HDU1525 Euclid's Game（博弈论找规律）

Euclid's Game

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

34 12
15 24
0 0

Sample Output

Stan wins
Ollie wins

题解

题意

正如题目所说，两个人对两个数进行轮流操作，只能在大数中减去较小数的倍数，且保证最后结果均为正整数。

思路

作为一个博弈论找规律的题目：

a，b两数可分为如下情况：

• a%b==0

• a>2*b

• b<a<2b

下面对3种情况进行分析：

1. a%b==0：就是a是b的倍数，那么是先手获胜。

2. a>2*b：那么当前人肯定知道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),那么先手获胜。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。

3. b<a<2*b：只能变成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

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;

#define REP(i,n) for(int i=0;i<(n);i++)

int main(){
	
	int a,b;
	while(~scanf("%d %d",&a,&b)){
		if(a==0&&b==0)	break;
		if(a<b)	swap(a,b);
		int win = 0;
		while(b){
			if(a%b==0||a>2*b)	break;
			a = a-b;
            swap(a,b);
            win ^= 1;
		}
		if(win==0)	printf("Stan wins\n");
		else	printf("Ollie wins\n");
	}
	return 0;
}