TZC 2414 Euclid's Game

 

Euclid's Game

http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=2414

时间限制(普通/Java):1000MS/10000MS     运行内存限制:65536KByte
总提交: 27            测试通过: 11

描述

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.

 

输入

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.

输出

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.

 

样例输入

 

34 12
15 24
0 0

 

样例输出

 

Stan wins
Ollie wins

代码：

#include<stdio.h> /* 分析：非原创[怎么就没想到 哈 悲剧 视乎前提是。。。。嘿嘿不可说] 观察后发现：假定后面的减法已经可以控制（已经确定），如果当前a，b(a>=b)两个数字，且a-b>=b , 不妨设a=n*b+c,那么当前的玩家就有胜负决定权，他可以决定减n*b使得a=c 或(n-1)*b使得a=b+c。 还有如果a-b<a 那么当前玩家是被动的，他只能选择a-1*b;(1是不可以选的) 既然当前步可以控制，所以以后的所有步都可以控制。 有了以上结论，得出只要谁得到两个数字a，b(a>=b) 那么他就获胜了。也有可能直到结束，都没有谁得到主动权，那就按照规则，谁可以减到0，谁获胜。 int gcd(int a,int b) { if(b==0) return a; int t = gcd(b,a%b); return t; } 上面这个是写的递归的欧几里德算法 */ int solve(int m,int n) { int tm,t=1; if(n>m){tm=m;m=n;n=tm;} while((m-2*n)<0&&m%n&&n) { tm=m%n;//相当于减 m=n; n=tm; t++; } return t; } int main() { int m,n,t; while(scanf("%d%d",&m,&n),n||m) { t = solve(m,n); if(!(t%2))printf("Ollie wins/n"); else printf("Stan wins/n"); } return 0; }