Codeforces Round ＃804 (Div. 2) Editorial(A-B)

C I really don't understand the solution of the question , Write first A,B

A — The Third Three Number Problem

The question

Give a number n , Find any three numbers a b c Satisfy

$(a\oplus b)+(b\oplus c)+(a\oplus c)=n$

Ideas

First of all, according to the
$a+b=a\oplus b+2\ast (a\&b)$
This formula shows that
$a\oplus b$ The parity of is with $a+b$ same
so $(a\oplus b)+(b\oplus c)+(a\oplus c)$ The parity of is with $(a+b)+(b+c)+(a+c)$ same , Is with the $2\ast (a+b+c)$ The parity of is the same , so It can be proved that n It must be an even number , Therefore, there is no solution in the case of odd numbers ;

And then the structure , According to the nature $a\oplus 0=a$ We Can construct 0 0 $\frac{n}{2}$ Three numbers

#include<bits/stdc++.h>
 
using namespace std;
 
int n,t; 
int main()
{
    
	cin>>t;
	while(t--)
	{
    
		cin>>n;
		
		if(n%2!=0) puts("-1");
		else cout<<"0 0 "<<n/2<<endl;
		
	}
}

B — Almost Ternary Matrix

The question

Matrix construction , There are exactly two different blocks that make each block connected

Ideas ：

hold n And m Get a little bigger and draw a picture to see the law

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;

int a,b;

int t;

int s[51][51];

int main()
{
    
	cin>>t;
	
	while(t--)
	{
    
		cin>>a>>b;
		
		s[1][1]=1;
		s[1][2]=0;
		s[2][1]=0;
		s[2][2]=1;// First construct  2*2  The smallest unit block of 
		
		for(int i=3;i<=b;i++) 
			if(i%2==0) s[1][i]=s[1][i-3];
			else s[1][i]=s[1][i-1];
		
		for(int i=3;i<=b;i++) 
			if(i%2==0) s[2][i]=s[2][i-3];
			else s[2][i]=s[2][i-1];// Construct the first two lines 
			
		for(int i=3;i<=a;i++)
		{
    
			if(i%2==0)
			{
    
				for(int j=1;j<=b;j++)
				{
    
					s[i][j]=s[i-3][j];
				}
			}
			else
			{
    
				for(int j=1;j<=b;j++)
				{
    
					s[i][j]=s[i-1][j];
				}
			}
		}// Construct the following line 
		
		for(int i=1;i<=a;i++)
		{
    
			for(int j=1;j<=b;j++)
			{
    
				cout<<s[i][j]<<" ";
			}
			puts("");
		}
				
	}// Output 

    return 0;
}