768EGame of Stones[Nim游戏][博弈]

E. Game of Stones time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output

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 si stones. The players make their moves alternatively. A move is considered as removal of some number of stones from a pile. Removal of 0stones 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 si (1 ≤ si ≤ 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)

Examples 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: 

题意：Nim游戏的规则，n堆石头，没一堆石头有ki个，在有一定策略的情况下，求先手能不能必胜。其中与nim游戏的差别是，从一堆石头中取出m个以后，则以后不能从同一堆石头里面取m个。

根据Nim游戏的结论，可以用对所有的k求异或，其每一个向量的运算都是基于ki原因是最大能够移动ki步，现在在题目要求的约束下，先求出每一堆石头的最大取出次数即可。

代码：

#include<iostream> using namespace std; const int maxn=1e6+5; int a[maxn]; int ans, n, x; int main() { cin>>n; for(int i=0; i<n; i++) { cin >> x; int j; for(j=1; j<=60; j++) { x-=j; if(x<=j)break; } a[i]=j;//最大步数 ans^=a[i]; } if(ans)cout << "NO"; else cout << "YES"; }