求出满足ax + by = m的m的最小值，并且m之后的所有值也满足

给定两个正整数’a’和’b’，它们表示方程ax + by = m的系数。对于x和y的任何正整数，找到满足方程式的m的最小值。在此最小值之后，该方程由m的所有(更大)值满足。如果不存在这样的最小值，则返回“ -1”。
例子：

Input: a = 4, b = 7
Output: 18
Explanation: 18 is the smallest value that can
             can be satisfied by equation 4x + 7y.         
             4*1 + 7*2 = 18

             And after 18 all values are satisifed 
             4*3 + 7*1 = 19
             4*5 + 7*0 = 20
             ... and so on.

这是Frobenius硬币问题的一种变体。在Frobenius硬币问题中，我们需要找到两个硬币无法表示的最大数字。面额为“ a”和“ b”的硬币的最大金额为a * b –(a + b)。因此可以使用两个硬币表示的最小数字，也可以表示之后的所有数字是a * b –(a + b)+ 1。
一个重要的情况是’a’和’b’的GCD不为1。例如，如果’a’= 4并且’b’= 6，则可以使用两个硬币表示的所有值都是偶数(或可以对方程进行分层的m是偶数。因此，所有不是2的倍数的值都不能满足该方程。在这种情况下，没有最小值，之后所有值都满足方程式。
下面是上述想法的实现：

C++

// C++ program to find the minimum value of m that satisfies
// ax + by = m and all values after m also satisfy
#include
using namespace std;
 
int findMin(int a, int b)
{
    // If GCD is not 1, then there is no such value,
    // else value is obtained using "a*b-a-b+1'
    return (__gcd(a, b) == 1)? a*b-a-b+1 : -1;
}
 
// Driver code
int main()
{
    int a = 4, b = 7;
    cout << findMin(a, b) << endl;
    return 0;
}

Java

// Java program to find the
// minimum value of m that
// satisfies ax + by = m
// and all values after m
// also satisfy
import java.io.*;
 
class GFG
{
// Recursive function to
// return gcd of a and b   
static int __gcd(int a, int b)
{
     
    // Everything divides 0
    if (a == 0 && b == 0)
    return 0;
 
    // base case
    if (a == b)
        return a;
 
    // a is greater$
    if (a > b)
        return __gcd(a - b, b);
    return __gcd(a, b - a);
}
 
static int findMin( int a, int b)
{
     
    // If GCD is not 1, then
    // there is no such value,
    // else value is obtained
    // using "a*b-a-b+1'
    return (__gcd(a, b) == 1)?
        a * b - a - b + 1 : -1;
}
 
// Driver code
public static void main (String[] args)
{
    int a = 4;
    int b = 7;
    System.out.println(findMin(a, b));
}
}
 
// This code is contributed
// by akt_mit

Python3

# Python3 program to find the minimum
# value of m that satisfies ax + by = m
# and all values after m also satisfy
 
# Recursive function to return
# gcd of a and b
def __gcd(a, b):
     
    # Everything divides 0
    if (a == 0 or b == 0):
        return 0;
 
    # base case
    if (a == b):
        return a;
 
    # a is greater
    if (a > b):
        return __gcd(a - b, b);
    return __gcd(a, b - a);
 
def findMin( a, b):
     
    # If GCD is not 1, then
    # there is no such value,
    # else value is obtained
    # using "a*b-a-b+1'
    if(__gcd(a, b) == 1):
        return (a * b - a - b + 1)
    else:
        return -1
 
# Driver code
a = 4;
b = 7;
print(findMin(a, b));
     
# This code is contributed by mits

C#

// C# program to find the minimum
// value of m that satisfies 
// ax + by = m and all values
// after m also satisfy
class GFG
{
// Recursive function to
// return gcd of a and b
static int __gcd(int a, int b)
{
     
    // Everything divides 0
    if (a == 0 && b == 0)
    return 0;
 
    // base case
    if (a == b)
        return a;
 
    // a is greater$
    if (a > b)
        return __gcd(a - b, b);
    return __gcd(a, b - a);
}
 
static int findMin( int a, int b)
{
     
    // If GCD is not 1, then
    // there is no such value,
    // else value is obtained
    // using "a*b-a-b+1'
    return (__gcd(a, b) == 1)?
           a * b - a - b + 1 : -1;
}
 
// Driver code
public static void Main()
{
    int a = 4;
    int b = 7;
    System.Console.WriteLine(findMin(a, b));
}
}
 
// This code is contributed
// by mits

PHP

 $b)
        return __gcd($a - $b, $b);
    return __gcd($a, $b - $a);
}
 
function findMin( $a, $b)
{
     
    // If GCD is not 1, then
    // there is no such value,
    // else value is obtained
    // using "a*b-a-b+1'
    return (__gcd($a, $b) == 1)?
           $a * $b - $a - $b + 1 : -1;
}
 
    // Driver code
    $a = 4; $b = 7;
    echo findMin($a, $b) ;
     
// This code is contributed by anuj_67.
?>

Javascript

 
    // Javascript program
    // to find the minimum
    // value of m that satisfies 
    // ax + by = m and all values
    // after m also satisfy
     
    // Recursive function to
    // return gcd of a and b
    function __gcd(a, b)
    {
 
        // Everything divides 0
        if (a == 0 && b == 0)
            return 0;
 
        // base case
        if (a == b)
            return a;
 
        // a is greater$
        if (a > b)
            return __gcd(a - b, b);
        return __gcd(a, b - a);
    }
 
    function findMin(a, b)
    {
 
        // If GCD is not 1, then
        // there is no such value,
        // else value is obtained
        // using "a*b-a-b+1'
        return (__gcd(a, b) == 1)?
                a * b - a - b + 1 : -1;
    }
     
    let a = 4;
    let b = 7;
    document.write(findMin(a, b));
 

输出：

18