A. XORwice (Thinking) Codeforces Round #676 (Div. 2)

Link to the original question: https://codeforces.com/contest/1421/problem/A

Test sample

input
6
6 12
4 9
59 832
28 14
4925 2912
1 1
output
10
13
891
18
6237
0

Note

For the first test case Sana can choose x=4 and the value will be$(6\oplus 4)+(12\oplus 4)=2+8=10$. It can be shown that this is the smallest possible value.

Question: Give two numbers $a$ and$b$ , you are asked to choose a number$x$ such that$(a\oplus x)+(b\oplus x)\; has$ the largest result. (⊕ stands for exclusive OR operation)

Problem-solving ideas: Let’s think about it. If the XOR operation is performed, the bits of the same bit are taken as $0$ , if the bit is different, take$1$ , then we hope to make the same bit as possible so that the number will become smaller after the operation, so how to choose?Combine two numbers $a$，$b$ , Hypothesis$i$ rank up$a$ is 1,$b$ is$0$ , then no matter what$x$ of the$i$ will not change the size for any value, only when they are all$1$ when$1$ . After XOR operation, this bit becomes$0$ , the sum will become smaller.So what we have to do is to build a $x$ , so that it can optimize the above operations. We have two ways to solve it. The first method was solved during the game, which is to take out the2 2 of thetwo numbers.$2$ decimal digits, and then compared to give$x$ . The second method is particularly good, is to refer to$cf$ big guy, that is, the best result is actually$a\oplus b$ . why? Because this is the same number of bits, take 0, and different ones take$1$ . The final addition is indeed different digits, one of which is 0 and one is 1, then the addition is 1. This effect can be achieved by performing an exclusive OR operation directly. It is equivalent to a formula derivation, cow.

AC code 1

/*
*邮箱：[email protected]
*blog:https://me.csdn.net/hzf0701
*注：文章若有任何问题请私信我或评论区留言，谢谢支持。
*
*/
#include<bits/stdc++.h>	//POJ不支持

#define rep(i,a,n) for (int i=a;i<=n;i++)//i为循环变量，a为初始值，n为界限值，递增
#define per(i,a,n) for (int i=a;i>=n;i--)//i为循环变量， a为初始值，n为界限值，递减。
#define pb push_back
#define IOS ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define fi first
#define se second
#define mp make_pair

using namespace std;

const int inf = 0x3f3f3f3f;//无穷大
const int maxn = 1e5;//最大值。
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll>  pll;
typedef pair<int, int> pii;
//*******************************分割线，以上为自定义代码模板***************************************//

int t,n;
int a,b;
void solve(){
    
    
	vector<int> a1;
	vector<int> b1;
	int temp1=a,temp2=b;
	while(a){
    
    
		a1.push_back(1&a);
		a=a>>1;
	}
	while(b){
    
    
		b1.push_back(1&b);
		b=b>>1;
	}
	int len=min(a1.size(),b1.size());
	int ans=0;
	rep(i,0,len-1){
    
    
		ans+=a1[i]*pow(2,i);
	}
	cout<<(ans^temp1)+(ans^temp2)<<endl;
}
int main(){
    
    
	//freopen("in.txt", "r", stdin);//提交的时候要注释掉
	IOS;
	while(cin>>t){
    
    
		while(t--){
    
    
			cin>>a>>b;
			solve();
		}
	}
	return 0;
}

AC code 2

/*
*邮箱：[email protected]
*blog:https://me.csdn.net/hzf0701
*注：文章若有任何问题请私信我或评论区留言，谢谢支持。
*
*/
#include<bits/stdc++.h>	//POJ不支持

#define rep(i,a,n) for (int i=a;i<=n;i++)//i为循环变量，a为初始值，n为界限值，递增
#define per(i,a,n) for (int i=a;i>=n;i--)//i为循环变量， a为初始值，n为界限值，递减。
#define pb push_back
#define IOS ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define fi first
#define se second
#define mp make_pair

using namespace std;

const int inf = 0x3f3f3f3f;//无穷大
const int maxn = 1e5;//最大值。
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll>  pll;
typedef pair<int, int> pii;
//*******************************分割线，以上为自定义代码模板***************************************//

int t,n;
int a,b;
int main(){
    
    
	//freopen("in.txt", "r", stdin);//提交的时候要注释掉
	IOS;
	while(cin>>t){
    
    
		while(t--){
    
    
			cin>>a>>b;
			cout<<(a^b)<<endl;
		}
	}
	return 0;
}