HDU 2177 取(2堆)石子游戏 Wythoff Game 求第一步方案

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题目大意：

就是Wythoff Game, 但是判断胜负之后还要输出方案

大致思路：

首先了解到Wythoff Game的性质：

可以参考 HDU1527

对于每一种方案讨论下一步可能走到的P点位置就行了, 细节问题看代码注释吧, 写的很详细了

代码如下：

Result : Accepted Memory : 1580 KB Time : 0 ms

/* * Author: Gatevin * Created Time:  2015/5/8 15:53:15 * File Name: Rin_Tohsaka.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e)#define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endlconst double gold1 = (1 + sqrt(5.)) / 2;const double gold2 = (sqrt(5.) - 1) / 2;int main(){    int a, b;    while(scanf("%d %d", &a, &b), a || b)    {        int k = floor(a*gold2);        if(floor((k + 1)*gold1) == a)//a = A[k + 1]            k = k + 1;        if(floor(k*gold1) == a)// a = A[k], b = B[k];            if(b == a + k)            {                puts("0");//输了                continue;            }         //否则就要输出方案了        puts("1");        //首先是两个都拿的情况, 必须变成A[b - a], A[b - a] + b - a        int t = floor((b - a)*gold1);//A[b - a]        if(t < a)//可以两个都拿            printf("%d %d\n", t, t + b - a);        //否则只拿一个                //拿b的那堆, a = A[k], 或者a = B[k]        //a = A[k]        if(floor(k*gold1) == a)            if(b > a + k)                printf("%d %d", a, a + k);        //a = B[k]        k = floor(a*gold2*gold2);        if(floor((k + 1)*gold1*gold1) == a) k++;        if(floor(k*gold1*gold1) == a)//a = B[k]            printf("%d %d\n", a - k, a);                //拿a的那堆一定要有b = B[k];        if(a != b)//防止相同的结果出现两次        {            k = floor(b*gold2*gold2);            if(floor((k + 1)*gold1*gold1) == b) k++;            if(floor(k*gold1*gold1) == b && b - k < a)                printf("%d %d\n", a - (b - k), b);        }            }    return 0;}

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