【Codeforces 1438 C】Engineer Artem，矩阵，奇偶性构造

problem

C. Engineer Artem
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value ai,j written in it.

If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.

Artem wants to increment the values in some cells by one to make a good.

More formally, find a good matrix b that satisfies the following condition —

For all valid (i,j), either bi,j=ai,j or bi,j=ai,j+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.

Input
Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤10). Description of the test cases follows.

The first line of each test case contains two integers n,m (1≤n≤100, 1≤m≤100) — the number of rows and columns, respectively.

The following n lines each contain m integers. The j-th integer in the i-th line is ai,j (1≤ai,j≤109).

Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is bi,j.

Example
inputCopy
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
outputCopy
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one.

solution

/*
题意：
+ 给出一个n*m的矩阵。可以对每个元素+1或不变，使得每个元素的相邻元素都不相同，输出最后满足条件的矩阵。
思路：
+ 构造形如"奇偶奇偶奇偶，偶奇偶奇偶奇"的矩阵。
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 600000+10;
const int mod = 1e9+7;

int main(){
	ios::sync_with_stdio(false);
	int T;  cin>>T;
	while(T--){
		int n, m;  cin>>n>>m;
		for(int i = 1; i <= n; i++){
			for(int j = 1; j <= m; j++){
				int x;  cin>>x;
				if((i+j)%2==0 && x%2==0)x++;
				if((i+j)%2==1 && x%2==1)x++;
				cout<<x<<" ";
			}
			cout<<"\n";
		}
	}
	return 0;
}

  

 

  
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文章来源: gwj1314.blog.csdn.net，作者：小哈里，版权归原作者所有，如需转载，请联系作者。

原文链接：gwj1314.blog.csdn.net/article/details/114071032