检查每个元素乘以整数时是否存在总和为1的子集

给定一个正数数组arr[] 。任务是检查是否存在任何大小的子集，以便在将子集的每个元素与任何整数相乘后，将子集的总和设为 1。

例子：

Input: arr[] = {12, 5, 9, 21}
Output: True
Explanation: Pick numbers 5 and 9. 5*2 + 9*(-1) = 1

Input: arr[] = {9, 29, 6, 10}
Output: True
Explanation: Pick numbers 29 and 10. 29*(-1) + 10*(3) = 1

方法：如果数组中任何一对数字的HCF为1 ，则返回True否则返回False 。如果gcd(a, b) = 1 ，方程ax + by = 1对x 和 y有解。无需检查所有可能对的 HCF，因为 GCD 是一个关联函数。

gcd(a0, a1, …, an – 1) = gcd(a0, gcd(a1, …, an-1) = gcd(gcd(a0, a1), gcd(a2, …, an-1)) = …

以此类推，包括所有可能的组合。所以只需遍历数组，直到找到1的 gcd。在另一个世界中，如果gcd(a0, a1, ..., an – 1) = 1，则存在#{ai0, ..., aik} = k, 1 <= k <= N的子序列aij ，其 gcd 为1 .请按照以下步骤解决问题：

• 将变量res初始化为arr[0]。
• 使用变量i遍历范围[1, N)并执行以下任务：
  • 将res的值设置为res和arr[i] 的 gcd。
  • 如果res等于1，则返回true。
• 执行上述步骤后，打印false作为答案。

下面是上述方法的实现：

C++

// C++ Program for the above approach
#include 
using namespace std;
 
// Function to find GCD
int gcd(int a, int b)
{
    if (a < b)
        gcd(b, a);
    if (a % b == 0)
        return b;
    else
        return gcd(b, a % b);
}
 
// Utility Function
bool IsArrayGood(int arr[], int N)
{
    int i, res = arr[0];
    for (i = 1; i < N; i++) {
        res = gcd(res, arr[i]);
        if (res == 1)
            return true;
    }
    return false;
}
 
// Driver Code
int main()
{
 
    int arr[] = { 12, 5, 9, 21 };
    int N = sizeof(arr) / sizeof(arr[0]);
    bool ver = IsArrayGood(arr, N);
 
    if (ver == true)
        cout << "True";
    else
        cout << "False";
 
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
 
  // Recursive function to return gcd of a and b
  static int gcd(int a, int b)
  {
    if (a < b)
      gcd(b, a);
    if (a % b == 0)
      return b;
    else
      return gcd(b, a % b);
  }
 
  // Utility Function
  static boolean IsArrayGood(int arr[], int N)
  {
    int i, res = arr[0];
    for (i = 1; i < N; i++) {
      res = gcd(res, arr[i]);
      if (res == 1)
        return true;
    }
    return false;
  }
 
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 12, 5, 9, 21 };
    int N = arr.length;
    boolean ver = IsArrayGood(arr, N);
 
    if (ver == true){
      System.out.println("True");
    }
    else{
      System.out.println("False");
    }
  }
}
 
// This code is contributed by hrithikgarg03188

Python3

# Python code for the above approach
 
# Function to find GCD
def gcd(a, b):
    if (a < b):
        gcd(b, a);
    if (a % b == 0):
        return b;
    else:
        return gcd(b, a % b);
 
# Utility Function
def IsArrayGood(arr, N):
    i = None
    res = arr[0];
    for i in range(1, N):
        res = gcd(res, arr[i]);
        if (res == 1):
            return True;
    return False;
 
# Driver Code
arr = [12, 5, 9, 21];
N = len(arr)
ver = IsArrayGood(arr, N);
 
if (ver == True):
    print("True");
else:
    print("False");
 
# This code is contributed by Saurabh Jaiswal

C#

/*package whatever //do not write package name here */
 
using System;
public class GFG {
 
  // Recursive function to return gcd of a and b
  static int gcd(int a, int b)
  {
    if (a < b)
      gcd(b, a);
    if (a % b == 0)
      return b;
    else
      return gcd(b, a % b);
  }
 
  // Utility Function
  static bool IsArrayGood(int[] arr, int N)
  {
    int i, res = arr[0];
    for (i = 1; i < N; i++) {
      res = gcd(res, arr[i]);
      if (res == 1)
        return true;
    }
    return false;
  }
 
 
  // Driver Code
  public static void Main()
  {
    int[] arr = { 12, 5, 9, 21 };
    int N = arr.Length;
    bool ver = IsArrayGood(arr, N);
 
    if (ver == true){
      Console.Write("True");
    }
    else{
      Console.Write("False");
    }
  }
}
 
// This code is contributed by gfgking

Javascript

       // JavaScript code for the above approach
 
       // Function to find GCD
       function gcd(a, b) {
           if (a < b)
               gcd(b, a);
           if (a % b == 0)
               return b;
           else
               return gcd(b, a % b);
       }
 
       // Utility Function
       function IsArrayGood(arr, N) {
           let i, res = arr[0];
           for (i = 1; i < N; i++) {
               res = gcd(res, arr[i]);
               if (res == 1)
                   return true;
           }
           return false;
       }
 
       // Driver Code
       let arr = [12, 5, 9, 21];
       let N = arr.length;
       let ver = IsArrayGood(arr, N);
 
       if (ver == true)
           document.write("True");
       else
           document.write("False");
 
 // This code is contributed by Potta Lokesh
   

输出

True

时间复杂度： O(N*log(D)) 其中 D 是数组中的最大元素
辅助空间： O(1)