C - Line——（扩展欧几里得算法）

传送门

C. Line

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

A line on the plane is described by an equation Ax + By + C = 0. You are to find any point on this line,
whose coordinates are integer numbers from  - 5·1018 to 5·1018 inclusive,
or to find out that such points do not exist.

Input

The first line contains three integers A, B and C ( - 2·109 ≤ A, B, C ≤ 2·109)
— corresponding coefficients of the line equation. It is guaranteed that A2 + B2 > 0.

Output

If the required point exists, output its coordinates, otherwise output -1.

Examples

input

2 5 3

output

6 -3

题目大意：

就是判断一下给定的三个数，a，b，c 是否符合 a*x + b*y + c == 0的方程，如果符合输出x 和 y的值，否者输出 -1

解题思路：

就是一个扩展欧几里得算法， 不是很难的，注意的是 将c 用 -c来代替剩下的也没啥了，扩展欧几里得是模板。。。

上代码：

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
void exgcd(LL a, LL b, LL &x, LL &y)
{
if(b == 0)
{
x = 1;
y = 0;
return;
}
LL x1, y1;
exgcd(b, a%b, x1, y1);
x = y1;
y = x1 - (a/b)*y1;
}
LL gcd(LL a, LL b)
{
if(b == 0)
return a;
return gcd(b, a%b);
}
int main()
{
LL a, b, c, x, y;
while(cin>>a>>b>>c)
{
c = -c;
LL d = gcd(a, b);
if(c % d)
puts("-1");
else
{
a /= d;
b /= d;
c /= d;
exgcd(a, b, x, y);
x *= c;
y *= c;
cout<<x<<" "<<y<<endl;
}
}
return 0;
}
</span>