A

注意到 \(a[i]\) 可以异或上任意多次 \(a[1 \to i - 1]\)，于是 求出 \(1\to i - 1\) 的线性基 \(V\)，能变成数的个数是 \(2^{|V|}\)。

评测记录

// Sparkle
#include <bits/stdc++.h>
#include <assert.h>
using namespace std;
typedef long long ll;
typedef double db;
#define ep emplace_back
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define fout freopen("out.out","w",stdout);
#define fin freopen("in.in","r",stdin);
#define dd(x) cerr << #x" = " << x << endl;
inline int read() {
	int x=0, v=1,ch=getchar();
	while('0'>ch||ch>'9') {
		if(ch=='-') v=0;
		ch=getchar();
	}while('0'<=ch&&ch<='9') {
		x=(x*10)+(ch^'0');
		ch=getchar();
	}return v?x:-x;
}
const int P = 998244353;
int n, p[35], sz;
void ins(int x) {
	for(int i = 30; i >= 0; -- i) {
		if(x >> i & 1) {
			if(p[i]) {
				x ^= p[i];
			}else {
				p[i] = x;
				++ sz;
				return;	
			}
		}
	}
}
int qpow(int x,int p) {
	int ret = 1;
	for(;p;p>>=1,x=1ll*x*x%P)if(p&1)ret=1ll*ret*x%P;
	return ret;
}
signed main() {
 	n = read();
	int ans = 1;
 	for(int i = 1; i <= n; ++ i) {
 		int x = read();
		ans = 1ll * ans * qpow(2, sz) % P;
		ins(x);
	}
	cout << ans;
	return 0;
}