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 - ) * B)
         {
             puts("-1");
             continue;
         }
         r = bsearch(l,t,m);
         printf("%lld\n",r);
     }

     return    ;
 }

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

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

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

     return    low - ;
 }