来自前 N 个自然数且余数至少为 K 的对数

给定两个正整数N和K ，任务是在[1, N]范围内找到对(a, b)的数量，使得a%b至少为 K 。

例子：

Input: N = 5, K = 2
Output: 7
Explanation: 
Following are the all possible pairs satisfying the given criteria:

1. (2, 3): The value of 2%3 = 2(>= K).
2. (5, 3): The value of 5%3 = 2(>= K).
3. (2, 4): The value of 2%4 = 2(>= K).
4. (3, 4): The value of 3%4 = 3(>= K).
5. (2, 5): The value of 2%5 = 2(>= K).
6. (3, 5): The value of 3%5 = 3(>= K).
7. (4, 5): The value of 4%5 = 4(>= K).

Therefore, the total count of pairs is 7.

Input: N = 6, K = 0
Output: 36

朴素方法：解决给定问题的最简单方法是在[1, N]范围内生成所有可能的对(a, b) ，如果a%b的值至少为 K ，则计算这对对。检查所有对后，打印获得的总对。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：上述方法也可以通过迭代范围[1, N]并固定对中的第二个数字，即b来优化。对于每个固定的b ，将有一个N/b周期，并且每个周期都可以与(b – K)个元素组合。因此，总共有(N/b)*(b – K)个元素。现在对于剩余的N%b元素，将有max(0, n%b – k + 1)对。请按照以下步骤解决问题：

• 如果K的值为0 ，则打印N ²作为有效对的结果数。
• 初始化变量，例如ans为0 ，它存储对的结果计数。
• 使用变量b遍历范围[K + 1, N]并执行以下步骤：
  • 将(N/b)*(b – K)的值添加到变量ans中。
  • 将(N % b – K + 1)的最大值或0添加到变量ans中。
• 执行上述步骤后，打印ans的值作为结果对数。

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 number of pairs
// (a, b) such that a%b is at least K
int countTotalPairs(int N, int K)
{
    // Base Case
    if (K == 0) {
        return N * N;
    }
 
    // Stores resultant count of pairs
    int ans = 0;
 
    // Iterate over the range [K + 1, N]
    for (int b = K + 1; b <= N; b++) {
 
        // Find the cycled elements
        ans += (N / b) * (b - K);
 
        // Find the remaining elements
        ans += max(N % b - K + 1, 0);
    }
 
    // Return the resultant possible
    // count of pairs
    return ans;
}
 
// Driver Code
int main()
{
    int N = 5, K = 2;
    cout << countTotalPairs(N, K);
 
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
 
class GFG {
 
  // Function to count the number of pairs
  // (a, b) such that a%b is at least K
  public static int countTotalPairs(int N, int K)
  {
 
    // Base case
    if (K == 0) {
      return N * N;
    }
 
    // Stores resultant count of pairs
    int ans = 0;
 
    // Iterate over the range [K + 1, N]
    for (int i = K + 1; i <= N; i++)
    {
 
      // Find the cycled element
      ans += (N / i) * (i - K);
      if ((N % i) - K + 1 > 0)
      {
 
        // Find the remaining element
        ans += (N % i) - K + 1;
      }
    }
 
    // Return the resultant possible
    // count of pairs
    return ans;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 5, K = 2;
    System.out.println(countTotalPairs(N, K));
  }
}
 
// This code is contributed by maddler.

Python3

# Python program for the above approach;
 
# Function to count the number of pairs
# (a, b) such that a%b is at least K
def countTotalPairs(N, K):
    # Base Case
    if (K == 0) :
        return N * N
     
 
    # Stores resultant count of pairs
    ans = 0
 
    # Iterate over the range [K + 1, N]
    for b in range(K + 1, N + 1) :
 
        # Find the cycled elements
        ans += (N // b) * (b - K)
 
        # Find the remaining elements
        ans += max(N % b - K + 1, 0)
     
 
    # Return the resultant possible
    # count of pairs
    return ans
 
 
# Driver Code
 
N = 5
K = 2
print(countTotalPairs(N, K))
 
# This code is contributed by _saurabh_jaiswal

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count the number of pairs
// (a, b) such that a%b is at least K
static int countTotalPairs(int N, int K)
{
    // Base Case
    if (K == 0) {
        return N * N;
    }
 
    // Stores resultant count of pairs
    int ans = 0;
 
    // Iterate over the range [K + 1, N]
    for (int b = K + 1; b <= N; b++) {
 
        // Find the cycled elements
        ans += (N / b) * (b - K);
 
        // Find the remaining elements
        ans += Math.Max(N % b - K + 1, 0);
    }
 
    // Return the resultant possible
    // count of pairs
    return ans;
}
 
// Driver Code
public static void Main()
{
    int N = 5, K = 2;
    Console.Write(countTotalPairs(N, K));
}
}
 
// This code is contributed by ipg2016107.

Javascript

 
        // JavaScript program for the above approach;
 
        // Function to count the number of pairs
        // (a, b) such that a%b is at least K
        function countTotalPairs(N, K) {
            // Base Case
            if (K == 0) {
                return N * N;
            }
 
            // Stores resultant count of pairs
            let ans = 0;
 
            // Iterate over the range [K + 1, N]
            for (let b = K + 1; b <= N; b++) {
 
                // Find the cycled elements
                ans += Math.floor((N / b) * (b - K));
 
                // Find the remaining elements
                ans += Math.max(N % b - K + 1, 0);
            }
 
            // Return the resultant possible
            // count of pairs
            return ans;
        }
 
        // Driver Code
 
        let N = 5, K = 2;
        document.write(countTotalPairs(N, K));
 
   // This code is contributed by Potta Lokesh
    

输出：

7

时间复杂度： O(N)
辅助空间： O(1)