CodeForce-807C Success Rate(二分数学)

Posted 2020-09-27 Ying_zx

 

Success Rate

CodeForces - 807C

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。

(0?≤?x?≤?y?≤?10⁹; 0?≤?p?≤?q?≤?10⁹; y?>?0; q?>?0)

解法：

 

(x+a)/(y+b)==p/q;

-->

x+a=np;

y+b=nq;

-->

a=np-x;

b=nq-y;

-->

二分n;

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
using namespace std;
#define _ ios::sync_with_stdio(false),cin.tie(0)

const int MAXN = 5010;
const int INF = 0xfffffff;
typedef long long ll;

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

int main()
{
    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;
}