如果任意一个乘以 M，则找到 M，其中 A、B、C 按给定顺序构成 AP

给定 3 个正整数A、B 和 C。选择一个正整数M ，并将 A、B 或 C 中的任何一个与 M 相乘。任务是决定 A、B 和 C 是否符合算术级数平均 (AP)表示执行上述操作一次后。 A、B、C 的顺序不能更改。

例子：

Input: A = 30, B = 5, C = 10
Output: YES
Explanation: B = 5 can be multiplied by M = 4.
Then 2 * B = A + C (2 * 20 = 30 + 10).

Input: A = 2, B = 1, C = 1
Output: NO
Explanation: A + C = 3 which is odd. B = 1 which is odd.
Both these values cannot be made even (because 2*B is always even so need for making both of them even) and equal simultaneously.

编程需要懂一点英语

方法：

如果三个整数在 AP 中

2 * B = A + C -(1)

This equation can also be written as

2 * B – C = A -(2)

and

2 * B – A = C -(3)

编程需要懂一点英语

以下是使用这种关系来确定给定的三个数字是否可以形成 AP 的步骤-

声明一个变量new_b并将其初始化为2 * B 。
如果A、B、C已经在 AP 中，则返回true 。它们中的任何一个都可以乘以 M=1，答案总是“是”。
如果 B 可以乘以任何数字以使其等于 A+C 即(A + C) % new_b = 0 ，则返回true 。无论M的值如何，余数始终为 0。
现在剩下的唯一选择是将A或C相乘以使它们的总和等于new_b 。
- 如果(A + C) > new_b，则返回false 。在这种情况下，将A或C与M相乘只会增加其值。答案永远是“不”。
- 否则，如果(new_b – C) % A = 0 ，则返回true 。在这种情况下， A可以乘以某个整数M 。在这种情况下， (new_b – C) 可以看作是A的倍数。因此，在(new_b – C)除以A时，无论 M 的值如何，余数始终为0。答案将是“是”（根据等式 (2) 的 a）。
- 否则，如果(new_b – A) % B = 0 ，则返回true 。在这种情况下， C可以乘以某个整数M 。在这种情况下， (new_b – A) 可以看作是C的倍数。因此，在(new_b – A)除以C时，余数将始终为0 ，而与M的值无关。答案将是“是”（根据等式（3））。
- 否则返回false 。

插图：

Input: A = 30, B = 5, C = 10
Output: YES
Explanation:
new_b = 2 * B
= 2 * 5
= 10
(A + C) % new_b
(30 + 10) % 10
40 % 10 = 0
This proves that the numbers 30 5 10 are in A.P. after multiplying with some number M.

编程需要懂一点英语

以下是上述方法的实现 -

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to check if A, B, C
// can be in A.P. after multiplying
// by any integer M
bool isAP(int A, int B, int C)
{
    int new_b = B * 2;
 
    // Check if A, B, C are already
    // in A.P.
    if (new_b == (A + C))
        return true;
 
    // Check if multiplying B will
    // give A.P.
    if ((A + C) % new_b == 0)
        return true;
 
    if ((A + C) > new_b)
        return false;
 
    // Check if multiplying A will
    // give A.P.
    if ((new_b - C) % A == 0)
        return true;
 
    // Check if multiplying C will
    // give A.P.
    if ((new_b - A) % C == 0)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    int A, B, C;
    A = 30;
    B = 5;
    C = 10;
 
    // Function call
    bool ans = isAP(A, B, C);
 
    // Displaying the answer
    if (ans)
        cout << "YES";
    else
        cout << "NO";
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to check if A, B, C
  // can be in A.P. after multiplying
  // by any integer M
  static Boolean isAP(int A, int B, int C)
  {
    int new_b = B * 2;
 
    // Check if A, B, C are already
    // in A.P.
    if (new_b == (A + C))
      return true;
 
    // Check if multiplying B will
    // give A.P.
    if ((A + C) % new_b == 0)
      return true;
 
    if ((A + C) > new_b)
      return false;
 
    // Check if multiplying A will
    // give A.P.
    if ((new_b - C) % A == 0)
      return true;
 
    // Check if multiplying C will
    // give A.P.
    if ((new_b - A) % C == 0)
      return true;
 
    return false;
  }
 
  // Driver code
  public static void main (String[] args) {
    int A, B, C;
    A = 30;
    B = 5;
    C = 10;
 
    // Function call
    Boolean ans = isAP(A, B, C);
 
    // Displaying the answer
    if (ans)
      System.out.print("YES");
    else
      System.out.print("NO");
  }
}
 
// This code is contributed by hrithikgarg03188

Python3

# Python code for the above approach
 
# Function to check if A, B, C
# can be in A.P. after multiplying
# by any integer M
def isAP(A, B, C):
    new_b = B * 2
 
    # Check if A, B, C are already
    # in A.P.
    if (new_b == (A + C)):
        return True
 
    # Check if multiplying B will
    # give A.P.
    if ((A + C) % new_b == 0):
        return True
 
    if ((A + C) > new_b):
        return False
 
    # Check if multiplying A will
    # give A.P.
    if ((new_b - C) % A == 0):
        return True
 
    # Check if multiplying C will
    # give A.P.
    if ((new_b - A) % C == 0):
        return True
 
    return False
 
# Driver code
A = 30
B = 5
C = 10
 
# Function call
ans = isAP(A, B, C)
 
# Displaying the answer
if (ans):
    print("YES")
else:
    print("NO")
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program to implement
// the above approach
using System;
class GFG
{
 
  // Function to check if A, B, C
  // can be in A.P. after multiplying
  // by any integer M
  static bool isAP(int A, int B, int C)
  {
    int new_b = B * 2;
 
    // Check if A, B, C are already
    // in A.P.
    if (new_b == (A + C))
      return true;
 
    // Check if multiplying B will
    // give A.P.
    if ((A + C) % new_b == 0)
      return true;
 
    if ((A + C) > new_b)
      return false;
 
    // Check if multiplying A will
    // give A.P.
    if ((new_b - C) % A == 0)
      return true;
 
    // Check if multiplying C will
    // give A.P.
    if ((new_b - A) % C == 0)
      return true;
 
    return false;
  }
 
  // Driver code
  public static int Main()
  {
    int A, B, C;
    A = 30;
    B = 5;
    C = 10;
 
    // Function call
    bool ans = isAP(A, B, C);
 
    // Displaying the answer
    if (ans)
      Console.Write("YES");
    else
      Console.Write("NO");
    return 0;
  }
}
 
// This code is contributed by Taranpreet

Javascript

输出

YES

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