priority = [ 2, 6, 1, 7, 10, 8, 5]
time = [30,200,20,70,120,60, 150]
# define number of tasks: don't hardwire this but rather use the len() command
n_tasks = len(priority)
max_time = 8*60
cur_time = 0
print ( f"TASK PRIORITY TIME TO COMPLETE")
for task in range (n_tasks):
# you need to add an appropriate for loop here and modify the next print statement so that you can print out all tasks
print (f" {task+1} {priority[task]} {time[task]}")
test_list = [ 4, 5, 3, 7, 11, 2]
max_value = max(test_list) # use function max() to find maximum value in list
max_index = test_list.index (max_value) # use method list.index () to find the index where this max occurs
print (f'the maximum output is {max_value} and the index where it occurs is {max_index}') # print out results in formatted statement
test_list = [ 4, 5, 3, 7, 11, 2]
time_list = [30, 45, 20, 120, 90, 70 ]
max_value = max(test_list)
max_index = test_list.index (max_value)
task_time = 90 # input the time it takes to complete this task
print (f'maximum output is {max_value}, the index where it occurs is {max_index} and the time it takes to complete this task is {task_time} minutes') #print out values as in Step #2 plus the time it takes to complete task
max_time = 6 * 60 # this is not the maximum time available for your example but rather for testing
time_used = 3 * 60
#
# Following 6 statements are from Step 3
test_list = [ 4, 5, 3, 7, 11, 2]
time_list = [30, 45, 20, 120, 90, 70 ]
max_value = max(test_list)
max_index = test_list.index (max_value)
task_time = 90
print (f'maximum output is {max_value}, the index where it occurs is {max_index} and the time it takes to complete this task is {task_time} minutes')
#
# Add conditional to see if this task can be completed in the allotted time
if ( time_used + task_time <= max_time) :
time_used = time_used + task_time
print (f'the task added is {test_list [1]}, time used is {time_used}')
# Input problem specific data and print it out from Step 1
priority = [ 2, 6, 1, 7, 10, 8, 5]
time = [30,200,20,70,120,60, 150]
n_tasks = len(priority)
max_time = 8*60
print ( f"TASK PRIORITY TIME TO COMPLETE")
#
# Initialize any variables before loop
cur_time = 0
cur_priority= 0
#
# add for loop to go through all tasks
#
for task in range (0,n_tasks) :
max_value = max(priority)
max_index = priority.index (max_value)
task_time = time[max_index]
if (cur_time + task_time <= max_time):
print (f'Add task with task {task+1} with time of {task_time}')
cur_time = cur_time + task_time
cur_priority = cur_priority + max_value
print (f'current time used is {cur_time} and current priority value is {cur_priority}')
#
# find highest priority left and task number it corresponds to from Step 3/4
#
# zero out this entry in the priority list so you won't find it again
#
# add conditional to check to see if task can be completed as in Step 4
#
# Print out the total time used for all tasks in day 1
# Input problem specific data and print it out from Step 1
priority = [ 2, 6, 1, 7, 10, 8, 5]
time = [30,200,20,70,120,60, 150]
n_tasks =
max_time =
print ( f"TASK PRIORITY TIME TO COMPLETE")
#
# Initialize any variables before loop
#
# add for loop to go through all tasks
#
for :
#
# find highest priority left and task number it corresponds to from Step 3/4
#
# zero out this entry in the priority list so you won't find it again
#
# add conditional to check to see if task can be completed as in Step 4
# if this conditional is satisfied then add another conditional to check if time_used = max_time; if so break out of loop
#
# Print out the total time used for all tasks in day 1
# Input problem specific data
old_priority = [ 2, 6, 1, 7, 10, 8, 5]
time = [30,200,20,70,120,60, 150]
# n_tasks =
#
# Create new priority list containing ratios
#
priority = [] # initialize the list which contains the new priorities which are original priority divided by
# time it takes to complete this task
# add for loop to create this list
# use method .append() to create elements of the new priority list
#
# Remainder of code should be the same; this is because we called the new list which contained the ratios the same as
# before.