A Multiplication Game - HDU 1517 博弈

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本文深入探讨了一款独特的数学游戏——乘法挑战赛，玩家通过连续将初始数值乘以2到9，以率先达到或超过指定目标数者为胜。文章揭示了游戏背后的先手必胜策略，以及如何通过数学分析预测游戏结果。

A Multiplication Game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4259 Accepted Submission(s): 2435

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：有一个数字，开始为1，每个人可以把它乘以2-9，谁先得到大于等于n的数字，谁就获胜，问先手胜负。

思路：我们可以得到2-9先手必胜，10-18先手必败，19-18*9先手必胜，以此类推。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
struct node
{
    ll a,b;
    bool type;
}arr[110];
ll n;
int main()
{
    int i,j,k=0;
    arr[0].a=arr[0].b=1;
    while(true)
    {
        k++;
        if(k&1)
        {
            arr[k].a=arr[k-1].b+1;
            arr[k].b=arr[k-1].b*9;
        }
        else
        {
            arr[k].a=arr[k-1].b+1;
            arr[k].b=arr[k-1].b*2;
        }
        if(arr[k].b>4294967295LL)
          break;
    }
    while(~scanf("%I64d",&n))
    {
        for(i=1;i<=15;i++)
           if(arr[i].a<=n && n<=arr[i].b)
              break;

        if(i&1)
          printf("Stan wins.\n");
        else
          printf("Ollie wins.\n");
    }
}