基本思路
[egin{array}{l}
k,n,a,b,假设能完成游戏,则有\
x_1+x_2=n\
x_1*a+x_2*b<k\
由于 quad a<b quad则有\
x_1*(a-b)+(x_1+x_2)*b<k\
x<frac{k-n*b}{a-b}
end{array}
复杂度O(1)
]
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll q,k,n,a,b;
int main()
{
cin>>q;
while(q--)
{
cin>>k>>n>>a>>b;
ll x;
if((k-b*n)<=0) cout<<-1<<endl;
else
{
x=(k-n*b)/(a-b);
if((k-n*b)%(a-b)==0)x--;
cout<<min(x,n)<<endl;
}
}
return 0;
}