Vlad and a Pair of Numbers

题目链接

题目描述：

Vlad found two positive numbers \(a\) and \(b(a,b>0)\). He discovered that \(a⊕b=\frac{a+b}{2}\), where \(⊕\) means the bitwise exclusive OR , and division is performed without rounding..

Since it is easier to remember one number than two, Vlad remembered only \(a⊕b\), let's denote this number as \(x\) . Help him find any suitable \(a\) and \(b\) or tell him that they do not exist.

输入描述：

The first line of the input data contains the single integer \(t(1≤t≤10^4)\) — the number of test cases in the test.

Each test case is described by a single integer \(x (1≤x≤2^{29})\) — the number that Vlad remembered.

输出描述：

Output \(t\) lines, each of which is the answer to the corresponding test case. As the answer, output \(a\) and \(b(0<a,b≤2^{32})\), such that \(x=a⊕b=\frac{a+b}{2}\)
. If there are several answers, output any of them. If there are no matching pairs, output \(-1\).

样例：

input:

6
2
5
10
6
18
36

output:

3 1
-1
13 7
-1
25 11
50 22

AC代码：

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

// 对于一个数x，某个位上是1的情况下
// 让a在对应位上为1，b在对应位上为0
// 而对于x某个位上是0的情况下
// 先贪心地让a，b对应位都为1
// 最后判断一下是否符合题意即可
void solve()
{
	int x;
	cin >> x;

	// a和b必是一个大于x，一个小于x
	int a = x;
	int b = 0;

	for(int i = 29; i >= 0; i --)
	{
		// 防止前面有0的情况，比如10 = (000...00001010)
		if(x & (1 << i) > 0)
			continue;

		// 在某个位是0的情况下，假设a，b对应位都为1，所以就加上两倍的(1 << i)
		// 只要小于等于2 * x就可以
		if(2 * x >= (2 << i) + a + b)
		{	
			a += (1 << i);
			b += (1 << i);
		}
	}

	if(a + b == 2 * x && (a ^ b) == x)
		cout << a << ' ' << b << '\n';
	else 
		cout << -1 << '\n';
}

int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

	int T;
	cin >> T;

	while(T --)
		solve();

	return 0;
}