poj 2348 Euclid's Game 題解

Euclid's Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9023   Accepted: 3691 Description Two players, Stan and Ollie, play, starting with two natural numbers. Stan, the first player, subtracts any positive multiple of the lesser of the two numbers from the greater of the two numbers, provided that the resulting number must be nonnegative. Then Ollie, the second player, does the same with the two resulting numbers, then Stan, etc., alternately, until one player is able to subtract a multiple of the lesser number from the greater to reach 0, and thereby wins. For example, the players may start with (25,7):  25 7 11 7 4 7 4 3 1 3 1 0 an Stan wins. Input The input consists of a number of lines. Each line contains two positive integers giving the starting two numbers of the game. Stan always starts. Output For each line of input, output one line saying either Stan wins or Ollie wins assuming that both of them play perfectly. The last line of input contains two zeroes and should not be processed. Sample Input 34 12 15 24 0 0 Sample Output Stan wins Ollie wins Source Waterloo local 2002.09.28

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————————————————————————我是分割線——————————————————————————

博弈論題目。

第一眼看見我竟覺得是nim遊戲 TAT~

其實當n/m>=2時即爲關鍵點，此時關鍵點掌控方能夠決定遊戲勝敗。

因此判斷誰掌握關鍵點便可。

 1 /*
 2     Problem:
 3     OJ:
 4     User:    S.B.S.
 5     Time:
 6     Memory:
 7     Length:
 8 */
 9 #include<iostream>
10 #include<cstdio>
11 #include<cstring>
12 #include<cmath>
13 #include<algorithm>
14 #include<queue>
15 #include<cstdlib>
16 #include<iomanip>
17 #include<cassert>
18 #include<climits>
19 #include<functional>
20 #include<bitset>
21 #include<vector>
22 #include<list>
23 #include<map>
24 #define F(i,j,k) for(int i=j;i<=k;i++)
25 #define M(a,b) memset(a,b,sizeof(a))
26 #define FF(i,j,k) for(int i=j;i>=k;i--)
27 #define maxn 10001
28 #define inf 0x3f3f3f3f
29 #define maxm 1001
30 #define mod 998244353
31 //#define LOCAL
32 using namespace std;
33 int read(){
34     int x=0,f=1;char ch=getchar();
35     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
36     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
37     return x*f;
38 }
39 int n,m;
40 inline int gcd(int a,int b)
41 {
42     return b ? gcd(b,a%b) : a;
43 }
44 int main()
45 {
46     std::ios::sync_with_stdio(false);//cout<<setiosflags(ios::fixed)<<setprecision(1)<<y;
47     #ifdef LOCAL
48     freopen("coin.in","r",stdin);
49     freopen("coin.out","w",stdout);
50     #endif
51     while(cin>>n>>m&&(n!=0||m!=0))
52     {
53         if(n<m) swap(n,m);
54         if(m==0) cout<<"Ollie wins"<<endl;
55         else if(n%m==0) cout<<"Stan wins"<<endl;
56         else{
57             long long cnt=0;
58             while(m!=0&&n/m==1){
59                 cnt++;
60                 n=n-m;swap(n,m);
61             }
62 //            cout<<cnt<<endl;
63             if(cnt%2==0) cout<<"Stan wins"<<endl;
64             else cout<<"Ollie wins"<<endl;
65         }
66     }
67     return 0;
68 }

poj 2348