具有给定约束的N * N矩阵中的最大个数

给定两个整数$n$和$x$ ， 在哪里$x <= n$ 。找出一个人的最大人数$n * n$二进制矩阵可以具有每个大小的子矩阵$x * x$至少有一个像元为零。

例子：

Input:5 3
Output: Maximum number of ones = 24
The matrix will be:
1 1 1 1 1
1 1 1 1 1 
1 1 0 1 1 
1 1 1 1 1
1 1 1 1 1 

Input:5 2
Output: Maximum number of ones = 21
The matrix will be:
1 1 1 1 1
1 0 1 0 1 
1 1 1 1 1 
1 0 1 0 1
1 1 1 1 1 

方法可以使用贪婪方法解决问题。在第一个正方形子矩阵的右下角放置一个零，即坐标为(1，1)和(x，x)的子矩阵，并对称地创建其余矩阵，我们可以得到最小值零个数，或者最大个数。因此，通过观察，可以得出一个共同的结论，即存在$\left \lfloor (\frac{n}{x})^2 \right \rfloor$零的数目(以最小排列方式)。可用单元总数为$n^2$在NxN矩阵中。

$max(ones) = n^2 - \left \lfloor (\frac{n}{x})^2 \right \rfloor$.

下面是上述方法的实现：

C++

// C++ program to get Maximum Number of
// ones in a matrix with given constraints
#include 
  
using namespace std;
  
// Function that returns the maximum 
// number  of ones 
int getMaxOnes(int n, int x)
{
    // Minimum number of zeroes
    int zeroes = (n / x);
    zeroes = zeroes * zeroes;
  
    // Totol cells = square of the size of the matrices
    int total = n * n;
  
    // Intialising the answer
    int ans =  total - zeroes;
  
    return ans; 
}
  
// Driver code
int main()
{
    // Intitalising the variables
    int n = 5;
    int x = 2;
  
      
    cout << getMaxOnes(n, x);
  
    return 0;
}

Java

// Java program to get Maximum
// Number of ones in a matrix 
// with given constraints
import java.io.*;
  
class GFG
{
      
// Function that returns 
// the maximum number of ones 
static int getMaxOnes(int n, 
                      int x)
{
    // Minimum number of zeroes
    int zeroes = (n / x);
    zeroes = zeroes * zeroes;
  
    // Totol cells = square of
    // the size of the matrices
    int total = n * n;
  
    // Intialising the answer
    int ans = total - zeroes;
  
    return ans; 
}
  
// Driver code
public static void main (String[] args) 
{
  
// Intitalising the variables
int n = 5;
int x = 2;
System.out.println(getMaxOnes(n, x));
}
}
  
// This code is contributed
// by akt_mit

Python3

# Python3 program to get 
# Maximum Number of ones 
# in a matrix with given
# constraints
  
# Function that returns 
# the maximum number of ones 
def getMaxOnes(n, x):
      
    # Minimum number
    # of zeroes
    zeroes = (int)(n / x);
    zeroes = zeroes * zeroes;
  
    # Totol cells = square of
    # the size of the matrices
    total = n * n;
  
    # Intialising 
    # the answer
    ans = total - zeroes;
  
    return ans; 
  
# Driver code
  
# Intitalising the variables
n = 5;
x = 2;
print(getMaxOnes(n, x));
  
# This code is contributed
# by mits

C#

// C# program to get Maximum
// Number of ones in a matrix 
// with given constraints
using System;
  
class GFG
{
      
// Function that returns 
// the maximum number of ones 
static int getMaxOnes(int n, 
                      int x)
{
    // Minimum number of zeroes
    int zeroes = (n / x);
    zeroes = zeroes * zeroes;
  
    // Totol cells = square of
    // the size of the matrices
    int total = n * n;
  
    // Intialising the answer
    int ans = total - zeroes;
  
    return ans; 
}
  
// Driver code
static public void Main ()
{
          
    // Intitalising the
    // variables
    int n = 5;
    int x = 2;
    Console.WriteLine(getMaxOnes(n, x));
}
}
  
// This code is contributed
// by ajit

输出：

21

时间复杂度： O(1)
辅助空间： O(1)