HDU 5973-Game of Taking Stones（威佐夫博弈-JAVA BigDecimal）

左手的ㄟ右手 2022-06-16 01:43 239阅读 0赞

# Game of Taking Stones #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)  
Total Submission(s): 459 Accepted Submission(s): 176**

Problem Description

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?

Input

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.

Output

For 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

Source

[2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）][2016ACM_ICPC_-]

威佐夫博弈的裸题，就是咳咳咳…数据量巨大，JAVA BigDecimal 可搞。

![20170502122639333][] 要精确到小数点后100位，精度要求太高了∑(っ °Д °;)っ

这个黄金分割数×(b-a)后如果等于a，那么为0，否则为1。(默认a<b)

import java.math.BigDecimal;
    import java.math.BigInteger;
    import java.util.Scanner;
    
    public class Main
    {
    
        public static BigDecimal setScaleRoundDown(int newScale, BigDecimal b)
        {
            return b.setScale(newScale, BigDecimal.ROUND_DOWN);
        }
    
        public static void main(String[] args)
        {
            Scanner cin=new Scanner(System.in);
            BigDecimal a,b,tmp,yi,er,x,tp,l;
            while(cin.hasNextBigDecimal())
            {
                a=cin.nextBigDecimal();
                b=cin.nextBigDecimal();
                l= new BigDecimal(String.valueOf("0"));
                if(a.compareTo(l)==0)
                {
                    if(b.compareTo(l)==0)
                    {
                        System.out.println("0");
                        continue;
                    }
                }
                if(a.compareTo(b)==0)
                {
                    System.out.println("1");
                    continue;
                }
                if(a.compareTo(b)>0)
                {
                    tp=a;
                    a=b;
                    b=tp;
                }
                x=b.subtract(a);
                tmp = new BigDecimal(String.valueOf("1.61803398874989484820458683436563811772030917980576286213544862270526046281890244970720720418939113748475408807"));
                tmp=tmp.multiply(x);
                //System.out.println(tmp);
                //tmp.setScale(0,BigDecimal.ROUND_DOWN);
                //System.out.println(setScaleRoundDown(0, tmp));
                if(setScaleRoundDown(0, tmp).compareTo((a))==0)
                    System.out.println("0");
                else
                {
                    System.out.println("1");
                }
            }
        }
    }

[2016ACM_ICPC_-]: http://acm.hdu.edu.cn/search.php?field=problem&key=2016ACM%2FICPC%D1%C7%D6%DE%C7%F8%B4%F3%C1%AC%D5%BE-%D6%D8%CF%D6%C8%FC%A3%A8%B8%D0%D0%BB%B4%F3%C1%AC%BA%A3%CA%C2%B4%F3%D1%A7%A3%A9&source=1&searchmode=source
[20170502122639333]: /images/20220616/eb20545d5b5549bc9732d1af543b8e6f.png