A multiplication game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6028   Accepted: 3013 Description Stan and Ollie play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Stan always starts with p =…

题目链接:http://poj.org/problem?id=2505 题目大意: 两个人轮流玩游戏,Stan先手,数字 p从1开始,Stan乘以一个2-9的数,然后Ollie再乘以一个2-9的数,直到谁先将p乘到p>=n时那个人就赢了,而且轮到某人时,某人必须乘以2-9的一个数. 解题思路: 这是一道博弈论的题目.不过这道题并没有用SG函数相关的知识.首先我们可以很快判断区间[2,9]必定是先手胜然后紧接着是区间[10,18]区间是后手胜然后是什么呢?[18,??]我们可以这样来理解:我们可以…

http://poj.org/problem?id=2505 感觉博弈论只有找规律的印象已经在我心中埋下了种子... 题目大意:两个人轮流玩游戏,Stan先手,数字 p从1开始,Stan乘以一个2-9的数,然后Ollie再乘以一个2-9的数,直到谁先将p乘到p>=n时那个人就赢了,而且轮到某人时,某人必须乘以2-9的一个数. 题目大意来源http://blog.csdn.net/jc514984625/article/details/71157698 因为谷歌翻译太难懂了,所以总是找题解找题目大…

思路:求必胜区间和必败区间! 1-9 先手胜 10-2*9后手胜 19-2*9*9先手胜 163-2*2*9*9后手胜 …… 易知右区间按9,2交替出现的,所以每次除以18,直到小于18时就可以直接判断了. 代码如下: #include<cstdio> int main() { double n; while(scanf("%lf",&n)!=EOF){ ) n/=; ) puts("Stan wins."); else puts("O…

A mutiplication game poj-2505 题目大意:给定一个数n和p,两个选手每次可以将p乘上[2,9].最先使得p大于n的选手胜利. 注释:$1\le n\le 4294967295$,$p=1$. 想法: 这个题比较新颖,我们可以直接推出必败态区间. 最后,附上丑陋的代码... ... #include <iostream> #include <cstdio> #include <cstring> #include <algorithm>…

Description Alice and Bob are in their class doing drills on multiplication and division. They quickly get bored and instead decide to play a game they invented. The game starts with a target integer N≥2N≥2 , and an integer M=1M=1. Alice and Bob take…

Description Alice and Bob are in their class doing drills on multiplication and division. They quickly get bored and instead decide to play a game they invented. The game starts with a target integer N≥2N≥2 , and an integer M=1M=1. Alice and Bob take…

背景: 昨天看了<最强大脑>,由于节目比较有争议性,不知为什么,作为一名感性的人,就想试一下如果自己理性分析会是怎样的呢? 过程是这样的: 中国队(3人)VS英国队(4人). 1:李建东(队长)出战,[并说中国队不胜就再不参加最强大脑]3局过后,打平,双方都没脑力进行下一轮,所以评委各得1分,结果:1V1. 2:苏XX(忘名了)出战,打败对手,结果:2V1. 3:申一帆出战,失败,结果2V2平(同时申一帆情绪失控离开节目现场,经节目组一番说辞后回归节目) 问题来了:最后一战,谁出站,在大屏幕播…

                             Matrix Multiplication Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 18173   Accepted: 3912 Description You are given three n × n matrices A, B and C. Does the equation A × B = C hold true? Input The first l…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Given two matrices A and B of size n×n, find the product of them. bobo hates big integers. So you are only asked to find t…

Euclid's Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9033   Accepted: 3695 Description Two players, Stan and Ollie, play, starting with two natural numbers. Stan, the first player, subtracts any positive multiple of the lesser o…

Game Theory Reveals the Future of Deep Learning Carlos E. Perez Deep Learning Patterns, Methodology and Strategy @ IntuitionMachine.com 译自:https://medium.com/intuitionmachine/game-theory-maps-the-future-of-deep-learning-21e193b0e33a#.2vjbrl5di 若你一直fo…

一些比较水的博弈论...(为什么都没有用到那什么SG呢....) TYVJ 1140  飘飘乎居士拯救MM 题解: 歌德巴赫猜想 #include <cmath> #include <cstdio> int n, a, b, ta, tb; inline bool isPrime(int x){ ) return true; ; i<=sqrt(x); i++) if (!(x%i)) return false; return true; } inline int getTi…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 820    Accepted Submission(s): 328 Problem Description Given two matrices A and B of size n×n, find the product of them. b…

A typical implementation Booth's algorithm can be implemented by repeatedly adding (with ordinary unsigned binary addition) one of two predetermined values A and S to a product P, then performing a rightward arithmetic shift on P. Let m and r be the…

C. The Game Of Parity time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n cities in Westeros. The i-th city is inhabited by ai people. Daenerys and Stannis play the following game:…

http://acm.hdu.edu.cn/showproblem.php?pid=4951 2014多校 第八题 1008 2014 Multi-University Training Contest 8 Multiplication table Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 438    Accepted Submi…

hdu4920 Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 568    Accepted Submission(s): 225 Problem Description Given two matrices A and B of size n×n, find the product o…

题目链接:http://poj.org/problem?id=1651 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the prod…

题目传送门 /* 题意:加上适当的括号,改变计算顺序使得总的计算次数最少 矩阵连乘积问题,DP解决:状态转移方程: dp[i][j] = min (dp[i][k] + dp[k+1][j] + p[i-1] * p[k] * p[j]) (i<=k<j) s[i][j] 记录断开的地方(即加括号的位置),回溯法输出结果 */ #include <cstdio> #include <cstring> #include <string> #include &l…

Matrix Chain Multiplication Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 834    Accepted Submission(s): 570 Problem DescriptionMatrix multiplication problem is a typical example of dynamical…

http://poj.org/problem?id=2234 博弈论真是博大精深orz 首先我们仔细分析很容易分析出来,当只有一堆的时候,先手必胜:两堆并且相同的时候,先手必败,反之必胜. 根据博弈论的知识(论文 张一飞:<由感性认识到理性认识——透析一类搏弈游戏的解答过程>) 局面可以分解,且结果可以合并. 局面均是先手 当子局面是 胜 和 败,那么局面则为胜 当子局面是 败 和 胜,那么局面则为胜 当子局面是 败 和 败,那么局面则为败 当子局面为 胜 和 胜,那么局面为不确定 而这些性质…

D - Matrix Multiplication Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description       Let us consider undirected graph G = {V; E} which has N vertices and M edges. Incidence matrix of this g…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1775    Accepted Submission(s): 796 Problem Description Given two matrices A and B of size n×n, find the product of them.…

Matrix Chain Multiplication Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 442 Appoint description:  System Crawler  (2015-08-25) Description   Suppose you have to evaluate an expression like A*B*C*D…

文章原地址:http://blog.csdn.net/zhangxiang0125/article/details/6174639 博弈论:是二人或多人在平等的对局中各自利用对方的策略变换自己的对抗策略,达到取胜目标的理论.博弈论是研究互动决策的理论.博弈可以分析自己与对手的利弊关系,从而确立自己在博弈中的优势,因此有不少博弈理论,可以帮助对弈者分析局势,从而采取相应策略,最终达到取胜的目的. 博弈论分类:(摘自百度百科) (一)巴什博奕(Bash Game):只有一堆n个物品,两个人轮流从这堆…

思路:首先用Tarjan算法找出树中的环,环为奇数变为边,为偶数变为点. 之后用博弈论的知识:某点的SG值等于子节点+1后的异或和. 代码如下: #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<vector> #include<cstring> using namespace std; int ans; vector<…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1404 一看就是博弈论的题目,但并没有什么思路,看了题解,才明白 就是求六位数的SG函数,暴力一遍,打表就OK. 具体的操作是先找P态,即最终无法移动的状态,可知无数可取是P态,0是N态,1是P态,然后从1开始进行暴力, 所有可以到!sg[i]的点标记为N态,暴力过程为标记一步可以到sg[i]的数,包括两类: 一类是仅某一位数字不同,提取方法比较巧妙: ; --i){ int m = x; ; ; j…

Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken…

这个题思路没有任何问题,但还是做了近三个小时,其中2个多小时调试 得到的经验有以下几点: 一定学会调试,掌握输出中间量的技巧,加强gdb调试的学习 有时候代码不对,得到的结果却是对的(之后总结以下常见错误) 能用结构体,就别用数组,容易出错(暂时还不知道为什么)=>现在知道申请的数组空间在运行期间被释放,除非用malloc去申请数组 代码要规范,空格该有就要有 有些不规范表达式,不同编译器出现不同结果,注意应避免使用这类语句 像这道题主要坑在了第三点上,以后要注意避免 以下是AC代码 第一次完成…