	}

	printf ("\nThe winner is %d\n", G_josephus_ring[s]);

	return 0;
}