HDU 1517 A Multiplication Game

A Multiplication Game

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6079    Accepted Submission(s): 3450

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

题意：2 个人玩游戏，给定一个数n,从 1 开始，轮流对数进行累乘一个数（2~9中取），
直到第一次等于或超过n为赢.
如果n是 2 ~ 9 ，Stan 必胜。
如果输入是 10~18 ，不管第一次Stan 乘的是什么，Stan肯定在 2 ~ 9 之间，
  无论stan乘以什么，Ollie乘以大于1的数都都能超过 10 ~ 18 中的任何一个数。Ollie 必胜。
如果输入是 19 ~ 162，那么这个范围是 Stan 的必胜态。
如果输入是 163 ~ 324 ，这是又是Ollie的必胜态。

#include <iostream> 
#include <algorithm> 
#include <cstring> 
#include <cstdio>
#include <vector> 
#include <queue> 
#include <cstdlib> 
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime> 
#include <map> 
#include <set> 
using namespace std; 
#define lowbit(x) (x&(-x)) 
#define max(x,y) (x>y?x:y) 
#define min(x,y) (x<y?x:y) 
#define MAX 100000000000000000 
#define MOD 1000000007
#define pi acos(-1.0) 
#define ei exp(1) 
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
double n;
int main()
{
    while(scanf("%lf",&n)!=EOF)
    {
        while(n>18)
        {
            n/=18;
        }
        puts(n<=9?"Stan wins.":"Ollie wins.");
    }
    return 0;
}