HDU 1525　Euclid's Game

最新推荐文章于 2020-10-24 10:38:46 发布

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本文介绍了一款基于数学策略的游戏，玩家Stan和Ollie轮流从两个正整数中减去较小数的倍数，直至某一方使其中一个数归零。通过分析游戏规则和给出的代码，本文揭示了获胜策略的关键因素。

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

题意

题目给出了两个正数a.b
每次操作，大的数减掉小的数的整数倍。一个数变为0 的时候结束。
谁先先把其中一个数减为0的获胜。问谁可以赢。Stan是先手。

题解

如果a%b==0.就是a是b的倍数，那么一定是先手获胜。
如果a>=2*b. 那么那个人肯定知道a%b,b是必胜态还是必败态。如果是必败态，先手将a,b变成a%b,b,那么先手肯定赢。如果是必胜态，先手将a,b变成a%b+b,b.那么对手只有将这两个数变成a%b,b,先手获胜。
如果是b

CODE:

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int main()
{
    long long a,b;
    while(scanf("%lld%lld",&a,&b)!=EOF&&a&&b)
    {
        bool flag=0;
        while(1)
        {
            if(a<b)
                swap(a,b);
            if(a%b==0||a>=2*b)
            {
                flag? printf("Ollie wins\n"):printf("Stan wins\n");
                break;
            }
            flag=!flag;
            a-=b;
        }
    }
    return 0;
}