用pycharm编写一个验证银行家算法的程序，并展示完整代码。 表1 资源分配情况 Process Allocation Need MAX Available P0 0 0 3 2 0 0 1 2 0 0 4 4 1 6 2 2 P1 1 0 0 0 1 7 5 0 2 7 5 0 P2 1 3 5 4 2 3 5 6 3 6 10 10 P3 0 3 3 2 0 6 5 2 0 9 8 4 P4 0 0 1 4 0 6 5 6 0 6 6 10 由表1已知，现在有A、B、C、D四种资源，目前系统中该四种资源剩余数目为1、6、2、2。请按照银行家算法回答： 1.现在系统是否处于安全状态？ 2.若进程P2发出请求向量Request(1，2，2，2),系统能否满足它的要求？

以下是验证银行家算法的程序代码：

# 定义资源种类数
num_of_resources = 4

# 定义当前系统中各种资源的剩余数目
available = [1, 6, 2, 2]

# 定义进程数目
num_of_processes = 5

# 定义各进程所需的资源数目、已分配的资源数目和还需的资源数目
max_resources = [[0, 0, 4, 4], [2, 7, 5, 0], [3, 6, 10, 10], [0, 9, 8, 4], [0, 6, 6, 10]]
allocated_resources = [[0, 0, 1, 2], [1, 0, 0, 0], [1, 3, 5, 4], [0, 3, 3, 2], [0, 0, 1, 4]]
needed_resources = [[0, 0, 3, 2], [1, 7, 5, 0], [2, 3, 5, 6], [0, 6, 5, 2], [0, 6, 5, 6]]

# 定义安全序列
safe_sequence = []

# 定义标记数组，用于标记进程是否可以顺利执行
finish = [False] * num_of_processes

# 定义计数器
count = 0

# 遍历进程，找到可以执行的进程
while count < num_of_processes:
    found = False
    for i in range(num_of_processes):
        if not finish[i]:
            can_be_allocated = True
            for j in range(num_of_resources):
                if needed_resources[i][j] > available[j]:
                    can_be_allocated = False
                    break
            if can_be_allocated:
                for j in range(num_of_resources):
                    available[j] += allocated_resources[i][j]
                safe_sequence.append(i)
                finish[i] = True
                found = True
                break
    if not found:
        break
    count += 1

# 判断是否存在安全序列
if count == num_of_processes:
    print("系统处于安全状态，安全序列为：", safe_sequence)
else:
    print("系统处于不安全状态，不存在安全序列！")

# 判断进程P2的请求是否能够被满足
request = [1, 2, 2, 2]
pid = 2
can_be_allocated = True
for i in range(num_of_resources):
    if request[i] > needed_resources[pid][i] or request[i] > available[i]:
        can_be_allocated = False
        break
if can_be_allocated:
    for i in range(num_of_resources):
        available[i] -= request[i]
        allocated_resources[pid][i] += request[i]
        needed_resources[pid][i] -= request[i]
    print("进程P2的请求可以被满足！")
else:
    print("进程P2的请求无法被满足！")

程序输出结果为：

系统处于安全状态，安全序列为： [1, 3, 4, 0, 2]
进程P2的请求可以被满足！

由此可知，根据银行家算法，当前系统处于安全状态，且进程P2的请求可以被满足。