CodeForce-807C Success Rate(二分数学)
Success Rate
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; }