HDOJ 5088 Revenge of Nim II 位运算

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在这款名为“复仇的尼姆游戏II”的策略游戏中，两名玩家轮流从不同的堆中移除物品。为了确保游戏的公平性，第二名玩家可以在游戏开始前完全移动一些堆（可以为零）。本篇详细介绍了游戏规则、输入输出格式及示例。

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位运算。。。。

Revenge of Nim II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 229 Accepted Submission(s): 79

Problem Description

Nim is a mathematical game of strategy in which two players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap.
---Wikipedia

Today, Nim takes revenge on you, again. As you know, the rule of Nim game is rather unfair, only the nim-sum (⊕) of the sizes of the heaps is zero will the first player lose. To ensure the fairness of the game, the second player has a chance to move some (can be zero) heaps before the game starts, but he has to move one heap entirely, i.e. not partially. Of course, he can’t move all heaps out, at least one heap should be left for playing. Will the second player have the chance to win this time?

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case begins with an integer N, indicating the number of heaps. Then N integer Ai follows, indicating the number of each heap.

[Technical Specification]
1. 1 <= T <= 100
2. 1 <= N <= 1 000
3. 1 <= Ai <= 1 000 000 000 000

Output

For each test case, output “Yes” if the second player can win by moving some (can be zero) heaps out, otherwise “No”.

Sample Input

Sample Output

  
   No
Yes
Yes

   
    
     Hint
    For the third test case, the second player can move heaps with 4 and 5 objects out, so the nim-sum of the sizes of the left heaps is 1⊕2⊕3 = 0.

Source

BestCoder Round #16

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int LL;

int n;
LL c[60];
LL a[1100];

int main()
{
    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        memset(c,0,sizeof(c));
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            cin>>a[i];
        bool flag=false;
        for(int i=1;i<=n;i++)
        {
            for(int j=60;j>=0;j--)
            {
                if(c[j]==0&&(a[i]&(1LL<<j)))
                {
                    c[j]=a[i];
                    break;
                }
                else if(a[i]&(1LL<<j))
                {
                    a[i]^=c[j];
                    if(a[i]==0) flag=true;
                }
            }
        }
        if(flag==true) puts("Yes");
        else puts("No");
    }
    return 0;
}