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

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

     return ;
 }

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

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

     return ;
 }