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see xia da an
want to know
我来看看
check ans..
6 and 6 then 3 and 3 then 1 and 1
5識答?
thank you
将12个乒乓球分成平均两份,称一次
* R( q# }* A4 l  X/ W- |: Ttvb now,tvbnow,bttvb将比较重的那份再平均分成两份成一次
0 ?) j# V" a% {/ R6 u0 n" E9 m6 Vtvb now,tvbnow,bttvb剩下3个随意拿两个来称,如果天平平衡就是第三个,如果不平衡就是重的那个
thank
Thanks!
第一次系一边四个,一边四个秤既
1 - compare 4 with 4tvb now,tvbnow,bttvb+ P! a$ f  c, I$ \
if  (4v4)= same that means the rest of 4 is got more weight
: x$ ^" P9 j: A. x8 R+ ]7 Itvb now,tvbnow,bttvb    then goto 2 - compare rest of(4)  - 2 with 2
% t4 `& E# S" q% S3 W2 `. {4 i- O5.39.217.77           if (left side is more weight)' l, H' v' B( @+ \( _' H
              then goto 3 - compare last (left side of ( 2) - 1 with 1 => get answer5.39.217.77/ a" ~: Q2 [+ X5 a
          else ( right side is more weight)
, w' j% ^7 R) C  b/ Z公仔箱論壇              then goto 3 - compare last (right side of (2) - 1 with 1 => get answer
  S& i8 Q5 w1 {1 {  b公仔箱論壇else ( the left 4) then compare with above  , or right 4 also can compare above method...
想了好久都不行
我的做法与4楼的做法一致,这样应该可以排除出来。但这个题目有个漏洞,只说重量异常,没说是重了,还是轻了。
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