POJ2348 Euclid's Game

Euclid's Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9011		Accepted: 3687

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):

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

看这样一个例子：

前两个数字分别为硬币数x，y，第三个数字为这是玩家n取完以后的结果

32 15
17 15 1
2 15 2
2 13 1
2 11 2
2 9 1
2 7 2
2 5 1
2 3 2
2 1 1
0 1 2

可以发现当x>=2*y时，此时操作的玩家必胜，因为他可以通过自由选择取走y的a倍数量的硬币，使得对方不得不采取己方想看到的行动。

此即为“控场”之术。

算法是几分钟就能想出来的，但是这题要注意细节，比如说2*y可能会越int界……

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 int a,b;
 9 int main(){
10     while(scanf("%d%d",&a,&b) && a && b){
11         int side=1;
12         while(a && b){
13             if(a<b)swap(a,b);
14 //            if(a>2*b)break; 错误的判断方式
15             if((long long)a>=(long long)2*b)break;
16             a=a%b;
17             if(!a)break;
18             side^=1;
19         }
20         if(side)printf("Stan wins
");
21         else printf("Ollie wins
");
22     }
23     return 0;
24 }