周赛 2 【拓展GCD&&欧几里得数】

B - 楼下水题
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Time Limit:1000MS
Memory Limit:262144KB
64bit IO Format:%I64d & %I64u
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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

分析：

这道题意思就是 给你 a b c 的值 ，然后求最小的x ，y 的值。

因为数据太大，这道题用暴力跟定超时，所有用拓展gcd来写。

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)
{
return b?gcd(b,a%b):a;
}
void extend_Euclid(ll a,ll b,ll &x,ll &y)
{
if(b==0)
{
x=1;
y=0;
return;
}
extend_Euclid(b,a%b,x,y);
ll tmp =x;
x=y;
y=tmp-(a/b)*y;
}

int main()
{
ll a,b,c,x,y;
while(cin>>a>>b>>c)
{
c=-c;
ll g=gcd(a,b);
if(c%g)
{
printf("-1\n");
continue;
}
a/=g;
b/=g;
c/=g;
extend_Euclid(a,b,x,y);
printf("%lld %lld\n",x*c,y*c);
}
return 0;
}

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