Codeforces 535C Tavas and Karafs

题目链接：CF - 535C

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 is si = 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 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.

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.

题目描述：给出一个等差数列，操作严格要求从最左边不为零的连续m个数减去1，最多执行t次后问离最左边最远的位置在哪里。

算法分析：我们可以二分位置，仔细想想这两个条件：max(hl，hl+1，hl+2，，，，hr)<=t && hl+(hl+1)+(hl+2)，，，，hr<=m*t

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std;
typedef long long LL;
const LL maxn=1000000+10;

LL A,B,n;
LL l,t,m;

LL getValue(LL u) {return A+(u-1)*B; }
LL getSeg(LL u,LL v)
{
    return (getValue(u)+getValue(v))*(v-u+1)/2;
}

int main()
{
    while (scanf("%I64d%I64d%I64d",&A,&B,&n)!=EOF)
    {
        while (n--)
        {
            scanf("%I64d%I64d%I64d",&l,&t,&m);
            LL left=A+(l-1)*B;
            if (left>t) printf("-1\n");
            else
            {
                LL L=l,R=(t-A)/B+1;
                while (L<=R)
                {
                    LL mid=(L+R)>>1;
                    if (getSeg(l,mid)<=m*t) L=mid+1;
                    else R=mid-1;
                }
                printf("%I64d\n",L-1);
            }
        }
    }
    return 0;
}