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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5.39.217.77:8898/ L+ O1 v3 f8 T% g+ \
将比较重的那份再平均分成两份成一次
& U; u+ {- F$ O/ v9 J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 4tvb now,tvbnow,bttvb- |7 l" D: o7 j- F
if  (4v4)= same that means the rest of 4 is got more weight6 A3 ?9 G3 c; d, T: I* b" s
    then goto 2 - compare rest of(4)  - 2 with 2 6 J9 B5 a( X/ n# k* {
           if (left side is more weight)公仔箱論壇- V. q* j+ D9 R- x6 G
              then goto 3 - compare last (left side of ( 2) - 1 with 1 => get answer
$ I) B$ S  c8 e/ g) M8 B) \6 ltvb now,tvbnow,bttvb          else ( right side is more weight)7 a$ H) ?% W# h3 ]
              then goto 3 - compare last (right side of (2) - 1 with 1 => get answertvb now,tvbnow,bttvb' V) f7 v; y9 N% m- U( u
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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