通过= n 找到满足 ax +的 x 和 y
给定 a、b 和 n，找到满足 ax + by = n 的 x 和 y，打印满足公式 的 x 和 y 中的任何一个，示例:
Input : n=7 a=2 b=3
Output : x=2, y=1
Explanation: here x and y satisfies the equation
Input : 4 2 7
Output : No solution
我们可以使用线性丢番图方程来检查是否存在任何解，但是这里我们需要找出这个方程的解，所以我们可以简单地迭代从 0 到 n 的所有可能的值，因为对于这个给定的方程，它不能超过 n。所以用纸和笔解这个方程得到 y=(n-ax)/b ，同样我们得到另一个数为 x=(n-by)/a 。如果没有一个值满足等式，最后我们打印“无解”
C++
// CPP program to find solution of ax + by = n
#include <bits/stdc++.h>
using namespace std;
// function to find the solution
void solution(int a, int b, int n)
{
// traverse for all possible values
for (int i = 0; i * a <= n; i++) {
// check if it is satisfying the equation
if ((n - (i * a)) % b == 0) {
cout << "x = " << i << ", y = "
<< (n - (i * a)) / b;
return;
}
}
cout << "No solution";
}
// driver program to test the above function
int main()
{
int a = 2, b = 3, n = 7;
solution(a, b, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find solution
// of ax + by = n
import java.io.*;
class GfG {
// function to find the solution
static void solution(int a, int b, int n)
{
// traverse for all possible values
for (int i = 0; i * a <= n; i++)
{
// check if it is satisfying the equation
if ((n - (i * a)) % b == 0)
{
System.out.println("x = " + i +
", y = " +
(n - (i * a)) / b);
return ;
}
}
System.out.println("No solution");
}
public static void main (String[] args)
{
int a = 2, b = 3, n = 7;
solution(a, b, n);
}
}
// This code is contributed by Gitanjali.
Python 3
# Python3 code to find solution of
# ax + by = n
# function to find the solution
def solution (a, b, n):
# traverse for all possible values
i = 0
while i * a <= n:
# check if it is satisfying
# the equation
if (n - (i * a)) % b == 0:
print("x = ",i ,", y = ",
int((n - (i * a)) / b))
return 0
i = i + 1
print("No solution")
# driver program to test the above function
a = 2
b = 3
n = 7
solution(a, b, n)
# This code is contributed by "Sharad_Bhardwaj".
C
// C# program to find solution
// of ax + by = n
using System;
class GfG {
// function to find the solution
static void solution(int a, int b, int n)
{
// traverse for all possible values
for (int i = 0; i * a <= n; i++)
{
// check if it is satisfying the
// equation
if ((n - (i * a)) % b == 0)
{
Console.Write("x = " + i +
", y = " +
(n - (i * a)) / b);
return ;
}
}
Console.Write("No solution");
}
// Driver code
public static void Main ()
{
int a = 2, b = 3, n = 7;
solution(a, b, n);
}
}
// This code is contributed by Vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find
// solution of ax + by = n
// function to find the solution
function solution($a, $b, $n)
{
// traverse for all possible values
for ($i = 0; $i * $a <= $n; $i++)
{
// check if it is satisfying
// the equation
if (($n - ($i * $a)) % $b == 0)
{
echo "x = " , $i , ", y = " ,
($n - ($i * $a)) / $b;
return;
}
}
echo "No solution";
}
// Driver Code
$a = 2; $b = 3; $n = 7;
solution($a, $b, $n);
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// JavaScript program to find solution
// of ax + by = n
// function to find the solution
function solution(a, b, n)
{
// traverse for all possible values
for (let i = 0; i * a <= n; i++)
{
// check if it is satisfying the equation
if ((n - (i * a)) % b == 0)
{
document.write("x = " + i +
", y = " +
(n - (i * a)) / b);
return ;
}
}
document.write("No solution");
}
// Driver code
let a = 2, b = 3, n = 7;
solution(a, b, n);
// This code is contributed by sanjoy_62.
</script>
输出:
x = 2, y = 1
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