289675283

呢题有D难,不过半个钟来谂,都係谂得到嘅~~7 z" S$ P6 u4 L2 w  }

2 J# C2 X" U2 j5.39.217.77将12个波分成3组: team A, team B, and team C.
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# N; G! O6 ~1 h* ]; V5 \第一次称:
5 X# l* C- i' X, F+ C8 M6 r* `3 B  将team A 和team B 分别放落天平两边, 如果天平平衡(最理想嘅情况,跟住就好简单了), 即係有问题嘅波喺team C度, 而team A同team B都係正常. 咁样第二次就喺team C度选2只波出来,同team A或者team B度嘅任意2只波称,如果天平仲係平衡,咁有问题嘅波就係team C选净嘅2只波度啦,咁第三次就大家都知点称啦...假如第二次唔平衡,即係有问题嘅波係team C选咗嘅呢2只波度,咁跟住第三次就好易啦~~以上都係建立喺第一次平衡嘅条件上,假如第一次天平係唔平衡,咁跟住落来就要有技巧咁称第二次啦...公仔箱論壇% H! s) s2 G, g! C+ M. s

6 u' w, b6 M  z+ p  Ttvb now,tvbnow,bttvb第二次称:
8 `. X) F! _! }% L1 a公仔箱論壇 假设第一次係team A轻过team B(其实边一边重D轻D都唔係问题), 咁第二次就係天平左边摆上AAB,右边亦摆上AAB(呢一部係就关键嘅!!谂到呢部就大功告成了!!) 假如第二次天平嘅左边轻咗,咁有问题嘅波就係左边嘅两个A或者係右边嘅B了(因为假设咗第一次team A轻过team B, 所以造成天平唔平衡嘅原因就係team A有波轻咗或者team B有波重咗),咁样只要第三次将有问题嘅2个A对称一次,咁其中轻咗果个波就係有问题啦!假如2个A对称完仲係平衡,咁有问题嘅波就係右边嘅B啦...!其实问题到呢度已经解决咗了,因为如果第二次称係右边轻咗,用同样嘅方法就可以揾到有问题嘅波...仲有就係假如第二次称两边係一样重,咁有问题嘅波就係team B选净咗果2个波啦,第三次只要将呢两个波对称一次.重咗果个就有问题啦...
& j- L& F# h+ a3 A& B: O9 N5.39.217.77公仔箱論壇: w2 S& q- v. p8 N* u6 U& b# ]
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希望大家睇得明我个答案啦...!唔明再稳我我!!

289675283

答啱咗有冇奖励咖```?

363990627

不用20秒就計到....要30分鐘??還是3分鐘才對呀...

289675283

佩服阁下嘅能耐!小弟甘拜下风~

nightrider

put 4 & 4 is a good start. Your answer is right when the problem ball exists in team A or team B. But when the problem ball exists in team C then your answer is wrong because you end up not knowing whether the problem is heavier or lighter so you still don't know which one is the problem ball.
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0 m# o, }) @% J- H+ kThe correct answer is after the first weight, (4 balls on each side and balance), you know the problem ball is among the remaining 4 balls (team C). Take 3 balls from team C, weight against 3 normal balls. If balance, then the problem is with the remaining 4th team C ball, just weight against any normal ball, then you know it's heavier or lighter. If not balance then note heavier or lighter. Just take 2 from the 3 and weight against each other, if balance it's the third ball and you already know it's heavier or lighter. If not balance then the problem ball is either the heavier ball or the lighter ball base on the previous weight.

jasonlam@work

good~~~

nightrider

Okay let's try again, first divide the 12 balls into 3 groups.....let's name them AAAA, BBBB & CCCC
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First weight....AAAA ^ BBBB, if balance, then problem in C balls, see my post above. If not balance, continue below.
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From the first weight, let's say AAAA is heavier then BBBB, then either the problem ball is heavier among AAAA, or lighter among BBBB, but all C balls are normal.
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2nd weight....AAAB ^ ACCC, there will be 3 scenarios:$ q8 X: T: \4 @2 z) V7 ^
(1) if AAAB is heavier then ACCC, for sure the problem ball is a heavier ball and it's among AAA. Take 2 of these and weight against each other, the heavier ball is the problem, if balance then it's the remaining ball.
. A' K. g5 [) F! B& |) gtvb now,tvbnow,bttvb(2) if ACCC is heavier then AAAB then weight the A ball in ACCC against a normal C ball, if balance then the problem ball is the B ball in AAAB and it's a lighter ball. If heavier then this is the problem ball. (note it cannot be lighter)
5 |1 u6 z  E* C, N9 ~  o(3) if balance then the problem ball is one of the 3 B balls not touched in the 2nd weight and it's a lighter ball. Take 2 of these B balls and weight against each other. If balance then the other B ball is the problem. If not balance then the lighter ball is the problem.

samw23

i wanna know how you do it

nightrider

I wanna tell you I use my brain.

289675283

如果第一次係平衡嘅话,跟住就好简单啦```都唔使讲出来点做啦```

289675283

画公仔唔使画出场...我只係讲最难果种情况,其他嘅情况一下就可以谂到,冇必要再加解释~

peiqun

wht

nightrider

Yeah Yeah! LMAO!

larcpo

answer plz...

fisheatcat

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