摘要: 2020牛客寒假算法基础训练营2 A.做游戏(思维) "传送门" 题意:两个人玩石头剪刀布,牛牛出石头a次,剪刀b次,布c次,牛可乐出石头x次,剪刀y次,布z次,问牛牛最多可以赢多少次 题解:对牛牛每次可以赢的出法取最小值即可 代码: c++ include include include incl 阅读全文
posted @ 2020-02-07 23:00 BT-7274 阅读(483) 评论(0) 推荐(0) 编辑