We were just children and grown up closeHow we made it this far only god knowsWe bend the rulesSmashed our carsPlayed with fire and broke our heartSomehow came out laughing through the tearsSure don't seem like twenty years Drove into the desert and…

标题套用了路飞的格式,其实我想说的是大多数都不相信你的梦想,直到你快走到了. 不废话了,介绍一下twenty: 这是基于CMS框架 zerojs打造一个博客.zerojs 的架构介绍在这里http://www.cnblogs.com/sskyy/p/3918129.html. 相比其他博客系统wordpress.ghost等等.twenty有以下优势: 技术架构更好!是的,你没看错.twenty 前端用angular,后端用sailsjs(基于express).对开发者来说几乎没有学习成本.基于…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9771   Accepted: 3220 Description Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2, 3, ..., and place N ch…

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…

Alice and BobTime Limit: 1 Sec  Memory Limit: 64 MBSubmit: 255  Solved: 43 Description Alice is a beautiful and clever girl. Bob would like to play with Alice. One day, Alice got a very big rectangle and wanted to divide it into small square pieces.…

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…

译者注: Bob大叔14年后再次谈论极限编程.极限编程经历了14年的风风雨雨后,Bob大叔将会给它怎么样的定义那? 在我手中拿着的一本白皮薄书,在14年前彻底的改变了软件世界.这本书的标题是解析极限编程,副标题是拥抱变化.作者是Kent Beck,出版时间为1999年. 这本书很薄,不到200页.排版很宽,间隔很远.写作风格即自由散漫又平易近人.章节不多,概念简单. 但是其影响却像地震一样,甚至至今震动仍未平息下来. 起始于第53页的章节10,列出了12项实践,引爆了行业内的大辩论.并催生了一场…

题目链接: Bob and Alice are playing numbers DESCRIPTION Bob and his girl friend are playing game together.This game is like this: There are nn numbers. If op = 11,Bob wants to find two numbers aiai and ajaj,that aiai & ajaj will become maximum value. If…

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…

Georgia and Bob Time Limit: 1000MS   Memory Limit: 10000KB   64bit IO Format: %I64d & %I64u Submit Status Description Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2,…

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…

Bob wants to pour water Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge There is a huge cubiod house with infinite height. And there are some spheres and some cuboids in the house. They do not intersect with others and the house.…

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…

Alice and Bob Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3716 Accepted Submission(s): 1179 Problem Description Alice and Bob's game never ends. Today, they introduce a new game. In this game…

十五 Twenty Questions Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1252 Appoint description:  System Crawler  (2015-08-25) Description   Consider a closed world and a set of features that are defined…

Bob’s Race Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Bob wants to hold a race to encourage people to do sports. He has got trouble in choosing the route. There are N houses and N - 1 roads…

Bob and math problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 401    Accepted Submission(s): 149 Problem Description Recently, Bob has been thinking about a math problem.There are N Digit…

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…

类似于石子合并的游戏,在黑板上写下N个数,每次只能将其中的一个数减1(结果为0自动消去),或者将某两个数消去,将其和写在黑板上. Alice先手,彼此都采用最优策略,将最后一个数消去者获胜. 思路:设s为石子总数,n为总堆数,分3种情况: (1).全1 :如果n能被3整除,先手败,否则先手胜. (2).有一个2,其余为1:如果n除以3余数为1,先手败,否则先手胜. (3)其他情况: a)1的个数为奇数,先手胜. b)1的个数为偶数,且s+n-1为奇数,先手胜. c)1的个数为偶数,且s+n-1为…

思路:每个数的SG值就是其质因子个数,在进行nim博弈 代码如下: #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> #include<set> #include<vector> #define ll long long #define M 5000005 #define inf 1e10 #d…

copy VS study 1.每堆部是1的时候,是3的倍数时输否则赢: 2.只有一堆2其他全是1的时候,1的堆数是3的倍数时输否则赢: 3.其他情况下,计算出总和+堆数-1,若为偶数,且1的堆数是偶数,则一定输: 4.不在上述情况下则赢. #include<stdio.h> int main() { ; int _case,i,n,x; int flag1,flag2,flag,sum; scanf("%d",&_case); while(_case--) { j…

题意: 一个无限大的棋盘,一开始在1,1,有三种移动方式,(x+1,y)(x,y+1) (x+k,y+k)最后走到nm不能走了的人算输.. 析:.我们看成一开始在(n,m),往1,1,走,所以自然可以从1,1,开始递推往出,那么打表程序就出来了.. 打出表以后我们观察到k等于1时稍有特殊,其他则与  (min(cx,cy)&1)^((n+m)&1)) 有关ps(其中cx=n/(k+1),cy=m/(k+1)) 那么就愉快的分类讨论外加试一试和表对照一下就好了.. 代码如下: #includ…

按照mahout官网https://cwiki.apache.org/confluence/display/MAHOUT/Twenty+Newsgroups的说法,我只用运行一条命令就可以完成这个算法的调用了,如下: mahout@ubuntu:~/mahout-d-0.7/examples/bin$ ./classify-20newsgroups.sh 但是,我首先运行就出错了,因为我不是root账户,所以先改下路径,打开classify-20newsgroups.sh,替换/tmp/maho…

题目链接:1484 - Alice and Bob's Trip 题意:BOB和ALICE这对狗男女在一颗树上走,BOB先走,BOB要尽量使得总路径权和大,ALICE要小,可是有个条件,就是路径权值总和必须在[L,R]之间,求终于这条路径的权值. 思路:树形dp,dp[u]表示在u结点的权值,往下dfs的时候顺带记录下到根节点的权值总和,然后假设dp[v] + w + sum 在[l,r]内,就是能够的,状态转移方程为 dp[u] = max{dp[v] + w }(bob) dp[u] = m…

Problem Description Alice and Bob's game never ends. Today, they introduce a new game. In this game, both of them have N different rectangular cards respectively. Alice wants to use his cards to cover Bob's. The card A can cover the card B if the hei…

组合游戏 Nim游戏的一个变形 题解请看金海峰的博客 以下为引用: 分析:我们把棋子按位置升序排列后,从后往前把他们两两绑定成一对.如果总个数是奇数,就把最前面一个和边界(位置为0)绑定. 在同一对棋子中,如果对手移动前一个,你总能对后一个移动相同的步数,所以一对棋子的前一个和前一对棋子的后一个之间有多少个空位置对最终的结果是没有影 响的.于是我们只需要考虑同一对的两个棋子之间有多少空位.我们把每一对两颗棋子的距离(空位数)视作一堆石子,在对手移动每对两颗棋子中靠右的那一颗 时,移动几位就相当于…