题解 UVA1500 Alice and Bob

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题解 UVA1500 Alice and Bob

题目传送门题目大意给出 \(n\) 堆石子,每次可以做以下两种操作之一: 将某两堆石子进行合并将某一堆石子抽走一个石子问谁必胜. 思路就nm很妙好么? 首先,我们需要考虑每堆石子大小都 \(>1\) 的情况,你发现判断条件就是判断石子总数加上堆数减一是否为奇数.因为每次合并实际上就相当于翻转先后手,然后如果两人足够聪明那么他们一定会用完合并的机会,因为如果自己当前必败就让对方走,反之对方就会让自己走. 我们再考虑存在大小 \(=1\) 的情况,你发现无非就多了几种操作: 将某两个大小为…

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Alice and Bob Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 147 Accepted Submission(s): 22 Problem Description As you know, Alice and Bob always play game together, and today they get a…

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Alice and Bob Time Limit: 1000ms Memory limit: 65536K 题目描述 Alice and Bob like playing games very much.Today, they introduce a new game. There is a polynomial like this: (a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1). Then Alice…

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Alice and Bob Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4174 Accepted Submission(s): 1310 Problem Description Alice and Bob's game never ends. Today, they introduce a new game. In thi…

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题目描述 Alice和Bob正在一棵树上玩游戏.这棵树有\(n\)个结点,编号由\(1\)到\(n\).他们一共玩\(q\)盘游戏. 在第\(i\)局游戏中,Alice从结点\(a_i\)出发,Bob从结点\(b_i\)出发.开始时,除了\(a_i\)和\(b_i\)这两个结点外,所有结点都没有染色.结点\(a_i\)被Alice染色,结点\(b_i\)被Bob染色. 接下来,两位玩家轮流移动,两位玩家移动步数之和为\(k_i\)步.Alice走第一步,Bob走第二步,Alice走第三步\(\c…