Revenge of Nim II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 443 Accepted Submission(s): 157
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
3 1 2 3 2 2 2 5 1 2 3 4 5
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.
题意：RT
思路：将每个数的每一位看成列，n个数看成n行，由于数值是<=1e12的，所以最多也就40列，如果行数大于40，那么一定可以用高斯消元得到至少有两个数是相等的（一个数是0也行）
如果n<=40那么直接高斯消元求解即可