用vector单向链表解决约瑟夫问题

约瑟夫问题：给定n个人，每个人按顺时针顺序从1开始报数，报到m数的人出局，出局人的下一个人从1开始接着报数，依此循环，最后仅剩的一个人即为最幸运的人。

解决约瑟夫问题，即对给定的n和m，找出最幸运的人。

程序代码：

//约瑟夫问题

#include <iostream> #include <vector> using namespace std; void select(vector<int> v, int n, int m) { int c = -1; while (n >= 2) { for (int i = 0; i < m; i++) { c++; if (c == n - 1 && c != i) { c = -1; } if (c == n) { c = 0; } } vector<int>::iterator it = v.begin(); for (int j = 0; j < c; j++, it++){} cout<<"Delete "<<v[c]<<endl; *v.erase(it); c--; n--; } cout<<"The lucky one: "<<v[0]<<endl; } void main(int argc, char *args[]) { vector<int> v; int i, n, m; cin>>n>>m; for (i = 1; i <= n; i++) { v.push_back(i); } select(v, n, m); }

注：解决约瑟夫问题用循环链表会更加简单！

测试数据1：

8 3

测试结果1：

Delete 3

Delete 6

Delete 1

Delete 5

Delete 2

Delete 8

Delete 4

The lucky one: 7

Press any key to continue

测试数据2：

19 3

测试结果2：

Delete 3

Delete 6

Delete 9

Delete 12

Delete 15

Delete 18

Delete 2

Delete 7

Delete 11

Delete 16

Delete 1

Delete 8

Delete 14

Delete 4

Delete 13

Delete 5

Delete 19

Delete 10

The lucky one: 17