Game of Taking Stones（威佐夫博奕 待整理）

Game of Taking Stones

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 477    Accepted Submission(s): 184

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）

 

Recommend

wange2014

 

本来就是一个简单的威佐夫博奕，但是因为大数！！！

#include <bits/stdc++.h>
using namespace std;

#define exp 1e-14

int main()
{
//    printf("%f\n", pow(10, 100));
    double gr = (sqrt(5.0) + 1) / 2;
    //printf("gr = %f\n", gr);
	double a, b, t;
	double k;
	while (~scanf("%lf%lf", &a, &b)) {
        if (a > b) {
            t = a;
            a = b;
            b = t;
        }
	    k = (b - a) * gr;
	    //printf("k = %f\n", k);
	    if ((-exp < k - a && k - a < exp) || (k > a && k - a < 1.0)) {
            printf("0\n");
	    } else {
            printf("1\n");
	    }
	}
    return 0;
}

因为10^100,所以黄金分割需要精确到100位

import java.math.*;
import java.math.BigDecimal;  
import java.util.Scanner;

public class Main {//类名要用Main
    public static void main(String[] args){
        BigDecimal TWO = BigDecimal.valueOf(2);  
        BigDecimal FIVE = BigDecimal.valueOf(5); 
        
        
        BigDecimal EPS = new BigDecimal(  
                "-0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001");  
        BigDecimal l = new BigDecimal("2.2360679774997"), r = new BigDecimal("2.2360679774998");  
        BigDecimal m = null;  
        while (l.subtract(r).compareTo(EPS) < 0)  
        {  
            m = l.add(r).divide(TWO);  
            if (m.multiply(m).subtract(FIVE).abs().compareTo(EPS.abs()) < 0)  
                break;  
            if (m.multiply(m).subtract(FIVE).compareTo(EPS) < 0)  
                l = m;  
            else  
                r = m;  
        }  
         BigDecimal gold = m.add(BigDecimal.ONE).divide(TWO);  
        
        BigDecimal a2, b2,k;
    	BigInteger a, b, t,k2;
    	Scanner sc=new Scanner(System.in);
    	while (sc.hasNextBigInteger()) {
    		a=sc.nextBigInteger();
            b=sc.nextBigInteger();
            if (a.compareTo(b) > 0) {
                t = a;
                a = b;
                b = t;
            }
            a2 = new BigDecimal(a);
            b2 = new BigDecimal(b);
            
    	    k = b2.subtract(a2).multiply(gold);
    	    k2 = k.toBigInteger();
    	    
    	    if (k2.compareTo(a) == 0) {
    	    	System.out.println("0");
    	    } else {
                System.out.println("1");
    	    }
    	}
        
    }
}