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FrontPage › ÀÛÀº³ª¹«/2008-03
Mar 27, 2008
\isqrt
Submitted by ÀÛÀº³ª¹« @ 03-27 [09:40 pm]
\newcount\x \newcount\nx \def\mysqrt#1{\x=#1 \nx=\x \isqrt{#1} \big\lfloor\sqrt{#1}\big\rfloor=\number\x} \def\isqrt#1{\nx=#1 \divide\nx by\x \advance\nx by\x \divide\nx by2 \ifnum\x=\nx \else\x=\nx \isqrt{#1}\fi} $\mysqrt{361}$
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