C - Crazy Calendar Time Limit:4000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1393 Description 2011 was a crazy year. Many people all over the world proposed on 11-11-11, married on 11-11-11, some even w…

思路:当移到右下角时,就不能移动了.所以与右下角的奇偶性相同的位置,都不能直接到达,先手必败! 只需考虑与右下角奇偶不同的位置,可以看成NIM博弈.最后NIM和不为0的胜,否者败!! 代码如下: #include<iostream> #include<cstdio> #include<algorithm> #define M 105 using namespace std; int main() { ,n,m,a; scanf("%d",&t…

LightOJ1253 :Misere Nim 时间限制:1000MS    内存限制:32768KByte   64位IO格式:%lld & %llu 描述 Alice and Bob are playing game of Misère Nim. Misère Nim is a game playing on k piles of stones, each pile containing one or more stones. The players alternate turns and…

Rabbit and Grass Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2145    Accepted Submission(s): 1622 Problem Description 大学时光是浪漫的,女生是浪漫的,圣诞更是浪漫的,但是Rabbit和Grass这两个大学女生在今年的圣诞节却表现得一点都不浪漫:不去逛商场,不去逛…

Rabbit and Grass Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3864    Accepted Submission(s): 2919 Problem Description 大学时光是浪漫的,女生是浪漫的,圣诞更是浪漫的,但是Rabbit和Grass这两个大学女生在今年的圣诞节却表现得一点都不浪漫:不去逛商场,不去逛…

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description   Input   Output   Sample Input   Sample Output   Hint   Description 大家常常感慨,要做好一件事情真的不容易,确实,失败比成功容易多了! 做好“一件”事情尚且不易,若想永远成功而总从不失败,那更是难上加难了,就像花钱总是比挣钱…

Climbing the Hill Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u Submit Status Description Alice and Bob are playing a game called "Climbing the Hill". The game board consists of cells arranged vertically, as the…

A - Matrix Game Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Description Given an m x n matrix, where m denotes the number of rows and n denotes the number of columns and in each cell a pile of stones is gi…

Fibonacci again and again Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5088    Accepted Submission(s): 2126 Problem Description 任何一个大学生对菲波那契数列(Fibonacci numbers)应该都不会陌生,它是这样定义的:F(1)=1;F(2)=2;…

ZYB's Game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 527    Accepted Submission(s): 430 Problem Description ZYB played a game named NumberBomb with his classmates in hiking:a host keeps a…

http://acm.hdu.edu.cn/showproblem.php? pid=1849 简单的尼姆博弈: 代码例如以下: #include <iostream> #include <cstdio> using namespace std; int main() { int m,n,t; while(cin>>m,m) { int ans=0; for(int i=0; i<m; i++) { cin>>n; ans^=n; } if(ans==…

题目思路: 对于尼姆博弈我们知道:op=a[1]^a[2]--a[n],若op==0先手必败 一个简单的数学公式:若op=a^b 那么:op^b=a: 对于第i堆a[i],op^a[i]的值代表其余各个堆值的亦或值. 我们现在希望将a[i]改变成某个更小的值使得,op^a[i]=0,反过来a[i]=op^0,输出它就好了 #include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h&…

题目大意:尼姆博弈,求先手必胜的情况数 题目思路:判断 ans=(a[1]^a[2]--^a[n]),求ans^a[i] < a[i]的个数. #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<stdio.h> #include<stdlib.h> #include<queue> #include<…

题目大意:尼姆博弈,判断是否先手必胜. 题目思路: 尼姆博弈:有n堆各a[]个物品,两个人轮流从某一堆取任意多的物品,规定每次至少取一个,多者不限,最后取光者得胜. 获胜规则:ans=(a[1]^a[2] --^a[n]),若ans==0则后手必胜,否则先手必胜. #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<stdio.h>…

Being a Good Boy in Spring Festival Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51 Accepted Submission(s): 45   Problem Description 一年在外 父母时刻牵挂春节回家 你能做几天好孩子吗寒假里尝试做做下面的事情吧 陪妈妈逛一次菜场悄悄给爸爸买个小礼物主动地…

Being a Good Boy in Spring Festival Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4894    Accepted Submission(s): 2930 Problem Description 一年在外 父母时刻牵挂 春节回家 你能做几天好孩子吗 寒假里尝试做做以下的事情吧 陪妈妈逛一次菜场 悄悄…

题型: 有3堆各若干个物品,两个人轮流从某一堆取任意多的物品,规定每次至少取1个,多者不限,最后取光者得胜. 思路 首先自己想一下,就会发现只要最后剩两堆物品一样多(不为零),第三堆为零,那面对这种局势的一方就必败 那我们用(a,b,c)表示某种局势,首先(0,0,0)显然是必败态,无论谁面对(0,0,0) ,都必然失败:第二种必败态是(0,n,n),自己在某一堆拿走k(k ≤ n)个物品,不论k为多少,对方只要在另一堆拿走k个物品,最后自己都将面临(0,0,0)的局势,必败.仔细分析一下,(1…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1907 Problem Description Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same c…

博弈论入门: 巴什博弈: 两个顶尖聪明的人在玩游戏,有一堆$n$个石子,每次每个人能取$[1,m]$个石子,不能拿的人输,请问先手与后手谁必败? 我们分类讨论一下这个问题: 当$n\le m$时,这时先手的人可以一次取走所有的: 当$n=m+1$时,这时先手无论取走多少个,后手的人都能取走剩下所有的: 当$n=k*(m+1)$时,对于每$m+1$个石子,先手取$i$个,后手一定能将剩下的$(m+1-i)$个都取走,因此后手必胜: 当$n=k*(m+1)+x(0<x<m+1)$时,先手可以先取$…

(一)巴什博奕(Bash Game):只有一堆n个物品,两个人轮流从这堆物品中取物,规定每次至少取一个,最多取m个.最后取光者得胜.若(m+1) | n,则先手必败,否则先手必胜.显然,如果n=m+1,那么由于一次最多只能取m个,所以,无论先取者拿走多少个,后取者都能够一次拿走剩余的物品,后者取胜.因此我们发现了如何取胜的法则:如果n=(m+1)r+s,(r为任意自然数,s≤m),那么先取者要拿走s个物品,如果后取者拿走k(≤m)个,那么先取者再拿走m+1-k个,结果剩下(m+1)(r-1)个,…

尼姆博弈(Nimm's Game) 题型 尼姆博弈模型,大致上是这样的: 有3堆各若干个物品,两个人轮流从某一堆取任意多的物品,规定每次至少取1个,多者不限,最后取光者得胜. 分析 1.首先自己想一下,就会发现只要最后剩两堆物品一样多(不为零),第三堆为零,那面对这种局势的一方就必败 那我们用(a,b,c)表示某种局势,首先(0,0,0)显然是必败态,无论谁面对(0,0,0) ,都必然失败:第二种必败态是(0,n,n),自己在某一堆拿走k(k ≤ n)个物品,不论k为多少,对方只要在另一堆拿走k…

题目链接: https://cn.vjudge.net/problem/POJ-2234 题目描述: Here is a simple game. In this game, there are several piles of matches and two players. The two player play in turn. In each turn, one can choose a pile and take away arbitrary number of matches fro…

HDU.1850 Being a Good Boy in Spring Festival (博弈论 尼姆博弈) 题意分析 简单的nim 博弈 博弈论快速入门 代码总览 #include <bits/stdc++.h> #define nmax 105 using namespace std; int a[nmax]; int main() { int m; while(scanf("%d",&m) != EOF && m){ int ans = 0,…

http://poj.org/problem?id=1704 我并不知道阶梯博弈是什么玩意儿,但是这道题的所有题解博客都写了这个标签,所以我也写了,百度了一下,大概是一种和这道题类似的能转换为尼姆博弈的博弈. 解法大概是配对之后的尼姆博弈,没看到一个格子只能放一个石头(以为可以直接把石头移动到前一个石头在的格子),所以莫名其妙wa了orz,总是因为题目被踩爆orz 其实博弈也是个很神奇的东西,现在这个东西给我的印象又变成配对找规律了(泪),虽然看上去从后往前两两配对的做法很有道理但是迫切需要更数…

http://poj.org/problem?id=2975 题目始终是ac的最大阻碍. 问只取一堆有多少方案可以使当前局面为先手必败. 显然由尼姆博弈的性质可以知道需要取石子使所有堆石子数异或和为0,那么将某一堆a个石子变为a^异或和即可. a1^a2^a3^...^an=y; a1^a2^a3^...^an^y=0; #include<cstdio> #include<cstring> #include<algorithm> #include<cmath>…

http://poj.org/problem?id=2234 尼姆博弈(Nimm's Game) 指的是这样一个博弈游戏:有任意堆物品,每堆物品的个数是任意的,双方轮流从中取物品,每一次只能从一堆物品中取物品,最少取一件,取到最后一件物品的人获胜. 结论:把每堆物品数全部异或起来,如果得到的值为0,那么先手必败,否则先手必胜. 直接判定即可. #include<cstdio> #include<cstring> #include<algorithm> #include&…

John Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 3762    Accepted Submission(s): 2127 Problem Description Little John is playing very funny game with his younger brother. There is one big bo…

题目链接: PKU:http://poj.org/problem? id=3480 HDU:http://acm.hdu.edu.cn/showproblem.php? pid=1907 Description Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to…

链接:https://www.nowcoder.com/acm/contest/86/E来源:牛客网 题目描述 坎为水,险阳失道,渊深不测:离为火,依附团结,光明绚丽. 坎卦:水洊至,习坎:君子以常德行,习教事.一轮明月照水中,只见影儿不见踪,愚夫当财下去取,摸来摸去一场空. 离卦:明两作,离,大人以继明照四方.官人来占主高升,庄农人家产业增,生意买卖利息厚,匠艺占之大亨通. 有一些石子堆,第 堆有 个石子.你和算卦先生轮流从任一堆中任取若干颗石子(每次只能取自一堆,并且不能不取),取到最后一颗…

John Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 3745    Accepted Submission(s): 2116 Problem Description Little John is playing very funny game with his younger brother. There is one big bo…