Codeforces Round #412 ( Div. 2) C

C. Success Rate
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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
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 haxiave 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：现在正确率时x/y，让你把它变成p/q。问你最少要做的题。
题解：变成数学式为 （x+a)/(y+b)==p/q(a<=b)
x+a==k*p; a=k*p-x;
y+b==k*q; b=k*q-y;
然后二分k就可以了。。。。。
代码：

#include<bits/stdc++.h>#define ll long longusing namespace std;const int N=1e6;ll x,y,q,p;int main(){    int t;    cin>>t;    while(t--)    {        cin>>x>>y>>p>>q;        ll l=0;        ll r=1e10;        ll ans=-1;        while(r>=l)        {            ll mid=(r+l)>>1;            ll a=mid*p-x;            ll b=mid*q-y;            if(b>=0&&a>=0&&b>=a)            {                ans=mid;                r=mid-1;            }            else                l=mid+1;        }        if(ans==-1) cout<<ans<<endl;        else            cout<<ans*q-y<<endl;    }}