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CodeForce-807C Success Rate(二分數學)

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Success Rate

CodeForces - 807C

給你4個數字 x y p q ,要求讓你求最小的非負整數b,使得 (x+a)/(y+b)==p/q,同時a為一個整數且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; }

CodeForce-807C Success Rate(二分數學)