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
題目大意:
有1到n個數字,兩個人輪流選1個數,並把它的所有約數擦去。擦去最後一個數的人贏,輸出先手必勝還是必敗。
解題思路:
如果後手能贏,也就是說後手有必勝策略,使得先手第一次無論取哪個石子,後手都能獲得最後的勝利。那麼現在假設先手取1,接下來
後手通過某種取法使自己進入必勝狀態,但是,先手第1次取得時候就可以和後手這次取的一樣,搶先進入必勝局面。於是除了0,其他都是
必勝。
代碼:
#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;
}