求 N 次操作的功率和 LCM 的乘积之和

给定一个整数N ，任务是在i最初为1时执行以下N操作后找到该值：

求i的值的(N – i)次幂，然后将其乘以i 和 N 的 lcm 。
将此值添加到最终答案中。
将i增加1 。

例子：

Input: 5
Output: 305
Explanation: Following are the steps of the operations:
pow(1, 5-1) * (lcm(1, 5)) = 1 * 5
pow(2, 5-2) * (lcm(2, 5)) = 8 * 10 = 80
pow(3, 5-3) * (lcm(3, 5)) = 9 * 15 = 135
pow(4, 5-4) * (lcm(4, 5)) = 4 * 20 = 80
pow(5, 5-5) * (lcm(5, 5)) = 1 * 5 = 5
Hence the final sum is: 5 + 80 + 135 + 80 + 5 = 305

Input: 6
Output: 612

编程需要懂一点英语

方法：这是一个简单的基于实现的问题。我们的想法是如前所述逐个执行步骤并找到最终总和。请按照以下步骤解决问题：

从i = 1开始遍历i到N
- 计算i和N的 LCM。
- 计算i的Ni次方的值。
- 继续将上述两步的值相乘，并将其添加到最终总和（最初总和将为 0）。
返回总和。

下面是上述方法的实现：

C++

// C++ program to implement the approach
 
#include 
using namespace std;
 
// GCD function
int gcd(int x, int y)
{
    if (x == 0)
        return y;
    return gcd(y % x, x);
}
 
// Function to find sum
int findsum(int N)
{
    int i, hcf, sum = 0, lcm;
 
    for (i = 1; i <= N; i++) {
 
        // Returning the gcd of two numbers
        hcf = gcd(i, N);
 
        // Calculating lcm using lcm*hcf=
        // value1*value2.
        lcm = (i * N) / hcf;
        sum += pow(i, N - i) * lcm;
    }
    return sum;
}
 
// Driver function
int main()
{
    int N = 5;
 
    // Function call
    cout << findsum(N) << endl;
    return 0;
}

Java

// Java program to implement the approach
import java.io.*;
 
class GFG
{
 
  // GCD function
  public static int gcd(int x, int y)
  {
    if (x == 0)
      return y;
    return gcd(y % x, x);
  }
 
  // Function to find sum
  public static int findsum(int N)
  {
    int i = 0, hcf = 0, sum = 0, lcm = 0;
 
    for (i = 1; i <= N; i++) {
 
      // Returning the gcd of two numbers
      hcf = gcd(i, N);
 
      // Calculating lcm using lcm*hcf=
      // value1*value2.
      lcm = (i * N) / hcf;
      sum += Math.pow(i, N - i) * lcm;
    }
    return sum;
  }
  public static void main(String[] args)
  {
    int N = 5;
 
    // Function call
    System.out.println(findsum(N));
  }
}
 
// This code is contributed by Rohit Pradhan

C#

// C# program to implement the approach
using System;
 
public class GFG{
 
  // GCD function
  public static int gcd(int x, int y)
  {
    if (x == 0)
      return y;
    return gcd(y % x, x);
  }
 
  // Function to find sum
  public static int findsum(int N)
  {
    int i = 0, hcf = 0, sum = 0, lcm = 0;
 
    for (i = 1; i <= N; i++) {
 
      // Returning the gcd of two numbers
      hcf = gcd(i, N);
 
      // Calculating lcm using lcm*hcf=
      // value1*value2.
      lcm = (i * N) / hcf;
      sum += (int)Math.Pow(i, N - i) * (int)lcm;
    }
    return sum;
  }
 
  // Driver Code
  static public void Main ()
  {
    int N = 5;
 
    // Function call
    Console.Write(findsum(N));
  }
}
 
// This code is contributed by hrithikgarg03188.

输出

时间复杂度： O(N * log N)
辅助空间： O(1)