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back to boardMath explanation We are given positive x and y. Let's go in the first loop. Let's do some method refactoring for better understanding: y0 = x*x+y; x0 = x*x+y0; y1 = sqrt(x0+(y0/labs(y0))*(-labs(y0))); for (j = 1; j <= 2*y1; j++) x0 = x0-y1; x = x0; y = y1; Let's go through the lines: y0 = x*x + y Next: x0 = x*x + y0 = 2*x*x + y As y0 > 0 (x, y are positive) => y0/labs(y0) = 1 So y1 = sqrt(x0+(y0/labs(y0))*(-labs(y0))) = sqrt(x0-labs(y0)) = sqrt(2*x*x + y - (x*x + y)) = sqrt(x*x) = x Next 2 lines equals to this: x0 = x0 - 2*y1*y1 = 2*x*x + y - 2*x*x = y So, x=y and y=x. x and y are swapped. After that you need to count amount of swaps and print appropriate answer Re: Math explanation Posted by D4nick 27 Mar 2020 22:02 Thank you very much. |
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