Hdu 1517 博弈论

A Multiplication Game

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3625    Accepted Submission(s): 2057

Problem Description

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.

 

Input

Each line of input contains one integer number n.

 

Output

For each line of input output one line either 

Stan wins. 

or 

Ollie wins.

assuming that both of them play perfectly.

 

Sample Input

162
17
34012226

 

Sample Output

Stan wins.
Ollie wins.
Stan wins.

 

题意就是一个人每次操作可以乘上2-9任意一个数字，谁先超过n就算赢。

题解：博弈论题目。一般先列举状态找下规律想一下。n=2-9，肯定先手必胜，n=9-18肯定后手必胜，然后第三个状态就值得商酌了。n=18-162先手必胜，为什么，想一下吧。然后就能找到规律了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
using namespace std;
#define LL __int64
//#define DEBUG
int a[10000000];
int main()
{
	a[1]=1;
	a[2]=9;
	LL s=9;
	LL t=3;
	LL n;
	int f=0,i;
	while (s<=4294967295)
	{
		if (f==0) s*=2;
		else s*=9;
		f=!f;
		a[t++]=s;
	}
	while (~scanf("%I64d",&n))
	{
		int pos=t;
		for (i=1;i<t;i++)
		{
			if (n<=a[i])
			{
				pos=i;
				break;
			}
		}
		if (pos % 2==0) printf("Stan wins.\n");
		else printf("Ollie wins.\n");
	}
	return 0;
}