C - Line——(扩展欧几里得算法)

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C - Line——(扩展欧几里得算法)

angel_imp 2016-03-23 19:05:00 浏览1158
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传送门

C. Line
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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 AB 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.

Examples
input
2 5 3
output
6 -3

题目大意:

就是判断一下给定的三个数,a,b,c 是否符合 a*x + b*y + c == 0的方程,如果符合输出x 和 y的值,否者输出 -1


解题思路:

就是一个扩展欧几里得算法, 不是很难的,注意的是 将c 用 -c来代替剩下的也没啥了,扩展欧几里得是模板。。。


上代码:

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
void exgcd(LL a, LL b, LL &x, LL &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return;
    }
    LL x1, y1;
    exgcd(b, a%b, x1, y1);
    x = y1;
    y = x1 - (a/b)*y1;
}
LL gcd(LL a, LL b)
{
    if(b == 0)
        return a;
    return gcd(b, a%b);
}
int main()
{
    LL a, b, c, x, y;
    while(cin>>a>>b>>c)
    {
        c = -c;
        LL d = gcd(a, b);
        if(c % d)
            puts("-1");
        else
        {
            a /= d;
            b /= d;
            c /= d;
            exgcd(a, b, x, y);
            x *= c;
            y *= c;
            cout<<x<<" "<<y<<endl;
        }
    }
    return 0;
}
</span>




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