zoj 3882 Help Bob（博弈）

题目链接

Help Bob

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Time Limit: 2 Seconds       Memory Limit: 65536 KB

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There is a game very popular in ZJU at present, Bob didn't meant to participate in it. But he decided to join it after discovering a lot of pretty girls playing it.

There are n stones on the ground and they are marked as 1 to n respectively. There will be 2 players in each competition. And the game rules are simple, A and B take turns to move. Each round, one of them can only take 1 number away, and then pick out all the divisors of the choosed number. When anyone who can not take away 1 number any longer, he will fail the whole game.

Input

There are multiple cases. Each case include an integer number n (0 ≤ n ≤ 100).

Output

For each case, A win, output "win". If not, output"fail".

Sample Input1

3
4

Sample Output1

win
win

题意：N个数1到N。两个人轮流操作，每次操作可以把一个数及其因子拿掉，不能拿的那个人输。问先手是的输赢。

题解：

若n为1先手赢

若n为0先手输

若n>1,若先手第一手不拿一，那么他所面临的状态和 先手拿一后后手所面临的状态是一样的。若第一手先手不拿一，一定必输。那么先手改变策勒拿一一定赢。若拿一必赢，则先手必赢。所以先手必赢。

代码如下：

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<string>
#include<string.h>
#include<stack>
#include<vector>
#include<set>
#include<map>
typedef long long LL;
typedef unsigned long long LLU;
double pi = acos(-1);
const int nn = 210;
const int inff = 0x3fffffff;
const LL mod = 1000000007;
using namespace std;
int n;
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
        {
            puts("fail");
        }
        else
            puts("win");

    }
    return 0;
}