Codeforcess 535C Tavas and Karafs【二分+思维】

C. 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.

Examples

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

题目大意：

现在给你一个以A为基，B为公差的等差数列（无现长），其中有N个询问。

对于N个询问，每个询问有三个元素，l，t，m；

表示我们现在有t次操作，每次可以选择m个数将其都-1.

现在问你能够使得以l为起点，r为终点最远的r，区间【l，r】所有数都减少为0.

思路：

1、显然这个终点r越远，需要的操作就越多，其具有单调性，那么我们二分终点。进行判断。

如果当前终点可行，那么让这个终点更远一些，否则就更近一些。

2、对于判定：

显然我们要满足两个条件才能叫可行：

①从起点到当前终点mid所有的数都小于等于t.其实就是判断a【mid】是否小于等于t.

②从起点到当前终点mid所有的数的和小于t*m.表示我们需要在t次内将所有数都减少为0.

3、题目并不难，读懂题最重要。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll __int64
ll A,B,n;
ll get(ll mid)
{
return A+(mid-1)*B;
}
ll getsum(ll l,ll r)
{
ll rr=(A+A+(r-1)*B)*r/2;
ll lll=(A+A+(l-2)*B)*(l-1)/2;
return rr-lll;
}
int main()
{
while(~scanf("%I64d%I64d%I64d",&A,&B,&n))
{
for(ll i=0;i<n;i++)
{
ll s,t,m;
scanf("%I64d%I64d%I64d",&s,&t,&m);
ll left=s;
ll right=1000000000;
ll ans=-1;
while(right-left>=0)
{
ll mid=(right+left)/2;
if(get(mid)>t)right=mid-1;
else
{
if(getsum(s,mid)>m*t)
{
right=mid-1;
}
else
{
ans=mid;
left=mid+1;
}
}
}
printf("%d\n",ans);
}
}
}