Ending a for loop, Python

Question

I'm writing a program that reads an integer number "n" from the user, and then add the numbers 1^2-2^2+3^2-4^2+...±n^2 together. For example if n = 7, then the program will return 28. I've done it like this:

n = int(input("n = "))
summ = 0

for number in range (1,n +1):
    square = (number**2)*((-1)**(number+1))
    summ += square
    print(square)

print("The loop ran",number,"times, the sum is", summ)

The problem is that I want the program to end before the sum reaches an input "k" from the user.

n = int(input("n = "))
k = int(input("k = "))

summ = 0

for number in range (1,n +1):
    square = (number**2)*((-1)**(number+1))
    summ += square
    print(square)
    if summ > k:
        break


print("The loop ran",number,"times, the sum is", summ)

If k = 6, the program returns "The loop ran 5 times, the sum is 15", but 15 is obviously over 6. The correct answer would be "The loop ran 4 times, the sum is -10". Does anyone know how to fix this? I've also tried putting the if statement right under the "for number" line, but that returns "The loop ran 6 times, the sum is 15".

Answer 1

You are updating the summ value and then checking the condition. You should instead check the total before actually adding the number to the sum.

for number in range(1, n+1):
    square = (number**2)*((-1)**(number+1))
    if summ + square > k:
        break
    summ += square
    ...

# this should work, assuming the rest of your code works.

Answer 2

Just put a condition before adding the square to summ

if summ + square > 7:
    break

n = int(input("n = "))
summ = 0

for number in range (1,n +1):
    square = (number**2)*((-1)**(number+1))
    if summ + square > 7:
        break
    summ += square

    print(square)

print("The loop ran",number,"times, the sum is", summ)