C++ 约瑟夫斯的问题

我一直在做作业的另一部分，我们要做的是给出两个变量m和n。m代表开始时的玩家，n代表玩家数量。我在计算m变量的位置时遇到了一个问题，这样程序就可以从正确的播放器开始，并跳过适当数量的空格

所以如果m是2，有6名球员，它将从球员2开始，传给球员3，消灭球员3，传给球员4，传给球员5，消灭他们，依此类推，直到剩下一名球员。这是我希望得到的结果：

相反，用我的代码，我得到

---

我有了替换int I=1的想法；我描述了约瑟夫的问题，或者提供了一个指向描述/@BenVoigt Oh的链接。我可能应该这么做。抱歉。@iKyriaki:在消除之间，您需要m个增量，第一个增量是it=eraseit；。您必须检查它==在每次增量之后结束，而不仅仅是在m之后。@iKyriaki：您有两个选项：在它周围放置另一个循环++；通过适当的环绕检查重复m-1次，或者使主循环只在第m次通过时移除某人。

Enter M and N: 2 6

Person removed: 3
Person removed: 6
Person removed: 4
Person removed: 2
Person removed: 5

*** AND THE WINNER IS 1!! ***

Person removed: 2
Person removed: 4
Person removed: 6 
Expression: list iterator not incrementable

Player removed: 3
Player removed: 5
Player removed: 2
Player removed: 6
*** AND THE WINNER IS 4!! ***

#include <iostream>
#include <list>

using namespace std;

int main() {
    int m, n;
    cout << "Enter M and N: ";
    cin >> m;
    cin >> n;
    list<int> players;
    for (int i = 1; i <= n; i++) {
        players.push_back(i); // adding players to the list
    }
    list<int>::iterator it = players.begin();
    while (players.size() > 1) {
        it++;
        if (it == players.end()) { // if the iterator reaches the end of the list
            it = players.begin();  // wrap around to beginning
        }
        cout << "Person removed: " << *it << endl;
        it = players.erase(it);
        if (players.size() == 1) {
            cout << "*** AND THE WINNER IS " << players.front() << "!! ***" << endl;
        }
    }

    system("pause");
    return 0;
}