[刷题]Codeforces Round #412(Div. 2)

PROBLEMSET里面神tm搜不到这题，很迷。所以标题就只好注明比赛出处而没法标明题号了。

Description

You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made y submissions, out of which x have been successful. Thus, your current success rate on Codeforces is equal to x / y.

Your favorite rational number in the [0;1] range is p / q. Now you wonder: what is the smallest number of submissions you have to make if you want your success rate to be p / q?

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

input43 10 1 27 14 3 820 70 2 75 6 1 1output4100-1

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.

Key

给你参加的总场次y与其中胜利的场次x，现要求胜率达到p/q，问还要再比多少场（每一场可胜可负）。
想明白了其实是个巨简单的题目。为了达到目标，比赛的总场次必为q的倍数（现设倍数为k），则k∗q>=y；在比了k∗q场次的情况下，必须胜利k∗p次，则k∗p>=x。胜利的场数肯定不大于总场数，则只要找到k∗p−x<=k∗q−y的最小k即获得了答案。
综合一下三个条件，找出满足以下条件的k的最小值：

k>=y/q;k>=x/p;k>=(y-x)/(q-p);

即k=max(y/q,x/p,(y−x)/(q−p))。其中除法结果若为小数则向上取整。
因此O(1)时间复杂度即可求出。
至于无法达到的情况，只有x/y不为0而p/q为0、或x/y不为1而p/q为1两种。一开始判断掉就行。

Code

#include<iostream>using namespace std;typedef long long LL;LL T, x, y, p, q;int main(){    cin >> T;    while (T--) {        cin >> x >> y >> p >> q;        if (x*q == p*y) {            cout << "0\n";            continue;        }        if (p == 0 || p == q) {            cout << "-1\n";            continue;        }        LL k = (y - x - 1) / (q - p) +1;        if (k*p < x) k = (x - 1) / p + 1;        if (k*q < y) k = (y - 1) / q + 1;        LL res = k*q - y;        cout << res << '\n';    }    return 0;}