POJ 2505 - A multiplication game （乘法博弈）

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A multiplication game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6826   Accepted: 3422 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. Source Waterloo local 2001.09.22

遇事不决先打表：

[1,9]  [10,18]  [19,162]  [163,324]  [325,2916]

A        B           A            B               A

发现规律：每个区间的有界  r = 9^x * 2 ^y；

对每个n进行判断即可

#pragma GCC optimize(2)
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
#define clr(a) memset(a,0,sizeof(a))
#define rep(i,x,y) for(int i=x;i<y;i++)
#define REP(i,x,y) for(int i=x;i<=y;i++)
#define pre(i,x,y) for(int i=x;i>=y;i--)
#define line cout<<"-----------------"<<endl;

typedef long long ll;
const int maxn = 1e5+10;
const int MAXN = 1e6+10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9+7;
const int N = 1010;


int main(){
	ll n;
	while(scanf("%lld", &n) != EOF){
		ll flag = 0;
		ll ans = 1;
		while(ans < n){
			if(flag & 1) ans *= 2;
			else ans *= 9;
			flag ++;
		}
		if(flag & 1) puts("Stan wins.");
		else puts("Ollie wins.");
	}
	return 0;
}