Baby Ming and Weight lifting

Problem Description
Baby Ming is fond of weight lifting. He has a barbell pole(the weight of which can be ignored) and two different kinds of barbell disks(the weight of which are respectively $a$ and $b$), the amount of each one being infinite.

Baby Ming prepare to use this two kinds of barbell disks to make up a new one weighted $C$(the barbell must be balanced), he want to know how to do it.

Input
In the first line contains a single positive integer $T$, indicating number of test case. For each test case: There are three positive integer $a, b$, and $C$. $1 \leq T \leq 1000, 0 < a, b, C \leq 1000, a \neq b$

Output
For each test case, if the barbell weighted $C$ can’t be made up, print Impossible. Otherwise, print two numbers to indicating the numbers of $a$ and $b$ barbell disks are needed. (If there are more than one answer, print the answer with minimum $a+b$)

Sample Input
2
1 2 6
1 4 5

Sample Output
2 2
Impossible

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

int main()
{
    int a,b,c,t;
    cin>>t;
    while(t--)
    {
        cin>>a>>b>>c;
        if(c%2!=0)
        {
            cout<<"Impossible"<<endl;
            continue;
        }
        c=c/2;
        int num1=-1,num2=-1;
        if(a<b)
        for(int i=0;i*a<=c;i++)
        {
            if((c-i*a)%b==0)
            {
                num2=(c-i*a)/b;
                num1=i;
                break;
            }
        }
        else
        {
            for(int i=0;i*b<=c;i++)
        {
            if((c-i*b)%a==0)
            {
                num1=(c-i*b)/a;
                num2=i;
                break;
            }
        }
        }
        if(num1>=0)
        {
            cout<<num1*2<<' '<<num2*2<<endl;
        }
        else cout<<"Impossible"<<endl;
    }
    return 0;
}