詹姆斯·乔伊斯的《尤利西斯》系列

詹姆斯·乔伊斯(James Joyce)的“尤利西斯”(Ulysses)序列表示N ^NN中的位数。
詹姆斯·乔伊斯(James Joyce)的《尤利西斯》(Ulysses)序列中的前几个词是

1, 2, 13, 155, 2185, 36306……..

给定整数N ，任务是打印詹姆斯·乔伊斯(James Joyce)的“尤利西斯”(Ulysses)序列的第N个项。
例子：

Input: 2 
Output: 2 
Explanation: 
Number of digits in 2²² is 2 since 2²² = 16
Input: 3 
Output: 13 
 

方法： Num中的位数由下式给出： $\lfloor (log_{10}(Num)) \rfloor + 1$ 。因此，N ^NN中的位数由下式给出：

$\lfloor (N^{N} * log_{10}(N)) \rfloor + 1$

要查看乔伊斯的所有数字，我们需要打印大约3.7亿个数字。假设每行100位和每页100行，这意味着需要写下约37卷(每1000页) $N^{N^{N}}$明确地。乔伊斯在《尤利西斯》中的估计为33卷；考虑到他在数学上的糟糕表现，还不错。
下面是上述方法的实现：

C++

// C++ implementation of
// the above approach
 
#include 
using namespace std;
 
// Function to return N-th term
// of James Joyce's "Ulysses" sequence.
int nthTerm(int n)
{
    return floor(pow(n, n) * log10(n)) + 1;
}
 
// Driver Code
int main()
{
    int n = 3;
    cout << nthTerm(n);
 
    return 0;
}

Java

// Java implementation of
// the above approach
class GFG{
 
// Function to return N-th term
// of James Joyce's "Ulysses" sequence.
static int nthTerm(int n)
{
    return (int)(Math.floor(Math.pow(n, n) *
                            Math.log10(n)) + 1);
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 3;
    System.out.print(nthTerm(n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the
# above approach
import math
 
# Function to return N-th term
# of James Joyce's "Ulysses" sequence.
def nthTerm(n):
     
    return (math.floor(math.pow(n, n) *
                       math.log10(n)) + 1);
 
# Driver Code
if __name__ == "__main__" :
     
    # Given number
    n = 3;
 
    # Function call
    print(nthTerm(n));
 
# This code is contributed by rock_cool

C#

// C# implementation of
// the above approach
using System;
class GFG{
 
// Function to return N-th term
// of James Joyce's "Ulysses" sequence.
static int nthTerm(int n)
{
    return (int)(Math.Floor(Math.Pow(n, n) *
                            Math.Log10(n)) + 1);
}
 
// Driver Code
public static void Main()
{
    int n = 3;
    Console.Write(nthTerm(n));
}
}
 
// This code is contributed by Nidhi_biet

Javascript

// Javascript implementation of
// the above approach
 
    // Function to return N-th term
    // of James Joyce's "Ulysses" sequence.
    function nthTerm( n)
    {
        return parseInt( (Math.floor(Math.pow(n, n) * Math.log10(n)) + 1));
    }
 
    // Driver Code
    let n = 3;
    document.write(nthTerm(n));
 
// This code is contributed by todaysgaurav

输出：

13

参考文献： https : //oeis.org/A054382