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到 1

给定一个由N 个正整数组成的数组arr[] ，任务是检查给定数组的任何子集的元素之和在将其所有元素乘以任何整数后是否可以减少到1 。如果不可能这样做，则打印“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

朴素方法：最简单的方法是生成给定数组的所有可能子集，如果数组中存在任何子集，使得其元素之和乘以任何整数后为 1，然后打印“是” 。否则打印“否” 。
时间复杂度： O(N * 2 ^N )
辅助空间： O(1)

高效方法：上述方法也可以通过使用 Bezout 的恒等式(Bezout 引理)进行优化，它指出如果任意两个整数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，要满足给定条件，必须存在任何两个GCD为1 的元素，则数组的 GCD 将等于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

Javascript

// javascript program for the above approach
    // Function to return gcd of a and b
    function 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
    function findGCDofArray(arr , N) {
 
        // Stores the GCD of array
        var g = 0;
 
        // Travese the array arr
        for (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
    function findSubset(arr , N) {
 
        // Calculate the gcd of the array
        var gcd = findGCDofArray(arr, N);
 
        // If gcd is 1, then prvar Yes
        if (gcd == 1) {
            document.write("Yes");
        }
 
        // Otherwise, prvar No
        else {
            document.write("No");
        }
    }
 
    // Driver code
     
 
        // Given array
        var arr = [ 29, 6, 4, 10 ];
 
        // length of the array
        var N = arr.length;
 
        // function call
        findSubset(arr, N);
 
// This code contributed by gauravrajput1

输出：

Yes

时间复杂度： O(N * log(M))，其中 M 是数组的最小元素。
辅助空间： O(1)

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