【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);
         }
     }
 }