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📜  构造两个具有相同索引元素的N长度数组作为互素,并且它们的和之差为N

📅  最后修改于: 2021-05-19 18:16:20             🧑  作者: Mango

给定一个正整数N ,任务是生成两个长度为N的数组,以使两个数组的相同索引元素都是互质的,并且数组元素之和之间的绝对差为N。

例子:

方法:这个想法是基于以下观察:两个连续的自然数始终是素数,并且它们之间的差是1 。请按照以下步骤解决问题:

  • 初始化大小为N的两个数组A []B []
  • 使用变量i遍历[1,2 * N]范围。对于范围内的每个元素,检查i是否可被2整除。如果发现为真,则将i插入数组A [] 。否则,将i插入数组B []中
  • 完成上述步骤后,打印两个阵列。

下面是上述方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function to generate two arrays
// satisfying the given conditions
void printArrays(int n)
{
    // Declare the two arrays A and B
    vector A, B;
 
    // Iterate from range [1, 2*n]
    for (int i = 1; i <= 2 * n; i++) {
 
        // Assign consecutive numbers to
        // same indices of the two arrays
        if (i % 2 == 0)
            A.push_back(i);
        else
            B.push_back(i);
    }
 
    // Print the first array
    cout << "{ ";
    for (int i = 0; i < n; i++) {
        cout << A[i];
 
        if (i != n - 1)
            cout << ", ";
    }
    cout << " }\n";
 
    // Print the second array, B
    cout << "{ ";
    for (int i = 0; i < n; i++) {
        cout << B[i];
 
        if (i != n - 1)
            cout << ", ";
    }
    cout << " }";
}
 
// Driver Code
int main()
{
    int N = 5;
 
    // Function Call
    printArrays(N);
 
    return 0;
}


Java
// Java program for the above approach
import java.io.*;
import java.util.*;
  
class GFG{
       
// Satisfying the given conditions
static void printArrays(int n)
{
     
    // Declare the two arrays A and B
    ArrayList A = new ArrayList();
    ArrayList B = new ArrayList();
     
    // Iterate from range [1, 2*n]
    for(int i = 1; i <= 2 * n; i++)
    {
         
        // Assign consecutive numbers to
        // same indices of the two arrays
        if (i % 2 == 0)
            A.add(i);
        else
            B.add(i);
    }
  
    // Print the first array
    System.out.print("{ ");
    for(int i = 0; i < n; i++)
    {
        System.out.print(A.get(i));
  
        if (i != n - 1)
            System.out.print(", ");
    }
    System.out.print(" }\n");
  
    // Print the second array, B
    System.out.print("{ ");
    for(int i = 0; i < n; i++)
    {
        System.out.print(B.get(i));
         
        if (i != n - 1)
            System.out.print(", ");
    }
    System.out.print(" }");
}
  
// Driver code
public static void main (String[] args)
{
    int N = 5;
     
    // Function Call
    printArrays(N);
}
}
 
// This code is contributed by sanjoy_62


Python3
# Python3 program for the above approach
 
# Function to generate two arrays
# satisfying the given conditions
def printArrays(n) :
     
    # Declare the two arrays A and B
    A, B = [], [];
 
    # Iterate from range [1, 2*n]
    for i in range(1, 2 * n + 1):
         
        # Assign consecutive numbers to
        # same indices of the two arrays
        if (i % 2 == 0) :
            A.append(i);
        else :
            B.append(i);
 
    # Print the first array
    print("{ ", end="");
    for i in range(n) :
        print(A[i], end="");
 
        if (i != n - 1) :
            print(", ", end="");
     
    print("}");
 
    # Print the second array, B
    print("{ ", end="");
     
    for i in range(n) :
        print(B[i], end="");
 
        if (i != n - 1) :
            print(",", end=" ");
             
    print(" }", end="");
 
# Driver Code
if __name__ == "__main__" :
     
    N = 5;
 
    # Function Call
    printArrays(N);
 
    # This code is contributed by AnkitRai01


C#
// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
   
// Satisfying the given conditions
static void printArrays(int n)
{
     
    // Declare the two arrays A and B
    List A = new List();
    List B = new List();
      
    // Iterate from range [1, 2*n]
    for(int i = 1; i <= 2 * n; i++)
    {
         
        // Assign consecutive numbers to
        // same indices of the two arrays
        if (i % 2 == 0)
            A.Add(i);
        else
            B.Add(i);
    }
   
    // Print the first array
    Console.Write("{ ");
    for(int i = 0; i < n; i++)
    {
        Console.Write(A[i]);
   
        if (i != n - 1)
            Console.Write(", ");
    }
    Console.Write(" }\n");
   
    // Print the second array, B
    Console.Write("{ ");
    for(int i = 0; i < n; i++)
    {
        Console.Write(B[i]);
          
        if (i != n - 1)
            Console.Write(", ");
    }
    Console.Write(" }");
}
   
// Driver Code
public static void Main()
{
    int N = 5;
     
    // Function Call
    printArrays(N);
}
}
 
// This code is contributed by susmitakundugoaldanga


输出:
{ 2, 4, 6, 8, 10 }
{ 1, 3, 5, 7, 9 }

时间复杂度: O(N)
辅助空间: O(N)