YLL討論網

#ed_op#DIV#ed_cl#你寫的我有點看不太懂 #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#首先,有兩節課不是迪姆教授的課#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#在這五節課內 C共上了四節 而迪姆的課只上了兩節#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以這兩節C一定有上.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#以相同方法得知A一定沒上這兩節#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#而B一定上了其中一節．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以現在出現了兩種組合[C]還有[BC]#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#再來看迪姆的課 #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#三人都上了兩節#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#先列出可能的出席狀況:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#一 [ABC][ABC][]...........................................出現重複的組合.....不合 #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#二 [ABC][XX][X]   (X表示三者其中之一)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#三 [AB][BC][AC]................................................[BC]為重複之組合....不合#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#從以上可以得知 必定是第二種情況#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#再來把所有的可能再次列出#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#一 [A][BC][ABC]....................................[BC]重複....不合#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#二 [AC][ABC]#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#三  [C][AB][ABC]....................................[C]重複.......不合#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#從以上可以輕易看出 [AC]為兩人出席的那堂課#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以犯人是B#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#我的表達怎麼樣`:)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl#