Find square difference: Difference between revisions
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=={{header|Python}}== |
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<lang python> |
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import math |
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print("working...") |
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limit1 = 6000 |
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limit2 = 1000 |
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oldSquare = 0 |
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newSquare = 0 |
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for n in range(limit1): |
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newSquare = n*n |
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if (newSquare - oldSquare > limit2): |
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print("Least number is = ") |
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#print(int(math.sqrt(oldSquare))) |
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print(int(math.sqrt(newSquare))) |
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break |
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oldSquare = n*n |
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print("done...") |
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print() |
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</lang> |
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{{out}} |
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<pre> |
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working... |
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Least number is = |
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501 |
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done... |
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</pre> |
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=={{header|Ring}}== |
=={{header|Ring}}== |
Revision as of 16:14, 18 November 2021
Find square difference is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
- Task
Find and show on this page the least integer number n, where diffrence of n*n and (n-1)*(n-1) greater than 1000
The result is 5001 because 501*501 - 500*500 = 251001 - 250000 = 1001 > 1000
Python
<lang python> import math print("working...") limit1 = 6000 limit2 = 1000 oldSquare = 0 newSquare = 0
for n in range(limit1):
newSquare = n*n if (newSquare - oldSquare > limit2): print("Least number is = ") #print(int(math.sqrt(oldSquare))) print(int(math.sqrt(newSquare))) break oldSquare = n*n
print("done...") print() </lang>
- Output:
working... Least number is = 501 done...
Ring
<lang ring> load "stdlib.ring" see "working..." + nl limit1 = 6000 limit2 = 1000 oldPrime = 0 newPrime = 0
for n = 1 to limit1
newPrime = n*n if newPrime - oldPrime > limit2 see "Latest number is = " + sqrt(newPrime) + nl exit ok oldPrime = n*n
next
see "done..." + nl </lang>
- Output:
working... Latest number is = 501 done...