CF Tavas and Karafs (二分)

Tavas and Karafs

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
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 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.

Sample test(s)

input

2 1 4
1 5 3
3 3 10
7 10 2
6 4 8

output

4
-1
8
-1

input

1 5 2
1 5 10
2 7 4

output

1
2

今天才发现比赛时候的算法是对的。。。只不过数据爆了，改成long long就过了，还重新找题解写了一次。。。。

不过新写的更快一点,130ms,二分查找，下界为L,二分查上界,上界的初始值可以算出来。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <ctime>
using    namespace    std;

long long    bsearch(long long,long long,long long);

long long    A,B;
int        main(void)
{
long long    l,t,m,r;
long long    n;

scanf("%lld%lld%lld",&A,&B,&n);
while(n --)
{
scanf("%lld%lld%lld",&l,&t,&m);
if(t < A + (l - 1) * B)
{
puts("-1");
continue;
}
r = bsearch(l,t,m);
printf("%lld\n",r);
}

return    0;
}

long long    bsearch(long long l,long long t,long long m)
{
long long    low = l;
long long    high = (t - A) / B + 1;

while(low <= high)
{
long    long    mid = (low + high) / 2;
long    long    box = (A + (l - 1) * B + A + (mid - 1) * B) * (mid - l + 1) / 2;

if(box <= m * t)
low = mid + 1;
else
high = mid - 1;
}

return    low - 1;
}