求 x 和 y 的最小值，使得 ax–by = 0
原文:https://www . geeksforgeeks . org/find-x 和 y 的最小值，即 ax 乘 0/
给定代表“ax–by = 0”中系数的两个值‘a’和‘b’，找到满足等式的 x 和 y 的最小值。也可以假设 x > 0，y > 0，a > 0，b > 0。
Input: a = 25, b = 35
Output: x = 7, y = 5
A 简单解法是从 1，1 开始尝试 x 和 y 的每一个可能值，当方程满足时停止。 一个直接解决方案是使用最小公倍数(LCM)。“a”和“b”的 LCM 代表可以使两边相等的最小值。我们可以用下面的公式找到 LCM。
LCM(a, b) = (a * b) / GCD(a, b)
最大公约数(GCD)可以使用欧几里德算法来计算。
C++
// C++ program to find the smallest values of x and y that
// satisfy "ax - by = 0"
#include <iostream>
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 语言(一种计算机语言，尤用于创建网站)
// 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.
Python 3
# 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.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP 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
$lcm = ($a * $b) / gcd($a, $b);
echo "x = ", $lcm/$a, "\ny = ", $lcm/$b;
}
// Driver Code
$a = 25;
$b = 35;
findSmallest($a, $b);
// This code is contributed by ajit
?>
java 描述语言
<script>
// 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);
</script>
输出:
x = 7
y = 5
以上 findSmallest()的代码可以简化:
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;
}
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