G. Gift Set（二分）

传送门

Polycarp has x of red and y of blue candies. Using them, he wants to make gift sets. Each gift set contains either a red candies and b blue candies, or a blue candies and b red candies. Any candy can belong to at most one gift set.

Help Polycarp to find the largest number of gift sets he can create.

For example, if x=10, y=12, a=5, and b=2, then Polycarp can make three gift sets:

• In the first set there will be 5 red candies and 2 blue candies;
• In the second set there will be 5 blue candies and 2 red candies;
• In the third set will be 5 blue candies and 2 red candies.

Note that in this example there is one red candy that Polycarp does not use in any gift set.

Input

The first line contains an integer t (1≤t≤1e4). Then t test cases follow.

Each test case consists of a single string containing four integers x, y, a, and b (1≤x,y,a,b≤1e9).

Output

For each test case, output one number — the maximum number of gift sets that Polycarp can make.

 

9
10 12 2 5
1 1 2 2
52 311 13 27
1000000000 1000000000 1 1
1000000000 1 1 1000000000
1 1000000000 1000000000 1
1 2 1 1
7 8 1 2
4 1 2 3

output

Copy

3
0
4
1000000000
1
1
1
5
0


这个题目是一个二分的题目，确实没有想到，首先先把题目转化为数学符号
设第一种有s个，第二中有t个，总数有c个

 

 

#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
ll x,y,a,b;
int judge(ll c){
    ll xx=x-a*c;
    ll yy=y-a*c;
    if(a==b){
        return ((xx>=0)&&(yy>=0));
    }
    else{
        if(xx<0||yy<0){
            return 0;
        }
        xx/=(b-a);
        yy/=(b-a);
        return (xx+yy>=c);
    }
}
int main(){
    int t;
    cin>>t;
    while(t--){
        cin>>x>>y>>a>>b;
        if (a>b) swap(a,b);
        ll l=0,r=1e9,ans;
        while(r>=l){
            ll mid=(l+r)/2;
            if(judge(mid)){
                ans=mid;
                l=mid+1;
            }
            else{
                r=mid-1;
            }
        }
        cout<<ans<<endl;
    }
}