hdu------(1525)Euclid's Game(博弈决策树)
Euclid's Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2074 Accepted Submission(s): 924
players, Stan and Ollie, play, starting with two natural numbers. Stan,
the first player, subtracts any positive multiple of the lesser of the
two numbers from the greater of the two numbers, provided that the
resulting number must be nonnegative. Then Ollie, the second player,
does the same with the two resulting numbers, then Stan, etc.,
alternately, until one player is able to subtract a multiple of the
lesser number from the greater to reach 0, and thereby wins. For
example, the players may start with (25,7):
25 7
11 7
4 7
4 3
1 3
1 0
an Stan wins.
input consists of a number of lines. Each line contains two positive
integers giving the starting two numbers of the game. Stan always
starts.
each line of input, output one line saying either Stan wins or Ollie
wins assuming that both of them play perfectly. The last line of input
contains two zeroes and should not be processed.
15 24
0 0
Ollie wins
#include<cstring>
#include<cstdio>
bool flag=false;
int cont=;
void botree(int n,int m){
if(n<m)n^=m^=n^=m;
if(n%m)
for(int i=;i*m<n;i++){
cont++;
botree(n-i*m,m);
}
else {
if(cont&) flag=true;
cont=;
}
}
int main(){
int n,m;
//freopen("test.in","r",stdin);
while(scanf("%d%d",&n,&m),n+m){
flag=false;
botree(n,m);
if(flag) printf("Stan wins\n");
else printf("Ollie wins\n");
}
return ;
}
然后进行了优化之后.....得到这样的代码:
/*Problem : 1525 ( Euclid's Game ) Judge Status : Accepted
RunId : 11528629 Language : C++ Author : huifeidmeng
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta*/ #include<cstring>
#include<cstdio>
bool flag=false;
int cont=,cc=;
void botree(int n,int m){
if(n<m) n^=m^=n^=m;
if(m>&&((n%m)&&n/m==)){
cont++;
botree(n%m,m);
}
else if(cont&) flag=true;
}
int main(){
int n,m;
//freopen("test.in","r",stdin);
while(scanf("%d%d",&n,&m),n+m){
flag=false;
cont=;
botree(n,m);
if(flag) printf("Stan wins\n");
else printf("Ollie wins\n");
}
return ;
}
hdu------(1525)Euclid's Game(博弈决策树)的更多相关文章
- HDU 1525 Euclid's Game 博弈
Euclid's Game Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Tot ...
- HDU 1525 Euclid's Game (博弈)
Euclid's Game Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Tot ...
- hdu 1525 Euclid's Game 博弈论
思路:两个数a和b,总会出现的一个局面是b,a%b,这是必然的,如果a>=b&&a<2*b,那么只有一种情况,直接到b,a%b.否则有多种情况. 对于a/b==1这种局面, ...
- HDU 1525 Euclid's Game
题目大意: 题目给出了两个正数a.b 每次操作,大的数减掉小的数的整数倍.一个数变为0 的时候结束. 谁先先把其中一个数减为0的获胜.问谁可以赢.Stan是先手. 题目思路: 无论a,b的值为多少,局 ...
- hdu 1525 Euclid's Game【 博弈论】
Two players, Stan and Ollie, play, starting with two natural numbers. Stan, the first player, subtra ...
- HDU 1525 Euclid Game
题目大意: 给定2个数a , b,假定b>=a总是从b中取走一个a的整数倍,也就是让 b-k*a(k*a<=b) 每人执行一步这个操作,最后得到0的人胜利结束游戏 (0,a)是一个终止态P ...
- HDU 1525 类Bash博弈
给两数a,b,大的数b = b - a*k,a*k为不大于b的数,重复过程,直到一个数为0时,此时当前操作人胜. 可以发现如果每次b=b%a,那么GCD的步数决定了先手后手谁胜,而每次GCD的一步过程 ...
- A - 无聊的游戏 HDU - 1525(博弈)
A - 无聊的游戏 HDU - 1525 疫情当下,有两个很无聊的人,小A和小B,准备玩一个游戏,玩法是这样的,从两个自然数开始比赛.第一个玩家小A从两个数字中的较大者减去两个数字中较小者的任何正倍数 ...
- HDU 1524 树上无环博弈 暴力SG
一个拓扑结构的图,给定n个棋的位置,每次可以沿边走,不能操作者输. 已经给出了拓扑图了,对于每个棋子找一遍SG最后SG和就行了. /** @Date : 2017-10-13 20:08:45 * @ ...
随机推荐
- Refresh / Updating a form screen in Oracle D2k Forms 6i
Refresh / Updating a form screen in Oracle D2k Forms 6i ProblemYou want to show number of records pr ...
- Run_Product Example Form - Oracle Forms 6i
I have already posted in my previous post Running Reports Using Run_Product to run reports in Oracle ...
- struts2--表单标签
struts2的表单标签可分为两类:form标签本身和包装HTML表单元素的其他标签.form标签本身的行为不同于它内部的元素. struts2表单标签包括: form.textfield.passw ...
- 服务器端验证--验证框架验证required.
struts2表单验证里field-validator type值一共可以取哪些?都什么含义? int 整数:double 实数:date 日期:expression 两数的关系比较: email E ...
- HDU 5831 Rikka with Parenthesis II(六花与括号II)
31 Rikka with Parenthesis II (六花与括号II) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536 ...
- Python基础学习笔记(八)常用字典内置函数和方法
参考资料: 1. <Python基础教程> 2. http://www.runoob.com/python/python-dictionary.html 3. http://www.lia ...
- Python基础学习笔记(一)入门
参考资料: 1. <Python基础教程> 2. http://www.runoob.com/python/python-chinese-encoding.html 3. http://w ...
- [转载] 一致性问题和Raft一致性算法
原文: http://daizuozhuo.github.io/consensus-algorithm/ raft 协议确实比 paxos 协议好懂太多了. 一致性问题 一致性算法是用来解决一致性问题 ...
- Centos7 PHP7 编译安装 开机自启动
1.PHP7.0.13下载 wget http://cn2.php.net/get/php-7.0.13.tar.gz/from/this/mirror 2.解压 .tar.gz 3. 进入目录 cd ...
- 闲谈--心态 (zhuan)
http://blog.csdn.net/marksinoberg/article/details/53261034 ***************************************** ...