题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5973
Game of Taking Stones
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)
#### 问题描述
> Two people face two piles of stones and make a game. They take turns to take stones. As game rules, there are two different methods of taking stones: One scheme is that you can take any number of stones in any one pile while the alternative is to take the same amount of stones at the same time in two piles. In the end, the first person taking all the stones is winner.Now,giving the initial number of two stones, can you win this game if you are the first to take stones and both sides have taken the best strategy?
输入
Input contains multiple sets of test data.Each test data occupies one line,containing two non-negative integers a andb,representing the number of two stones.a and b are not more than 10^100.
输出
For each test data,output answer on one line.1 means you are the winner,otherwise output 0.
样例输入
2 1
8 4
4 7
样例输出
0
1
0
题解
威佐夫博弈+大数
对于a,b(a<b),如果有floor((b-a)*((sqrt(5)+1)/2))==a，则为必败态。
import java.awt.peer.SystemTrayPeer;
import java.math.BigDecimal;
import java.text.DecimalFormat;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
BigDecimal two=new BigDecimal(2);
BigDecimal thr=new BigDecimal(3);
BigDecimal fiv=new BigDecimal(5);
BigDecimal l=two,r=thr;
for(int i=0;i<500;i++){
BigDecimal mid=r.add(l).divide(two);
if(mid.multiply(mid).compareTo(fiv)<0){
l=mid;
}else{
r=mid;
}
}
BigDecimal gold=l.add(BigDecimal.ONE).divide(two);
BigDecimal a,b;
Scanner cin=new Scanner(System.in);
while(cin.hasNext()){
a=cin.nextBigDecimal();
b=cin.nextBigDecimal();
if(a.compareTo(b)>0){
BigDecimal tmp=a;
a=b;
b=tmp;
}
BigDecimal zero=a.subtract(b.subtract(a).multiply(gold).setScale(0,BigDecimal.ROUND_FLOOR));
//100位
BigDecimal sma=new BigDecimal("0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001");
if(zero.abs().compareTo(sma)<0){
System.out.println("0");
}else{
System.out.println("1");
}
}
}
}