C. Line（扩展欧几里得求不定方程的解）

最新推荐文章于 2023-03-20 16:42:43 发布

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最新推荐文章于 2023-03-20 16:42:43 发布

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分类专栏： CodeForce 数论文章标签：扩展欧几里得不定方程求解

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本文介绍了一个算法问题，即如何找到一条由方程 Ax + By + C = 0 定义的直线上，坐标为整数且位于特定范围内的任意一点。通过扩展欧几里得算法确定该点是否存在并计算其坐标。

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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#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

typedef long long int ll;

ll ext_gcd(ll a,ll b,ll& s,ll& t){
	
	if(b==0){
		s = 1;
		t = 0;
		return a;
	}
	ll ans = ext_gcd(b,a%b,s,t);
	ll tmp = s;
	s = t;
	t = tmp-a/b*s;
	
	return ans;
}


int main(){
	
	
	ll a,b,c;
	scanf("%I64d %I64d %I64d",&a,&b,&c);
	ll s=0,t=0;
	ll d = ext_gcd(a,b,s,t); 
	if(c%d!=0)
		printf("-1\n");
	else
		printf("%I64d %I64d\n",-1*c/d*s,-1*c/d*t);
	return 0;
}