Codeforces Round #299 (Div. 1) A. Tavas and Karafs 水题

Tavas and Karafs

Time Limit: 1 Sec Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/536/problem/A

Description

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.

$\text{[math]}$

Each Karafs has a positive integer height. Tavas has an infinite 1-based sequence of Karafses. The height of the i-th Karafs is s_i = 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 that l ≤ r and sequence s_l, s_l + 1, ..., s_r can be eaten by performing m-bite no more than t times or print -1 if there is no such number r.

Input

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.

Output

For each query, print its answer in a single line.

Sample Input

Sample Output

4
-1
8
-1

HINT

题意

给你一个等差数列，然后10e6次询问

问你一次可以吃m个，最多吃t次，问你最多往右吃多少个

题解：

傻逼题，只要每个数都小于等于t，前缀和小于等于t*m就好了

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff;   //无限大
const int inf=0x3f3f3f3f;
/*

int buf[10];
inline void write(int i) {
  int p = 0;if(i == 0) p++;
  else while(i) {buf[p++] = i % 10;i /= 10;}
  for(int j = p-1; j >=0; j--) putchar('0' + buf[j]);
  printf("
");
}
*/
//**************************************************************************************
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
ll A,B,q,t,m,l;
ll ans(ll x)
{
    return A+(x-1)*B;   
}
ll sum(ll l,ll r)
{
    ll x1=A+(l-1)*B,x2=A+(r-1)*B;
    return 1ll*(x1+x2)*(r-l+1)/2;
}
int main()
{
    scanf("%d%d%d",&A,&B,&q);
    while(q--)
    {
        int l,t,m;
        scanf("%d%d%d",&l,&t,&m);
        int lef=l,rig=inf,mid;
        while(lef<=rig)
        {
            mid=lef+rig>>1;
            if(ans(mid)<=t && sum(l,mid)<=1ll*t*m)lef=mid+1;
            else rig=mid-1;
        }
        if(rig==l-1)printf("-1
");
        else printf("%d
",rig);
    }
    return 0;
}