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 ;
}