CF 535c Tavas and Karafs

Tavas and Karafs

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 535C

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.

Output

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

Sample Input

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

定理  序列h1,h2,...,hn 可以在t次时间内(每次至多让m个元素减少1)  全部减小为0  当且仅当  

max(h1, h2, ..., hn) <= t  &&  h1 + h2 + ... + hn <= m*t   

然后用二分来做

网上的代码

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 12345678
#define M 1234

long long a,b,n,l,t,m;

int main()
{
    while(~scanf("%lld%lld%lld",&a,&b,&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%lld%lld%lld",&l,&t,&m);
            if(a+(l-1)*b>t)
            {
                puts("-1");
                continue;
            }
            long long ll=l,r=(t-a)/b+1,mid;

            while(ll<=r)
            {
                mid=(ll+r)/2;
                if( (2*a+(mid+l-2)*b)*(mid-l+1)/2  <=t*m)
                {
                    ll=mid+1;
                }
                else
                {
                    r=mid-1;
                }
            }
            cout<<ll-1<<endl;
        }
    }

    return 0;
}

我的代码

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 12345678
#define M 1234

long long a,b,n,l,t,m;

int main()
{
    while(~scanf("%lld%lld%lld",&a,&b,&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%lld%lld%lld",&l,&t,&m);
            if(a+(l-1)*b>t)
            {
                puts("-1");
                continue;
            }
            long long ll=l,r=(t-a)/b+1,mid;

            while(ll<=r)
            {
                mid=(ll+r)/2;
                if( (2*a+(mid+l-2)*b)*(mid-l+1)/2  <=t*m
                    &&  (2*a+(mid+1+l-2)*b)*(mid+1-l+1)/2  >t*m)
                                        //这里wa了一次，  把 > 写成了 >= ,
                {
                    break;
                }
                else if( (2*a+(mid+l-2)*b)*(mid-l+1)/2  <=t*m)
                {
                    ll=mid+1;
                }
                else
                {
                    r=mid-1;
                }
            }
            cout<<mid<<endl;
        }
    }

    return 0;
}