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I just started taking classes of python and I am trying to make a code in where the value of the first month = 1 and multiply its value by 2 next month, and then by 3 next month, and 2 next month and so on. Until it reaches 6 months. I am using this code but it only gives me Month 1 = 1 which is the initial value.
P = 1
count = 12
print ("month 1: ",P)
for month in range(count-1):
if month %2 == 0:
P = P*2
else:
P = P*3
print:("month", month+2 ,":",P)
Change
print:("month", month+2 ,":",P)
to
print("month", month+2 ,":",P)
I'm not sure why python didn't complain about the colon. You can actually put anything there
weird: ("month", month+2 ,":",P)
And it won't complain. Awesome mistake, thanks!
Remember that range(count-1) will return numbers between 0 and 11 (the last number is non-inclusive)
Your logic could be like this:
(a) A counter that starts at 1 (cnt)
(b) A loop that watches for the counter to reach 6, then exits
(c) A running total (month_val or some such)
cnt = 1
month_val = 1
while cnt < 7:
month_val = month_val * cnt
print(month_val)
cnt += 1
The above assumes that you retain the new value of month_num -- but re-reading your question, you might just want to print the values 1 to 6, in which case the month_num value should just remain 1 all the time:
cnt = 1
month_val = 1
while cnt < 7:
print(month_val * cnt)
cnt += 1
You can define a dictionary for your problem like
month_values={}
for i in range(1,7):
if i == 1:
month_values['Month'+str(i)]=1
elif i%2 == 0:
month_values['Month'+str(i)]=i*2
elif i%2 == 1:
month_values['Month'+str(i)]=i*3
print(month_values)
prints
{'Month1': 1, 'Month2': 4, 'Month3': 9, 'Month4': 8, 'Month5': 15, 'Month6': 12}
Related
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if __name__ == '__main__':
n = int(input())
for i in range (0,n):
result = i**2
print(result)
#input : 5
#output : 0 1 4 9 16
range 0,n = 0,1,2,3,4 and
we gave i**2 but how we got 0,1,4,9,16 as output?
range(start, stop, step)
** = square of Number
start Optional. An integer number specifying at which position to
start. Default is 0
stop Required. An integer number specifying at which position to
stop (not included).
step Optional. An integer number specifying the incrementation. Default is 1
you are passing required parameter as 5 which will not be included in the loop. so as per your calculation it will start from 0
result = 0**2 = 0
result = 1**2 = 1
result = 2**2 = 4
result = 3**2 = 9
result = 4**2 = 16
'i' will not reach 5 because of non-inclusive nature of range() operator.
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In a python code, how to skip a few lines of code?
For example, this is the code:
if value == 10:
<skip_the_next_seven_lines_of_code>
if value == 15:
<skip_the_next_eight_lines_of_code>
#code continues...
Note that this can be done using a lot of workarounds, like putting the code in the else statement, but I want to avoid that as it would require a lot of elses as I have a lot of ifs.
I am asking whether there is a function which is like:
skiplines(7)
Then the next 7 lines are skipped.
Also, this question has been downvoted many times, but I do not understand why. If there is no such way to accomplish this it can be answered in that way...
I don't think you can handle by the number of lines.
A closest I could think of is to define a function of the corresponding code block as below.
value = 10 # or set other value
task = lambda: print("this is original task")
if value == 10:
task = lambda: None
task() # if value==10, this does nothing.
This is the original answer to the original question (see question edits):
if value != 10:
<some_seven_lines>
<the_rest_of_the_code>
The updated question shows a slightly more complicated pattern. Let's say something like this (I have to create an example since OP does not provide specific code/problem that he/she is trying to solve):
a = 0 # some variable that lines of code will modify
if value == 10:
goto L1 # same as "skip next 7 lines"
if value == 15: # line 1 to "skip"
goto L2 # line 2 to "skip"; same as "skip next 8 lines"
a += 1 # line 3 (or 1 from second "if") to "skip"
a += 10 # line 4 (or 2 from second "if") to "skip"
a += 100 # line 5 (or 3 from second "if") to "skip"
a += 1000 # line 6 (or 4 from second "if") to "skip"
a += 10000 # line 7 (or 5 from second "if") to "skip"
L1: a += 100000 # line - (or 6 from second "if") to "skip"
a += 1000000 # line - (or 7 from second "if") to "skip"
a += 10000000 # line - (or 8 from second "if") to "skip"
L2: a += 100000000 # line 11 (or 13 from second "if")
This could be done in the following way:
a = 0
def f1(a):
a += 1
a += 10
a += 100
a += 1000
a += 10000
return a
def f2(a):
a += 100000
a += 1000000
a += 10000000
return a
if value == 10:
a = f2(a)
elif value == 15:
pass
else: # or elif value != 15 and skip the branch above
a = f1(a)
a = f2(a)
a += 100000000
Of course, specific answer would depend on the question with a specific example, which, as of now, is missing.
Also, see https://stackoverflow.com/a/41768438/8033585
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I am making a game where two players take turns adding a number 1 to 10 to the running total. The player who adds to get 100 wins. I stored the inputs of the players in a list, and at the end of the game, I want to display the total after every turn. The list is in the format:
allTurns = [player1move1, player2move1, player1move2, player2move2, ...]
How can I add only the first 2 elements and display them, then add only the first 4 elements and display them, and then the first 6, and so on?
for i in range(0, len(list)):
if i % 2 == 0:
print(sum(list[:i+1]))
Or
sums = []
for i in range(0, len(list)):
if i % 2 == 0:
sums.append(sums[-1] + list[i-1] + list[i])
Try this:
allTurns = [1,2,3,6,12,23]
out = []
for n in range(len(allTurns)):
if n % 2 is 1:
out.append(allTurns[n] + allTurns[n-1])
if len(allTurns) % 2 is 1:
out.append(allTurns[-1])
print(out)
I've coded it so that it could work for any situation, and not just yours.
Try this:
allTurns = [1,2,3,6,12,23,4]
out = []
for n in allTurns:
if allTurns.index(n) % 2 is 0:
i = 0
for a in range(allTurns.index(n)+2):
try:
i=i+allTurns[a]
except IndexError:
pass
out.append(i)
print(out)
This is probably as simple as it can get.
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This is for the Third Problem on Google's Codejam Qualification Round 2020
The Judging System says this solution gives a wrong answer, but I could not figure out why. Any insights would be much appreciated.
num_test_cases = int(input())
def notOverlap(activity, arr):
# returns true if we have no overlapping activity in arr
for act in arr:
if not (act[0] >= activity[1] or act[1] <= activity[0]):
return False
return True
def decide(act, num_act):
C, J = [], []
result = [None]*num_act
for i in range(num_act):
if notOverlap(act[i], C):
C.append(act[i])
result[i] = "C"
elif notOverlap(act[i], J):
J.append(act[i])
result[i] = "J"
else:
return "IMPOSSIBLE"
return "".join(result)
for i in range(num_test_cases):
num_act = int(input())
act = []
for _ in range(num_act):
act.append(list(map(int, input().split())))
print("Case #" + str(i+1) + ": " + decide(act, num_act))
You implemented a brute force way to solve it. Your code runs slow, Time complexity O(N^2), but you can do it in O(N*log(N))
Instead of check with notOverlap(activity, arr), sort the array and check with the last ending time activity of C or J. ( Greedy Way to solve it )
You have to store the index of activity before sorting the array.
Here is my solution, but before reading the solution try to implement it yourself
for testcasei in range(1, 1 + int(input())):
n = int(input())
acts = []
for index in range(n):
start, end = map(int, input().split())
acts.append((start, end, index)) # store also the index
acts.sort(reverse=True) # sort by starting time reversed
# so the first activity go to the last
d = ['']*n # construct the array for the answer
cnow = jnow = 0 # next time C or J are available
impossible = False # not impossible for now
while acts: # while there is an activity
start_time, end_time, index = acts.pop()
if cnow <= start_time: # C is available to do the activity
cnow = end_time
d[index] = 'C'
elif jnow <= start_time:
jnow = end_time
d[index] = 'J'
else: # both are'nt available
impossible = True
break
if impossible:
d = 'IMPOSSIBLE'
else:
d = ''.join(d) # convert the array to string
print("Case #%d: %s" % (testcasei, d))
I hope you find this informative and helped you to understand, and keep the hard work.
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I am given a user entered string 'aaabbbccaa'.
I want to find the duplicates and print the string back as 'a3b3c2a2'
Maybe with this way:
from itertools import groupby
s = "aaabbbccaa"
# group by characters
groups = groupby(s)
# process result
result = "".join([label + str(len(list(group))) for label, group in groups])
print(result)
Output:
a3b3c2a2
def process_string(source):
new = ''
while source:
counter = 0
first_char = source[0]
while source and source[0] == first_char:
counter += 1
source = source[1:]
new += f'{first_char}{counter}'
return new
print(process_string('aaabbbccaa'))
'a3b3c2a2'
this kind of solution could mabye solve it, it does what you specify, however if you are able to put it into your context, no idea :)
hope it helps!
c = 0
foo = "aaabbbccaa"
bar = ""
prev = None
for counter, index in enumerate(foo):
print(c)
if prev == None:
print("first")
elif prev == index:
print("second")
elif prev != index:
c = 0
c += 1
prev = index
try:
if index in foo[counter+1]:
print("third")
else:
print("fourth")
bar += index + str(c)
except:
print("fifth")
bar += index + str(c)
print("foo is {}".format(foo)) # will output aaabbbccaa
print("bar is {}".format(bar)) # will output a3b3c2a2