HDU 1525 博弈
来源:互联网 发布:网络购物诈骗特点 编辑:程序博客网 时间:2024/05/04 15:06
Euclid's Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1880 Accepted Submission(s): 825
Problem Description
Two 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.
25 7
11 7
4 7
4 3
1 3
1 0
an Stan wins.
Input
The input consists of a number of lines. Each line contains two positive integers giving the starting two numbers of the game. Stan always starts.
Output
For 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.
Sample Input
34 1215 240 0
Sample Output
Stan winsOllie wins
题意:给出两个正数。
两个人轮流进行操作:
较大的数减掉较小的数的倍数,得到两个非负数。
谁先先把其中一个数减为0的获胜。问谁可以赢。Stan是先手。
假设两个数为a,b(a>=b)
如果a==b.那么肯定是先手获胜。一步就可以减为0,b
如果a%b==0.就是a是b的倍数,那么也是先手获胜。
如果a>=2*b. 那么 那个人肯定知道a%b,b是必胜态还是必败态。如果是必败态,先手将a,b变成a%b,b,那么先手肯定赢。如果是必胜态,先手将a,b变成a%b+b,b.那么对手只有将这两个数变成a%b,b,先手获胜。
如果是b<a<2*b 那么只有一条路:变成a-b,b (这个时候0<a-b<b).这样一直下去看谁先面对上面的必胜状态。
代码:
/* ***********************************************Author :rabbitCreated Time :2014/7/4 13:47:33File Name :3.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <iostream>#include <algorithm>#include <sstream>#include <stdlib.h>#include <string.h>#include <limits.h>#include <string>#include <time.h>#include <math.h>#include <queue>#include <stack>#include <set>#include <map>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8#define pi acos(-1.0)typedef long long ll;int gcd(int a,int b){ if(a<b) //求出谁大 swap(a,b); if(b==0) //终止 必输状态 return 0; if(a>2*b) //必胜 return 1; return gcd(b,a%b)^1; //和拿了之后的输赢情况相反}int main(){ //freopen("data.in","r",stdin); //freopen("data.out","w",stdout); int a,b; while(cin>>a>>b){ if(!a&&!b)break; int ans=gcd(a,b);// cout<<ans<<endl;if(ans)puts("Stan wins");else puts("Ollie wins"); } return 0;}
0 0
- hdu 1525 博弈
- HDU 1525 博弈
- 博弈 hdu 3032 、hdu 1525
- hdu 1525找规律博弈
- HDU博弈
- HDU 1525 Euclid's Game (博弈)
- hdu 1525 Euclid's Game 博弈
- [简单博弈] hdu 1525 Euclid's Game
- hdu 1525 Euclid's Game 博弈
- HDU 1525 Euclid's Game 博弈
- HDU 1525 Euclid's Game (博弈)
- HDU 1525 Euclid's Game 博弈
- hdu 1525 Euclid's Game 博弈
- HDU 1525 Eculid's Game(博弈)
- HDU 1525 Euclid's Game(博弈)
- hdu 1525 Euclid's Game (博弈规律)
- HDU 1525 Euclid's Game (博弈)
- HDU 1525 Euclid's Game (博弈)
- 一次简单的AIX磁盘未正常激活导致RAC无法启动的故障排除
- Transform 学习和应用
- IOS AFNetWork 附件上传
- 海量数据面试题整理
- 消费分类
- HDU 1525 博弈
- 部分常见ORACLE面试题以及SQL注意事项(转载)
- Spring security初探
- Apple 企业开发者账号&邓白氏码申请记录
- java设计模式示例
- CSS框架
- 文件下载(已修改)
- Linux下编写选择排序(C语言)
- USB无线网卡配置ICS失败,你可以连接到SoftAP,但可能无法使用internet服务,