给定GCD的顺序增加

给定两个整数n和g ，任务是生成n个整数的递增序列，使得：

1. 序列中所有元素的gcd为g 。
2. 并且，所有元素的总和在所有可能的序列中最小。

例子：

Input: n = 6, g = 5
Output: 5 10 15 20 25 30

Input: n = 5, g = 3
Output: 3 6 9 12 15

方法：当序列由以下元素组成时，序列的总和将最小：
g，2 * g，3 * g，4 * g，…..，n * g 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to print the required sequence
void generateSequence(int n, int g)
{
    for (int i = 1; i <= n; i++)
        cout << i * g << " ";
}
  
// Driver Code
int main()
{
    int n = 6, g = 5;
    generateSequence(n, g);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
  
    // Function to print the required sequence 
    static void generateSequence(int n, int g) 
    { 
        for (int i = 1; i <= n; i++) 
            System.out.print(i * g + " ");; 
    } 
      
    // Driver Code
    public static void main(String []args)
    {
        int n = 6, g = 5; 
        generateSequence(n, g); 
      
    }
}
  
// This code is contributed by Rituraj Jain

Python3

# Python3 implementation of the approach
  
# Function to print the required sequence 
def generateSequence(n, g): 
  
    for i in range(1, n + 1): 
        print(i * g, end = " ")
  
# Driver Code
if __name__ == "__main__":
  
    n, g = 6, 5
    generateSequence(n, g) 
  
# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System ;
  
class GFG
{
  
    // Function to print the required sequence 
    static void generateSequence(int n, int g) 
    { 
        for (int i = 1; i <= n; i++) 
            Console.Write(i * g + " "); 
    } 
      
    // Driver Code
    public static void Main()
    {
        int n = 6, g = 5; 
        generateSequence(n, g); 
    }
}
  
// This code is contributed by Ryuga

输出：

5 10 15 20 25 30