递归程序为n个整数的GCD打印公式

给定函数gcd(a，b)来找到两个数字的GCD(最大公约数)。还已知可以通过gcd(a，gcd(b，c))找到三个元素的GCD，类似地，对于四个元素，也可以通过gcd(a，gcd(b，gcd(c，d))找到GCD。 )。给定正整数n 。工作是使用给定的gcd()函数打印公式以找到n个整数的GCD。
例子：

Input : n = 3
Output : gcd(int, gcd(int, int))

Input : n = 5
Output : gcd(int, gcd(int, gcd(int, gcd(int, int))))

方法：想法是使用递归来打印单行命令。现在，要编写一个递归函数，例如recursiveFun(n)，所需的字符串由gcd(int，+ recursiveFun(n – 1)+)组成。这意味着recursiveFun(n)应该返回一个包含对自身的调用的字符串，并且为了评估该值，递归函数将再次从n – 1开始。这将依次返回另一个字符串，其中包含对n – 1，直到n == 1为止，递归函数改为返回字符串“ int”。
下面是上述方法的实现：

C++

// CPP Program to print single line command
// to find the GCD of n integers
#include 
using namespace std;
 
// Function to print single line command
// to find GCD of n elements.
string recursiveFun(int n)
{
    // base case
    if (n == 1)
        return "int";
 
    // Recursive Step
    return "gcd(int, " + recursiveFun(n - 1) + ")";
}
 
// Driver Program
int main()
{
    int n = 5;
 
    cout << recursiveFun(n) << endl;
 
    return 0;
}

Java

// Java Program to print
// single line command to
// find the GCD of n integers
class GFG
{
     
// Function to print single
// line command to find GCD
// of n elements.
static String recursiveFun(int n)
{
    // base case
    if (n == 1)
        return "int";
 
    // Recursive Step
    return "gcd(int, " +
            recursiveFun(n - 1) + ")";
}
 
// Driver Code
public static void main(String [] arg)
{
    int n = 5;
 
    System.out.println(recursiveFun(n));
}
}
 
// This code is contributed
// by Smitha

Python3

# Python 3 Program to print single line
# command to find the GCD of n integers
 
# Function to print single line command
# to find GCD of n elements.
def recursiveFun(n):
     
    # base case
    if (n == 1):
        return "int"
 
    # Recursive Step
    return "gcd(int, " + recursiveFun(n - 1) + ")"
 
# Driver Code
if __name__ == '__main__':
    n = 5
    print(recursiveFun(n))
 
# This code is contributed
# by PrinciRaj1992

C#

// C# Program to print single
// line command to find the
// GCD of n integers
using System;
class GFG
{
     
// Function to print single
// line command to find GCD
// of n elements.
static String recursiveFun(int n)
{
    // base case
    if (n == 1)
        return "int";
 
    // Recursive Step
    return "gcd(int, " +
            recursiveFun(n - 1) + ")";
}
 
// Driver Code
public static void Main()
{
    int n = 5;
 
    Console.Write(recursiveFun(n));
}
}
 
// This code is contributed
// by smitha

Javascript

// javascript Program to prvar
// single line command to
// find the GCD of n integers   
// Function to prvar single
    // line command to find GCD
    // of n elements.
    function recursiveFun(n)
    {
     
        // base case
        if (n == 1)
            return "int";
 
        // Recursive Step
        return "gcd(int, " + recursiveFun(n - 1) + ")";
    }
 
    // Driver Code
        var n = 5;
 
        document.write(recursiveFun(n));
 
// This code is contributed by gauravrajput1

输出：

gcd(int, gcd(int, gcd(int, gcd(int, int))))

时间复杂度： O(N)，其中N是给定的数字。