【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。
1 import java.util.*; 2 import java.io.BufferedInputStream; 3 import java.math.*; 4 public class Main { 5 public static void main(String[] args){ 6 Scanner cin = new Scanner(new BufferedInputStream(System.in)); 7 BigDecimal L = new BigDecimal(1.6180339887), R = new BigDecimal(1.6180339888); 8 BigDecimal eps = new BigDecimal(0.1).pow(101); 9 while( R.subtract(L).subtract(eps).compareTo(BigDecimal.ZERO) > 0){ 10 BigDecimal M = L.add(R).divide(BigDecimal.valueOf(2)); 11 BigDecimal ret = M.multiply(BigDecimal.valueOf(2)).subtract(BigDecimal.ONE).pow(2); 12 if(ret.compareTo(BigDecimal.valueOf(5)) > 0) 13 R = M; 14 else 15 L = M; 16 } 17 BigDecimal t = L;//黄金分割数精确到1e-100 18 while(cin.hasNext()){ 19 BigInteger a = cin.nextBigInteger(), b = cin.nextBigInteger(); 20 int com = a.compareTo(b), ccom = 0; 21 if(com == 0){ 22 ccom = a.compareTo(BigInteger.ZERO); 23 if(ccom == 0) 24 System.out.println(0); 25 else 26 System.out.println(1); 27 continue ; 28 } 29 if(com == -1){ 30 BigInteger c = a; 31 a = b; 32 b = c; 33 } 34 BigInteger k = a.subtract(b); 35 BigDecimal kk = new BigDecimal(k).multiply(t).setScale(0, BigDecimal.ROUND_DOWN); 36 k = kk.toBigInteger(); 37 ccom = k.compareTo(b); 38 if(ccom == 0) 39 System.out.println(0); 40 else 41 System.out.println(1); 42 } 43 } 44 }
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