求x和y的最小值，使得ax – by = 0

给定代表“ ax – by = 0”中系数的两个值“ a”和“ b”，找到满足方程式的x和y的最小值。还可以假设x> 0，y> 0，a> 0和b> 0。

Input: a = 25, b = 35
Output: x = 7, y = 5

一个简单的解决方案是尝试从1、1开始的每个可能的x和y值，并在满足方程式时停止。
一个直接的解决方案是使用最小公倍数(LCM)。 “ a”和“ b”的LCM表示可以使双方相等的最小值。我们可以使用以下公式找到LCM。

LCM(a, b) = (a * b) / GCD(a, b) 

可以使用Euclid算法计算最大公约数(GCD)。

C++

// C++ program to find the smallest values of x and y that
// satisfy "ax - by = 0"
#include 
using namespace std;
 
// To find GCD using Eculcid's algorithm
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return (gcd(b, a % b));
}
 
// Prints smallest values of x and y that
// satisfy "ax - by = 0"
void findSmallest(int a, int b)
{
    // Find LCM
    int lcm = (a * b) / gcd(a, b);
 
    cout << "x = " << lcm / a
         << "\ny = " << lcm / b;
}
 
// Driver program
int main()
{
    int a = 25, b = 35;
    findSmallest(a, b);
    return 0;
}

Java

// Java program to find the smallest values of
// x and y that satisfy "ax - by = 0"
class GFG {
 
    // To find GCD using Eculcid's algorithm
    static int gcd(int a, int b)
    {
 
        if (b == 0)
            return a;
        return (gcd(b, a % b));
    }
 
    // Prints smallest values of x and y that
    // satisfy "ax - by = 0"
    static void findSmallest(int a, int b)
    {
 
        // Find LCM
        int lcm = (a * b) / gcd(a, b);
 
        System.out.print("x = " + lcm / a
                         + "\ny = " + lcm / b);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a = 25, b = 35;
        findSmallest(a, b);
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python program to find the
# smallest values of x and y that
# satisfy "ax - by = 0"
 
# To find GCD using Eculcid's algorithm
def gcd(a, b):
    if (b == 0):
        return a
    return(gcd(b, a % b))
 
# Prints smallest values of x and y that
# satisfy "ax - by = 0"
def findSmallest(a, b):
 
    # Find LCM
    lcm = (a * b)/gcd(a, b)
    print("x =", lcm / a, "\ny = ", lcm / b)
 
# Driver code
a = 25
b = 35
findSmallest(a, b)
 
# This code is contributed
# by Anant Agarwal.

C#

// C# program to find the smallest
// values of x and y that
// satisfy "ax - by = 0"
using System;
 
class GFG {
 
    // To find GCD using
    // Eculcid's algorithm
    static int gcd(int a, int b)
    {
 
        if (b == 0)
            return a;
        return (gcd(b, a % b));
    }
 
    // Prints smallest values of x and
    // y that satisfy "ax - by = 0"
    static void findSmallest(int a, int b)
    {
 
        // Find LCM
        int lcm = (a * b) / gcd(a, b);
 
        Console.Write("x = " + lcm / a + "\ny = " + lcm / b);
    }
 
    // Driver code
    public static void Main()
    {
        int a = 25, b = 35;
        findSmallest(a, b);
    }
}
 
// This code is contributed by Sam007.

Javascript

 
    // Javascript program to find the smallest
    // values of x and y that
    // satisfy "ax - by = 0"
     
    // To find GCD using
    // Eculcid's algorithm
    function gcd(a, b)
    {
   
        if (b == 0)
            return a;
        return (gcd(b, a % b));
    }
   
    // Prints smallest values of x and
    // y that satisfy "ax - by = 0"
    function findSmallest(a, b)
    {
   
        // Find LCM
        let lcm = parseInt((a * b) / gcd(a, b), 10);
   
        document.write("x = " + parseInt(lcm / a, 10) +
        "</br>y = " + parseInt(lcm / b, 10));
    }
     
    let a = 25, b = 35;
      findSmallest(a, b);
     

输出：

x = 7
y = 5

上面的findSmallest()代码可以减少：

Since ax - by = 0,
ax = by, which means x/y = b/a
So we can calculate gcd and directly do as -

Value of x = b / gcd;
Value of y = a / gcd; 

C

// Prints smallest values of x and y that
// satisfy "ax - by = 0"
void findSmallest(int a, int b)
{
    // Find GCD
    int g = gcd(a, b);
 
    cout << "x = " << b / g
         << "\ny = " << a / g;
}