异或= =

最新推荐文章于 2025-12-27 16:49:17 发布

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文章描述了一个关于异或操作的问题，其中N个戴着标有非负整数红围巾的猫学习了异或操作并计算除了自己围巾上的数以外其他所有数的异或值。给定这些异或值，需要恢复出每只猫围巾上的原始整数。解题的关键在于利用异或一个数偶数次结果为零的性质，通过对初始输入数组进行异或求和，然后逐个与输入数组元素异或来恢复每个元素的值。

简介：

异或，英文为exclusive OR，缩写成xor。

运算规则：相同为1，不同为0，

1 ^ 0 == 1
1 ^ 1 == 0
0 ^ 0 == 0
0 ^ 1 == 0
// 通过观察可知，异或的结果是加法

性质：

a ^ a = 0;
a ^ 0 = a; // 二进制中加0相当于不加

题目：

There are NN Snuke Cats numbered 1,2,…,N1,2,…,N, where NN is even.

Each Snuke Cat wears a red scarf, on which his favorite non-negative integer is written.

Recently, they learned the operation called xor (exclusive OR).

What is xor? For example, .

They wanted to use this operation quickly, so each of them calculated the xor of the integers written on their scarfs except his scarf.

We know that the xor calculated by Snuke Cat ii, that is, the xor of the integers written on the scarfs except the scarf of Snuke Cat ii is aiai. Using this information, restore the integer written on the scarf of each Snuke Cat.

Constraints

All values in input are integers.
2≤N≤2000002≤N≤200000
NN is even.
0≤ai≤1090≤ai≤109
There exists a combination of integers on the scarfs that is consistent with the given information.

Input

Input is given from Standard Input in the following format:

Output

Print a line containing NN integers separated with space.

The ii-th of the integers from the left should represent the integer written on the scarf of Snuke Cat ii.

If there are multiple possible solutions, you may print any of them.

Sample 1

Inputcopy	Outputcopy
4 20 11 9 24	26 5 7 22

5 xor 7 xor 22=205 xor 7 xor 22=20
26 xor 7 xor 22=1126 xor 7 xor 22=11
26 xor 5 xor 22=926 xor 5 xor 22=9
26 xor 5 xor 7=2426 xor 5 xor 7=24

Thus, this output is consistent with the given information.

题意：输入一个n，然后输入n个数字，然后请输出n个数字，让他们的数字XOR的值能组成原来的数组

解题思路：其实这道题利用到了同一个数XOR偶数次等于0，这个原理，根据XOR的原理，位相同为0，不同为1，我们可以将开始输入的n个数XOR起来，然后输出的时候分别再和输入的元素XOR，就能得到新的元素。

AC代码：

#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 10;
typedef long long LL;

int a[N];

int main()
{
	int n;
	scanf("%d",&n);
	int sum = 0;
	for(int i = 0 ; i < n ; i ++)
	{
		scanf("%d",&a[i]);
		sum ^= a[i];
	}
	
	for(int i = 0 ; i < n ; i ++)  printf("%d ",sum^a[i]);
	
	return 0;
}