(codeforce1699) a & B (construction)

A. The Third Three Number Problem

Topic link ：Problem - A - Codeforces

The sample input ：

5
4
1
12
2046
194723326

  Sample output ：

3 3 1
-1
2 4 6
69 420 666
12345678 87654321 100000000

The question ： Multiple sets of samples , One sample per group n, Ask if we can find three numbers that satisfy the equation

(a⊕b)+(b⊕c)+(a⊕c)=n(a⊕b)+(b⊕c)+(a⊕c)=n, If you can find it, output a,b,c, Otherwise output -1

analysis ：
Let's first look at what kind of n It must not be found a,b,c Satisfied with the question

First of all, there is ：

Odd numbers and odd numbers are XOR to get an even number

An even number is obtained by exclusive or of an even number

An odd number that is exclusive or of odd and even numbers

So when a,b,c When all three numbers are odd or even ,n It must be an even number

When a,b,c When two of the three numbers are odd , Might as well set a,b Is odd , Then there are n For me + p. + p. = accidentally

When a,b,c When two of the three numbers are even , Might as well set a,b It's even , Then there are n For me + p. + p. = accidentally

So no matter a,b,c It is impossible to get an odd number by taking any value , So when n When it is an odd number, it must be -1

When n For even when , We can make a=b=n>>1,c=0, This is a set of solutions that meet the meaning of the problem

Here's the code ：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const int N=1e5+10;
int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		long long n;
		scanf("%lld",&n);
		if(n&1) puts("-1");
		else
			printf("%lld %lld 0\n",n>>1,n>>1);
	}
	return 0;
} 

B. Almost Ternary Matrix

Topic link ：Problem - B - Codeforces

The sample input ：

3
2 4
2 2
4 4

Sample output ：

1 0 0 1
0 1 1 0
1 0
0 1
1 0 1 0
0 0 1 1
1 1 0 0
0 1 0 1

The question ： Multi group input , One sample per group n,m（n and m Is an even number ） Represents that we need to construct a n*m Matrix , Every number in the matrix is 0 perhaps 1, Meet each 0 There is 2 individual 1, Every 1 There is 2 individual 0, The surrounding of each grid refers to the four adjacent grids up, down, left and right .

analysis ： We can directly simulate the structure , I follow the example 1 Constructed in that way , namely ：

When i In an odd number of , We put i*2-1 Row sum i*2 Rows are constructed as follows  

0 1 1 0 0 1 1 0……

1 0 0 1 1 0 0 1……

When i For even when , We put i*2-1 Row sum i*2 Rows are constructed as follows  

1 0 0 1 1 0 0 1……

0 1 1 0 0 1 1 0……

i from 1 Traversing n/2 that will do .

Here is the code ：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const int N=1e2+10;
int s[N][N];
int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		int n,m,t;
		scanf("%d%d",&n,&m);
		for(int i=1;i*2<=n;i++)
		{
			if(i&1) t=1;
			else t=0;
			for(int j=1;j*2<=m;j++)
			{
				if((j+t)&1)
				{
					s[i*2-1][j*2-1]='0';
					s[i*2-1][j*2]='1';
					s[i*2][j*2-1]='1';
					s[i*2][j*2]='0';
				}
				else
				{
					s[i*2-1][j*2-1]='1';
					s[i*2-1][j*2]='0';
					s[i*2][j*2-1]='0';
					s[i*2][j*2]='1';
				}
			}
		}
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=m;j++)
				printf("%c ",s[i][j]);
			puts("");
		}
	}
	return 0;
}