Codeforces807 C. Success Rate 2017-05-08 23:27 91人阅读 评论(0) 收藏
2 seconds
256 megabytes
standard input
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?
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.
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.
4
3 10 1 2
7 14 3 8
20 70 2 7
5 6 1 1
4
10
0
-1
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.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <vector>
using namespace std;
#define inf 0x3f3f3f3f
#define LL long long LL x,y,p,q; bool ok(LL n)
{
if(q*n-y>=p*n-x&&q*n>=y&&p*n>=x)
return 1;
return 0;
} int main()
{
int t; scanf("%d",&t);
while(t--)
{
scanf("%lld%lld%lld%lld",&x,&y,&p,&q); LL l=1,r=1e9; LL ans=-1;
while(l<=r)
{
int mid=(l+r)/2;
if(ok(mid))
{
r=mid-1;
ans=mid;
}
else
{
l=mid+1;
}
}
if(ans==-1)
printf("-1\n");
else
printf("%lld\n",ans*q-y);
}
return 0;
}
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