计算元素在大小为N * N的矩阵中出现的次数，以使每个元素等于其索引|的乘积。套装2

给定两个正整数N和X ，任务是计算给定整数X在生成的N长度方阵中的出现次数，以使矩阵的每个元素等于其行索引和列索引的乘积(从1开始索引)。

例子：

Input: N = 5, X = 6
Output: 2
Explanation:
The 2D array formed is equal to the :
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
There are 2 occurrences of the element X(= 6) in the generated array.

Input: N = 7, X = 12
Output: 4

朴素的方法：请参考本文中最简单的方法来解决此问题，方法是通过将行和列的索引相乘以获得每个矩阵元素并计算X的出现次数来构造给定的矩阵。

时间复杂度： O(N ² )
辅助空间： O(N ² )

有效的方法：这个想法是基于观察到^第i^个 行包含范围[1，N]中i的所有倍数。因此， X出现在第i^个 当且仅当X可以被i整除并且X / i应该小于或等于N时，才行。如果发现是真的，则将计数加1。请按照以下步骤解决问题：

• 初始化一个变量，例如count ，以将X的出现次数存储在生成的矩阵中。
• 使用变量i遍历[1，N]范围，并执行以下步骤：
  • 如果X可被i整除，则将X / i的商存储在变量中，例如b 。
  • 如果b的值落在[1，N]范围内，则将计数增加1 。
• 完成上述步骤后，打印count的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count the occurrences
// of X in the generated square matrix
int countOccurrences(int n, int x)
{
    // Store the required result
    int count = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i <= n; i++) {
 
        // Check if x is a
        // multiple of i or not
        if (x % i == 0) {
 
            // Check if the other multiple
            // exists in the range or not
            if (x / i <= n)
                count++;
        }
    }
 
    // Print the result
    cout << count;
}
 
// Driver Code
int main()
{
    int N = 7, X = 12;
    countOccurrences(N, X);
 
    return 0;
}

Java

// Java program for the above approach
 
class GFG{
 
// Function to count the occurrences
// of X in the generated square matrix
static void countOccurrences(int n, int x)
{
    // Store the required result
    int count = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i <= n; i++) {
 
        // Check if x is a
        // multiple of i or not
        if (x % i == 0) {
 
            // Check if the other multiple
            // exists in the range or not
            if (x / i <= n)
                count++;
        }
    }
 
    // Print the result
    System.out.print(count);
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 7, X = 12;
    countOccurrences(N, X);
 
}
}
 
// This code contributed by shikhasingrajput

Python3

# Python3 program for the above approach
 
# Function to count the occurrences
# of X in the generated square matrix
def countOccurrences(n, x):
     
    # Store the required result
    count = 0
 
    # Iterate over the range [1, N]
    for i in range(1, n + 1):
 
        # Check if x is a
        # multiple of i or not
        if (x % i == 0):
 
            # Check if the other multiple
            # exists in the range or not
            if (x // i <= n):
                count += 1
 
    # Print the result
    print(count)
 
# Driver Code
if __name__ == "__main__":
     
    N = 7
    X = 12
     
    countOccurrences(N, X)
 
# This code is contributed by ukasp

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to count the occurrences
  // of X in the generated square matrix
  static void countOccurrences(int n, int x)
  {
    // Store the required result
    int count = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i <= n; i++) {
 
      // Check if x is a
      // multiple of i or not
      if (x % i == 0) {
 
        // Check if the other multiple
        // exists in the range or not
        if (x / i <= n)
          count++;
      }
    }
 
    // Print the result
    Console.WriteLine(count);
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int N = 7, X = 12;
    countOccurrences(N, X);
  }
}
 
// This code is contributed by splevel62.

输出：

4

时间复杂度： O(N)
辅助空间： O(1)