洛谷 P5656

#include 
#include 
 
using namespace std;
 
typedef long long ll;
 
inline long long read()
{
    long long s=0,f=1;        
    char ch=getchar();        
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}        
    while(isdigit(ch)) {s=s*10+ch-'0'; ch=getchar(); }        
    return s*f;        
}//快读
inline void write(long long x)        
{        
    if(x<0) {putchar('-');x=-x;}
	if(x>9) write(x/10);        
    putchar(x%10+'0');        
    return;        
}//快写 
 
void exgcd(ll a,ll b,ll &x,ll &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return ;
    }
    exgcd(b,a%b,x,y);
    ll temp=x;
    x=y;
    y=temp-a/b*y;
}
ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}
 
int main(){
	int t;
	cin>>t;
	while(t--){
		ll a,b,c,x,y;
		a=read();
		b=read();
		c=read();
		ll gcdd = gcd(a, b);
		if(c % gcdd != 0){
			puts("-1");
			continue;
		}
		a /= gcdd;
		b /= gcdd;
		c /= gcdd;//此时gcd(a, b) = 1 
		exgcd(a, b, x, y);//求ax + by = 1的解 
		x *= c;
		y *= c;//这样x,y就是ax + by = c 的解 
		//以下都是对a(x + tb) + b(y - ta) = c的应用 
		if(x <= 0){//如果x<=0那么求x的最小正整数值 
			ll t = abs(x) / b + 1;
			x = x % b + b;
			y -= t * a;
			if(y <= 0)//x是最小正整数解，y<=0,说明有整数解没有正整数解，输出x的最小正整数值，y的最小正整数值。 
			{
				write(x);
				printf(" ");
				y = y % a + a;
				write(y);
				printf("\n");
				continue;
			}
		}
		if(y <= 0){//同上 
			ll t = abs(y) / a + 1;
			y = y % a + a;
			x -=  t * b;
			if(x <= 0){
				x = x % b + b;
				write(x);
				printf(" ");
				write(y);
				printf("\n");
				continue;	
			}
		}
		
		ll maxx, minx, maxy, miny, p, q;
		
		p = x / b;//计算x的最小值，同时y取最大值 
		x %= b;
		if(!x) x += b,p--;
		y += p * a;
		minx = x;
		maxy = y;
		
		q = y/a;
		y %= a;
		if(!y) y+=a,q--;
		x += q * b;
		miny = y;
		maxx = x;	
		
		printf("%d %d %d %d %d\n", (maxx - minx) / b + 1, minx, miny, maxx, maxy);
	}
	return 0;
}