给定Number数组和两个值A和B ，任务是检查以下条件：
- 数组中是否存在两个数字。
- 如果是，则它们的算术平均值和谐波平均值也存在或不存在于同一数组中。
- 如果满足所有条件，则打印两个数字的几何平均值。
各个数字的均值可以表述为：
例子：
Input: arr[] = {1.0, 2.0, 2.5, 3.0, 4.0, 4.5, 5.0, 6.0}, A = 3, B = 6.
Output : GM = 4.24
Explanation:
A = 3, B = 6 are present in the array.
AM = 4.5, HM = 4 are also present in the array.
So, GM = 4.24
Input: arr = {1.0, 2.0, 2.5, 3.0, 4.0, 4.5, 5.0, 6.0}, A = 4, B = 6.
Output: AM and HM not found
方法：
- 这个想法是使用Hashing ，通过它我们可以将数组元素简单地存储在Hash容器中，并使用恒定时间O(1)操作来查找和跟踪数字及其均值。最后，如果满足所有条件，则通过观察简单关系AM * HM = GM 2来计算几何平均值。
- 上述方法的分步实施如下：
- 定义了一个散列容器来存储数组元素。
- 基于该公式计算算术和谐波均值。
- 简单的条件语句用于在恒定时间内查找Hash容器中的元素。
- 在满足所有条件的前提下，根据上述关系式计算出GM。
下面是上述方法的实现：
C++
// C++ program to check if two numbers
// are present in an array then their
// AM and HM are also present. Finally,
// find the GM of the numbers
#include
using namespace std;
// Function to find the Arithmetic Mean
// of 2 numbers
float ArithmeticMean(float A, float B)
{
return (A + B) / 2;
}
// Function to find the Harmonic Mean
// of 2 numbers
float HarmonicMean(float A, float B)
{
return (2 * A * B) / (A + B);
}
// Following function checks and computes the
// desired results based on the means
void CheckArithmeticHarmonic(float arr[],
float A,
float B, int N)
{
// Calculate means
float AM = ArithmeticMean(A, B);
float HM = HarmonicMean(A, B);
// Hash container (Set) to store elements
unordered_set Hash;
// Insertion of array elements in the Set
for (int i = 0; i < N; i++)
{
Hash.insert(arr[i]);
}
// Conditionals to check if numbers
// are present in array by Hashing
if (Hash.find(A) != Hash.end()
&& Hash.find(B) != Hash.end()) {
// Conditionals to check if the AM and HM
// of the numbers are present in array
if (Hash.find(AM) != Hash.end()
&& Hash.find(HM) != Hash.end()) {
// If all conditions are satisfied,
// the Geometric Mean is calculated
cout << "GM = ";
printf("%0.2f", sqrt(AM * HM));
}
else
{
// If numbers are found but the
// respective AM and HM are not
// found in the array
cout << "AM and HM not found";
}
}
else
{
// If none of the conditions are satisfied
cout << "Numbers not found";
}
}
int main()
{
float arr[] = {1.0, 2.0, 2.5, 3.0, 4.0,
4.5, 5.0, 6.0};
int N = sizeof(arr)/sizeof(arr[0]);
float A = 3.0;
float B = 6.0;
CheckArithmeticHarmonic(arr, A, B, N);
return 0;
} Java
// Java program to check if two numbers
// are present in an array then their
// AM and HM are also present. Finally,
// find the GM of the numbers
import java.util.*;
class GFG{
// Function to find the Arithmetic Mean
// of 2 numbers
static Double ArithmeticMean(Double A, Double B)
{
return (A + B) / 2;
}
// Function to find the Harmonic Mean
// of 2 numbers
static Double HarmonicMean(Double A, Double B)
{
return (2 * A * B) / (A + B);
}
// Following function checks and computes the
// desired results based on the means
static void CheckArithmeticHarmonic(Double arr[],
Double A,
Double B, int N)
{
// Calculate means
Double AM = ArithmeticMean(A, B);
Double HM = HarmonicMean(A, B);
// Hash container (HashMap) to store elements
HashMap Hash = new HashMap();
// Insertion of array elements in the Set
for(int i = 0; i < N; i++)
{
Hash.put(arr[i], 1);
}
// Conditionals to check if numbers
// are present in array by Hashing
if (Hash.get(A) != 0 &&
Hash.get(B) != 0)
{
// Conditionals to check if the AM and HM
// of the numbers are present in array
if (Hash.get(AM) != 0 &&
Hash.get(HM) != 0)
{
// If all conditions are satisfied,
// the Geometric Mean is calculated
System.out.print("GM = ");
System.out.format("%.2f", Math.sqrt(AM * HM));
}
else
{
// If numbers are found but the
// respective AM and HM are not
// found in the array
System.out.print("AM and HM not found");
}
}
else
{
// If none of the conditions are satisfied
System.out.print("numbers not found");
}
}
// Driver code
public static void main(String args[])
{
Double arr[] = { 1.0, 2.0, 2.5, 3.0,
4.0, 4.5, 5.0, 6.0};
int N = (arr.length);
Double A = 3.0;
Double B = 6.0;
CheckArithmeticHarmonic(arr, A, B, N);
}
}
// This code is contributed by Stream_Cipher Python3
# Python3 program to check if two numbers
# are present in an array then their
# AM and HM are also present. Finally,
# find the GM of the numbers
from math import sqrt
# Function to find the arithmetic mean
# of 2 numbers
def ArithmeticMean(A, B):
return (A + B) / 2
# Function to find the harmonic mean
# of 2 numbers
def HarmonicMean(A, B):
return (2 * A * B) / (A + B)
# Following function checks and computes the
# desired results based on the means
def CheckArithmeticHarmonic(arr, A, B, N):
# Calculate means
AM = ArithmeticMean(A, B)
HM = HarmonicMean(A, B)
# Hash container (set) to store elements
Hash = set()
# Insertion of array elements in the set
for i in range(N):
Hash.add(arr[i])
# Conditionals to check if numbers
# are present in array by Hashing
if (A in Hash and B in Hash):
# Conditionals to check if the AM and HM
# of the numbers are present in array
if (AM in Hash and HM in Hash):
# If all conditions are satisfied,
# the Geometric Mean is calculated
print("GM =", round(sqrt(AM * HM), 2))
else:
# If numbers are found but the
# respective AM and HM are not
# found in the array
print("AM and HM not found")
else:
# If none of the conditions are satisfied
print("Numbers not found")
# Driver Code
if __name__ == '__main__':
arr = [ 1.0, 2.0, 2.5, 3.0,
4.0, 4.5, 5.0, 6.0 ]
N = len(arr)
A = 3.0
B = 6.0
CheckArithmeticHarmonic(arr, A, B, N)
# This code is contributed by SamarthC#
// C# program to check if two numbers
// are present in an array then their
// AM and HM are also present. Finally,
// find the GM of the numbers
using System;
using System.Collections.Generic;
class GFG{
// Function to find the Arithmetic Mean
// of 2 numbers
static Double ArithmeticMean(Double A, Double B)
{
return (A + B) / 2;
}
// Function to find the Harmonic Mean
// of 2 numbers
static Double HarmonicMean(Double A, Double B)
{
return (2 * A * B) / (A + B);
}
// Following function checks and computes the
// desired results based on the means
static void CheckArithmeticHarmonic(Double []arr,
Double A,
Double B, int N)
{
// Calculate means
Double AM = ArithmeticMean(A, B);
Double HM = HarmonicMean(A, B);
// Hash container (Set) to store elements
// HashMap Hash = new HashMap();
Dictionary Hash = new Dictionary();
// Insertion of array elements in the Set
for(int i = 0; i < N; i++)
{
Hash[arr[i]] = 1;
}
// Conditionals to check if numbers
// are present in array by Hashing
if (Hash.ContainsKey(A) &&
Hash.ContainsKey(B))
{
// Conditionals to check if the AM and HM
// of the numbers are present in array
if (Hash.ContainsKey(AM) &&
Hash.ContainsKey(HM))
{
// If all conditions are satisfied,
// the Geometric Mean is calculated
Console.Write("GM = ");
Console.Write(Math.Round(
Math.Sqrt(AM * HM), 2));
}
else
{
// If numbers are found but the
// respective AM and HM are not
// found in the array
Console.WriteLine("AM and HM not found");
}
}
else
{
// If none of the conditions are satisfied
Console.WriteLine("numbers not found");
}
}
// Driver code
public static void Main()
{
Double []arr = { 1.0, 2.0, 2.5, 3.0,
4.0, 4.5, 5.0, 6.0 };
int N = (arr.Length);
Double A = 3.0;
Double B = 6.0;
CheckArithmeticHarmonic(arr, A, B, N);
}
}
// This code is contributed by Stream_Cipher 输出：
GM = 4.24
复杂度分析：
以上程序的总体时间复杂度基于用户定义的输入处数组元素的初始迭代。与Set关联的find操作都是O(1)恒定时间操作。因此，程序的复杂度为O(N) ，其中N是数组的大小。
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