B - 楼下水题(扩展欧几里德)

B - 楼下水题

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 7C

Description

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.

Sample Input

Input

2 5 3

Output

6 -3
题解：
模版欧几里德；Ax+By=gcd;
找到x,y;分式两边同乘以（-c/gcd）也就是x*(-c/gcd),y*(-c/gcd)；就好了；
代码：

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 typedef long long LL;
 using namespace std;
 void e_gcd(LL &x,LL &y,LL &d,LL a,LL b){
     if(!b){
         d=a;
         x=;
         y=;//y可以为任意数；因为b为0
     }
     else{
         e_gcd(x,y,d,b,a%b);//欧几里德；
         LL temp=x;
         x=y;
         y=temp-a/b*y;
     }
 }
 int main(){
     LL a,b,c,x,y,d;
     while(~scanf("%lld%lld%lld",&a,&b,&c)){
         e_gcd(x,y,d,a,b);
         //printf("%d %d %d\n",x,y,d);
         if(c%d!=)puts("-1");
         else printf("%lld %lld\n",x*(-c/d),y*(-c/d));
     }
     return ;
 }