【hdu5973】高精度威佐夫博弈

题意：输入a, b表示两堆石头数目，威佐夫博弈，问：先手胜负？

a, b <= 1e100.

高精度。当a > b时， a = (a-b)*黄金分割比 时是先手败状态。因为a, b <= 1e100，所以黄金分割比至少要保留小数点后100位。

BigDecimal没有开方操作，所以二分算出黄金分割比。

import java.util.*;
import java.io.BufferedInputStream;
import java.math.*;
public class Main {
    public static void main(String[] args){
        Scanner cin = new Scanner(new BufferedInputStream(System.in)); 
        BigDecimal L = new BigDecimal(1.6180339887), R = new BigDecimal(1.6180339888);
        BigDecimal eps = new BigDecimal(0.1).pow(101);
        while( R.subtract(L).subtract(eps).compareTo(BigDecimal.ZERO) > 0){
            BigDecimal M = L.add(R).divide(BigDecimal.valueOf(2));
            BigDecimal ret = M.multiply(BigDecimal.valueOf(2)).subtract(BigDecimal.ONE).pow(2);
            if(ret.compareTo(BigDecimal.valueOf(5)) > 0)
                R = M;
            else 
                L = M;
        }
        BigDecimal t = L;//黄金分割数精确到1e-100
        while(cin.hasNext()){
            BigInteger a = cin.nextBigInteger(), b = cin.nextBigInteger();
            int com = a.compareTo(b), ccom = 0;
            if(com == 0){
                ccom = a.compareTo(BigInteger.ZERO);
                if(ccom == 0)
                    System.out.println(0);
                else 
                    System.out.println(1);
                continue ;
            }
            if(com == -1){
                BigInteger c = a;
                a = b;
                b = c;
            }
            BigInteger k = a.subtract(b);
            BigDecimal kk = new BigDecimal(k).multiply(t).setScale(0,  BigDecimal.ROUND_DOWN);
            k = kk.toBigInteger();
            ccom = k.compareTo(b);
            if(ccom == 0)
                System.out.println(0);
            else 
                System.out.println(1);
        }
    }
}