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 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.
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
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 - ;
}