扩展欧几里得算法

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<stdio.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<string.h>
#include<map>
#include<math.h>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define MAX 50005
#define ll long long
using namespace std;
/*
对于不完全为0的非负整数a，b，gcd(a, b)表示a, b的最大
公约数，必定存在整数对x，y，满足a*x+b*y==gcd(a, b)
求解不定方程；如a*x+b*y=c; 已知a, b, c的值求x和y的值
a*x+b*y=gcd(a, b)*c/gcd(a, b);
最后转化为 a*x/(c/gcd(a, b))+b*y/(c/gcd(a, b))=gcd(a, b);
最后求出的解x0，y0乘上c/gcd(a, b)就是最终的结果了
x1=x0*c/gcd(a, b);
y1=y0*c/gcd(a, b);
*/
//求一组 x 和 y 使得：a*x + b*y = gcd(a,b)
ll egcd(ll a,ll b,ll &x,ll &y)//扩展欧几里得算法，返回的值为a,b最大公约数
{
if(b==0)
{
x=1;
y=0;
return a;
}
ll ans=egcd(b,a%b,x,y);
ll temp=x;
x=y;
y=temp-a/b*y;
return ans;
}
int main()
{
ll a,b,c,x,y,t;
while(cin>>a>>b>>c)
{
t=egcd(a,b,x,y);
//t=gcd(a,b);
if(c%t!=0)
printf("-1");
else
printf("%lld %lld\n",-x*c/t,-y*c/t);
}
}