【codeforces 807C】Success Rate

【题目链接】:http://codeforces.com/contest/807/problem/C

【题意】

给你4个数字
x y p q
要求让你求最小的非负整数b;
使得
(x+a)/(y+b)==p/q
同时a为一个整数且0<=a<=b

【题解】

/*
    (x+a)/(y+b)==p/q;
    则
    x+a=np
    y+b=nq
    (以上结论只在p和q是互质的情况下有效,当然题目有说p和q互质)
    a=np-x
    b=nq-y
    a<=b
    因为p<q所以n越大b会越大;
    又p和q都大于等于0,所以n越大,a和b都是不下降的
    二分n
*/

【Number Of WA】

3

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 110;

int t;
LL x,y,p,q;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not use
    cin >> t;
    while (t--)
    {
        cin >> x >> y >> p >> q;
        LL l = 0,r = 1e9,ans = -1;
        while (l<=r)
        {
            LL mid = (l+r)>>1;
            LL a=mid*p-x,b=mid*q-y;
            if (a>=0 && b>=0 && a<=b)
            {
                ans = mid;
                r = mid-1;
            }
            else
                l = mid+1;
        }
        if (ans==-1)
            cout <<-1<<endl;
        else
            cout <<ans*q-y<<endl;
    }
    return 0;
}