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最新推荐文章于 2026-01-04 16:35:21 发布

原创最新推荐文章于 2026-01-04 16:35:21 发布 · 697 阅读

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#integer #input #output #numbers #each #算法

joj博弈问题专栏收录该内容

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本文介绍了一个名为“乘法游戏”的博弈论问题，通过玩家交替将整数与2到9之间的数字相乘来达到目标值。文章提供了示例输入输出，并分享了C++实现代码，展示了如何判断赢家。

1889: A multiplication game

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	362	139	Standard

Stan and Ollie play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Stan always starts with p = 1, does his multiplication, then Ollie multiplies the number, then Stan and so on. Before a game starts, they draw an integer 1 < n < 4294967295 and the winner is who first reaches p >= n. Each line of input contains one integer number n. For each line of input output one line either

Stan wins.

Ollie wins.

assuming that both of them play perfectly. There should be one more blank line at the end of your output.

Sample input

162
17
34012226

Output for sample input

Stan wins.
Ollie wins.
Stan wins.

Submit / Problem List / Status / Discuss

这个题目的关键是了解博弈算法当中的关于最大最小策略。这既是一个博弈论的问题又是一个离散数学的问题。

#include <iostream>
#include<stdio.h>
using namespace std;

int main()
{
int n;
while(scanf("%d",&n)==1)
{
int p=1;
while(p<n)
{
p=p*9;
if(p>=n)
{
cout<<"Stan wins."<<endl;
break;
}
p=p*2;
if(p>=n)
{
cout<<"Ollie wins."<<endl;;
break;
}
}
}
cout<<endl;
return 0;
}