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

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：

Input: arr[] = {29, 6, 4, 10}
Output: Yes
Explanation:
Choose a subset {29, 6, 10} and multiply each corresponding element by {1, -3, -1}.
Therefore, sum of the subset = 29 * (1) + 6 * (-3) + 10 * (-1) = 29 – 18 – 10 = 1.
Therefore, print “Yes”.

Input: arr[] = {6, 3, 9}
Output: No

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a和b的GCD等于d ，则存在整数x和y ，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GCD为1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)