找出对数(a，b)使得a％b = K

给定两个整数N和K ，其中N，K> 0 ，任务是找到对数(a，b) ，其中1≤a，b≤N，这样a％b = K。

例子：

Input: N = 4, K = 2
Output: 2
Only valid pairs are (2, 3) and (2, 4).

Input: N = 11, K = 5
Output: 7

天真的方法：从1到n运行两个循环，并计算所有对(i，j) ，其中i％j = K。这种方法的时间复杂度将是O(n ² ) 。

高效的方法：最初的总计数= N – K，因为该范围内所有大于K的数字都将除以K作为余数。此后，对于所有i> K ，都有(N-K)/ i个正好数字，除以i后，余数将为K。

以下是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count
// of required pairs
int CountAllPairs(int N, int K)
{
  
    int count = 0;
  
    if (N > K) {
  
        // Initial count
        count = N - K;
        for (int i = K + 1; i <= N; i++)
            count = count + ((N - K) / i);
    }
  
    return count;
}
  
// Driver code
int main()
{
    int N = 11, K = 5;
  
    cout << CountAllPairs(N, K);
  
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
class GFG {
  
    // Function to return the count
    // of required pairs
    static int CountAllPairs(int N, int K)
    {
  
        int count = 0;
  
        if (N > K) {
  
            // Initial count
            count = N - K;
            for (int i = K + 1; i <= N; i++)
                count = count + ((N - K) / i);
        }
  
        return count;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int N = 11, K = 5;
        System.out.println(CountAllPairs(N, K));
    }
}

Python3

# Python3 implementation of the approach
import math
  
# Function to return the count 
# of required pairs
def CountAllPairs(N, K):
    count = 0
    if( N > K):
          
        # Initial count
        count = N - K
        for i in range(K + 1, N + 1):
            count = count + ((N - K) // i)
              
    return count
      
# Driver code
N = 11
K = 5
print(CountAllPairs(N, K))

C#

// C# implementation of the approach
using System;
class GFG {
  
    // Function to return the count
    // of required pairs
    static int CountAllPairs(int N, int K)
    {
  
        int count = 0;
  
        if (N > K) {
  
            // Initial count
            count = N - K;
            for (int i = K + 1; i <= N; i++)
                count = count + ((N - K) / i);
        }
  
        return count;
    }
  
    // Driver code
    public static void Main()
    {
        int N = 11, K = 5;
        Console.WriteLine(CountAllPairs(N, K));
    }
}

PHP

 $K){
          
        // Initial count
        $count = $N - $K;
        for($i = $K+1; $i <= $N ; $i++)
        {
                $x = ((($N - $K) / $i));
                $count = $count + (int)($x);
        }
    }
  
    return $count;
}
  
    // Driver code
    $N = 11; 
    $K = 5;
    echo(CountAllPairs($N, $K));
  
?>

输出：

7