【bzoj1982】【Spoj2021】【Moving Pebbles】【博弈论】

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

题解：博弈论的题都好神奇。

考虑如果这些石子能够两两配对配成n/2组(n为偶数),则是先手必败的。

因为这种情况下无论先手如何操作,后手都可以让局面回到等价的状态。

其余情况下先手必胜,因为其余情况下先手都有办法将局面变成先手必败。

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,a[100010];
int main(){
cin>>n;
if (n&1){cout<<"first player"<<endl;return 0;}
for (int i=1;i<=n;i++)cin>>a[i];
sort(a+1,a+n+1);
for (int i=1;i<=n;i+=2)
if (a[i]!=a[i+1]){
cout<<"first player"<<endl;
return 0;
}
cout<<"second player"<<endl;
}