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Title: Give you a set of data, let you choose as many numbers as possible, and choose the number between the two binary representation of different digits to be greater than 1.
Thought: Reverse thinking, if there is only one difference between the two numbers, then we will build an edge between the two numbers. So that is the problem of finding a maximum independent set.
theorem

But this problem requires not only the size, but also the output of the selected number.
So we can first find the minimum point coverage of this bipartite graph, and then the rest is the points of the largest independent set.
Finding the Minimum Point Covering of Bipartite Graphs

For this problem, we require the largest independent set, so the opposite is to go to the left marked point and the right unmarked point.

Here, when we construct a bipartite graph, we can divide left and right nodes according to the parity of the number of 1 under the binary of each number.

#include<bits/stdc++.h>
using namespace std;
const int N = 5010,M = N*N;
int n;
int a[N];
int head[N], to[M], nextt[M], cnt = 2;
int vis[N], match[N],sign[N];
int L[N], R[N], tot1, tot2;
void add(int u, int v) {
	to[cnt] = v, nextt[cnt] = head[u], head[u] = cnt++;
}
bool solve(int x, int y) {
	bitset<32>a(x), b(y);
	int ans = 0;
	for (int i = 0; i < 32; i++)if (a[i] != b[i])ans++;
	return ans == 1;
}

bool dfs(int x) {
	vis[x] = 1;
	for (int i = head[x], y; i; i = nextt[i]) {
		if (!vis[y = to[i]]) {
			vis[y] = 1;
			if (!match[y] || dfs(match[y])) {
				match[y] = x; match[x] = y; return true;
			}
		}
	}
	return false;
}

int main() {
	scanf("%d",&n);
	for (int i = 1; i <= n; i++)scanf("%d",a+i);
	for (int i = 1; i <= n; i++) {
		bitset<32>b(a[i]);
		if (b.count() % 2)L[++tot1] = i;
		else R[++tot2] = i;
	}

	for (int i = 1; i <= tot1; i++) {
		for (int j = 1; j <= tot2; j++) {
			if (solve(a[L[i]], a[R[j]]))add(L[i], R[j]);
		}
	}

	int ans = 0;
	for (int i = 1; i <= tot1; i++) {
		memset(vis,0,sizeof(vis));
		if (dfs(L[i]))ans++;
	}
	
	memset(vis, 0, sizeof(vis));
	for (int i = 1; i <= tot1; i++) {
		if (match[L[i]])continue;
		dfs(L[i]);
	}

	printf("%d\n",n-ans);
	for (int i = 1; i <= tot1; i++) {
		if (vis[L[i]])printf("%d ",a[L[i]]);
	}
	for (int i = 1; i <= tot2; i++) {
		if (vis[R[i]]==0)printf("%d ", a[R[i]]);
	}
	return 0;
}