矩阵中等于 X 的单元格计数构造为行和列的总和

给定整数N表示方阵的大小，其每个元素是其行和列的总和，即mat[i][j] = i + j (0 < i, j, ≤ N) 和整数X ，任务是找出有多少单元格具有值X 。

例子：

Input: N = 4, X = 2
Output: 1
Explanation: The matrix we get after using M(a, b) = a + b is:
2 3 4 5 
3 4 5 6 
4 5 6 7
5 6 7 8
At M[1][1] = X There is only value equals to 2.

Input: N = 3, X = 4
Output: 3
Explanation: The matrix we get after using M(a, b) = a + b is:
2 3 4
3 4 5
4 5 6
At M[1][3], M[2][2], M[3][1] = X There is 3 Times x appears.

Input: N = 1, X = 3
Output: 0
Explanation: The matrix we get after using M(a, b) = a + b is:
The matrix is { 2 }. So no value 3

天真的方法：最简单的方法是创建矩阵并计算具有值X的单元格。

下面是上述方法的实现。

C++

// C++ code of the above approach.
 
#include 
using namespace std;
 
// Function to find the number of times
// X is present in the matrix
long long int solve(long long int n,
                    long long int x)
{
    long long int total = 0;
 
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++)
 
            if (i + j == x)
                total++;
    }
 
    return total;
}
 
// Driver code
int main()
{
 
    int N = 4;
    int X = 2;
 
    cout << solve(N, X);
    return 0;
}

Java

// Java code of the above approach.
import java.io.*;
 
class GFG {
 
  // Function to find the number of times
  // X is present in the matrix
  static long  solve(long  n,
                     long  x)
  {
    long  total = 0;
 
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++)
 
        if (i + j == x)
          total++;
    }
 
    return total;
  }
 
  // Driver code
  public static void main (String[] args) {
    int N = 4;
    int X = 2;
 
    System.out.println(solve(N, X));
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python3 code of the above approach.
 
# Function to find the number of times
# X is present in the matrix
def solve(n, x):
    total = 0;
    for i in range(1, n + 1):
        for j in range(1, n + 1):
            if (i + j == x):
                total += 1;
    return total;
 
# Driver code
if __name__ == '__main__':
    N = 4;
    X = 2;
 
    print(solve(N, X));
 
# This code is contributed by gauravrajput1

C#

// C# code of the above approach.
using System;
 
class GFG {
 
    // Function to find the number of times
    // X is present in the matrix
    static long solve(long n, long x)
    {
        long total = 0;
 
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++)
 
                if (i + j == x)
                    total++;
        }
 
        return total;
    }
 
    // Driver code
    public static void Main()
    {
        int N = 4;
        int X = 2;
 
        Console.WriteLine(solve(N, X));
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript code for the above approach
 
// Function to find the number of times
// X is present in the matrix
function solve(n, x)
{
    let total = 0;
 
    for (let i = 1; i <= n; i++) {
        for (let j = 1; j <= n; j++)
 
            if (i + j == x)
                total++;
    }
 
    return total;
}
 
    // Driver Code
    let N = 4;
    let X = 2;
 
    document.write(solve(N, X));
     
    // This code is contributed by code_hunt.

输出

1

时间复杂度： O(N ² )
辅助空间： O(N ² )

有效方法：可以使用以下数学观察有效地解决问题：

The numbers across the secondary diagonal (top-right to bottom-left) and parallel to those will be the same.
The value in the secondary diagonal will be (1 + N). 
The minimum value in a diagonal will be 2 and maximum will be N+N = 2*N.

So there can be two cases:

When X < N+1:
    => It will be on the upper half of the secondary diagonal.
    => X can be formed by [1 + (X-1)], [2 + (X-2)], [3 + (X-3)] + . . . + [(X-1)+1].
    => From the above arrangement it can be seen that the X will be in the cells (1, X-1), (2, X-2), (3, X-3), . . ., (X-1, 1).
    => So total occurrences are is the total number of values in range [1, X-1] = X – 1.

When X > N+1:
    => It will be on the lower half of the secondary diagonal.
    => X can be formed by [(X-N) + N], [(X-N+1) + N-1], [(X-N+2) + (N-2)] + . . . + [N + (X-N)].
    => From the above arrangement it can be seen that the X will be in the cells (X-N, N), (X-N+1, N-1), ( X-N+2, N-2), . . ., (N, X-N).
    => So total occurrences are is the total number of values in range [X-N, N] = N – (X – N) + 1 = 2*N – X + 1.

(N+1) can be considered in any of the cases and that will give result as N.

按照下面提到的步骤来实现上述观察：

• 检查 X 是否大于 N+1。
• 然后使用上面推导的公式。

下面是上述方法的实现。

C++

// C++ code for the above approach.
 
#include 
using namespace std;
 
// Function to find
// the number of occurrences of X
long long int solve(long long int N,
                    long long int X)
{
    if (X == 0 || X > 2 * N)
        return 0;
 
    if (X <= N + 1)
        return X - 1;
    else
        return N + N + 1 - X;
}
 
// Driver code
int main()
{
 
    int N = 4;
    int X = 2;
 
    cout << solve(N, X);
    return 0;
}

Java

// Java Program of the above approach.
import java.util.*;
class GFG {
 
  // Function to find
  // the number of occurrences of X
  static int solve(int N, int X)
  {
    if (X == 0 || X > 2 * N)
      return 0;
 
    if (X <= N + 1)
      return X - 1;
    else
      return N + N + 1 - X;
  }
 
  // Driver Code
  public static void main(String args[])
  {
    int N = 4;
    int X = 2;
 
    System.out.print(solve(N, X));
  }
}
 
// This code is contributed by code_hunt.

C＃

// C# Program of the above approach.
using System;
 
class GFG {
 
    // Function to find
    // the number of occurrences of X
    static int solve(int N, int X)
    {
        if (X == 0 || X > 2 * N)
            return 0;
 
        if (X <= N + 1)
            return X - 1;
        else
            return N + N + 1 - X;
    }
 
    // Driver Code
    public static void Main()
    {
        int N = 4;
        int X = 2;
 
        Console.Write(solve(N, X));
    }
}
 
// This code is contributed by ukasp.

Javascript

    // JavaScript code for the above approach.
 
    // Function to find
    // the number of occurrences of X
    const solve = (N, X) => {
        if (X == 0 || X > 2 * N)
            return 0;
 
        if (X <= N + 1)
            return X - 1;
        else
            return N + N + 1 - X;
    }
 
    // Driver code
    let N = 4;
    let X = 2;
    document.write(solve(N, X));
 
// This code is contributed by rakeshsahni
 

输出

1

时间复杂度： O(1)
辅助空间： O(1)