我一直在使用Josephus problem(作业的另一部分),我们应该做的是提供两个变量m
和n
。 m
代表它开始的玩家,n
代表玩家的数量。我在确定放置m
变量的位置时遇到问题,以便程序在正确的播放器上启动并跳过适当数量的空格。
因此,如果m
为2并且有6名玩家,它将从玩家2开始,传递给玩家3,消除玩家3并转到玩家4,传递给玩家5并消除它们,等等,直到一名球员离开。这是我希望得到的输出:
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!! ***
相反,使用我的代码,我得到了
Person removed: 2 Person removed: 4 Person removed: 6 Expression: list iterator not incrementable
我想到用int i = 1; i <= n; i++
替换int i = m; i <= n; i++
,以正确的数字开始i
,但这给了我:
Player removed: 3 Player removed: 5 Player removed: 2 Player removed: 6 *** AND THE WINNER IS 4!! ***
我也知道为什么它只增加了五名球员,但我不确定m
可以去哪里,以便增加正确数量的球员。任何想法/提示将不胜感激。
#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;
}