Codeforces Round #299 (Div. 1) A. Tavas and Karafs 水题

Tavas and Karafs

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/536/problem/A

Description

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 s_i = 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 s_l, s_l + 1, ..., s_r 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.

1000000000.

Output

For each query, print its answer in a single line.

Sample Input

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

Sample Output

4
-1
8
-1

HINT

题意

给你一个等差数列，然后10e6次询问

问你一次可以吃m个，最多吃t次，问你最多往右吃多少个

题解：

傻逼题，只要每个数都小于等于t，前缀和小于等于t*m就好了

代码：

 

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 200001

#define mod 10007

#define eps 1e-9

//const int inf=0x7fffffff;   //无限大

const int inf=0x3f3f3f3f;

/*



int buf[10];

inline void write(int i) {

  int p = 0;if(i == 0) p++;

  else while(i) {buf[p++] = i % 10;i /= 10;}

  for(int j = p-1; j >=0; j--) putchar('0' + buf[j]);

  printf("\n");

}

*/

//**************************************************************************************

inline ll read()

{

    int x=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

ll A,B,q,t,m,l;

ll ans(ll x)

{

    return A+(x-1)*B;   

}

ll sum(ll l,ll r)

{

    ll x1=A+(l-1)*B,x2=A+(r-1)*B;

    return 1ll*(x1+x2)*(r-l+1)/2;

}

int main()

{

    scanf("%d%d%d",&A,&B,&q);

    while(q--)

    {

        int l,t,m;

        scanf("%d%d%d",&l,&t,&m);

        int lef=l,rig=inf,mid;

        while(lef<=rig)

        {

            mid=lef+rig>>1;

            if(ans(mid)<=t && sum(l,mid)<=1ll*t*m)lef=mid+1;

            else rig=mid-1;

        }

        if(rig==l-1)printf("-1\n");

        else printf("%d\n",rig);

    }

    return 0;

}