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

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 ;
}
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