[51Nod 1185 威佐夫游戏 V2]Wythoff Game+乘法模拟

题目链接：[51Nod 1185 威佐夫游戏 V2]
题意描述：有2堆石子。A B两个人轮流拿，A先拿。每次可以从一堆中取任意个或从2堆中取相同数量的石子，但不可不取。拿到最后1颗石子的人获胜。假设A B都非常聪明，拿石子的过程中不会出现失误。给出2堆石子的数量，问最后谁能赢得比赛。
例如：2堆石子分别为3颗和5颗。那么不论A怎样拿，B都有对应的方法拿到最后1颗。
注意：这个题目每堆石子的数目是在[1,1018]范围内。
解题思路：根据Wythoff 博弈定理：当且仅当a=⌊k∗5√+12⌋,b=a+k时，先手失败；否则后手失败。由于题目给定的石子数目很大，所以不能用浮点数直接进行运算。[tips:5√+12=1.6180339887498948482045868343656...]，这里用到了乘法模拟的方法，来增大精度。
这题还有一个Java大数的做法， 请参考另外一篇博客《 [hdu 5973 Game of Taking Stones] Wythoff Game+大数运算 》

#include <map>#include <set>#include <queue>#include <cmath>#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <algorithm>using namespace std;//#pragma comment(linker, "/STACK:1024000000,1024000000")#define FIN             freopen("input.txt","r",stdin)#define FOUT            freopen("output.txt","w",stdout)#define fst             first#define snd             second#define rep(i, f, t)    for(int i = (f); i <= (t); i++)//typedef __int64 LL;typedef long long LL;typedef unsigned long long ULL;typedef pair<int, int> PII;const ULL MOD = 1000000000; /// 1e9int T;ULL A, B;ULL X[] = {618033988, 749894848, 204586834};ULL Y[2], Z[4];bool check(ULL x, ULL y) {    if(x > y) swap(x, y);    /// 转化为1e9进制数的模拟乘法,计算k*618033988749894848204586834,结果保存在Z[]中    ULL k = y - x, temp = 0;    Y[0] = k / MOD, Y[1] = k % MOD;    Z[3] = (temp = X[2] * Y[1] + temp) % MOD, temp /= MOD;    Z[2] = (temp = X[1] * Y[1] + X[2] * Y[0] + temp) % MOD, temp /= MOD;    Z[1] = (temp = X[0] * Y[1] + X[1] * Y[0] + temp) % MOD, temp /= MOD;    Z[0] = (temp = X[0] * Y[0] + temp) % MOD, temp /= MOD;    return x == temp * MOD + k + Z[0];}int main() {#ifndef ONLINE_JUDGE    FIN;#endif // ONLINE_JUDGE    scanf("%d", &T);    while(T --) {        cin >> A >> B;        puts(check(A, B) ? "B" : "A");    }    return 0;}

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补充一个Java代码：

import java.math.BigDecimal;import java.util.Scanner;public class Main {    static BigDecimal sqrt_five, G;    static BigDecimal two, five;    public static void init() {        two = new BigDecimal(2);        five = new BigDecimal(5);        BigDecimal low, high, mid;        low = new BigDecimal(0);        high = new BigDecimal(5);        mid = new BigDecimal(0);        for(int i = 0; i < 100; i++) {            mid = low.add(high).divide(two);            if(mid.multiply(mid).compareTo(five) > 0) high = mid;            else low = mid;        }        sqrt_five = mid;        // G = (sqrt(5) + 1) / 2;        G = sqrt_five.add(BigDecimal.ONE).divide(two);    }    public static void main(String[] argv) {        BigDecimal a, b, c, left, right;        init();        int T;        Scanner input = new Scanner(System.in);        T = input.nextInt();        while ((T --) != 0) {            a = input.nextBigDecimal();            b = input.nextBigDecimal();            if (a.compareTo(b) > 0) {                c = a;                a = b;                b = c;            }            left = b.subtract(a).multiply(G);            right = a;            // 向下取整。统一精度。            left = left.setScale(0, BigDecimal.ROUND_DOWN);            right = right.setScale(0, BigDecimal.ROUND_DOWN);            if(left.compareTo(right) == 0) System.out.println("B");            else System.out.println("A");        }        return;    }}