【hdu 1517】A Multiplication Game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5536 Accepted Submission(s): 3151

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.

【题目链接】:http://acm.hdu.edu.cn/showproblem.php?pid=1517

【题解】

用sg函数容易推出来小数据的答案；
如下”()”表示先手赢而”{}”表示先手输

一开始l = 2,r = 9;
递推的话
l = r+1;
如果这次是先手赢
则接下来是先手输的态
r = r*2;
如果这次是先手输
则接下来是先手赢的态
r = r*9;
(感性理解:让先手输比较难)

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

LL n;

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    while (~scanf("%I64d",&n))
    {
        int now = 1;
        LL l = 2,r = 9;
        while (true)
        {
            if (l<=n && n <= r)
                break;
            l = r+1;
            if (now==1)
                r = r*2;
            else
                r = r*9;
            now ^= 1;
        }
        if (now)
            puts("Stan wins.");
        else
            puts("Ollie wins.");
    }
    return 0;
}