计算给定两个数组的所有成对和的 GCD

给定两个大小为N和M的数组A[]和B[]计算所有成对和的 GCD (A[i]+B[j]) 1<=i<=N 和 1<=j<=M。

例子：

Input: A[] = {1, 7, 25, 55}, B[] = {1, 3, 5}
Output: 2
Explaination: The GCD of all pairwise sum of (A[i]+B[j]) is equals to
GCD(1+1, 1+3, 1+5, 7+1, 7+3, 7+5, 25+1, 25+3, 25+5, 55+1, 55+3, 55+5)
GCD(2, 4, 6, 8, 10, 12, 26, 28, 30, 56, 58, 60) = 2

Input: A[] = {8, 16, 20}, B[] = {12, 24}
Output: 4

编程需要懂一点英语

朴素方法：这个问题的简单方法是计算所有成对的和，然后计算它们的 GCD。

下面是这种方法的实现。

C++

// C++ code to implement the approach
 
#include 
using namespace std;
 
// Function to calculate gcd
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to calculate the GCD
// of all pairwise sums
int calculateGCD(vector& a,
                 vector& b,
                 int N, int M)
{
    int ans = 0;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
 
            // Pairwise sum of all elements
            int sum = a[i] + b[j];
 
            // Finding gcd of the elements
            ans = gcd(ans, sum);
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    int N = 4, M = 3;
    // Initialization of the vector
    vector A = { 1, 7, 25, 55 };
    vector B = { 1, 3, 5 };
 
    // output
    cout << calculateGCD(A, B, N, M);
    return 0;
}

Java

// Java code to implement the approach
import java.util.*;
 
class GFG {
 
  // Function to calculate gcd
  static int gcd(int a, int b)
  {
    if (b == 0)
      return a;
    return gcd(b, a % b);
  }
 
  // Function to calculate the GCD
  // of all pairwise sums
  static int calculateGCD(int a[],
                          int b[],
                          int N, int M)
  {
    int ans = 0;
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < M; j++) {
 
        // Pairwise sum of all elements
        int sum = a[i] + b[j];
 
        // Finding gcd of the elements
        ans = gcd(ans, sum);
      }
    }
    return ans;
  }
 
  // Driver code
  public static void main (String[] args) {
    int N = 4, M = 3;
 
    // Initialization of the vector
    int A[] = { 1, 7, 25, 55 };
    int B[] = { 1, 3, 5 };
 
    // output
    System.out.print(calculateGCD(A, B, N, M));
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python code to implement the approach
# Function to calculate gcd
def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)
   
# Function to calculate the GCD
# of all pairwise sums
def calculateGCD(a, b, N, M):
    ans = 0
    for i in range(N):
        for j in range(M):
           
            # Pairwise sum of all elements
            sum = a[i]+b[j]
             
            # Finding gcd of the elements
            ans = gcd(ans, sum)
    return ans
 
# Driver code
N = 4
M = 3
A = [1, 7, 25, 55]
B = [1, 3, 5]
print(calculateGCD(A, B, N, M))
 
'''This Code is contributed by Rajat Kumar'''

C#

// C# code to implement the approach
using System;
using System.Numerics;
using System.Collections.Generic;
 
public class GFG {
 
  // Function to calculate gcd
  static int gcd(int a, int b)
  {
    if (b == 0)
      return a;
    return gcd(b, a % b);
  }
 
  // Function to calculate the GCD
  // of all pairwise sums
  static int calculateGCD(int[] a, int[] b, int N, int M)
  {
    int ans = 0;
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < M; j++) {
 
        // Pairwise sum of all elements
        int sum = a[i] + b[j];
 
        // Finding gcd of the elements
        ans = gcd(ans, sum);
      }
    }
    return ans;
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    int N = 4, M = 3;
     
    // Initialization of the vector
    int[] A = { 1, 7, 25, 55 };
    int[] B = { 1, 3, 5 };
 
    // output
    Console.WriteLine(calculateGCD(A, B, N, M));
  }
}
 
// This code is contributed by phasing17

Javascript

C++

// C++ code to implement the approach
 
#include 
using namespace std;
 
// Function for calculating gcd
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to calculate the required GCD
int calculateGCD(vector& a,
                 vector& b, int N, int M)
{
    int ans = 0;
 
    // Sorting the arrays
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
 
    // Calculating the gcd of all the elements
    // of type (i, j) i>0 and j>=0
    // Using the property
    // gcd(a[0]+b[j], a[i]+b[j])
    // =gcd(a[0]+b[j], a[i]-a[0])
    for (int i = 1; i < N; i++) {
        ans = gcd(ans, a[i] - a[0]);
    }
 
    // Calculating the gcd of the remaining
    // elements of the type (a[0]+b[j])
    for (int i = 0; i < M; i++) {
        ans = gcd(ans, a[0] + b[i]);
    }
    return ans;
}
 
// Driver code
int main()
{
    int N = 4, M = 3;
    // Initialization of the array
    vector A = { 1, 7, 25, 55 };
    vector B = { 1, 3, 5 };
 
    // Function call
    cout << calculateGCD(A, B, N, M);
    return 0;
}

Python3

# Python3 code to implement the approach
 
# Function for calculating gcd
def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)
 
# Function to calculate the required GCD
def calculateGCD(a, b, N, M):
    ans = 0
     
    # sorting the arrays
    a.sort()
    b.sort()
     
    # calculating the gcd of all the elements
    # of type (i, j) i>0 and j>=0
    # Using the property
    # gcd(a[0]+b[j], a[i]+b[j])
    # =gcd(a[0]+b[j], a[i]-a[0])
    for i in range(1, N):
        ans = gcd(ans, a[i] - a[0])
         
    # Calculating the gcd of the remaining
    # elements of the type (a[0]+b[j])
    for i in range(M):
        ans = gcd(ans, a[0] + b[i])
    return ans
 
# Driver code
N, M = 4, 3
A = [1, 7, 25, 55]
B = [1, 3, 5]
 
# function all
print(calculateGCD(A, B, N, M))
 
# This code is contributed by phasing17.

输出

时间复杂度：O(N*M)
辅助空间：O(N)

有效方法：可以根据以下数学观察有效地解决问题：

观察：

By Euclidian algorithm it can be said that GCD( a, b) = GCD(a-b, b) where(a>b) .
Then for any j:
GCD(A[0] + B[j], A[1] + B[j], . . ., A[N-1] + B[j]) = GCD(A[0]+B[j], A[1]-A[0], A[2]-A[0], . . ., A[N]-A[0]).
The term GCD(A[1]-A[0], A[2]-A[0], . . ., A[N]-A[0]) is common for all j from 0 to M-1 (say this is X)
This leaves us with only the elements of type (A[0] + B[j]) for which will calculate their GCD and will get the desired GCD upon calculating their GCD with X.

编程需要懂一点英语

请按照以下步骤解决此问题：

首先对两个数组进行排序。
然后遍历i = 1 到 N-1并计算所有 A[i] -A[0] 对的 GCD（比如X ）。
然后遍历i = 1 到 M-1并计算所有 A[0] +B[i] 对的 GCD（比如Y ）。
返回最终答案，即 (X 和 Y) 的 GCD。

下面是这种方法的实现。

C++

// C++ code to implement the approach
 
#include 
using namespace std;
 
// Function for calculating gcd
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to calculate the required GCD
int calculateGCD(vector& a,
                 vector& b, int N, int M)
{
    int ans = 0;
 
    // Sorting the arrays
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
 
    // Calculating the gcd of all the elements
    // of type (i, j) i>0 and j>=0
    // Using the property
    // gcd(a[0]+b[j], a[i]+b[j])
    // =gcd(a[0]+b[j], a[i]-a[0])
    for (int i = 1; i < N; i++) {
        ans = gcd(ans, a[i] - a[0]);
    }
 
    // Calculating the gcd of the remaining
    // elements of the type (a[0]+b[j])
    for (int i = 0; i < M; i++) {
        ans = gcd(ans, a[0] + b[i]);
    }
    return ans;
}
 
// Driver code
int main()
{
    int N = 4, M = 3;
    // Initialization of the array
    vector A = { 1, 7, 25, 55 };
    vector B = { 1, 3, 5 };
 
    // Function call
    cout << calculateGCD(A, B, N, M);
    return 0;
}

Python3

# Python3 code to implement the approach
 
# Function for calculating gcd
def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)
 
# Function to calculate the required GCD
def calculateGCD(a, b, N, M):
    ans = 0
     
    # sorting the arrays
    a.sort()
    b.sort()
     
    # calculating the gcd of all the elements
    # of type (i, j) i>0 and j>=0
    # Using the property
    # gcd(a[0]+b[j], a[i]+b[j])
    # =gcd(a[0]+b[j], a[i]-a[0])
    for i in range(1, N):
        ans = gcd(ans, a[i] - a[0])
         
    # Calculating the gcd of the remaining
    # elements of the type (a[0]+b[j])
    for i in range(M):
        ans = gcd(ans, a[0] + b[i])
    return ans
 
# Driver code
N, M = 4, 3
A = [1, 7, 25, 55]
B = [1, 3, 5]
 
# function all
print(calculateGCD(A, B, N, M))
 
# This code is contributed by phasing17.

输出

时间复杂度：O(N*logN + M*logM)
辅助空间：O(1)