CF1944B 题解

异或的性质

1. $x$ 异或 $x$ 等于 $0$；
2. $0$ 异或 $x$ 是 $x$。

思路

注意题目中说每个数恰好出现 $2$ 次，所以每个数可能在两边各出现一个，这种情况要两边一起选，使结果里多这个数；也可能在同一边出现两次，这样需要把它们两个数一起选，但对这个数不起作用。

代码

# include <bits/stdc++.h>
using namespace std;
int t, n, x, k, v1[50005], v2[50005], tot1, tot2;
bitset <50005> vis1, vis2;
int main () {
	ios::sync_with_stdio (0);
	cin.tie (0);
	cout.tie (0);
	cin >> t;
	while (t --) {
		cin >> n >> k;
		for (int i = 1; i <= n; ++ i)
			vis1[i] = vis2[i] = 0;
		for (int i = 1; i <= n; ++ i)
			cin >> x, vis1[x] = 1;
		for (int i = 1; i <= n; ++ i)
			cin >> x, vis2[x] = 1;
		tot1 = tot2 = 0;
		for (int i = 1; i <= n; ++ i)
			if (vis1[i] && ! vis2[i])
				v1[++ tot1] = i;
			else if (vis2[i] && ! vis1[i])
				v2[++ tot2] = i; 
		if (tot1 >= k) {
			for (int i = 1; i <= k; ++ i)
				cout << v1[i] << ' ' << v1[i] << ' ';
			cout << '\n';
			for (int i = 1; i <= k; ++ i)
				cout << v2[i] << ' ' << v2[i] << ' ';
			cout << '\n';
		} else {
			for (int i = 1; i <= n; ++ i)
				if (vis1[i] && vis2[i])
					++ tot2, v1[tot2] = v2[tot2] = i;
			tot2 = k * 2 - tot1;
			for (int i = 1; i <= tot1; ++ i)
				cout << v1[i] << ' ' << v1[i] << ' ';
			for (int i = tot1 + 1; i <= tot2; ++ i)
				cout << v1[i] << ' ';
			cout << '\n';
			for (int i = 1; i <= tot1; ++ i)
				cout << v2[i] << ' ' << v2[i] << ' ';
			for (int i = tot1 + 1; i <= tot2; ++ i)
				cout << v2[i] << ' ';
			cout << '\n';
		}
	}
	return 0;
}