题目链接：
http://poj.org/problem?id=2505
|
A multiplication game
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 |
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题目意思：
给一个n,两个人轮流玩游戏，从p=1开始，每次可以选择乘以2或者9，谁第一次到达n赢，假设两人都足够聪明。
1 < n < 4294967295
解题思路：
找P必败点，N必赢点
代码解释的很详细：
//#include<CSpreadSheet.h>
#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
int n;
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(~scanf("%d",&n))
{
int ans=0; //必输
int cur=n-1;
while(cur>=1)
{
if(!ans) //后面是必输的状态
cur/=9; //在cur/9+1~cur内的人肯定选择*9 获得必胜
else //后面是必赢
cur/=2; //至少要选择*2
// printf("ans:%d cur:%d\n",ans,cur);
//system("pause");
ans^=1;
}
if(ans)
printf("Stan wins.\n");
else
printf("Ollie wins.\n");
}
return 0;
}
[求PN点] poj 2505 A multiplication game,布布扣,bubuko.com
[求PN点] poj 2505 A multiplication game
原文：http://blog.csdn.net/cc_again/article/details/25904445