CF Round #412（ Div.2） Success Rate

C. Success Rate

time limit per test 2
seconds
memory limit per test256
megabytes

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

Your favorite rational number in the[0;1]
range is p / q.
Now youwonder: what is the smallest number of submissions you have to make if you wantyour success rate to bep / q?
Input

The first line contains a single integert
(1 ≤ t ≤ 1000) —
thenumber of test cases.

Each of the nextt
lines containsfour integers x,y,p
and q
(0 ≤ x ≤ y ≤ 109;0 ≤ p ≤ q ≤ 109;y > 0;q > 0).

It is guaranteed thatp / q
is anirreducible fraction.

Hacks.
For hacks, anadditional 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 rateto be equal to your favorite rational number, or-1
if this isimpossible 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 4successful submissions. Your success rate will be equal to7 / 14,
or 1 / 2.

In the second example, you have to make2 successful and 8 unsuccessful submissions. Your success rate will be equal
to9 / 24,
or 3 / 8.

In the third example, there is no needto make any new submissions. Your success rate is already equal to20 / 70,
or 2 / 7.
In the fourth example, the only unsuccessful
submissionbreaks your hopes of having the success rate equal to 1.

题意：现在有一个AC Ratio（分数形式，AC题数/总提交题数），问是否能够通过最小的提交数（每次提交可以是正确，也可以是错误）使AC
Patio 达到一个最简分数

这道题开始想要暴力搞，果断超时，后来列出等式（x+a）/(y+a+b)=p/q,化简得a=(py-qx+pb)/(q-b),求出最小的a，再加上b即可，我利用中国剩余定理来求最小的a，然而不是RE就是TLE。。。后来发现不用那么复杂

法一：直接求需要把第二个分数扩大的倍数

#include <bits/stdc++.h>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define f(i,a,b) for(int i=(a);i<(b);++i)
#define ll long long
const int maxn = 1005;
const int mod = 475;
const ll INF = 0x3f3f3f3f;
const double eps = 1e-6;
#define rush() int T;scanf("%d",&T);while(T--)

int main()
{
ll x,y,p,q;
rush()
{
scanf("%I64d%I64d%I64d%I64d",&x,&y,&p,&q);
if(p==q)
{
if(x==y)
puts("0");
else  puts("-1");
continue;
}
if(p==0)
{
if(x==0)
puts("0");
else puts("-1");
continue;
}
ll a=q-p;
ll b=y-x;
ll cnt=max((x+p-1)/p,max((y+q-1)/q,(b+a-1)/a));
ll ans=cnt*q-y;
printf("%I64d\n",ans);
}
return 0;
}

法二：二分

#include <bits/stdc++.h>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define f(i,a,b) for(int i=(a);i<(b);++i)
#define ll long long
const int maxn = 1005;
const int mod = 475;
const ll INF = 0x3f3f3f3f;
const double eps = 1e-6;
#define rush() int T;scanf("%d",&T);while(T--)

int main()
{
ll x,y,p,q;
rush()
{
scanf("%I64d%I64d%I64d%I64d",&x,&y,&p,&q);
ll l=0,r=INF;
ll m;
while(l<r)
{
m=(l+r)/2;
if(q*m<y||p*m<x)
{
l=m+1;
}
else if(q*m-y>=p*m-x)
{
r=m;
}
else l=m+1;
}
if(l==INF)
puts("-1");
else printf("%I64d\n",q*l-y);
}
return 0;
}