HDU1525 Euclid's Game (找规律博弈)

Euclid's Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3434    Accepted Submission(s): 1599

Problem 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 1215 240 0

 

Sample Output

Stan winsOllie wins

 

Source

University of Waterloo Local Contest 2002.09.28

 

思路：题目让求的是用大的数减去小的数，能够使得其中的一个数变成0的人为胜者。首先来分析下，假设a>=b，如果a==b 或者a%b==0则先手为胜，当a>=2*b时，则可以判断（a%b，b）是必胜点或者必败点，如果是必败点则可以走到这点，如果是必胜点则可以走到（a%b+b,b），当a < 2*b,则可以通过 a-=b计算

#include <cstdio>#include <algorithm>using namespace std;int a,b;int main(){while(~scanf("%d%d",&a,&b)){if(a==0&&b==0)break;if(a < b)swap(a,b);int flag = 0;while(b){if(a%b == 0 || a / b >= 2)break;a -= b;swap(a,b);flag = !flag;}if(flag)printf("Ollie wins\n");elseprintf("Stan wins\n");}return 0;}