全RE求调

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define end {puts("-1");return;}
ll a,b,c,x,y;
ll exgcd(ll at,ll bt)
{
    if(!bt)
    {
        x=1,y=0;
        return at;
    }
    ll t=exgcd(bt,at%bt);
    ll tmp=x;
    x=y;
    y=tmp-at/bt*y;
    return t;
}
void work()
{
    // x=y=0;
    scanf("%lld%lld%lld",&a,&b,&c);
    ll d=exgcd(a,b);
    if(c%d) end;
    x*=(c/d);
    y*=(c/d);
    ll tx=b/d,ty=a/d,k=(long long)ceil((1.0-x)/tx);
    x+=tx*k; //xmin
    y-=ty*k; //ymax
    if(y<=0)
    {
        ll ky=(long long)ceil((1.0-y)/ty);
        ll ymin=y+ty*ky;
        printf("%lld %lld\n",x,ymin); //这里x是xmin
    }
    else
    {
        printf("%lld ",(y-1)/ty+1); //正整数解的个数，通过y的可减区间来求！
        printf("%lld ",x); //xmin
        printf("%lld ",(y-1)%ty+1); //ymin
        printf("%lld ",x+(y-1)/ty*tx); //xmax
        printf("%lld\n",y); //ymax
    }
    return;
}
signed main()
{
    ll t;
    scanf("%lld",t);
    while(t--) work();
    return 0;
}