codeforces807div2 C.Success Rate[二分][数学]

C. Success Rate

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

43 10 1 27 14 3 820 70 2 75 6 1 1

output

4100-1

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;}