C. Engineer Artem

最新推荐文章于 2021-09-24 11:05:38 发布

原创最新推荐文章于 2021-09-24 11:05:38 发布 · 320 阅读

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该博客讨论了一个编程问题，涉及到创建一个新的矩阵，使得相邻元素不相等。作者给出了一个C++代码示例，该代码通过递增某些元素值来确保矩阵变为'好'矩阵，即没有相邻元素相等。每个测试用例中，矩阵的大小和初始值都是输入的，然后输出修改后的矩阵。示例展示了如何满足条件并确保解决方案的存在性和唯一性。

https://codeforces.ml/contest/1438/problem/C

Artem is building a new robot. He has a matrix aa consisting of nn rows and mm columns. The cell located on the ii-th row from the top and the jj-th column from the left has a value ai,jai,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 aa good.

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

For all valid (i,ji,j), either bi,j=ai,jbi,j=ai,j or bi,j=ai,j+1bi,j=ai,j+1.

For the constraints of this problem, it can be shown that such a matrix bb 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 tt (1≤t≤101≤t≤10). Description of the test cases follows.

The first line of each test case contains two integers n,mn,m (1≤n≤1001≤n≤100, 1≤m≤1001≤m≤100) — the number of rows and columns, respectively.

The following nn lines each contain mm integers. The jj-th integer in the ii-th line is ai,jai,j (1≤ai,j≤1091≤ai,j≤109).

Output

For each case, output nn lines each containing mm integers. The jj-th integer in the ii-th line is bi,jbi,j.

Example

input

Copy

output

Copy

Note

In all the cases, you can verify that no two adjacent cells have the same value and that bb is the same as aa with some values incremented by one.

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=5e5+5;
const int inf=0x3f3f3f3f;
ll n,t,m,s;
ll a[101][101];
int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        s=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                cin>>a[i][j];
                if((i+j)%2==a[i][j]%2)
                {
                    a[i][j]++;
                }
                cout<<a[i][j]<<" ";
            }
            cout<<endl;
        }
    }
    return 0;
}