HDOJ 1525 Euclid's Game 博弈论

最新推荐文章于 2018-12-20 17:55:54 发布

kdwycz

最新推荐文章于 2018-12-20 17:55:54 发布

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分类专栏：【OJ】其他算法文章标签： HDOJ 1525 Euclids Game 博弈论入门题

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Euclid's Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1700 Accepted Submission(s): 751

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.

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

Sample Output

  
  
   
   Stan wins
Ollie wins

Source

University of Waterloo Local Contest 2002.09.28

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分两种情况讨论：

如果输入的两个数a>b有 a>2b 则先手必胜

小于的话b 和 a-b 在进行如上的判断，先后手转换。

用递归解决，非常清晰明了。

#include <iostream>
using namespace std;
int main()
{
	ios::sync_with_stdio(false);
	int a,b,flag;
	while (cin>>a>>b && (a || b)){
		flag=1;
		while (1){
			if (a<b) swap(a,b);
			if ((a/b>=2) || (a%b==0) || (b==0)){
				if (flag)
					cout<<"Stan wins\n";
				else
					cout<<"Ollie wins\n";
				break;
			}
			a-=b;	
			flag^=1;
		}
	}
	return 0;
}

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