研究生复试含面试环节，某院研究生面试采用多对一的面试方式，对1个学生的面试称为一场面试。安排每场面试需要5位教师，有10间教室可作为面试场，每间教室安排的老师固定不变。每个面试场之间的面试进程相对独立，一位学生面试完毕后，可以马上进行下一位学生的面试。该院可参加面试的教师共有88人，参加面试的学生有102人，每位教师均可担任专业面试教师。 一、根据以往经验，10间面试场每场的面试时间（单位：分钟）已确定，见下表：面试场编号 1 2 3 4 5 6 7 8 9 10面试时间 24.2 29.3 13.0 25.4 23.3 16.8 20.3 23.4 36.3 33.1 请设计每间面试场的面试人数使整个面试能尽早结束。分别使用python和matlab对上述进行解答

Python代码实现：

import numpy as np

# 面试场次
num_rooms = 10

# 面试时间
interview_time = np.array([24.2, 29.3, 13.0, 25.4, 23.3, 16.8, 20.3, 23.4, 36.3, 33.1])

# 面试教师数
num_teachers = 5

# 每个面试场的教师数量
teachers_per_room = np.full((num_rooms, num_teachers), num_teachers)

# 每个面试场的面试时间
time_per_room = np.full(num_rooms, 0)

# 每个面试场的面试人数
num_students_per_room = np.full(num_rooms, 0)

# 每个教师能参加的面试场次数
num_rooms_per_teacher = np.full(88, 0)

# 每个学生的面试状态，0表示未面试，1表示已面试
interview_status = np.full(102, 0)

# 计算每个教师能参加的面试场次数
for i in range(88):
    num_rooms_per_teacher[i] = int(num_rooms / 88)
    if i < num_rooms % 88:
        num_rooms_per_teacher[i] += 1

# 随机分配教师到面试场
teachers_index = np.arange(88)
np.random.shuffle(teachers_index)

for i in range(num_rooms):
    # 随机选择教师
    selected_teachers = np.random.choice(teachers_index, num_teachers, replace=False)
    
    # 更新每个面试场的教师数量和面试时间
    teachers_per_room[i] = selected_teachers
    time_per_room[i] = np.sum(interview_time[i] / selected_teachers)
    
    # 更新每个教师的面试场次数
    for j in selected_teachers:
        num_rooms_per_teacher[j] -= 1
    
    # 分配学生到面试场
    for j in range(102):
        if interview_status[j] == 0:
            # 找到一个教师没有超过参加面试场次数限制的面试场
            for k in range(num_teachers):
                if num_rooms_per_teacher[selected_teachers[k]] > 0:
                    interview_status[j] = 1
                    num_rooms_per_teacher[selected_teachers[k]] -= 1
                    num_students_per_room[i] += 1
                    break

# 输出每个面试场的面试人数和面试时间
print("每个面试场的面试人数：", num_students_per_room)
print("每个面试场的面试时间：", time_per_room)

Matlab代码实现：

% 面试场次
num_rooms = 10;

% 面试时间
interview_time = [24.2, 29.3, 13.0, 25.4, 23.3, 16.8, 20.3, 23.4, 36.3, 33.1];

% 面试教师数
num_teachers = 5;

% 每个面试场的教师数量
teachers_per_room = repmat(num_teachers, num_rooms, num_teachers);

% 每个面试场的面试时间
time_per_room = zeros(num_rooms, 1);

% 每个面试场的面试人数
num_students_per_room = zeros(num_rooms, 1);

% 每个教师能参加的面试场次数
num_rooms_per_teacher = zeros(88, 1);
num_rooms_per_teacher(:) = floor(num_rooms / 88);
num_rooms_per_teacher(1:mod(num_rooms, 88)) = num_rooms_per_teacher(1:mod(num_rooms, 88)) + 1;

% 每个学生的面试状态，0表示未面试，1表示已面试
interview_status = zeros(102, 1);

% 计算每个教师能参加的面试场次数
for i = 1:88
    num_rooms_per_teacher(i) = floor(num_rooms / 88);
    if i <= mod(num_rooms, 88)
        num_rooms_per_teacher(i) = num_rooms_per_teacher(i) + 1;
    end
end

% 随机分配教师到面试场
teachers_index = randperm(88);

for i = 1:num_rooms
    % 随机选择教师
    selected_teachers = teachers_index((i - 1) * num_teachers + 1:i * num_teachers);
    
    % 更新每个面试场的教师数量和面试时间
    teachers_per_room(i, :) = selected_teachers;
    time_per_room(i) = sum(interview_time(i) ./ selected_teachers);
    
    % 更新每个教师的面试场次数
    for j = 1:num_teachers
        num_rooms_per_teacher(selected_teachers(j)) = num_rooms_per_teacher(selected_teachers(j)) - 1;
    end
    
    % 分配学生到面试场
    for j = 1:102
        if interview_status(j) == 0
            % 找到一个教师没有超过参加面试场次数限制的面试场
            for k = 1:num_teachers
                if num_rooms_per_teacher(selected_teachers(k)) > 0
                    interview_status(j) = 1;
                    num_rooms_per_teacher(selected_teachers(k)) = num_rooms_per_teacher(selected_teachers(k)) - 1;
                    num_students_per_room(i) = num_students_per_room(i) + 1;
                    break;
                end
            end
        end
    end
end

% 输出每个面试场的面试人数和面试时间
disp("每个面试场的面试人数：");
disp(num_students_per_room');
disp("每个面试场的面试时间：");
disp(time_per_room');