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 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 test(s) input

2 5 3

output

6 -3

暫時不是特別明白運算的原理，不過還是要先記住...........

#include<cstdio>
typedef long long ll;
ll gcd(ll a,ll b)
{
	if(b==0)
	{
		return a;
	}
	return gcd(b,a%b);
}
void extgcd(ll a,ll b,ll& d,ll& x,ll& y)
{
	if(!b)
	{
		d=a;x=1;y=0;
	}
	else
	{
		extgcd(b,a%b,d,y,x);
		y-=x*(a/b);
	}
} 

int main()
{
	ll a,b,c;
	while(~scanf("%lld%lld%lld",&a,&b,&c))
	{
		ll tp=gcd(a,b);
		if(c%tp!=0)//不能整除的
		{
			printf("-1\n");
		}
		else
		{
			ll x,y,d;
			extgcd(a,b,d,x,y);
			printf("%lld %lld\n",-x*c/tp,-y*c/tp);
		}
	}
	return 0;
}
 

2016年6月22日7:43

之前的做法雖然能ac，但是總感覺心裡不踏實，只有這樣找出所有解的情況，再進行判斷，這樣才更嚴謹

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const ll maxn=5e18;
ll abs(ll x)
{
	return x<0?-x:x;
}
ll gcd(ll a,ll b)
{
	if(b==0)
	{
		return a;
	}
	return gcd(b,a%b);
}
void extgcd(ll a,ll b,ll &m,ll &x,ll &y)
{
	if(b==0)
	{
		m=a;x=1;y=0;
		return;
	}
	extgcd(b,a%b,m,y,x);
	y-=(a/b)*x;
}
bool judge(ll a,ll b)//在範圍內 
{
	return a<=maxn&&a>=-maxn && b<=maxn&&b>=-maxn;
}
void slove(ll a,ll b,ll c)
{
	ll x,y,m;
	extgcd(a,b,m,x,y);
	c=-c;
	if(c%m!=0)
	{
		printf("-1\n");
		return;
	}
	x*=c/m;y*=c/m;
	if(a==0||b==0)
	{
		printf("%I64d %I64d\n",x,y);
		return;
	}
	ll lcm=a/m*b;
	ll tx=lcm/a,ty=lcm/b;
	ll tp=abs((x+maxn)/lcm+1);
	x=x-tp*(lcm/a);
	y=y+tp*(lcm/b);
	while(x<=maxn)
	{
		if(judge(x,y))
		{
			printf("%I64d %I64d\n",x,y);
			//printf("%lld\n",a*x+b*y);
			return;
		}
		x+=tx;y-=ty;
	}
	printf("-1\n");
}
int main()
{
	ll a,b,c;
	while(~scanf("%I64d%I64d%I64d",&a,&b,&c))
	{
		slove(a,b,c);
	}
	return 0;
}