Three cups of water

Determines whether a number is an integer: if (A==int (a))

The description gives three cups, varying in size, and only the largest cup of water is filled, the remaining two are empty cups . Three cups of water are poured between each other, and the cups are not identified and can only be calculated according to the volume of the cups given . You are now asked to write a program that outputs the minimum number of times that the initial state reaches the target State .

Input

The

first line an integer n (0<n<50) represents the n set of test data
Next, each set of test data has two lines, the first line gives three integers V1 V2 V3 (v1>v2>v3 v1<100 v3>0) represents the volume of three cups.
The second line gives three integers E1 E2 E3 (volume less than equal to the corresponding Cup volume) represents the final state we need

Output

each row outputs the minimum number of water pours for the corresponding test data. If the target status output is not reached-1

Sample input

26 3 14 1 19 3 27 1 1

Sample output

3-1

First: output n for loop output three cups a[0],b[0],d[0] open to 50//or record output with a[]
Second: The premise: v1=e1+e2+e3 (according to this equation V1 known if E2 and E3 can meet E1 will be able to find out)!!! 
     The final state of the V3 can only be full or empty.
     The final state of the V2 is either full or V3 multiples or empty (full itself is also a multiple of V3)//But when full, consider different methods.
     The final state of the V1 is full or empty or!!!!!!! The wrong idea
  The right thought: as long as the above situation V1 must be able to achieve the final state!!!!
The third E3 is 0 E2 0 0 times

    E3 is 0  E2 is a multiple of V3 V1 to give V3 a V3 to give V2 multiples times  + 1 times!!!!! Errors such as  5 3 2 and 2 3 0 only once
           Correct:  v2/v3 is not an integer when calculating multiples n1 from V1 to V3 to V2  to 2xn1 times
                  V2/v3 is an integer when calculating multiples n2= (v2-e2)/v3  from V1 to V2 to V3  to 2n2 +1  compared to 2n1
                                                                  
    E3==v3 e2=0      1 times
    E3=v3  E2 is a multiple of V3  V1 give V3 a V3 to V2 fold several times V1 again pour to V3 a  multiplier + 2 times/Error
        That's right

Misunderstanding: Both directions can be poured from the V3 to V2 can also be poured from the V2 to V3  things to the bidirectional nature
    

# include<stdio.h>
int main ()
{
int n,i,n1,n2,t,a[50]={0};
int v1,v2,v3,e1,e2,e3;
scanf ("%d", &n);
for (i=0;i<n;i++)
{
scanf ("%d%d%d", &v1,&v2,&v3);
scanf ("%d%d%d", &e1,&e2,&e3);
if (V1==E1+E2+E3)
{
if (e3==0 && e2==0)
{
a[i]=0;
}
else if (e3==v3 && e2==0)
{
A[i]=1;
}
else if (e3==0 && e2*1.0/v3 = = Int (e2*1.0/v3))
{
if (V2*1.0/v3==int (V2*1.0/V3))
{
N1= E2/v3; N2= (V2-E2)/v3
if (N1&GT;N2)
{t=n1;n1=n2;n2=t;}
A[I]=N1;
}
Else
{
A[i]=e2*1.0/v3 +1;
}
}
else if (E3 ==v3 && e2*1.0/v3 = = Int (e2*1.0/v3))
{

if (V2*1.0/v3==int (V2*1.0/V3))
{
N1= E2/v3; N2= (V2-E2)/v3
if (N1&GT;N2)
{t=n1;n1=n2;n2=t;}
A[I]=N1;
}
Else
{
A[i]=e2*1.0/v3 +2;
}

}

Else
{
A[i]=-1;
}
}
Else
A[i]=-1;
}
for (i=0;i<n;i++)
{
printf ("%d\n", A[i]);
}
return 0;
}

Three cups of water