从前N个自然数开始计算三元组(a，b，c)的数量，使得a * b + c = N

给定一个整数N ，任务是从前N个自然数开始计算三元组( a，b，c )，以便a * b + c = N。

例子：

Input: N = 3
Output: 3 
Explanation: 
Triplets of the form a * b + c = N are { (1, 1, 2), (1, 2, 1), (2, 1, 1) } 
Therefore, the required output is 3.

Input: N = 100
Output: 473

方法：可以通过以下观察来解决问题：

For every possible pairs (a, b), If a * b < N, then only c exists. Therefore, count the pairs (a, b) whose product is less than N.

请按照以下步骤解决问题：

• 初始化一个变量，例如cntTriplets ，以存储满足给定条件的前N个自然数的三元组的计数。
• 使用变量i遍历[1，N – 1]范围，并检查N％i == 0与否。如果发现是真的，则更新cntTriplets + =(N / i)– 1 。
• 否则，更新cntTriplets + =(N / i)。
• 最后，打印cntTriplets的值。

下面是上述方法的实现。

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find the count of
// triplets (a, b, c) with a * b + c = N
int findCntTriplet(int N)
{
    // Stores count of triplets of 1st
    // N natural numbers which are of
    // the form a * b + c = N
    int cntTriplet = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i < N; i++) {
 
        // If N is divisible by i
        if (N % i != 0) {
 
            // Update cntTriplet
            cntTriplet += N / i;
        }
        else {
 
            // Update cntTriplet
            cntTriplet += (N / i) - 1;
        }
    }
    return cntTriplet;
}
 
// Driver Code
int main()
{
    int N = 3;
    cout << findCntTriplet(N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG
{
 
  // Function to find the count of
  // triplets (a, b, c) with a * b + c = N
  static int findCntTriplet(int N)
  {
 
    // Stores count of triplets of 1st
    // N natural numbers which are of
    // the form a * b + c = N
    int cntTriplet = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i < N; i++)
    {
 
      // If N is divisible by i
      if (N % i != 0)
      {
 
        // Update cntTriplet
        cntTriplet += N / i;
      }
      else
      {
 
        // Update cntTriplet
        cntTriplet += (N / i) - 1;
      }
    }
    return cntTriplet;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 3;
    System.out.println(findCntTriplet(N));
  }
}
 
// This code is contributed by susmitakundugoaldanga

Python3

# Python program to implement
# the above approach
 
# Function to find the count of
# triplets (a, b, c) with a * b + c = N
def findCntTriplet(N):
   
    # Stores count of triplets of 1st
    # N natural numbers which are of
    # the form a * b + c = N
    cntTriplet = 0;
 
    # Iterate over the range [1, N]
    for i in range(1, N):
 
        # If N is divisible by i
        if (N % i != 0):
 
            # Update cntTriplet
            cntTriplet += N // i;
        else:
 
            # Update cntTriplet
            cntTriplet += (N // i) - 1;
 
    return cntTriplet;
 
# Driver code
if __name__ == '__main__':
    N = 3;
    print(findCntTriplet(N));
 
# This code is contributed by 29AjayKumar

C#

// C# program to implement
// the above approach
using System;
class GFG
{
 
  // Function to find the count of
  // triplets (a, b, c) with a * b + c = N
  static int findCntTriplet(int N)
  {
 
    // Stores count of triplets of 1st
    // N natural numbers which are of
    // the form a * b + c = N
    int cntTriplet = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i < N; i++)
    {
 
      // If N is divisible by i
      if (N % i != 0)
      {
 
        // Update cntTriplet
        cntTriplet += N / i;
      }
      else
      {
 
        // Update cntTriplet
        cntTriplet += (N / i) - 1;
      }
    }
    return cntTriplet;
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int N = 3;
    Console.WriteLine(findCntTriplet(N));
  }
}
 
// This code is contributed by 29AjayKumar

输出：

3

时间复杂度： O(N)
辅助空间： O(1)