Sum of 1 3 5... n until the sum exceeds 100-CodePudding

Then the sum and the last added number and the number of numbers added must be printed.

I am currently stuck, I managed to get the sum part working. The last added number output is printed "23" but should be "21". And lastly, how can I print the number of numbers added?

Output goal: 121 21 11

Here is my code:

n = int()
sum = 0
k = 1
while sum <= 100:
  if k%2==1:
    sum = sum   k
  k = k   2
print('Sum is:', sum)
print("last number:", k)

CodePudding user response：

You can do the following:

from itertools import count

total = 0
for i, num in enumerate(count(1, step=2)):
    total  = num
    if total > 100:
        break

print('Sum is:', total)
print("last number:", 2*i   1)

To avoid the update on k, you can also use the follwoing idiom

while True:
    total  = k  # do not shadow built-in sum
    if total > 100:
        break

Or in Python >= 3.8:

while (total := total   k) <= 100:
    k  = 2

CodePudding user response：

Instead of

k = k   2

say

if (sum <= 100):
  k = k  2

...because that is, after all, the circumstance under which you want to add 2.

To also count the numbers, have another counter, perhasp howManyNumbers, which starts and 0 and you add 1 every time you add a number.

CodePudding user response：

Change your while loop so that you test and break before the top:

k=1
acc=0
while True:
  if acc k>100:
    break
  else:
    acc =k
    k =2

>>> k
21
>>> acc 
100

And if you want the accumulator to be 121 just add k before you break:

k=1
acc=0
while True:
  if acc k>100:
    acc =k
    break
  else:
    acc =k
    k =2

CodePudding user response：

Based on your code, this would achieve your goal:

n = 0
summed = 0
k = 1

while summed <= 100:
   n  = 1
   summed = summed   k
   if summed <= 100:
       k = k   2

print(f"Sum is: {summed}")
print(f"Last number: {k}") 
print(f"Loop count: {n}")

CodePudding user response：

This will solve your problem without changing your code too much:

n = int()
counter_sum = 0
counter = 0
k = 1
while counter_sum <= 100:  
  k = 2
  counter_sum =counter_sum  k
  counter =1
print('Sum is:', counter_sum)
print("last number:", k) 
print("number of numbers added:", counter)

CodePudding user response：

Note, that

1   3   5   ...   2 * n - 1 == n**2
<-----    n items    ----->

So far so good you to get n all yu have to do is to compute square root:

n = sqrt(sum)

in case of 100 we can find n when sum reach 100 as

n = sqrt(100) == 10

So when n == 10 then sum == 100, when n = 11 (last item is 2 * n - 1 == 2 * 11 - 1 == 21) the sum exceeds 100: it will be

n*n == 11**2 == 121

Code:

def solve(sum):
    n = round(sum ** 0.5 - 0.5)   1;
    
    print ('Number if items: ', n);
    print ('Last item:       ', 2 * n - 1)
    print ('Sum of items:    ', n * n)
    
solve(100)

We have no need in loops here and can have O(1) time and space complexity solution

CodePudding user response：

If you have the curiosity to try a few partial sums, you immediately recognize the sequence of perfect squares. Hence, there are 11 terms and the last number is 21.

print(121, 21, 11)