題目鏈接：http://poj.org/problem?id=1067

威佐夫博弈 (Wythoff game): 當兩堆石子的數量符合特定關系時，先行者必贏。兩堆石子分別有a，b個石子時(a<=b), floor( (b-a) * ( (sqrt(5)+1) /2 ) ) == a 時，先行者必輸。代碼如下：

 1 #include <iostream>
 2 #include <math.h>
 3 #include <stdio.h>
 4 #include <cstdio>
 5 #include <algorithm>
 6 #include <string
 
.h>
 7 #include <string>
 8 #include <sstream>
 9 #include <cstring>
10 #include <queue>
11 #include <vector>
12 #include <functional>
13 #include <cmath>
14 #include <set>
15 #define SCF(a) scanf("%d", &a)
16 #define IN(a) cin>>a
17 #define
 
 FOR(i, a, b) for(int i=a;i<b;i++)
18 #define Infinity 999999999
19 #define NInfinity -999999999
20 #define PI 3.14159265358979323846
21 #define MAXN 1000000005
22 #define RATIO (sqrt(5.0)+1.0)/2.0
23 typedef long long Int;
24 using namespace std;
25 
26 int main()
27 {
28     int a, b;
29     while (scanf("%d %d
 
", &a, &b) != EOF)
30     {
31         if (a > b)
32         {
33             int temp = b;
34             b = a;
35             a = temp;
36         }
37         if (floor((b - a) * RATIO) == a)
38         {
39             printf("0\n");
40         }
41         else
42             printf("1\n");
43 
44     }
45     return 0;
46 }

POJ 1067 取石子遊戲