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📜  将N个自然数分成两个总和大于1的GCD

📅  最后修改于: 2021-06-26 15:49:10             🧑  作者: Mango

给定一个整数N ,任务是创建两组从1到N的截然不同的元素,使它们各自总和的gcd大于1。打印各个集合。如果无法进行此类拆分,请打印-1。

例子:

方法1:一种简单的方法是将直到N的所有偶数均分成一组,而将所有奇数均分成另一组。

下面的代码是该方法的实现:

C++
// C++ program to split N natural numbers
// into two sets having GCD
// of their sums greater than 1
 
#include 
using namespace std;
 
// Function to create
// and print the two sets
void createSets(int N)
{
    // No such split
    // possible for N <= 2
    if (N <= 2) {
        cout << "-1" << endl;
        return;
    }
 
    // Print the first set
    // consisting of even elements
    for (int i = 2; i <= N; i += 2)
        cout << i << " ";
    cout << "\n";
    // Print the second set
    // consisting of odd ones
    for (int i = 1; i <= N; i += 2) {
        cout << i << " ";
    }
}
// Driver Code
int main()
{
 
    int N = 6;
    createSets(N);
 
    return 0;
}


Java
// Java program to split N natural numbers
// into two sets having GCD
// of their sums greater than 1
class GFG{
   
// Function to create
// and print the two sets
static void createSets(int N)
{
    // No such split
    // possible for N <= 2
    if (N <= 2)
    {
        System.out.println("-1" );
        return;
    }
 
    // Print the first set
    // consisting of even elements
    for (int i = 2; i <= N; i += 2)
        System.out.print(i + " ");
    System.out.print("\n") ;
   
    // Print the second set
    // consisting of odd ones
    for (int i = 1; i <= N; i += 2)
    {
        System.out.print(i + " ");
    }
}
// Driver Code
public static void main(String[] args)
{
 
    int N = 6;
    createSets(N);
}
}
 
// This code is contributed by shivanisinghss2110


Python3
# Python3 program to split N natural numbers
# into two sets having GCD
# of their sums greater than 1
 
# Function to create
# and print the two sets
def createSets(N):
   
    # No such split
    # possible for N <= 2
    if (N <= 2):
        print("-1");
        return;
 
    # Print the first set
    # consisting of even elements
    for i in range(2, N + 1, 2):
        print(i, end=" ");
    print("");
 
    # Print the second set
    # consisting of odd ones
    for i in range(1, N + 1, 2):
        print(i, end = " ");
 
# Driver Code
if __name__ == '__main__':
    N = 6;
    createSets(N);
 
# This code is contributed by gauravrajput1


C#
// C# program to split N natural numbers
// into two sets having GCD
// of their sums greater than 1
using System;
class GFG{
 
// Function to create
// and print the two sets
static void createSets(int N)
{
    // No such split
    // possible for N <= 2
    if (N <= 2)
    {
        Console.WriteLine("-1" );
        return;
    }
 
    // Print the first set
    // consisting of even elements
    for (int i = 2; i <= N; i += 2)
        Console.Write(i + " ");
    Console.Write("\n") ;
 
    // Print the second set
    // consisting of odd ones
    for (int i = 1; i <= N; i += 2)
    {
        Console.Write(i + " ");
    }
}
// Driver Code
public static void Main(String[] args)
{
 
    int N = 6;
    createSets(N);
}
}
 
// This code is contributed by shivanisinghss2110


Javascript


C++
// C++ program to split N natural numbers
// into two sets having GCD
// of their sums greater than 1
 
#include 
using namespace std;
 
#define ll long long
 
// Function to create
// and print the two sets
void createSets(ll n)
{
    // For n <= 2 such sets
    // can never be formed
    if (n <= 2) {
        cout << "-1";
        return;
    }
 
    else {
 
        // Check if N is even or odd
        // and store the element
        // which divides the sum of N
        // natural numbers accordingly
        ll x = (n % 2 == 0) ? (n / 2)
                            : ((n + 1) / 2);
 
        // First set
        cout << x << endl;
 
        // Print elements of second set
        for (ll i = 1; i <= n; i++) {
 
            if (i == x)
                continue;
 
            cout << i << " ";
        }
    }
    return;
}
 
// Driver code
int main()
{
    ll N = 7;
 
    createSets(N);
 
    return 0;
}


Java
// Java program to split N natural numbers
// into two sets having GCD
// of their sums greater than 1
class GFG{
 
// Function to create
// and print the two sets
static void createSets(long n)
{
     
    // For n <= 2 such sets
    // can never be formed
    if (n <= 2)
    {
        System.out.print("-1");
        return;
    }
    else
    {
         
        // Check if N is even or odd
        // and store the element
        // which divides the sum of N
        // natural numbers accordingly
        long x = (n % 2 == 0) ? (n / 2) :
                          ((n + 1) / 2);
 
        // First set
        System.out.print(x + "\n");
 
        // Print elements of second set
        for(int i = 1; i <= n; i++)
        {
            if (i == x)
                continue;
 
            System.out.print(i + " ");
        }
    }
    return;
}
 
// Driver code
public static void main(String[] args)
{
    long N = 7;
 
    createSets(N);
}
}
 
// This code is contributed by PrinciRaj1992


Python3
# Python3 program to split N natural numbers
# into two sets having GCD
# of their sums greater than 1
 
# Function to create
# and print the two sets
def createSets(n):
     
    # For n <= 2 such sets
    # can never be formed
    if (n <= 2):
        print("-1");
        return;
    else:
 
        # Check if N is even or odd
        # and store the element
        # which divides the sum of N
        # natural numbers accordingly
        x = (n // 2) if(n % 2 == 0) else (
            (n + 1) // 2);
 
        # First set
        print(x, " ");
 
        # Print elements of second set
        for i in range(1, n + 1):
            if (i == x):
                continue;
 
            print(i, end = " ");
 
    return;
 
# Driver code
if __name__ == '__main__':
     
    N = 7;
 
    createSets(N);
 
# This code is contributed by 29AjayKumar


C#
// C# program to split N natural numbers
// into two sets having GCD
// of their sums greater than 1
using System;
class GFG{
 
// Function to create
// and print the two sets
static void createSets(long n)
{
     
    // For n <= 2 such sets
    // can never be formed
    if (n <= 2)
    {
        Console.Write("-1");
        return;
    }
    else
    {
         
        // Check if N is even or odd
        // and store the element
        // which divides the sum of N
        // natural numbers accordingly
        long x = (n % 2 == 0) ? (n / 2) :
                          ((n + 1) / 2);
 
        // First set
        Console.Write(x + "\n");
 
        // Print elements of second set
        for(int i = 1; i <= n; i++)
        {
            if (i == x)
                continue;
 
            Console.Write(i + " ");
        }
    }
    return;
}
 
// Driver code
public static void Main(String[] args)
{
    long N = 7;
 
    createSets(N);
}
}
 
// This code is contributed by sapnasingh4991


Javascript


输出:
2 4 6 
1 3 5

方法2:基于N,可以将该方法分为2种情况:

  1. 如果N为奇数:
    • 在这种情况下,N个自然数的总和可被(N + 1)/ 2整除。
    • 因此,我们只需要将(N + 1)/ 2放入第一组,其余数字放入第二组。
    • 这将自动使GCD的总和大于1。
  2. 如果N是偶数:
    • 在这种情况下,N个自然数的总和可被N / 2整除。
    • 因此,我们只需要将N / 2放在第一组中,其余的数字放在第二组中。
    • 这将自动使GCD的总和大于1。

下面是上述方法的实现:

C++

// C++ program to split N natural numbers
// into two sets having GCD
// of their sums greater than 1
 
#include 
using namespace std;
 
#define ll long long
 
// Function to create
// and print the two sets
void createSets(ll n)
{
    // For n <= 2 such sets
    // can never be formed
    if (n <= 2) {
        cout << "-1";
        return;
    }
 
    else {
 
        // Check if N is even or odd
        // and store the element
        // which divides the sum of N
        // natural numbers accordingly
        ll x = (n % 2 == 0) ? (n / 2)
                            : ((n + 1) / 2);
 
        // First set
        cout << x << endl;
 
        // Print elements of second set
        for (ll i = 1; i <= n; i++) {
 
            if (i == x)
                continue;
 
            cout << i << " ";
        }
    }
    return;
}
 
// Driver code
int main()
{
    ll N = 7;
 
    createSets(N);
 
    return 0;
}

Java

// Java program to split N natural numbers
// into two sets having GCD
// of their sums greater than 1
class GFG{
 
// Function to create
// and print the two sets
static void createSets(long n)
{
     
    // For n <= 2 such sets
    // can never be formed
    if (n <= 2)
    {
        System.out.print("-1");
        return;
    }
    else
    {
         
        // Check if N is even or odd
        // and store the element
        // which divides the sum of N
        // natural numbers accordingly
        long x = (n % 2 == 0) ? (n / 2) :
                          ((n + 1) / 2);
 
        // First set
        System.out.print(x + "\n");
 
        // Print elements of second set
        for(int i = 1; i <= n; i++)
        {
            if (i == x)
                continue;
 
            System.out.print(i + " ");
        }
    }
    return;
}
 
// Driver code
public static void main(String[] args)
{
    long N = 7;
 
    createSets(N);
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to split N natural numbers
# into two sets having GCD
# of their sums greater than 1
 
# Function to create
# and print the two sets
def createSets(n):
     
    # For n <= 2 such sets
    # can never be formed
    if (n <= 2):
        print("-1");
        return;
    else:
 
        # Check if N is even or odd
        # and store the element
        # which divides the sum of N
        # natural numbers accordingly
        x = (n // 2) if(n % 2 == 0) else (
            (n + 1) // 2);
 
        # First set
        print(x, " ");
 
        # Print elements of second set
        for i in range(1, n + 1):
            if (i == x):
                continue;
 
            print(i, end = " ");
 
    return;
 
# Driver code
if __name__ == '__main__':
     
    N = 7;
 
    createSets(N);
 
# This code is contributed by 29AjayKumar

C#

// C# program to split N natural numbers
// into two sets having GCD
// of their sums greater than 1
using System;
class GFG{
 
// Function to create
// and print the two sets
static void createSets(long n)
{
     
    // For n <= 2 such sets
    // can never be formed
    if (n <= 2)
    {
        Console.Write("-1");
        return;
    }
    else
    {
         
        // Check if N is even or odd
        // and store the element
        // which divides the sum of N
        // natural numbers accordingly
        long x = (n % 2 == 0) ? (n / 2) :
                          ((n + 1) / 2);
 
        // First set
        Console.Write(x + "\n");
 
        // Print elements of second set
        for(int i = 1; i <= n; i++)
        {
            if (i == x)
                continue;
 
            Console.Write(i + " ");
        }
    }
    return;
}
 
// Driver code
public static void Main(String[] args)
{
    long N = 7;
 
    createSets(N);
}
}
 
// This code is contributed by sapnasingh4991

Java脚本


输出:
4
1 2 3 5 6 7

时间复杂度: O(N)
辅助空间复杂度: O(1)