J - Success Rate

CodeForces - 773A

Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each of the next t lines contains four integers x, y, p and q (0 ≤ x ≤ y ≤ 109; 0 ≤ p ≤ q ≤ 109; y > 0; q > 0).

It is guaranteed that p / q is an irreducible fraction.

Hacks. For hacks, an additional constraint of t ≤ 5 must be met.

Output
For each test case, output a single integer equal to the smallest number of submissions you have to make if you want your success rate to be equal to your favorite rational number, or -1 if this is impossible to achieve.

Example

Input
4
3 10 1 2
7 14 3 8
20 70 2 7
5 6 1 1

Output
4
10
0
-1

Note
In the first example, you have to make 4 successful submissions. Your success rate will be equal to 7 / 14, or 1 / 2.

In the second example, you have to make 2 successful and 8 unsuccessful submissions. Your success rate will be equal to 9 / 24, or 3 / 8.

In the third example, there is no need to make any new submissions. Your success rate is already equal to 20 / 70, or 2 / 7.

In the fourth example, the only unsuccessful submission breaks your hopes of having the success rate equal to 1.

解法：通分k倍后边的数使 q-y>p-x，k*q-y为结果，重点是求k
k为 max(ceil(x*1.0/p),ceil((y-x)*1.0/(q-p)))

代码：

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

int main()
{
    int s;
    cin>>s;
    for(int i=0;i<s;i++){
        long long x,y,p,q;
        cin>>x>>y>>p>>q;
        long long flag=-1;
        if(p==q)
            flag=x==y?0:-1;
        else if(p==0)
            flag=x==0?0:-1;
        else{
            long long k=max(ceil(x*1.0/p),ceil((y-x)*1.0/(q-p)));
            flag = q*k-y;
        }
        cout<<flag<<endl;
    }
}