求系列 1、4、27、16、125、36、343 的第 n 项……。

给定一个数 n，找出系列 1、4、27、16、125、36、343……中的第 n 项。
例子：

Input : 5
Output : 125

Input : 6
Output : 36

天真的方法：在这个序列中，如果第 N 项是偶数，那么第 N 项是 N 的平方，否则第 N 项是 N 的立方。

N = 5 
In this case, N is odd 
N = N * N * N 
N = 5 * 5 * 5 
N = 125
Similarly, 
N = 6 
N = N * N 
N = 6 * 6 
N = 36 and so on.. 
 

上述方法的实现如下：

C++

// CPP code to generate first 'n' terms
// of Logic sequence
#include 
using namespace std;
 
// Function to generate a fixed \number
int logicOfSequence(int N)
{
    if (N % 2 == 0)
        N = N * N;   
    else
        N = N * N * N;  
    return N;
}
 
// Driver Method
int main()
{
    int N = 6;
    cout << logicOfSequence(N) << endl;
    return 0;
}

Java

// JAVA code to generate first
// 'n' terms of Logic sequence
 
class GFG {
 
// Function to generate
// a fixed \number
public static int logicOfSequence(int N)
{
    if (N % 2 == 0)
        N = N * N;
    else
        N = N * N * N;
    return N;
}
 
// Driver Method
public static void main(String args[])
{
    int N = 6;
    System.out.println(logicOfSequence(N));
}
}
 
// This code is contributed by Jaideep Pyne

Python 3

# Python code to generate first
# 'n' terms of Logic sequence
 
# Function to generate a fixed
# number
def logicOfSequence(N):
    if(N % 2 == 0):
        N = N * N
    else:
        N = N * N * N
    return N
 
N = 6
print (logicOfSequence(N))
 
# This code is contributed by
# Vishal Gupta

C#

// C# code to generate first
// 'n' terms of Logic sequence
using System;
 
class GFG {
 
// Function to generate
// a fixed number
public static int logicOfSequence(int N)
{
    if (N % 2 == 0)
        N = N * N;
    else
        N = N * N * N;
    return N;
}
 
// Driver Method
public static void Main()
{
    int N = 6;
    Console.Write(logicOfSequence(N));
}
}
 
// This code is contributed by nitin mittal

Javascript

// JavaScript code to generate first 'n' terms
// of Logic sequence
 
// Function to generate a fixed \number
function logicOfSequence( N)
{
    if (N % 2 == 0)
        N = N * N;   
    else
        N = N * N * N;  
    return N;
}
 
// Driver Function
 
    let N = 6;
    document.write(logicOfSequence(N));
     
// This code contributed by Rajput-Ji
 

输出：

36

时间复杂度：O(1)