codeforces 807C Success Rate 数论

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 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.

转自：http://blog.csdn.net/dormousenone/article/details/71404361

构造。推公式。 求最小的n。也就是求能满足两个情况的n。因为情况只有两种。

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>using namespace std;long long x, y, p, q;int main(){    int T;    scanf("%d",&T);    int x,y,p,q;    while(T--)    {        cin>>x>>y>>p>>q;        if(p==q)        {            printf("%d\n",x==y?0:-1 );            continue;        }        if(p==0)        {            printf("%d\n",x==0?0:-1 );            continue;        }        long long GCD=__gcd(p,q);        p/=GCD;q/=GCD;        long long n=max(ceil(x*1.0/p),ceil((y-x)*1.0/(q-p)));        printf("%lld\n",n*q-y );    }}