约瑟夫环问题

约瑟夫环是一个数学的应用问题：已知n个人（以编号1，2，3...n分别表示）围坐在一张圆桌周围。

从编号为k的人開始报数，数到m的那个人出列；他的下一个人又从1開始报数。数到m的那个人又出列；依此规律反复下去。直到圆桌周围的人所有出列。求剩下最后的一个人的编号。

#include "stdafx.h"
#include<vector>
#include<iostream>
using namespace std;

template <class InputIterator, class Distance>
void advance(vector<int>& i, int n);

vector<int>::iterator do_once(vector<int> &cycle, vector<int>::iterator it, int m)
{
	int k = cycle.end() - it-1;
	if (k >= m)
	{
		it = cycle.erase(it + m);
		if (it == cycle.end())
			it = cycle.begin();
		return it;
	}
	else
	{
		int h = m%cycle.size() - k;
		if (h>0)
		{
			it = cycle.erase(cycle.begin() + m%cycle.size() - k - 1);
			if (it == cycle.end())
				it = cycle.begin();
			return it;
		}
		else
		{
			it = cycle.erase(it + m%cycle.size());
			if (it == cycle.end())
				it = cycle.begin();
			return it;
		}
	}
}

int Joseph_problem(int n, int k, int m)
{
	_ASSERTE(k <= n);
	vector<int>cycle;
	vector<int>::iterator it;
	for (int i = 0; i < n; i++)
	{
		cycle.push_back(i);
	}
	it = cycle.begin();
	advance(it, k-1);
	while (cycle.size() != 1)
	{
		it = do_once(cycle, it, m - 1);
	}
	return *it+1;
}


int _tmain(int argc, _TCHAR* argv[])
{
	/*vector<int>cycle;
	vector<int>::iterator it;
	for (int i = 0; i < 4; i++)
	{
		cycle.push_back(i);
	}
	it = cycle.begin();
	advance(it, 3);
	cout << *(it) << endl;
	
	cout << *(cycle.erase(it))<< endl;*/
	
	cout << 4 % 4 << endl;
	cout << Joseph_problem(4, 2, 5) << endl;
	
	system("pause");
	return 0;
}