计算三元组，使得两个数字与第三个数字相加的乘积为 N

给定一个正整数N ，任务是找到三元组(A, B, C)的数量，其中A , B , C是正整数，使得两个数字与第三个数字相加的乘积为N ，即A * B + C = N。

例子：

Input: N = 3
Output: 3
Explanation:
Following are the possible triplets satisfying the given criteria:

(1, 1, 2): The value of 1*1 + 2 = 3.
(1, 2, 1): The value of 1*2 + 1 = 3.
(2, 1, 1): The value of 2*1 + 1 = 3.

Therefore, the total count of such triplets is 3.

Input: N = 5
Output: 8

方法：给定的问题可以通过将方程A * B + C = N重新排列为A * B = N – C来解决。现在，满足上述方程的唯一可能值A和B是N – C的除数。例如，如果N – C = 18的值，有 6 个除数，即 1, 2, 3, 6, 9, 18。因此，满足上述等式的A、B的值是： (1, 18), ( 2, 9), (3, 6), (6, 3), (9, 2), (18, 1) 。因此，对于N – C = 18的值， A和B的可能值为6 ，即N – C(= 18)的除数数。请按照以下步骤解决给定的问题：

初始化一个变量，比如count为0 ，它存储可能的三元组的总数。
使用变量i在[1, N – 1]范围内迭代一个循环，对于每个值，我找到(N – i)的除数的总数并将其添加到变量count中。
完成上述步骤后，将count的值打印为三元组的结果数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the divisors of
// the number (N - i)
int countDivisors(int n)
{
    // Stores the resultant count of
    // divisors of (N - i)
    int divisors = 0;
    int i;
 
    // Iterate over range [1, sqrt(N)]
    for (i = 1; i * i < n; i++) {
        if (n % i == 0) {
            divisors++;
        }
    }
    if (i - (n / i) == 1) {
        i--;
    }
    for (; i >= 1; i--) {
        if (n % i == 0) {
            divisors++;
        }
    }
    // Return the total divisors
    return divisors;
}
 
// Function to find the number of triplets
// such that A * B - C = N
int possibleTriplets(int N)
{
    int count = 0;
 
    // Loop to fix the value of C
    for (int i = 1; i < N; i++) {
 
        // Adding the number of
        // divisors in count
        count += countDivisors(N - i);
    }
 
    // Return count of triplets
    return count;
}
 
// Driver Code
int main()
{
    int N = 10;
    cout << possibleTriplets(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG
{
   
      // Function to find the divisors of
    // the number (N - i)
    static int countDivisors(int n)
    {
       
        // Stores the resultant count of
        // divisors of (N - i)
        int divisors = 0;
        int i;
 
        // Iterate over range [1, sqrt(N)]
        for (i = 1; i * i < n; i++) {
            if (n % i == 0) {
                divisors++;
            }
        }
        if (i - (n / i) == 1) {
            i--;
        }
        for (; i >= 1; i--) {
            if (n % i == 0) {
                divisors++;
            }
        }
 
        // Return the total divisors
        return divisors;
    }
 
    // Function to find the number of triplets
    // such that A * B - C = N
    static int possibleTriplets(int N)
    {
        int count = 0;
 
        // Loop to fix the value of C
        for (int i = 1; i < N; i++) {
 
            // Adding the number of
            // divisors in count
            count += countDivisors(N - i);
        }
 
        // Return count of triplets
        return count;
    }
 
    // Driver Code
    public static void main (String[] args) {
       
        int N = 10;
        System.out.println(possibleTriplets(N));
    }
}
 
// This code is contributed by Dharanendra L V.

Python3

# Python program for the above approach
import math
 
# function to find the divisors of
# the number (N - i)
def countDivisors(n):
 
    # Stores the resultant count of
    # divisors of (N - i)
    divisors = 0
 
    # Iterate over range [1, sqrt(N)]
    for i in range(1, math.ceil(math.sqrt(n))+1):
        if n % i == 0:
            divisors = divisors+1
 
        if (i - (n / i) == 1):
            i = i-1
 
    for i in range(math.ceil(math.sqrt(n))+1, 1, -1):
        if (n % i == 0):
            divisors = divisors+1
 
     # Return the total divisors
    return divisors
 
    # def to find the number of triplets
    # such that A * B - C = N
 
 
def possibleTriplets(N):
    count = 0
 
    # Loop to fix the value of C
    for i in range(1, N):
 
        # Adding the number of
        # divisors in count
        count = count + countDivisors(N - i)
 
        # Return count of triplets
    return count
 
    # Driver Code
# Driver Code
if __name__ == "__main__":
    N = 10
    print(possibleTriplets(N))
 
# This code is contributed by Potta Lokesh

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the divisors of
// the number (N - i)
static int countDivisors(int n)
{
    // Stores the resultant count of
    // divisors of (N - i)
    int divisors = 0;
    int i;
 
    // Iterate over range [1, sqrt(N)]
    for (i = 1; i * i < n; i++) {
        if (n % i == 0) {
            divisors++;
        }
    }
    if (i - (n / i) == 1) {
        i--;
    }
    for (; i >= 1; i--) {
        if (n % i == 0) {
            divisors++;
        }
    }
    // Return the total divisors
    return divisors;
}
 
// Function to find the number of triplets
// such that A * B - C = N
static int possibleTriplets(int N)
{
    int count = 0;
 
    // Loop to fix the value of C
    for (int i = 1; i < N; i++) {
 
        // Adding the number of
        // divisors in count
        count += countDivisors(N - i);
    }
 
    // Return count of triplets
    return count;
}
 
// Driver Code
public static void Main()
{
    int N = 10;
    Console.Write(possibleTriplets(N));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

输出：

时间复杂度： O(N*sqrt(N))
辅助空间： O(1)