Help Bob
Time Limit: 2 Seconds Memory Limit: 65536 KB
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
winwin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;
}
