HDOJ 1525 Euclid's Game 博弈论
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Euclid's Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1700 Accepted Submission(s): 751
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.
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
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分两种情况讨论:
如果输入的两个数a>b有 a>2b 则先手必胜
小于的话b 和 a-b 在进行如上的判断,先后手转换。
用递归解决,非常清晰明了。
#include <iostream>using namespace std;int main(){ios::sync_with_stdio(false);int a,b,flag;while (cin>>a>>b && (a || b)){flag=1;while (1){if (a<b) swap(a,b);if ((a/b>=2) || (a%b==0) || (b==0)){if (flag)cout<<"Stan wins\n";elsecout<<"Ollie wins\n";break;}a-=b;flag^=1;}}return 0;}
kdwycz的网站: http://kdwycz.com/
kdwyz的刷题空间:http://blog.csdn.net/kdwycz
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