// ====================方法1====================

int LastRemaining_Solution1(unsigned int n, unsigned int m)
{
    if(n < 1 || m < 1)
        return -1;

    unsigned int i = 0;

    list<int> numbers;
    for(i = 0; i < n; ++ i)
        numbers.push_back(i);

    list<int>::iterator current = numbers.begin();
    while(numbers.size() > 1)
    {
        for(int i = 1; i < m; ++ i)
        {
            current ++;
            if(current == numbers.end())
                current = numbers.begin();
        }

        list<int>::iterator next = ++ current;
        if(next == numbers.end())
            next = numbers.begin();

        -- current;
        numbers.erase(current);
        current = next;
    }

    return *(current);
}

// ====================方法2====================

int LastRemaining_Solution2(unsigned int n, unsigned int m)
{
    if(n < 1 || m < 1)
        return -1;

    int last = 0;
    for (int i = 2; i <= n; i ++) 
        last = (last + m) % i;

    return last;
}