我之前做过一些博弈的题目,以为博弈都是DP,结果被坑了很多次,其实博弈有很多种,在此,把我见过的类型都搬上来. 1,HDU3951(找规律) 题意:把n枚硬币围成一个圆,让Alice和Bob两个人分别每人每次拿k(1<=k<=m)枚连续的硬币,谁能拿到最后谁赢: 思路:找规律,A拿了之后,B只要把剩下的分成偶数块,B就能赢,找到的规律就是除了m=1 && n&1是A赢,其余全是B赢,即B能够分成偶数块: #include <cstdio> #include…

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…

Description Alice和Bob在玩游戏.有n个节点,m条边(0<=m<=n-1),构成若干棵有根树,每棵树的根节点是该连通块内编号最 小的点.Alice和Bob轮流操作,每回合选择一个没有被删除的节点x,将x及其所有祖先全部删除,不能操作的人输 .注:树的形态是在一开始就确定好的,删除节点不会影响剩余节点父亲和儿子的关系.比如:1-3-2 这样一条链 ,1号点是根节点,删除1号点之后,3号点还是2号点的父节点.问有没有先手必胜策略.n约为10w. 显然只要算出每颗子树的sg值就可以…

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 as…

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2608 Alice and Bob Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 Alice and Bob like playing games very much.Today, they introduce a new game. There is a polynomial li…

Alice and Bob Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 5   Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Alice and Bob's game nev…

原题: ZOJ 3666 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3666 博弈问题. 题意:给你1~N个位置,N是最终点,1~N-1中某些格子能够移石头到另外一些指定的格子,1~N-1上有M个石头,位置不定,现在Alice和Bob要把这些石头全部移到N点,谁不能移则输,问先手必胜还是后手必胜. 做法:求出每个位置的SG函数值,然后将放石头的M个位置的SG函数值做异或,异或为0则Alice赢.这里讲坐标反转,1~N…

Alice and Bob Time Limit:3000MS     Memory Limit:128000KB     64bit IO Format:%lld & %llu Submit Status Practice ACdream 1112 Description Here  is Alice and Bob again ! Alice and Bob are playing a game. There are several numbers. First, Alice choose…

题目传送门 /* 题意: 求(a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1) 式子中,x的p次方的系数 二进制位运算:p = 2 ^ i + 2 ^ j + 2 ^ k + ...,在二进制表示下就是1的出现 例如:10 的二进制 为1010,10 = 2^3 + 2^1 = 8 + 2,而且每一个二进制数都有相关的a[i],对p移位运算,累计取模就行了 */ #include <cstdio> #include…

Alice and Bob Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 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). T…