Bzoj 1982: [Spoj 2021]Moving Pebbles 博弈论

1982: [Spoj 2021]Moving Pebbles

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 130  Solved: 88
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Description

2021. Moving Pebbles Two players play the following game. At the beginning of the game they start with n (1<=n<=100000) piles of stones. At each step of the game, the player chooses a pile and remove at least one stone from this pile and move zero or more stones from this pile to any other pile that still has stones. A player loses if he has no more possible moves. Given the initial piles, determine who wins: the first player, or the second player, if both play perfectly. 给你N堆Stone，两个人玩游戏. 每次任选一堆，首先拿掉至少一个石头,然后移动任意个石子到任意堆中. 谁不能移动了，谁就输了...

Input

Each line of input has integers 0 < n <= 100000, followed by n positive integers denoting the initial piles. 

Output

For each line of input, output "first player" if first player can force a win, or "second player", if the second player can force a win. 

Sample Input

3 2 1 3

Sample Output

first player

HINT

鸣谢lqp18_31..

Source

 题解：

在纸上画画就可以得出必败态为：n为偶数且可以分成n/2组两两相同的石子堆。

例如：

n=8 

石子为：1 1 6 6 8 8 8 8

博弈的题都不太好想，要从小到大一个一个去尝试。

但。。。

代码。。。

#include<bits/stdc++.h>
using namespace std;
int a[100010];
int main()
{
    int n,i;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)scanf("%d",&a[i]);
        sort(a+1,a+n+1);
        if(n%2==0)
        {
            for(i=1;i<=n;i+=2)if(a[i]!=a[i+1])break;
            if(i>n){printf("second player\n");continue;}
        }
        printf("first player\n");
    }
    return 0;
}