Playing games

Subject:
https://ac.nowcoder.com/acm/contest/295/H

There are n numbers, choose as many numbers as possible so that the exclusive or sum is $0$ .
$1\le n\le 500000,0\le a_i\le 500000$
Idea: The
problem is equivalent to choosing as few numbers as possible so that the XOR sum is the XOR sum of all numbers.$val$ . According to the linear basis idea, it can be deduced that the exclusive OR of the entire set can be no more than$19$ XOR set representation. So the answer does not exceed$19$ . So the answer can be dichotomy.
So how to judge themid midCan the number of$m$$i$$d$ be XORed into$val$ , you can use the generating function$A={\sum }_{i=0}^{500000}{a}_i{x}^i$，$a_i=1$ DisplayiiThe number$i$ exists and is$0$ means not present,$A^k$ means choose$k$ , but here is the exclusive OR of the polynomial, use$FWT$ , and calculate$B=A^k$ , you can directly calculate$FWT(B)=FWT(A{)}^k$ , so just put$A$ becomes$FWT\left(A\right)$ , then countkkfor each bit$k\; to$ the power, then$FWT\left(B\right)$ becomes$B$ is fine, similar to point value notation.
Also note that this will select repetitions, so it is impossible to directly judge whether you can just select$The$ number of$m$$i$$d$ , but the repeated XOR will cancel out, which is equivalent to the number selected is less than$mid$ , because it is a binary judgment, it is the second best to judge whether you can use less than$The$ number of$m$$i$$d$ , then a function$a_0=1.$ Representatives may not choose.

Summary:
The convolution form of XOR is similar to the convolution obtained by addition, and the generating function can also be used.