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.
There are multiple cases. Each case include an integer number n (0 ≤ n ≤ 100).
For each case, A win, output "win". If not, output"fail".
3 4
winwin
题目大意:
有1到n个数字,两个人轮流选1个数,并把它的所有约数擦去。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,使得先手第一次无论取哪个石子,后手都能获得最后的胜利。那么现在假设先手取1,接下来
后手通过某种取法使自己进入必胜状态,但是,先手第1次取得时候就可以和后手这次取的一样,抢先进入必胜局面。于是除了0,其他都是
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; }