[刷题]Codeforces Round #412(Div. 2) - C. Success Rate

最新推荐文章于 2022-08-21 01:05:54 发布

蟹恼板

最新推荐文章于 2022-08-21 01:05:54 发布

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分类专栏：刷题 Codeforces 文章标签： codeforces

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本文介绍了一个简单的算法问题，旨在计算为了将Codeforces平台上的成功提交率调整到特定比率p/q，参赛者还需要进行多少次提交。通过确定提交次数的最小增量，确保成功率能够精确匹配指定值。

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PROBLEMSET里面神tm搜不到这题，很迷。所以标题就只好注明比赛出处而没法标明题号了。

Description

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

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.

Key

给你参加的总场次 $y$ 与其中胜利的场次 $x$ ，现要求胜率达到 $p/q$ ，问还要再比多少场（每一场可胜可负）。
想明白了其实是个巨简单的题目。为了达到目标，比赛的总场次必为 $q$ 的倍数（现设倍数为 $k$ ），则 $k*q>=y$ ；在比了 $k*q$ 场次的情况下，必须胜利 $k*p$ 次，则 $k*p>=x$ 。胜利的场数肯定不大于总场数，则只要找到 $k*p-x<=k*q-y$ 的最小k即获得了答案。
综合一下三个条件，找出满足以下条件的k的最小值：

k>=y/q;
k>=x/p;
k>=(y-x)/(q-p);

即 $k=max(y/q, x/p, (y-x)/(q-p))$ 。其中除法结果若为小数则向上取整。
因此 $O(1)$ 时间复杂度即可求出。
至于无法达到的情况，只有 $x/y$ 不为0而 $p/q$ 为0、或 $x/y$ 不为1而 $p/q$ 为1两种。一开始判断掉就行。

Code

#include<iostream>
using namespace std;
typedef long long LL;
LL T, x, y, p, q;

int main()
{
    cin >> T;
    while (T--) {
        cin >> x >> y >> p >> q;
        if (x*q == p*y) {
            cout << "0\n";
            continue;
        }
        if (p == 0 || p == q) {
            cout << "-1\n";
            continue;
        }
        LL k = (y - x - 1) / (q - p) +1;
        if (k*p < x) k = (x - 1) / p + 1;
        if (k*q < y) k = (y - 1) / q + 1;
        LL res = k*q - y;
        cout << res << '\n';
    }
    return 0;
}