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回復 #1 MagicAndy 的帖子

i wan to know
thx.~
thank you for sharing
look at it
i noe de ans cauz 1 do b4 already ^^
  H# l4 x3 ]/ ]+ C1st u divide the ball into two group (i group 6 ball)
$ X( c4 K# X0 t' C公仔箱論壇then 称 them c which side is the heavy
5 d* c9 B2 ~6 p; S( D公仔箱論壇then use the heavy side divide into 3 group 1group contain 2 ball 称 the 1st n 2nd group if its balance use the 3th group
: ^1 ?7 ]" j" E0 x2 H8 P# ?, a7 Iif not balance take the heavy and then tvb now,tvbnow,bttvb, [- E0 s( ~: _1 d  @8 \
take the heavy group divided into 2 group i group contain 1 ball than 称 it 公仔箱論壇, G- o% ]  {; ~- m
than the heavy ball is the ball u wan

回復 #1 MagicAndy 的帖子

how??????????
Oh..i did the same way as the first answer~am i wrong??Let me check~:onion18:
分三次算的话,每次分四粒来算,意思说当有一边较重时,只需要随意将其中六粒拿出来就知道那一粒是异常的了:onion14: 4 G( K* I9 L  Z1 V
(不知各位知不知道小弟说的解答,因为小弟不擅长解说)
thx 4 sharing
顶......

回復 #1 MagicAndy 的帖子

顶.......
简单到。。分3份。。每份4个。。第一次称。。必然知道3份中哪份包括异常的。。第二次..把包括的那份2:2的称。。又知道异常的一边; G3 i6 g2 M4 [, C# k6 m5 w# I+ W
然后。。1:1。。就知道了··
看解答!!!!!!!!!!!!!!
cccccc
idk..show me
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