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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
! m0 b" T7 M6 `+ Q将比较重的那份再平均分成两份成一次
/ j& E* f, D& k5.39.217.77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 4# C, h: I8 [: W: {5 v( S/ w3 ^
if  (4v4)= same that means the rest of 4 is got more weight5.39.217.77. T: m( v+ }) T1 w* e. e2 e4 ]
    then goto 2 - compare rest of(4)  - 2 with 2 : t- R& W+ W- B/ h4 R6 ^
           if (left side is more weight), p. m4 G8 `/ m$ z9 V& z
              then goto 3 - compare last (left side of ( 2) - 1 with 1 => get answertvb now,tvbnow,bttvb' C0 l- w: t. u9 [* V9 [0 E  P$ d
          else ( right side is more weight)
6 u' X- |) X/ q" o/ H" p3 _! }tvb now,tvbnow,bttvb              then goto 3 - compare last (right side of (2) - 1 with 1 => get answer8 R, K+ J/ h6 U. F/ W% n0 _
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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