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📜  通过将数组的所有元素乘以任何值,将数组的任何子集的总和减少为1

📅  最后修改于: 2021-05-17 04:43:09             🧑  作者: Mango

给定一个由N个正整数组成的数组arr [] ,任务是检查在将给定数组的所有子集的所有元素乘以任何整数之后,是否可以将其总和减小为1 。如果无法这样做,请打印“否” 。否则,打印“是”

例子:

天真的方法:最简单的方法是生成给定数组的所有可能的子集,并且如果数组中存在任何子集,以使其元素的总和乘以任何整数后得出1,则打印“是” 。否则,打印“否”
时间复杂度: O(N * 2 N )
辅助空间: O(1)

高效的方法:上述方法也可以通过使用Bezout的身份(Bezout引理)进行优化,该身份表示如果任意两个整数ab的GCD等于d ,则存在整数xy ,使得a * x + b * y = d

因此,我们的想法是检查给定数组arr []的GCD是否可以设为1 。因此,为了满足给定条件,必须存在GCD1的任何两个元素,然后数组的GCD等于1 。因此,打印“是” 。否则,打印“否”

下面是上述方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function to return gcd of a and b
int gcd(int a, int b)
{
    // Base Case
    if (a == 0)
        return b;
 
    // Find the GCD recursively
    return gcd(b % a, a);
}
 
// Function to calculate the GCD
// of the array arr[]
int findGCDofArray(int arr[], int N)
{
    // Stores the GCD of array
    int g = 0;
 
    // Travese the array arr[]
    for (int i = 0; i < N; i++) {
 
        // Update gcd of the array
        g = gcd(g, arr[i]);
 
        // If gcd is 1, then return 1
        if (g == 1) {
            return 1;
        }
    }
 
    // Return the resultant GCD
    return g;
}
 
// Function to check if a subset satisfying
// the given condition exists or not
void findSubset(int arr[], int N)
{
 
    // Calculate the gcd of the array
    int gcd = findGCDofArray(arr, N);
 
    // If gcd is 1, then print Yes
    if (gcd == 1) {
        cout << "Yes";
    }
 
    // Otherwise, print No
    else {
        cout << "No";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 29, 6, 4, 10 };
    int N = sizeof(arr) / sizeof(arr[0]);
    findSubset(arr, N);
 
    return 0;
}


Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{
 
  // Function to return gcd of a and b
  static int gcd(int a, int b)
  {
 
    // Base Case
    if (a == 0)
      return b;
 
    // Find the GCD recursively
    return gcd(b % a, a);
  }
 
  // Function to calculate the GCD
  // of the array arr[]
  static int findGCDofArray(int arr[], int N)
  {
 
    // Stores the GCD of array
    int g = 0;
 
    // Travese the array arr[]
    for (int i = 0; i < N; i++)
    {
 
      // Update gcd of the array
      g = gcd(g, arr[i]);
 
      // If gcd is 1, then return 1
      if (g == 1) {
        return 1;
      }
    }
 
    // Return the resultant GCD
    return g;
  }
 
  // Function to check if a subset satisfying
  // the given condition exists or not
  static void findSubset(int arr[], int N)
  {
 
    // Calculate the gcd of the array
    int gcd = findGCDofArray(arr, N);
 
    // If gcd is 1, then print Yes
    if (gcd == 1) {
      System.out.println("Yes");
    }
 
    // Otherwise, print No
    else {
      System.out.println("No");
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    // Given array
    int arr[] = { 29, 6, 4, 10 };
 
    // length of the array
    int N = arr.length;
 
    // function call
    findSubset(arr, N);
  }
}
 
// This code is contributed by Kingash.


Python3
# Python3 program for the above approach
 
# Function to return gcd of a and b
def gcd(a, b):
     
    # Base Case
    if (a == 0):
        return b
 
    # Find the GCD recursively
    return gcd(b % a, a)
 
# Function to calculate the GCD
# of the array arr[]
def findGCDofArray(arr, N):
     
    # Stores the GCD of array
    g = 0
 
    # Travese the array arr[]
    for i in range(N):
 
        # Update gcd of the array
        g = gcd(g, arr[i])
 
        # If gcd is 1, then return 1
        if (g == 1):
            return 1
 
    # Return the resultant GCD
    return g
 
# Function to check if a subset satisfying
# the given condition exists or not
def findSubset(arr, N):
 
    # Calculate the gcd of the array
    gcd = findGCDofArray(arr, N)
 
    # If gcd is 1, then prYes
    if (gcd == 1):
        print("Yes")
     
    # Otherwise, prNo
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
    arr = [29, 6, 4, 10]
 
    N = len(arr)
 
    findSubset(arr, N)
 
    # This code is contributed by mohit kumar 29.


C#
// C# program for the above approach
using System;
class GFG
{
 
  // Function to return gcd of a and b
  static int gcd(int a, int b)
  {
 
    // Base Case
    if (a == 0)
      return b;
 
    // Find the GCD recursively
    return gcd(b % a, a);
  }
 
  // Function to calculate the GCD
  // of the array arr[]
  static int findGCDofArray(int[] arr, int N)
  {
 
    // Stores the GCD of array
    int g = 0;
 
    // Travese the array arr[]
    for (int i = 0; i < N; i++) {
 
      // Update gcd of the array
      g = gcd(g, arr[i]);
 
      // If gcd is 1, then return 1
      if (g == 1) {
        return 1;
      }
    }
 
    // Return the resultant GCD
    return g;
  }
 
  // Function to check if a subset satisfying
  // the given condition exists or not
  static void findSubset(int[] arr, int N)
  {
 
    // Calculate the gcd of the array
    int gcd = findGCDofArray(arr, N);
 
    // If gcd is 1, then print Yes
    if (gcd == 1) {
      Console.Write("Yes");
    }
 
    // Otherwise, print No
    else {
      Console.Write("No");
    }
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int[] arr = { 29, 6, 4, 10 };
    int N = arr.Length;
    findSubset(arr, N);
  }
}
 
// This code is contributed by shivani


输出:
Yes

时间复杂度: O(N * log(M)),其中M是数组中最小的元素
辅助空间: O(1)