codeforces807div2 C.Success Rate[二分][数学]

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本文介绍了一个有趣的问题：如何通过增加成功的或不成功的提交来调整Codeforces平台上的个人成功率，使其达到一个特定的理想比率。文中提供了一种有效的算法实现，通过二分搜索找到达到目标成功率所需的最小提交次数。

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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

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.

题意：现在的正确率是 a/b , 要达到c/d ,可以进行任意次的正确与非正确提交，求最小的提交次数能够得到c/d, 如果不能得到 pirnt -1

思路：

假设要进行p次正确提交，q-p次非正确提交，则 (a+p)/(b+q) == c/d

由题意知c、d互质，所以 (a+p)/(b+q)应该是 c/d的非最简分数或最简分数，即 (a+p)/(b+q) == nc/nd ，其中 0<=p<=q n>=1 且都是整数

所以 p = nc-a, q = nd-b p、q都是n的递增函数，所以二分n寻找最小的n满足条件，无满足输出-1

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while(t--)
    {
        int a, b, c, d;
        cin >> a >> b >> c >> d;
        LL l = 1, r = 1e9;
        LL ans = -1;
        while(l<=r)
        {
            LL mid = (l+r) >> 1;
            //cout << mid << endl;
            LL p = mid*c-a, q = mid*d-b;
            if(p>=0 && q>=0 && p<=q)
            {
                r = mid-1;
                ans = q;
            }
            else
                l = mid+1;
        }
        cout << ans << endl;
    }
    return 0;
}