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的人开始报数，数到m的那个人出列；他的下一个人又从1开始报数，数到m的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。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;}