A Simple Game

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 1487    Accepted Submission(s): 939

Problem Description
Agrael likes play a simple game with his friend Animal during the classes. In this Game there are n piles of stones numbered from 1 to n, the 1st pile has M1 stones, the 2nd pile has M2 stones, ... and the n-th pile contain Mn stones. Agrael and Animal take turns to move and in each move each of the players can take at most L1stones from the 1st pile or take at most L2 stones from the 2nd pile or ... or take Ln stones from the n-th pile. The player who takes the last stone wins.

After Agrael and Animal have played the game for months, the teacher finally got angry and decided to punish them. But when he knows the rule of the game, he is so interested in this game that he asks Agrael to play the game with him and if Agrael wins, he won't be punished, can Agrael win the game if the teacher and Agrael both take the best move in their turn?

The teacher always moves first(-_-), and in each turn a player must takes at least 1 stones and they can't take stones from more than one piles.

 
Input
The first line contains the number of test cases. Each test cases begin with the number n (n ≤ 10), represent there are n piles. Then there are n lines follows, the i-th line contains two numbers Mi and Li (20 ≥ Mi > 0, 20 ≥ Li > 0). 
 
Output
Your program output one line per case, if Agrael can win the game print "Yes", else print "No". 
 
Sample Input
2
1
5 4
2
1 1
2 2
 
Sample Output
Yes
No
 
Author
Agreal@TJU
 
Source
 
Recommend
lcy   |   We have carefully selected several similar problems for you:  1852 1854 1858 1730 1857 
 
简单的SG函数题,结果为每个子游戏的sg值的异或,每个子游戏都是简单的巴什博弈
 #include<stdio.h>
int main()
{
int t,n,m,l;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
int sg=;
for(int i=;i<n;i++){
scanf("%d%d",&m,&l);
sg ^= m%(l+);
}
if(sg)
puts("No");
else
puts("Yes");
}
return ;
}
 

hdu 1851(A Simple Game)(sg博弈)的更多相关文章

  1. HDU 5795 A Simple Nim (博弈 打表找规律)

    A Simple Nim 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5795 Description Two players take turns ...

  2. HDU 1851 A Simple Game

    典型的尼姆博弈,在n对石子中,告诉你每堆的数目和每次从该堆最多可以取的数目,求最终谁将其取完. 题解:SG(i)=mi%(li+1),求异或值即可. #include <cstdio> i ...

  3. hdu 1851 尼姆+巴什博弈

    先在每堆中进行巴什博弈,然后尼姆 #include<stdio.h> int main() { int T; int i,n; int ans,m,l; scanf("%d&qu ...

  4. hdu 1851 A Simple Game 博弈论

    简单博弈问题(巴什博弈-Bash Game) 巴什博弈:只有一堆n个物品,两个人轮流从这对物品中取物,规定每次至少取一个,最多取m个,最后取光着得胜. 很容易想到当n%(m+1)!=0时,先取者必胜, ...

  5. HDU 1851 (N个BASH博弈子游戏)

    题意:n堆石子,分别有M1,M2,·······,Mn个石子,各堆分别最多取L1,L2,·····Ln个石头,两个人分别取,一次只能从一堆中取,取走最后一个石子的人获胜.后选的人获胜输出Yes,否则输 ...

  6. HDU 1848(sg博弈) Fibonacci again and again

    Fibonacci again and again Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Jav ...

  7. HDU 5795 A Simple Nim(简单Nim)

    p.MsoNormal { margin: 0pt; margin-bottom: .0001pt; text-align: justify; font-family: Calibri; font-s ...

  8. HDU 5794 - A Simple Chess

    HDU 5794 - A Simple Chess题意: 马(象棋)初始位置在(1,1), 现在要走到(n,m), 问有几种走法 棋盘上有r个障碍物, 该位置不能走, 并规定只能走右下方 数据范围: ...

  9. HDU 4974 A simple water problem(贪心)

    HDU 4974 A simple water problem pid=4974" target="_blank" style="">题目链接 ...

  10. hdu 4972 A simple dynamic programming problem(高效)

    pid=4972" target="_blank" style="">题目链接:hdu 4972 A simple dynamic progra ...

随机推荐

  1. 北京Uber优步司机奖励政策(12月27日)

    滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:http://www.cnblogs.com/mfry ...

  2. JS学习 函数的理解

    ECMAScript 的函数实际上是功能完整的对象. 函数的理解 用 Function 类直接创建函数,格式如下.可理解为Function构造器. var function_name = new Fu ...

  3. 92套AE抖音快闪模板(精品)

    包含很多场景和类型,直接用即可,下载地址:百度网盘,https://pan.baidu.com/s/1bRFql1zFWyfpTAwa6MhuPA 内容截图:    

  4. 165. Merge Two Sorted Lists【LintCode by java】

    Description Merge two sorted (ascending) linked lists and return it as a new sorted list. The new so ...

  5. [Clr via C#读书笔记]Cp13接口

    Cp13接口 类和接口继承 接口只提供签名,不提供实现:等效于契约:凡事能使用具名接口的地方都能够使用实现了的接口. 定义接口 定义很简单,FCL也提供了大量的现成接口供使用: 继承接口 类不能多继承 ...

  6. 1053 Path of Equal Weight (30 分)(树的遍历)

    题目大意:给出树的结构和权值,找从根结点到叶子结点的路径上的权值相加之和等于给定目标数的路径,并且从大到小输出路径 #include<bits/stdc++.h> using namesp ...

  7. poj 2155 (二维树状数组 区间修改 求某点值)

    Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 33682   Accepted: 12194 Descript ...

  8. mysql下分组取关联表指定提示方法,类似于mssql中的cross apply

    转至:https://stackoverflow.com/questions/12113699/get-top-n-records-for-each-group-of-grouped-results ...

  9. lsscsi命令详解

    基础命令学习目录首页 lsscsi包默认是不安装的.lsscsi包安装完之后,lsscsi命令就可以使用了.lsscsi命令(lsscsi -t -L)能很方便的看出哪些是固态硬盘(SSD),哪些是S ...

  10. redis集群sentinel哨兵模式的搭建与实际应用

    参考资料:https://blog.csdn.net/men_wen/article/details/72724406 之前环境使用的keepalived+redis vip集群模式,现在我们服务切换 ...