使用第二个数组使所有元素相等的最小操作

给定两个数组A []和B []都具有N个元素。查找使A的所有元素等于第二个数组B的最小操作数。一个操作包括：

A[i] = A[i] - B[i], 0 <= i <= n-1 

注意：如果不可能使数组元素等于print -1。
例子：

Input: A[] = {5, 7, 10, 5, 15}, B[] = {2, 2, 1, 3, 5} 
Output: 8 
Explanation: 
Following are the operations: 
1) Choosing index 1 -> 5 5 10 5 15 
2) Choosing index 2 -> 5 5 9 5 15 
3) Choosing index 2 -> 5 5 8 5 15 
4) Choosing index 2 -> 5 5 7 5 15 
5) Choosing index 2 -> 5 5 6 5 15 
6) Choosing index 2 -> 5 5 5 5 15 
7) Choosing index 4 -> 5 5 5 5 10 
8) Choosing index 4 -> 5 5 5 5 5

Input: A[] = {3, 5, 8, 2}, B[] = {1, 2, 1, 1} 
Output: 12

方法：

• 可以观察到，如果可以使A的所有元素相等，则最大可能值是A的最小元素。
• 因此，当A的所有元素都相等时，我们可以遍历最终值x。使用上述观察结果– 0 <= x <= min(A [i])。对于每个x，遍历A并检查是否可以使用B [i]使A [i]等于x：

  A[i] - k*B[i] = x
  Take mod with B[i] on both sides
  A[i] %B[i] = x %B[i]  ->  
  must be satisfied for A[i] to be converted to x
  

  如果满足条件，则此元素所需的操作数=(A [i] – x)/ B [i]

下面是上述方法的实现：

C++

// C++ implementation to find the
// minimum operations make all elements 
// equal using the second array
  
#include 
  
using namespace std;
  
// Function to find the minimum 
// operations required to make
// all elements of the array equal
int minOperations(int a[], int b[], int n)
{
    // Minimum element of A[]
    int minA = *min_element(a, a + n);
  
    // Traverse through all final values
    for (int x = minA; x >= 0; x--) {
          
        // Variable indicating 
        // whether all elements
        // can be converted to x or not
        bool check = 1;
  
        // Total operations
        int operations = 0;
          
        // Traverse through 
        // all array elements
        for (int i = 0; i < n; i++) {
            if (x % b[i] == a[i] % b[i]) {
                operations += 
                    (a[i] - x) / b[i];
            }
  
            // All elements can't 
            // be converted to x
            else {
                check = 0;
                break;
            }
        }
        if (check)
            return operations;
    }
    return -1;
}
  
// Driver Code
int main()
{
    int N = 5;
    int A[N] = { 5, 7, 10, 5, 15 };
    int B[N] = { 2, 2, 1, 3, 5 };
  
    cout << minOperations(A, B, N);
    return 0;
}

Java

// Java implementation to find the
// minimum operations make all elements 
// equal using the second array
import java.util.*;
class GFG{
  
// Function to find the minimum 
// operations required to make
// all elements of the array equal
static int minOperations(int a[], int b[], int n)
{
    // Minimum element of A[]
    int minA = Arrays.stream(a).min().getAsInt();
  
    // Traverse through all final values
    for (int x = minA; x >= 0; x--) 
    {
          
        // Variable indicating 
        // whether all elements
        // can be converted to x or not
        boolean check = true;
  
        // Total operations
        int operations = 0;
          
        // Traverse through 
        // all array elements
        for (int i = 0; i < n; i++) 
        {
            if (x % b[i] == a[i] % b[i]) 
            {
                operations += (a[i] - x) / b[i];
            }
  
            // All elements can't 
            // be converted to x
            else 
            {
                check = false;
                break;
            }
        }
        if (check)
            return operations;
    }
    return -1;
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 5;
    int A[] = { 5, 7, 10, 5, 15 };
    int B[] = { 2, 2, 1, 3, 5 };
  
    System.out.print(minOperations(A, B, N));
}
}
  
// This code is contributed by AbhiThakur

Python3

# Python3 implementation to find the
# minimum operations make all elements
# equal using the second array
  
# Function to find the minimum
# operations required to make
# all elements of the array equal
def minOperations(a, b, n):
      
    # Minimum element of A
    minA = min(a);
  
    # Traverse through all final values
    for x in range(minA, -1, -1):
  
        # Variable indicating
        # whether all elements
        # can be converted to x or not
        check = True;
  
        # Total operations
        operations = 0;
  
        # Traverse through
        # all array elements
        for i in range(n):
            if (x % b[i] == a[i] % b[i]):
                operations += (a[i] - x) / b[i];
  
            # All elements can't
            # be converted to x
            else:
                check = False;
                break;
  
        if (check):
            return operations;
  
    return -1;
  
# Driver Code
if __name__ == '__main__':
      
    N = 5;
    A = [ 5, 7, 10, 5, 15 ];
    B = [ 2, 2, 1, 3, 5 ];
  
    print(int(minOperations(A, B, N)));
  
# This code is contributed by amal kumar choubey

C#

// C# implementation to find the
// minimum operations make all elements 
// equal using the second array
using System;
using System.Linq;
  
class GFG{
  
// Function to find the minimum 
// operations required to make
// all elements of the array equal
static int minOperations(int []a, int []b, int n)
{
      
    // Minimum element of A[]
    int minA = a.Max();
  
    // Traverse through all final values
    for(int x = minA; x >= 0; x--) 
    {
         
       // Variable indicating 
       // whether all elements
       // can be converted to x or not
       bool check = true;
         
       // Total operations
       int operations = 0;
         
       // Traverse through 
       // all array elements
       for(int i = 0; i < n; i++) 
       {
          if (x % b[i] == a[i] % b[i]) 
          {
              operations += (a[i] - x) / b[i];
          }
            
          // All elements can't 
          // be converted to x
          else
          {
              check = false;
              break;
          }
       }
       if (check)
           return operations;
    }
    return -1;
}
  
// Driver Code
public static void Main(string[] args)
{
    int N = 5;
    int []A = { 5, 7, 10, 5, 15 };
    int []B = { 2, 2, 1, 3, 5 };
  
    Console.WriteLine(minOperations(A, B, N));
}
}
  
// This code is contributed by SoumikMondal

输出：

8

时间复杂度： O(N * min(A _i ))