hdu1525&&poj2348

最新推荐文章于 2025-09-10 22:21:01 发布

转载最新推荐文章于 2025-09-10 22:21:01 发布 · 104 阅读

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原文链接：http://www.cnblogs.com/lxm940130740/p/3281158.html

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#java

本文详细解析了欧几里得游戏的胜负规律，通过寻找关键态来预测玩家的胜利情况，涉及数学逻辑和博弈论知识。

Euclid's Game

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 0 Accepted Submission(s) : 0

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):

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

方法一：

找规律，谁先遇到关键态，谁就获胜，关键态为：a>=2*b;

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
int main()
{
int a,b;
while(~scanf("%d%d",&a,&b)&&(a||b))
{
if(a<b)swap(a,b);
int flag=1;
while(1)
{
if(a==b||a/b>=2)
break;
a=a-b;
swap(a,b);
flag=!flag;
}
if(flag)
printf("Stan wins\n");
else
printf("Ollie wins\n");
}
return 0;
}

方法二：

给两堆石子（题目中是数，配合一下上文这里说石子），两人依次取石子，规则是：

每次从石子数较多的那堆取（两堆石子数目相等时任选一堆），取的数目只能为石子少的

那一堆的正整数倍。最后取完一堆石子者胜。问给定情况下先手胜负情况。

多列几项就会发现一个熟悉的身影：斐波拉契数列。最后的结论是如果两个数相等，或者两数之比大于斐

波拉契数列相邻两项之比的极限（(sqrt(5)+1)/2），则先手胜，否则后手胜。

#include<stdio.h>
#include <math.h>
int main()
{
int a,b;
double c=(1.0+sqrt(5.0))/2.0;
while(scanf("%d%d",&a,&b)!=EOF && (a ||b))
{
if(a==b)
{
printf("Stan wins\n");
continue;
}
if(a<b)
{
a=a^b;
b=a^b;
a=a^b;
}
if((double)a/b>c)
printf("Stan wins\n");
else
printf("Ollie wins\n");
}
}

转载于:https://www.cnblogs.com/lxm940130740/p/3281158.html