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?
InputInput 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.
OutputFor each test data,output answer on one line.1 means you are the winner,otherwise output 0.
Sample Input
2 1
8 4
4 7
Sample Output
0
1
0
給兩堆石子，兩個人輪流拿，可以從其中的一堆拿任意多數量，也可以從兩堆中拿相同的數量，拿到最後一顆爲勝利者
l這是一個威佐夫博弈的裸題，對於a，b（a<b）若有floor((b-a)*((sqrt(5)+1)/2))==a則是必敗，但由於是大數，所以用java就比較簡單了，但是這面sqrt（5）不能直接算，所以用二分查找，找到近似值
l（sqrt（5）+1）/2是黃金分割數
import java.math.BigDecimal;
import java.util.Scanner;
public class Main{
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
BigDecimal two=new BigDecimal(2);
BigDecimal three= new BigDecimal(3);
BigDecimal five=new BigDecimal(5);
BigDecimal l=two,r=three;
for(int i=0;i<500;i++){
BigDecimal mid=l.add(r).divide(two);
if(mid.multiply(mid).compareTo(five)<0)
l=mid;
else
r=mid;
}
BigDecimal gold=l.add(BigDecimal.ONE).divide(two);//黃金分割數
BigDecimal a,b;
while(sc.hasNext()){
a=sc.nextBigDecimal();
b=sc.nextBigDecimal();
if(a.compareTo(b)>0){
BigDecimal tmp=a;
a=b;b=tmp;
}
b=b.subtract(a).multiply(gold);
b=b.setScale(0,BigDecimal.ROUND_DOWN);
if(a.compareTo(b)==0)
System.out.println("0");
else
System.out.println("1");
}
}
}
