zoj 3882 Help Bob(zoj 7月月赛)

Posted 2020-09-30 mfmdaoyou

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


题目大意：
    有1到n个数字，两个人轮流选1个数。并把它的全部约数擦去。擦去最后一个数的人赢，输出先手必胜还是必败。

解题思路：
    如果后手能赢。也就是说后手有必胜策略，使得先手第一次不管取哪个石子，后手都能获得最后的胜利。那么如今如果先手取1，接下来
后手通过某种取法使自己进入必胜状态，可是，先手第1次取得时候就能够和后手这次取的一样，抢先进入必胜局面。于是除了0。其它都是
必胜。

代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==0)
        printf("fail\n");
        else
        printf("win\n");
    }
    return 0;
}