2 seconds
256 megabytes
standard input
standard output
Karafs is some kind of vegetable in shape of an 1 × h rectangle. Tavaspolis people love Karafs and they use Karafs in almost any kind of food. Tavas, himself, is crazy about Karafs.
Each Karafs has a positive integer height. Tavas has an infinite 1-based sequence of Karafses. The height of the i-th Karafs issi = A + (i - 1) × B.
For a given m, let's define an m-bite operation as decreasing the height of at most m distinct not eaten Karafses by 1. Karafs is considered as eaten when its height becomes zero.
Now SaDDas asks you n queries. In each query he gives you numbers l, t and m and you should find the largest number r such thatl ≤ r and sequence sl, sl + 1, ..., sr can be eaten by performing m-bite no more than t times or print -1 if there is no such number r.
The first line of input contains three integers A, B and n (1 ≤ A, B ≤ 106, 1 ≤ n ≤ 105).
Next n lines contain information about queries. i-th line contains integers l, t, m (1 ≤ l, t, m ≤ 106) for i-th query.
For each query, print its answer in a single line.
2 1 4
1 5 3
3 3 10
7 10 2
6 4 8
4
-1
8
-1
1 5 2
1 5 10
2 7 4
1
2
题意:给一个等差数列,基为a,差为b,每次能对m个值减一,给出范围l,m,和次数t,求最能是得有边为0的最大值;
题解:二分找到符合的位置;
#include<iostream>
#include<cstdio>
#include<cstring>
#include<fstream>
#include<cmath>
#include<string>
#include<vector>
#include<algorithm>
#include<string>
#define ll long long
using namespace std;
ll a,b,n,m,l,t;
ll slove(ll x)
{
ll ans=(a*+(l+x-)*b)*(x-l+)/;
return ans;
}
bool judge(ll x)
{
if(a+(x-)*b>t)return true;
if(slove(x)>t*m)return true;
return false;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie();
cin>>a>>b>>n;
while(n--)
{
cin>>l>>t>>m;
if(a+(l-)*b>t)
{
cout<<-<<endl;continue;
}
ll L=l,R=1e9;
while(R-L>=)
{
ll mid=(R+L)>>;
if(judge(mid))
{
R=mid-;
}
else
{
L=mid+;
}
}
cout<<L-<<endl;
}
return ;
}