Codeforces 807C(二分)

深藏阁楼爱情的钟 2022-06-01 12:08 88阅读 0赞

问题描述:

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

题目题意:给我们一个p/q,x/y让我们在给分母加上最小的数字m,于此同时我们可以给分子加上n (0<=n<=m)使得p/q==(x+n)/(y+m)

题目分析:我们二分倍数使得q\*base>=y   p\*base>=x,(p\*base-x)<=(q\*base-y),自己画一下就知道了。

p/q=0和1的情况特判，比较特殊。因为极限只能接近不能到达。

代码如下:

#include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #define ll long long
    using namespace std;
    
    ll x,y,p,q;
    bool check(int base)
    {
        if (base*q<y) return false;
        if (x>p*base) return false;
        if ((p*base-x)>(q*base-y)) return false;
        return true;
    }
    int main()
    {
        int t;
        scanf("%d",&t);
        while (t--) {
            scanf("%lld%lld%lld%lld",&x,&y,&p,&q);
            if (p==0) {
                if (x!=0) {
                    printf("%d\n",-1);
                    continue;
                }
            }
            if (p==q) {
                if (x!=y) {
                    printf("%d\n",-1);
                    continue;
                }
            }
            ll left=1,right=1e9,mid;
            while (right>=left) {
                mid=(right+left)/2;
                if (check(mid)) right=mid-1;
                else left=mid+1;
            }
            printf("%lld\n",left*q-y);
        }
        return 0;
    }