Computer Science Canada HELP ON TURING PERFECT SQUARE 
Author:  dinktsr [ Mon Apr 26, 2004 5:05 pm ] 
Post subject:  HELP ON TURING PERFECT SQUARE 
i struggle so much at this it's not even funny... i need a program that keep inputting integers until a perfect square (ex. 81, 9) between 40 and 100 PLEASE HELP ME, WITHOUT THIS IM DEAD, THANKS IN ADVANCE 
Author:  Tony [ Mon Apr 26, 2004 5:08 pm ]  
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eh? what seems to be the problem?

Author:  Paul [ Mon Apr 26, 2004 5:17 pm ]  
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wha...? here's how I would do it, it might be sucky, but i always like to showoff my newly learned knowledge (strintok)

Author:  Tony [ Mon Apr 26, 2004 5:18 pm ] 
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hah, while we at it, might also use if sqrt(num) mod 1 = 0 to test for a perfect square 
Author:  Cervantes [ Mon Apr 26, 2004 5:29 pm ]  
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jeeze, thats the way i did it. I made the prog but then just as i was about to post my "do your own homework" rant followed by "this is how in english, not in code", I scroll down and there's tony's very blatant answer staring at me. anywho, I don't understand this bit.
if you take the sqrt of a perfect square and you get an integer. round it and you get an integer. square it and you get the first number but take the sqrt of a nonperfect square number and you get a num with decimals, round that and you get an integer, square that and you get the number you started with, yes? but obviously if you input 10 into your prog it doesn't exit, but if you input 9 it does. 
Author:  Tony [ Mon Apr 26, 2004 5:38 pm ] 
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Cervantes  since round drops the decimal point, when you square the number it will not be the same as the one you squared. It is basically a test to see if the squareroot of a number is an integer (perfect square) or not. That's why I later posted with mod 1 method, as non integer root will have a .something remainder. 10 (non perfect square) sqrt = 3.16 round(3.16) = 3 3 * 3 = 9 9 != 10 
Author:  Cervantes [ Mon Apr 26, 2004 5:49 pm ] 
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i have an incredibly strong urge to edit my above post and erase all my moronic ramblings 
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