ZOJ 月赛 Help Bob-优快云博客

探讨了在特定游戏规则下，如何通过选择石子编号数及其因数，实现先手必胜的目标。通过数学归纳法证明了n=0时先手必输，其余情况下先手必胜的结论。

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
win
win

题意：给n个石子，标记为1，2，……，n;双方人选一个存在的石子编号数，去掉该数的全部因数编号石子。问先手是否必胜。

这个题虽然很多人过了，但是却都不知道怎么过的，n = 0时，先手必输，其余情况先手必胜。

这里我严格证明这个答案的正确性：
先枚举找规律：
n = 0 : 为{}， sg = 0;

n = 1 : 可选1，得{}，其 sg = 1;

n = 2：可选1，得{2}，sg = 1;
可选2，得{}，sg = 0;故最终sg = 1;

n = 3: 可选1，得{2， 3}
可选2，得{3}；
可选3，得{2}；
而对{2，3}，可选2，得{3}；也可选3，得{2}。
这里，留意{2， 3} 推出的两个子集恰与n = 3推出的剩余两个集合相同。若要sg[n = 3] = 1，则必须有其推出的一个集合sg值为0,而由{2，3}的sg值由{2}、{3}确定，若{2}、{3}的sg值为1，则{2，3}的sg值为0，则可得sg[n=3] = 1;而若{2}、{3}中有一个sg值为0, 则也可得sg[n = 3] = 1. 故sg[n = 3] = 1;

……
你将发现，这不仅对于n=3时，sg = 1比成立，n > 3时，sg = 1同样成立。
O(∩_∩)O~ 可能没讲得太清楚，大家可以留言给我交流。

#include <stdio.h>
int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        if (n == 0) printf("win\n");
        else printf("fail\n");
    }
}