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

Copy
2 5 3
output

Copy
6 -3

ax+by+c=0,化为ax+by=-c/gcd(a,b)*gcd(a,b),

套拓展欧几里得就可以解出了

 #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + ;
const int mod = 1e9 + ;
typedef long long LL;
LL exgcd(LL a, LL b, LL &x, LL &y) {
if (b == ) {
x = , y = ;
return a;
}
LL g = exgcd(b, a % b, x, y);
LL t;
t = x, x = y, y = t - (a / b) * y;
return g;
}
int main() {
LL a, b, c, x, y;
cin >> a >> b >> c;
LL t = exgcd(a, b, x, y);
if (c % t == ) printf("%lld %lld\n", -x * c / t, -y * c / t);
else printf("-1\n");
return ;
}

C. Line (扩展欧几里得)的更多相关文章

  1. Line(扩展欧几里得)

    题意:本题给出一个直线,推断是否有整数点在这条直线上: 分析:本题最重要的是在给出的直线是不是平行于坐标轴,即A是不是为0或B是不是为0..此外.本题另一点就是C输入之后要取其相反数,才干进行扩展欧几 ...

  2. Codeforces7C 扩展欧几里得

    Line Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status ...

  3. POJ2115(扩展欧几里得)

    C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 23700   Accepted: 6550 Descr ...

  4. Root(hdu5777+扩展欧几里得+原根)2015 Multi-University Training Contest 7

    Root Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Su ...

  5. UVA 10090 Marbles(扩展欧几里得)

    Marbles Input: standard input Output: standard output I have some (say, n) marbles (small glass ball ...

  6. [ACM] hdu 3923 Invoker (Poyla计数,高速幂运算,扩展欧几里得或费马小定理)

    Invoker Problem Description On of Vance's favourite hero is Invoker, Kael. As many people knows Kael ...

  7. Root(hdu5777+扩展欧几里得+原根)

    Root                                                                          Time Limit: 30000/1500 ...

  8. Gym100812 L 扩展欧几里得

    L. Knights without Fear and Reproach time limit per test 2.0 s memory limit per test 256 MB input st ...

  9. Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) C.Ray Tracing (模拟或扩展欧几里得)

    http://codeforces.com/contest/724/problem/C 题目大意: 在一个n*m的盒子里,从(0,0)射出一条每秒位移为(1,1)的射线,遵从反射定律,给出k个点,求射 ...

  10. UVA 12169 Disgruntled Judge 枚举+扩展欧几里得

    题目大意:有3个整数 x[1], a, b 满足递推式x[i]=(a*x[i-1]+b)mod 10001.由这个递推式计算出了长度为2T的数列,现在要求输入x[1],x[3],......x[2T- ...

随机推荐

  1. 利用nohup后台运行jar文件包程序

    Linux 运行jar包命令如下: 方式一: java -jar XXX.jar特点:当前ssh窗口被锁定,可按CTRL + C打断程序运行,或直接关闭窗口,程序退出 那如何让窗口不锁定? 方式二 j ...

  2. vs_code 快捷键

    一般的Ctrl+Shift+P,F1显示命令面板按Ctrl+P快速打开,到文件.Ctrl + Shift + N新窗口/实例Ctrl + Shift + W /关闭窗口实例Ctrl +.用户设置Ctr ...

  3. eos教程如何创建eos测试账号并且使用scatter插件

    EOS代币租赁平台 --- Chintai平台已经在Jungle测试网络上部署了,欢迎大家来体验. 地址见: Chintai 公测版 官网是: Chintai 目前测试网络上面需要用到Scatter插 ...

  4. POJ 2653 Pick-up sticks(线段判交)

    Description Stan has n sticks of various length. He throws them one at a time on the floor in a rand ...

  5. Simple Expression

    Description You probably know that Alex is a very serious mathematician and he likes to solve seriou ...

  6. 【android】实现手指滑动来切换activity(转)

    http://code.eoe.cn/115 1.jpg外部引用 原始文档 MainActivity.java外部引用 原始文档 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...

  7. <Android>tab选项卡

    1.继承TabActivity实现 a)         在布局文件中使用FrameLayout列出Tab组件及Tab中的内容组件 b)        Activity要继承TabActivity c ...

  8. setsockopt 设置socket 详细用法

    1.closesocket(一般不会立即关闭而经历TIME_WAIT的过程)后想继续重用该socket:BOOL bReuseaddr=TRUE;setsockopt(s,SOL_SOCKET ,SO ...

  9. vi/sed等遵循的搜索正则语法

    转自:http://blog.csdn.net/lanxinju/article/details/5731843 一.查找 查找命令 /pattern<Enter> :向下查找patter ...

  10. phpcms 模型