给定一个数字，我们的任务是找到该数字的超级阶乘。

将 1 到给定数字的连续整数相乘的结果，其中每个整数都以其自身为幂，这就是所谓的数字 超级阶乘。

H(n)= 1 ^ 1 * 2 ^ 2 * 3 ^ 3 * . . . . . * n ^ n

示例：

> 输入： 2

> 输出： 4

>

>

>

>

>

> 输入： 4

> 输出： 27648

> H(4) = 1^1 2^2 3^3 * 4^4 = 27648

一种朴素的方法是使用两个循环，一个用于计算 i^i 的总和，另一个用于计算 i^i。但这种方法的时间复杂度将是 O(n2)。
一种高效的方法是使用内置的 pow() 函数或 O(log n) 方法 来计算 i^i，然后将其相加。

下面是上述方法的实现。

C++


“/// C++ program to find the hyperfactorial

// of a number

#include

using namespace std;

#define ll long long

// function to calculate the value of hyperfactorial

ll boost_hyperfactorial(ll num)

{

// initialise the val to 1

ll val = 1;

for (int i = 1; i <= num; i++) {

val = val * pow(i,i);

}

// returns the hyperfactorial of a number

return val;

}

// Driver code

int main()

{

int num = 5;

cout << boost_hyperfactorial(num);

return 0;

}

`

            Java

    `

// Java program to find the

// hyperfactorial of a number

// function to calculate the

// value of hyperfactorial

class GFG

{

static long boost_hyperfactorial(long num)

{

// initialise the val to 1

long val = 1;

for (int i = 1; i <= num; i++)

{

val = val * (long)Math.pow(i, i);

}

// returns the hyperfactorial

// of a number

return val;

}

// Driver code

public static void main(String args[])

{

int num = 5;

System.out.println(boost_hyperfactorial(num));

}

}

// This code is contributed

// by chandan_jnu

`

            Python3

    `

# Python3 program to find the

hyperfactorial of a number

function to calculate the

value of hyperfactorial

def boost_hyperfactorial(num):

# initialise the

# val to 1

val = 1;

for i in range(1, num + 1):

val = val * pow(i, i);

# returns the hyperfactorial

# of a number

return val;

Driver code

num = 5;

print(boost_hyperfactorial(num));

This code is contributed

by mits

`

            C#

    `

// C# program to find the

// hyperfactorial of a number

using System;

class GFG

{

// function to calculate the

// value of hyperfactorial

static long boost_hyperfactorial(long num)

{

// initialise the val to 1

long val = 1;

for (long i = 1; i <= num; i++)

{

val = val * (long)Math.Pow(i, i);

}

// returns the hyperfactorial

// of a number

return val;

}

// Driver code

public static void Main()

{

int num = 5;

Console.WriteLine(boost_hyperfactorial(num));

}

}

// This code is contributed

// by chandan_jnu

`

            PHP

    `

<?php

// PHP program to find the

// hyperfactorial of a number

// function to calculate the

// value of hyperfactorial

function boost_hyperfactorial($num)

{

// initialise the

// val to 1

$val = 1;

for ($i = 1; $i <= $num; $i++)

{

$val = $val * pow($i, $i);

}

// returns the hyperfactorial

// of a number

return $val;

}

// Driver code

$num = 5;

echo boost_hyperfactorial($num);

// This code is contributed

// by Akanksha Rai(Abby_akku)

?>

`

            JavaScript

    `

// Javascript program to find the

// hyperfactorial of a number

// function to calculate the

// value of hyperfactorial

function boost_hyperfactorial(num)

{

// initialise the

// val to 1

let val = 1;

for (let i = 1; i <= num; i++)

{

val = val * Math.pow(i, i);

}

// returns the hyperfactorial

// of a number

return val;

}

// Driver code

let num = 5;

document.write(boost_hyperfactorial(num));

// This code is contributed

// by gfgking

`

**输出：**

86400000

 

**时间复杂度：** O(N * log N)
**辅助空间：** O(1)

由于数字的超级阶乘可能非常巨大，数字将会溢出。我们可以使用 [C++ 中的 boost 库](https://www.geeksforgeeks.org/cpp/advanced-c-boost-library/) 或 [Java 中的 BigInteger](https://www.geeksforgeeks.org/java/biginteger-class-in-java/) 来存储数字 N 的超级阶乘。

            C++

    `

// C++ program to find the hyperfactorial

// of a number using boost libraries

#include

#include

using namespace boost::multiprecision;

using namespace std;

// function to calculate the value of hyperfactorial

int1024t boosthyperfactorial(int num)

{

// initialise the val to 1

int1024_t val = 1;

for (int i = 1; i <= num; i++) {