检查两个数字是否在白银比例中

给定两个数字A和B ，任务是检查A和B的银比。

Silver Ratio: Two numbers are said to be in silver ratio if the ratio of the sum of the smaller and twice the larger number to the larger number is the same as the ratio of the larger one to the smaller one. Below is the representation of the silver ratio:

$\frac{2*A+B}{A} = \frac{A}{B} = \delta_{\varsigma} = {1+ \sqrt{2}} = 2.414$
 

for A > 0, B > 0

 

例子：

Input: A = 2.414, B = 1 
Output: Yes 
Explanation:

$\frac{A}{B} = 2.414 \;\;\text{as well as}\;\; \frac{2*A + B}{A} = \frac{2.414}{1} = 2.414$
 

Input: A = 1, B = 0.414 
Output No 
Explanation: Ratio of A to B do not form a golden ratio

 

方法：想法是找到两个比率并检查它们是否等于银比率(2.414)。

//这里A代表较大的数字$\frac{A}{B} = \frac{2*A + B}{A} = 2.414$

下面是上述方法的实现：

C++

// C++ implementation to check
// whether two numbers are in
// silver ratio with each other
#include
using namespace std;
 
// Function to check that two
// numbers are in silver ratio
bool checksilverRatio(float a, float b)
{
     
    // Swapping the numbers such
    // that A contains the maximum
    // number between these numbers
    if(a < b)
        swap(a, b);
     
    // First Ratio
    float ratio1 = ((a / b) * 1000.0) / 1000.0;
     
    // Second Ratio
    float ratio2 = (int)(((2 * a + b) /
                          a) * 1000);
    ratio2 = ratio2 / 1000;
     
    // Condition to check that two
    // numbers are in silver ratio
    if (ratio1 == ratio2 &&
       (int)(ratio1 - 2.414) == 0)
    {
        cout << "Yes\n";
        return true;
    }
    else
    {
        cout << "No\n";
        return false;
    }
}
 
// Driver Code
int main()
{
    float a = 2.414;
    float b = 1;
     
    // Function call
    checksilverRatio(a, b);
}
 
// This code is contributed by ishayadav181

Java

// Java implementation to check
// whether two numbers are in
// silver ratio with each other
import java.util.*;
import java.lang.*;
 
class GFG{
 
// Function to check that two
// numbers are in silver ratio
static boolean checksilverRatio(double a,
                                double b)
{
 
    // Swapping the numbers such
    // that A contains the maximum
    // number between these numbers
    if (a < b)
    {
        a = a + b;
        b = a - b;
        a = a - b;
    }
 
    // First Ratio
    double ratio1 = ((a / b) * 1000) / 1000;
 
    // Second Ratio
    double ratio2 = (int)(((2 * a + b) /
                           a) * 1000);
    ratio2 = ratio2 / 1000;
 
    // Condition to check that two
    // numbers are in silver ratio
    if (ratio1 == ratio2 &&
       (int)(ratio1 - 2.414) == 0)
    {
        System.out.println("Yes");
        return true;
    }
    else
    {
        System.out.println("No");
        return false;
    }
}
 
// Driver Code
public static void main(String[] args)
{
    double a = 2.414;
    double b = 1;
 
    // Function call
    checksilverRatio(a, b);
}
}
 
// This code is contributed by jana_sayantan

Python3

# Python3 implementation to check
# whether two numbers are in
# silver ratio with each other
 
# Function to check that two
# numbers are in silver ratio
def checksilverRatio(a, b):
     
    # Swapping the numbers such
    # that A contains the maximum
    # number between these numbers
    a, b = max(a, b), min(a, b)
     
    # First Ratio
    ratio1 = round(a / b, 3)
     
    # Second Ratio
    ratio2 = round((2 * a + b)/a, 3)
    # Condition to check that two
    # numbers are in silver ratio
    if ratio1 == ratio2 and\
       ratio1 == 2.414:
        print("Yes")
        return True
    else:
        print("No")
        return False
         
# Driver Code
if __name__ == "__main__":
    a = 2.414
    b = 1
     
    # Function Call
    checksilverRatio(a, b)

C#

// C# implementation to check
// whether two numbers are in
// silver ratio with each other
using System;
 
class GFG{
 
// Function to check that two
// numbers are in silver ratio
static bool checksilverRatio(double a,
                             double b)
{
 
    // Swapping the numbers such
    // that A contains the maximum
    // number between these numbers
    if (a < b)
    {
        a = a + b;
        b = a - b;
        a = a - b;
    }
 
    // First Ratio
    double ratio1 = ((a / b) * 1000) / 1000;
 
    // Second Ratio
    double ratio2 = (int)(((2 * a + b) /
                        a) * 1000);
    ratio2 = ratio2 / 1000;
 
    // Condition to check that two
    // numbers are in silver ratio
    if (ratio1 == ratio2 &&
       (int)(ratio1 - 2.414) == 0)
    {
        Console.WriteLine("Yes");
        return true;
    }
    else
    {
        Console.WriteLine("No");
        return false;
    }
}
 
// Driver Code
public static void Main()
{
    double a = 2.414;
    double b = 1;
 
    // Function call
    checksilverRatio(a, b);
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
// Javascript Program to check
// whether two numbers are in
// silver ratio with each other
 
// Function to check that two
// numbers are in silver ratio
function checksilverRatio(a, b)
{
   
    // Swapping the numbers such
    // that A contains the maximum
    // number between these numbers
    if (a < b)
    {
        a = a + b;
        b = a - b;
        a = a - b;
    }
   
    // First Ratio
    let ratio1 = ((a / b) * 1000) / 1000;
   
    // Second Ratio
    let ratio2 = Math.floor(((2 * a + b) /
                           a) * 1000);
    ratio2 = ratio2 / 1000;
   
    // Condition to check that two
    // numbers are in silver ratio
    if (ratio1 == ratio2 &&
       (ratio1 - 2.414) == 0)
    {
        document.write("Yes");
        return true;
    }
    else
    {
        document.write("No");
        return false;
    }
}
 
// Driver Code
  let a = 2.414;
    let b = 1;
   
    // Function call
    checksilverRatio(a, b);
     
    // This code is contributed by chinmoy1997pal.

输出：

Yes

参考： https : //en.wikipedia.org/wiki/Silver_ratio