http://codeforces.com/contest/347/problem/C 这道题就是求出n个数的最大公约数,求出n个数的最大值,总共有max1/gcd-n个回合.然后判断如果回合数%2==0 输出Bob,否则输出Alice. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int main() { ; int n; int m; while(sca…

Alice and Bob decided to eat some fruit. In the kitchen they found a large bag of oranges and apples. Alice immediately took an orange for herself, Bob took an apple. To make the process of sharing the remaining fruit more fun, the friends decided to…

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…