推荐回答(5个)
不对上面还是需要14场来完成这样的情况,因为要到达5人,满足10场,其中4人都各自需要离开一场,所以最少需要14场,即使推前也满足不了,也就是你抓14场可以满足5人都第10个第1场12345 第2场12345 第3场12345 第4场12345 第5场12345 第6场2345 第7场1345 第8场1245 第9场1234 第10场12345 1队长第11场 23451 第12场34512 第13场45123 第14场54123 前头那个数字代表队长的人
你开5个号 拍好顺序 你1号 后面4个号一次是2号 3号 4号 5号 领第一个鬼 5号离队 领第二个鬼 5号和4号离队 领第三个鬼 5号和3号离队 第四个鬼5号和2号离队 第5个鬼3号和4号离队 第6个鬼4号离队
这是不可能的,最多就是能让5个人都都在10个鬼时不重叠,先5人去领任务,过后一人退队,4人杀,这样来5次,接下来就5人杀,保证每个人都带一次对,这样10个鬼下来就得5个东西!
第一个抓鬼4次第二个抓鬼3次以此类推 开始麻烦 在组队抓鬼 接下来的就一直 有顺序了
我帮你计算下好了
这里五个人名字按1到5 名字是1的
在第5个的时候
离开队伍名字是2的
第6个离开名字是3的
第7个离开名字是4的,第8个离开名字是5个,第9个离开
也就是第1个鬼的人员是1.2.3.4.5第2个鬼是
1.2
.3
.4
.5第3个鬼1.2.3.4.5第4个鬼
1.2.
3.4.5第5个鬼.2.3.4.5第6个鬼1.3.4.5第7个鬼1.2.4.5第8个鬼1.2.3.5第9个鬼1.2.3.4第10个鬼全部带队
有1当队长第11个
第2个当队长12个
第3个队长13个
第4个队长14个
第5个队长这样是满足了,但是需要抓14个。
如果以上方法在名字是1的
在第2个鬼的时候离开
就可以满足10个鬼
满足你的条件
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