zoj3882 Help Bob 博弈

题意：两个人博弈，地上有n个标号从1~n的石子，然后每人依次拿走一个石头，在拿走一个石头的同时，它的因数石子也会被同时去掉，0<=n<=60。最后石子都被拿光后，没的石子拿的那个人输掉。给你不同的n，让你求先手胜还是负。

解题思路：想明白了很简单，想不通就做不出来。

假设存在一种n并且先手选2~n，是先手必败，那么后手就是必胜了，此时1已经被取走了，因为1是所有数的因数。那么如此来看，先手如果不选2~n，而选1，那么必败态就留给后手了，所以无论如何主动权都是在先手的人，先手可以选择是否先拿走1，所以先手必胜。当然，n=0除外。.

Help Bob

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Time Limit: 2 Seconds       Memory Limit: 65536 KB

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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

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Author:  ZHANG, Yunxiao
Source:  ZOJ Monthly, July 2014

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <stack>
using namespace std;
typedef long long LL;
const int INF=0x7fffffff;
const int MAX_N=10000;

int a;
int main(){
    while(cin>>a){
        if(a==0)printf("fail\n");
        else printf("win\n");
    }
    return 0;
}