矩阵之和，其中每个元素是其行号和列号的绝对差

给定一个正整数n 。考虑一个n行n列的矩阵，其中每个元素都包含其行数和行数的绝对差。任务是计算矩阵的每个元素的总和。
例子 ：

Input : n = 2
Output : 2
Matrix formed with n = 2 with given constraint:
0 1
1 0
Sum of matrix = 2.

Input : n = 3
Output : 8
Matrix formed with n = 3 with given constraint:
0 1 2
1 0 1
2 1 0
Sum of matrix = 8.

方法1（蛮力）：
只需构造一个 n 行 n 列的矩阵，并用其对应的行号和列号的绝对差来初始化每个单元格。现在，找到每个单元格的总和。
以下是上述想法的实现：

C++

// C++ program to find sum of matrix in which each
// element is absolute difference of its corresponding
// row and column number row.
#include
using namespace std;
 
// Return the sum of matrix in which each element
// is absolute difference of its corresponding row
// and column number row
int findSum(int n)
{
    // Generate matrix
    int arr[n][n];
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            arr[i][j] = abs(i - j);
 
    // Compute sum
    int sum = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            sum += arr[i][j];
 
    return sum;
}
 
// Driven Program
int main()
{
    int n = 3;
    cout << findSum(n) << endl;
    return 0;
}

Java

// Java program to find sum of matrix
// in which each element is absolute
// difference of its corresponding
// row and column number row.
import java.io.*;
 
public class GFG {
 
// Return the sum of matrix in which
// each element is absolute difference
// of its corresponding row and column
// number row
static int findSum(int n)
{
     
    // Generate matrix
    int [][]arr = new int[n][n];
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            arr[i][j] = Math.abs(i - j);
 
    // Compute sum
    int sum = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            sum += arr[i][j];
 
    return sum;
}
 
    // Driver Code
    static public void main (String[] args)
    {
        int n = 3;
        System.out.println(findSum(n));
    }
}
 
// This code is contributed by vt_m.

Python3

# Python3 program to find sum of matrix
# in which each element is absolute
# difference of its corresponding
# row and column number row.
 
# Return the sum of matrix in which each
# element is absolute difference of its
# corresponding row and column number row
def findSum(n):
 
    # Generate matrix
    arr = [[0 for x in range(n)]
              for y in range (n)]
    for i in range (n):
        for j in range (n):
            arr[i][j] = abs(i - j)
 
    # Compute sum
    sum = 0
    for i in range (n):
        for j in range(n):
            sum += arr[i][j]
 
    return sum
 
# Driver Code
if __name__ == "__main__":
 
    n = 3
    print (findSum(n))
     
# This code is contributed by ita_c

C#

// C# program to find sum of matrix
// in which each element is absolute
// difference of its corresponding
// row and column number row.
using System;
 
public class GFG {
 
// Return the sum of matrix in which
// each element is absolute difference
// of its corresponding row and column
// number row
static int findSum(int n)
{
     
// Generate matrix
    int [,]arr = new int[n, n];
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            arr[i,j ] = Math.Abs(i - j);
  
    // Compute sum
    int sum = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            sum += arr[i, j];
  
    return sum;
}
 
    // Driver Code
    static public void Main(String[] args)
    {
        int n = 3;
        Console.WriteLine(findSum(n));
    }
}
 
// This code is contributed by vt_m.

Javascript

// Javascript program to find sum of matrix
// in which each element is absolute
// difference of its corresponding
// row and column number row.
     
    // Return the sum of matrix in which
    // each element is absolute difference
    // of its corresponding row and column
    // number row
    function findSum(n)
    {
     
        // Generate matrix
        let arr=new Array(n);
        for(let i = 0; i < n; i++)
        {
            arr[i] = new Array(n);
            for(let j = 0; j < n; j++)
            {
                arr[i][j] = 0;
            }
        }
         
        for (let i = 0; i < n; i++)
            for (let j = 0; j < n; j++)
                arr[i][j] = Math.abs(i - j);
       
        // Compute sum
        let sum = 0;
        for (let i = 0; i < n; i++)
            for (let j = 0; j < n; j++)
                sum += arr[i][j];
       
        return sum;
    }
     
    // Driver Code
    let n = 3;
    document.write(findSum(n));
     
    // This code is contributed by avanitrachhadiya2155
     

输出：

8

方法2（O（n））：
考虑 n = 3，形成的矩阵将是：
0 1 2
1 0 1
2 1 0
请注意，主对角线始终为 0，因为所有 i 都等于 j。正上方和正下方的对角线始终为 1，因为在每个单元格中，要么 i 比 j 大 1，要么 j 比 i 大 1，依此类推。
按照该模式，我们可以看到矩阵中所有元素的总和将是，对于从 0 到 n 的每个 i，添加 i*(ni)*2。
以下是上述想法的实现：

C++

// C++ program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
#include
using namespace std;
 
// Return the sum of matrix in which each
// element is absolute difference of its
// corresponding row and column number row
int findSum(int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += i*(n-i);
    return 2*sum;
}
 
// Driven Program
int main()
{
    int n = 3;
    cout << findSum(n) << endl;
    return 0;
}

Java

// Java program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
import java.io.*;
 
class GFG {
 
// Return the sum of matrix in which each
// element is absolute difference of its
// corresponding row and column number row
static int findSum(int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += i * (n - i);
    return 2 * sum;
}
 
    // Driver Code
    static public void main(String[] args)
    {
        int n = 3;
        System.out.println(findSum(n));
    }
}
 
// This code is contributed by vt_m.

C＃

// C# program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
using System;
 
class GFG {
 
// Return the sum of matrix in which each
// element is absolute difference of its
// corresponding row and column number row
static int findSum(int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += i * (n - i);
    return 2 * sum;
}
 
    // Driver Code
    static public void Main(String[] args)
    {
        int n = 3;
        Console.WriteLine(findSum(n));
    }
}
 
// This code is contributed by vt_m.

Python3

# Python 3 program to find sum
# of matrix in which each element
# is absolute difference of its
# corresponding row and column
# number row.
 
# Return the sum of matrix in
# which each element is absolute
# difference of its corresponding
# row and column number row
def findSum(n):
    sum = 0
    for i in range(n):
        sum += i * (n - i)
    return 2 * sum
 
# Driver code
n = 3
print(findSum(n))
 
# This code is contributed by Shrikant13

Javascript

// Java  script program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
// Return the sum of matrix in which each
// element is absolute difference of its
// corresponding row and column number row
function findSum( n)
{
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += i * (n - i);
    return 2 * sum;
}
 
    // Driver Code   
        let n = 3;
        document.write(findSum(n));
 
// This code is contributed by mohan pavan
 

输出：

8

方法3（技巧）：
考虑 n = 3，形成的矩阵将是：
0 1 2
1 0 1
2 1 0
所以，总和 = 1 + 1 + 1 + 1 + 2 + 2。
在重新排列时， 1 + 2 + 1 + 2 + 2 = 1 + 2 + 1 + 2 ² 。
因此，在每种情况下，我们都可以重新排列矩阵之和，以便答案始终是第一个 n – 1 个自然数的和和第一个 n – 1 个自然数的平方和。

Sum of first n natural number = ((n)*(n + 1))/2.
Sum of first n natural number = ((n)*(n + 1)*(2*n + 1)/6.

以下是上述想法的实现：

C++

// C++ program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
#include
using namespace std;
 
// Return the sum of matrix in which each element
// is absolute difference of its corresponding
// row and column number row
int findSum(int n)
{
    n--;
    int sum = 0;
    sum += (n*(n+1))/2;
    sum += (n*(n+1)*(2*n + 1))/6;
    return sum;
}
 
// Driven Program
int main()
{
    int n = 3;
    cout << findSum(n) << endl;
    return 0;
}

Java

// Java program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
import java.io.*;
 
public class GFG {
     
// Return the sum of matrix in which each element
// is absolute difference of its corresponding
// row and column number row
static int findSum(int n)
{
    n--;
    int sum = 0;
    sum += (n * (n + 1)) / 2;
    sum += (n * (n + 1) * (2 * n + 1)) / 6;
    return sum;
}
 
    // Driver Code
    static public void main (String[] args)
    {
        int n = 3;
        System.out.println(findSum(n));
    }
}
 
// This code is contributed by vt_m.

Python3

# Python 3 program to find sum of matrix
# in which each element is absolute
# difference of its corresponding row
# and column number row.
 
# Return the sum of matrix in which
# each element is absolute difference
# of its corresponding row and column
# number row
def findSum(n):
    n -= 1
    sum = 0
    sum += (n * (n + 1)) / 2
    sum += (n * (n + 1) * (2 * n + 1)) / 6
    return int(sum)
 
# Driver Code
n = 3
print(findSum(n))
 
# This code contributed by Rajput-Ji

C＃

// C# program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
using System;
 
public class GFG {
     
// Return the sum of matrix in which each element
// is absolute difference of its corresponding
// row and column number row
static int findSum(int n)
{
    n--;
    int sum = 0;
    sum += (n * (n + 1)) / 2;
    sum += (n * (n + 1) * (2 * n + 1)) / 6;
    return sum;
}
 
    // Driver Code
    static public void Main(String[] args)
    {
        int n = 3;
        Console.WriteLine(findSum(n));
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// Java script program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
     
// Return the sum of matrix in which each element
// is absolute difference of its corresponding
// row and column number row
function findSum( n)
{
    n--;
    let sum = 0;
    sum += (n * (n + 1)) / 2;
    sum += (n * (n + 1) * (2 * n + 1)) / 6;
    return sum;
}
 
    // Driver Code
   let n = 3;
       document.write(findSum(n));
    
// This code is contributed by mohan pavan
 

输出 ：

8