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To print first n even numbers in reverse order using while loop
#4 print first n even numbers in reverse order n= int(input()) num= 0 x= 0 listt= [] while n>x: if num%2==0: listt.append(str(num)) x+=1 num+=1 listt= listt.reverse() #for loop is working fine but using while loop it is creating a problem #sample input- 5 #desired output- 8,6,4,2,0
11 Answers
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n= int(input())
num= 2 #debug
listt= []
while n>0:
listt.append(num)
n-=1
num+=2
listt.reverse()
print(listt)
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This just indicates that the `reverse` method does not return anything. It only modifies the existing list. It does not return an updated list.
Don't bother doing listt = listt.reverse(). If you just want to reverse a list, do listt.reverse() to modify your listt.
listt.reverse()
print(listt)
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Now solve this problem without the extra variable num ☺️
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Thanks to all for your help 🙏🙏
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Brian, great example with bitwise operation 👍, but it's easy ☺️.
Try all the same to print the n-th number of even numbers.
Example:
input 5
output 10,8,6,4,2
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Shaurya Agarwal the description says to print them in reverse order, but the #desired output shows them in forward order. Which is correct?
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n = int(input())
# If n is odd, force n to next lower even.
if n&1: n -= 1
#now only even numbers will print
while n>0:
print(n, end=',')
n -= 2
print(n) #final 0 without comma
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Solo oops, I missed that detail.
Okay, it is easily corrected with the same loop, only a different adjustment to n.
#print the first n even Natural numbers in reverse order
n = int(input())
n = 2*(n - 1)
while n>0:
print(n, end=',')
n -= 2
print(n) #final 0 without comma
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Solo thanks bro!
I know there is debate about whether 0 is even, so I included it only because the OP did.
There is also debate about whether 0 belongs in the Natural number set. I decided to adopt it in that set in my comment because it is the only difference that would distinguish Natural numbers from Counting numbers. On the other hand, 0 is most unnatural!
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Brian corrected
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Brian, yes, you did it. 👏👏👏
Except 0 is not an even number ☺️.