使用双向循环链表解决约瑟夫问题

#include<iostream>
using namespace std;

class monkey {
public:
int No;
monkey *next, * pre;

monkey( int info,  monkey* preValue = 0,  monkey *nextValue = 0) {
No = info;
pre = preValue;
next = nextValue;
}
monkey() {};

};

class MyList{
//private:
public:
monkey *head;
monkey *setPos(const int i);

MyList(monkey* h) {
//monkey tmp(0);
head = h;
head->No = 0;
head->next = 0;
head->pre = 0;
}
//~MyList();
//bool isEmpty();
//void clear();
//int length();
bool append(int value, monkey* tmp);
bool remove(monkey* tmp);
int getValue(const int i);
};
//find the ith componet of the myList;
monkey* MyList::setPos(const int i) {
int myCount = 0;
if (i == -1) return head;
monkey* p = head->next;
while (p != 0 && myCount < i) {
p = p->next;
myCount++;
}
return p;
}
bool MyList::append(int value, monkey* tmp) {
tmp = new monkey;
tmp->No = value;
//if this is a empty list, then add value after head;
if (head->next == 0) {

tmp->next = head;
tmp->pre = head;
head->next = tmp;
head->pre = tmp;
//cout << tmp->pre->pre->No << endl;
//cout << "tmp=" << tmp << "  head=" << head << endl;
}
else {
//if this list isn't empty, then add value between head and head->next;
tmp->pre = head->pre;

tmp->next = head;
head->pre->next = tmp;
head->pre = tmp;
//cout << head->next->No << endl;
//cout << "tmp=" << tmp << "  head=" << head << endl;
}

return 1;
}

int MyList::getValue(int i) {
return setPos(i)->No;
}
bool MyList::remove(monkey* tmp) {
tmp->next->pre = tmp->pre;
tmp->pre->next = tmp->next;
tmp->next = 0;
tmp->pre = 0;
free(tmp);
return true;
}

int main() {
int n = 0, m = 0;
cin >> n >> m;
monkey mhead(0);
MyList MonkeyList(&mhead);
monkey tmp(0);
for (int i = 1; i < n+1; i++) {

MonkeyList.append(i,&tmp);
}

int myCount = 0;
monkey* curPoint = MonkeyList.setPos(0);

while (1) {
//if curPoint's next point is head, then there is only one point left, done
if (curPoint->next->No == 0 && curPoint->pre->No==0) {
cout << curPoint->No << endl;
break;
}
//if the curPoint is the head, then skip to next point;
if (curPoint->No == 0) {
curPoint = curPoint->next;
continue;
}
myCount++;
if (myCount == m) {
int curNo = curPoint->No;
monkey* nextPoint = curPoint->next;
MonkeyList.remove(curPoint);
curPoint = nextPoint;
myCount = 0;
}
else {
curPoint = curPoint->next;
}
}
return 0;

}