Codeforces Round #399 (Div. 1 + Div. 2, combined) （博弈，Nim游戏）E.Game of Stones

Sam has been teaching Jon the Game of Stones to sharpen his mind and help him devise a strategy to fight the white walkers. The rules of this game are quite simple:

• The game starts with n piles of stones indexed from 1 to n. The i-th pile contains s_i stones.
• The players make their moves alternatively. A move is considered as removal of some number of stones from a pile. Removal of 0 stones does not count as a move.
• The player who is unable to make a move loses.

Now Jon believes that he is ready for battle, but Sam does not think so. To prove his argument, Sam suggested that they play a modified version of the game.

In this modified version, no move can be made more than once on a pile. For example, if 4 stones are removed from a pile, 4 stones cannot be removed from that pile again.

Sam sets up the game and makes the first move. Jon believes that Sam is just trying to prevent him from going to battle. Jon wants to know if he can win if both play optimally.

Input

First line consists of a single integer n (1 ≤ n ≤ 10⁶) — the number of piles.

Each of next n lines contains an integer s_i (1 ≤ s_i ≤ 60) — the number of stones in i-th pile.

Output

Print a single line containing "YES" (without quotes) if Jon wins, otherwise print "NO" (without quotes)

Example

Input

1
5

Output

NO

Input

2
1
2

Output

YES

Note

In the first case, Sam removes all the stones and Jon loses.

In second case, the following moves are possible by Sam: 

In each of these cases, last move can be made by Jon to win the game as follows: 

 

第一次见到Nim游戏的博弈题目。状态数为 n(n+1)/2 <=x 的最大x，之后就进行异或和即可。

参考词条：   Nim游戏  SG函数 Sprague–Grundy theorem

 1 #include <iostream>
 2 //#include<bits/stdc++.h>
 3 #include <stack>
 4 #include <queue>
 5 #include <map>
 6 #include <set>
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <algorithm>
10 using namespace std;
11 typedef long long ll;
12 typedef unsigned long long ull;
13 const int MAX=1e5+5;
14 int n;
15 int a[66];
16 int main()
17 {
18     int i,j,st=0,an=0;
19     a[0]=0;
20     for(i=1;i<=10;i++)
21     {
22         for(j=1;j<=i+1;j++)
23         {
24             a[++st]=i;
25         }
26     }
27     scanf("%d",&n);
28     while(n--)
29     {
30         scanf("%d",&st);
31         an^=a[st];
32     }
33     if(an==0)
34         printf("YES\n");
35     else
36         printf("NO\n");
37 }

 

转载于:https://www.cnblogs.com/quintessence/p/6423700.html