博弈

http://acm.hdu.edu.cn/showproblem.php?pid=2516

斐波那契博弈
判断这个数是不是斐波那契额数就可以，如果是，则后手必赢，如果不是可以分解成多个斐波那契数，则先手必赢


import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        long[]f = new long[50];
        f[0] = 1;
        f[1] = 1;
        for(int i = 2;i<50;i++)
            f[i] = f[i-1]+f[i-2];
        int a;
        while(true){
            a = cin.nextInt();
            if(a==0)
                break;
            boolean flag = false;
            for(int i = 0;i<50;i++)
                if(a==f[i]){
                    flag = true;
                    break;
                }
            if(flag)
                System.out.println("Second win");
            else 
                System.out.println("First win");
        }
    }
}

/***
 * Wythoff博弈入门
 * 这个博弈关键是在与判断，m,n是不是奇异局势
 * 如果是奇异局势的话，那么先手必赢
 * 所谓奇异局势也就是说当j = m*(sqrt(5)-1)/2
 * 且 a == j*(sqrt(5)+1）/2
 * 如果m == a&&n ==a+j的话，这是一个奇异局势
 * 或则当a = (j+1)*(sqrt(5)+1)/2时
 *  m==a&&n==a+j+1时，也是奇异局势
 */
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int m, n;
        double eps1 = (Math.sqrt(5) - 1) / 2;
        double eps2 = (Math.sqrt(5) + 1) / 2;
        while (cin.hasNext()) {
            m = cin.nextInt();
            n = cin.nextInt();
            if (m > n) {
                int c = m;
                m = n;
                n = c;
            }
            int j = (int) (m * eps1);
            int a = (int)(eps2 * j);
            int b = (int)(eps2 * (j+1));
            boolean flag = false;
            if (m == a && n == a + j)
                flag = true;
            if (m == b && n == b + j + 1)
                flag = true;
            if (flag)
                System.out.println("0");
            else
                System.out.println("1");
        }
    }
}