Northcott Game Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4465    Accepted Submission(s): 2047 Tom和Jerry正在玩一种Northcott游戏,可是Tom老是输,因此他怀疑这个游戏是不是有某种必胜策略,郁闷的Tom现在向你求救了,你能帮帮他么?  游戏规则是这样的:  如图所示…

题解: 转化成求Nim-sum 每行黑白棋的初始间距作为每堆石子个数 假设当前为P态,则无论当前选手如何操作,下一个选手都能使其操作后的局面又变为P态. Nim-sum = 0,即P态. #include<cmath> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() { int n, m; while(~scanf("%d%…

Northcott Game Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3101    Accepted Submission(s): 1355 Problem Description Tom和Jerry正在玩一种Northcott游戏,可是Tom老是输,因此他怀疑这个游戏是不是有某种必胜策略,郁闷的Tom现在向你求救了,你能帮帮他…

Tom和Jerry正在玩一种Northcott游戏,可是Tom老是输,因此他怀疑这个游戏是不是有某种必胜策略,郁闷的Tom现在向你求救了,你能帮帮他么? 游戏规则是这样的: 如图所示,游戏在一个n行m列(1 ≤ n ≤ 1000且2 ≤ m ≤ 100)的棋盘上进行,每行有一个黑子(黑方)和一个白子(白方).执黑的一方先行,每次玩家可以移动己方的任何一枚棋子到同一行的任何一个空格上,当然这过程中不许越过该行的敌方棋子.双方轮流移动,直到某一方无法行动为止,移动最后一步的玩家获胜.Tom总是先下(…

Northcott Game Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4549    Accepted Submission(s): 2092 Problem Description Tom和Jerry正在玩一种Northcott游戏,可是Tom老是输,因此他怀疑这个游戏是不是有某种必胜策略,郁闷的Tom现在向你求救了,你能帮帮他…

You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove th…

http://codeforces.com/problemset/problem/662/A 题目大意: 给定n(n <= 500000)张卡片,每张卡片的两个面都写有数字,每个面都有0.5的概率是在正面,各个卡牌独立.求把所有卡牌来玩Nim游戏,先手必胜的概率. (⊙o⊙)-由于本人只会在word文档里写公式,所以本博客是图片格式的. Code #include <cstdio> #include <cstring> #include <algorithm> u…

A Simple Nim Problem Description   Two players take turns picking candies from n heaps,the player who picks the last one will win the game.On each turn they can pick any number of candies which come from the same heap(picking no candy is not allowed)…

Problem: You are playing the following Nim Game with your friend: There to stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones. Both of you are very clever and have optimal strategies for t…

题意 \(K(1 \le K \le 10^9)\)堆石子,每堆石子个数不超过\(L(2 \le 50000)\),问Nim游戏中先手必败局面的数量,答案对\(10^9+7\)取模. 分析 容易得到\(f(i, k) = \sum_{j=0}^{n-1} f(i-1, j) f(i-1, k^j), f(1, i(2 \le i \le L))=1\),其中\(n=min(2^i, 2^i > L)\).发现其实这就是操作为\(xor\)的卷积.于是用鬼畜的fwt做就行了. 题解 然后fwt+快…