zoj 3882 Help Bob(zoj 2015年7月月赛)

标签：acm   zoj   algorithm   算法   编程   

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;
}


    

zoj 3882 Help Bob(zoj 2015年7月月赛)

标签：acm   zoj   algorithm   算法   编程   

原文地址：http://blog.csdn.net/caduca/article/details/47099775