题意：两个玩家玩一个游戏，从 p = 1，开始，然后依次轮流选择一个2 - 9的数乘以 p，问你谁先凑够 p >= n。
析：找规律，我先打了一下SG函数的表，然后就找到规律了
我找到的是：
1 - 9 Stan wins. 1 ~ 9
10 - 18 Ollie wins. 9+1 ~ 9*2
19 - 162 Stan wins. 9*2+1 ~ 9*2*9
163 - 324 Ollie wins. 9*2*9+1 ~ 9*2*9*2
325 - 2916 Stan wins. 9*2*9*2+1 ~ 9*2*9*2*9
2917 - 5832 Ollie wins. 9*2*9*2*9+1 ~ 9*2*9*2*9*2
规律就很明显了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 10;
const int maxm = 100 + 2;
const LL mod = 100000000;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
int main(){
unsigned int n;
while(scanf("%d", &n) == 1){
LL p = 1;
while(1){
p *= 9;
if(p >= n){ puts("Stan wins."); break; }
p *= 2;
if(p >= n){ puts("Ollie wins."); break; }
}
}
return 0;
}