Description:
A line on the plane is described by an equation . You are to find any point on this line, whose coordinates are integer numbers from to inclusive, or to find out that such points do not exist.
Input
The first line contains three integers A, B and C ( ) — corresponding coefficients of the line equation. It is guaranteed that
Output
If the required point exists, output its coordinates, otherwise output .
Examples
Input
2 5 3
Output
6 -3
求
中
的值。
扩展欧几里得模板。
AC代码:
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", (n))
#define pdd(n, m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, n, a) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
int ret = 0, sgn = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
sgn = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
ret = ret * 10 + ch - '0';
ch = getchar();
}
return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
if (a > 9)
Out(a / 10);
putchar(a % 10 + '0');
}
ll gcd(ll a, ll b)
{
return b == 0 ? a : gcd(b, a % b);
}
ll lcm(ll a, ll b)
{
return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
if (a >= mod)
a = a % mod + mod;
ll ans = 1;
while (b)
{
if (b & 1)
{
ans = ans * a;
if (ans >= mod)
ans = ans % mod + mod;
}
a *= a;
if (a >= mod)
a = a % mod + mod;
b >>= 1;
}
return ans;
}
// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
return qpow(a, p - 2, p);
}
///扩展欧几里得
ll exgcd(ll a, ll b, ll &x, ll &y)
{
if (b == 0)
{
x = 1, y = 0;
return a;
}
ll d = exgcd(b, a % b, x, y);
ll z = x;
x = y;
y = z - y * (a / b);
return d;
}
ll a, b, c;
int main()
{
slddd(a, b, c);
ll x, y;
ll res = exgcd(a, b, x, y);
if (c % res == 0)
pldd(-x * c / res, -y * c / res);
else
puts("-1");
return 0;
}