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：
 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 typedef long long LL;
 6 using namespace std;
 7 void e_gcd(LL &x,LL &y,LL &d,LL a,LL b){
 8     if(!b){
 9         d=a;
10         x=1;
11         y=0;//y可以为任意数；因为b为0 
12     }
13     else{
14         e_gcd(x,y,d,b,a%b);//欧几里德；
15         LL temp=x;
16         x=y;
17         y=temp-a/b*y;
18     }
19 }
20 int main(){
21     LL a,b,c,x,y,d;
22     while(~scanf("%lld%lld%lld",&a,&b,&c)){
23         e_gcd(x,y,d,a,b);
24         //printf("%d %d %d\n",x,y,d);
25         if(c%d!=0)puts("-1");
26         else printf("%lld %lld\n",x*(-c/d),y*(-c/d));
27     }
28     return 0;
29 }

 

posted @ 2015-10-24 22:00  handsomecui  阅读(294)  评论(0)  编辑  收藏  举报