返回列表 回復 發帖
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
$ w+ X6 P9 h* e5.39.217.76将比较重的那份再平均分成两份成一次
! G$ K$ L. Q. @! H) L公仔箱論壇剩下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
9 e: |8 e9 O5 G5.39.217.76if  (4v4)= same that means the rest of 4 is got more weight5.39.217.760 i' V/ S  Z) }" U0 n
    then goto 2 - compare rest of(4)  - 2 with 2 ' n3 T# h0 H4 t
           if (left side is more weight)
. [$ ?/ ^* o: x7 b! g- c8 _/ F( @- D) `$ u5.39.217.76              then goto 3 - compare last (left side of ( 2) - 1 with 1 => get answer
: C: ?+ ?7 |5 w' ], M9 _7 p5.39.217.76          else ( right side is more weight)5.39.217.76% `4 c  y( o" M# g
              then goto 3 - compare last (right side of (2) - 1 with 1 => get answer  z  Y0 j+ }! G3 L+ {* j( T
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。
返回列表