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 41 5 33 3 107 10 26 4 8

Output

4-18-1

Input

1 5 21 5 102 7 4

Output

12

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

 1 #include<queue>
 2 #include<math.h>
 3 #include<stdio.h>
 4 #include<string.h>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;
 8 #define N 12345678
 9 #define M 1234
10
11 long long a,b,n,l,t,m;
12
13 int main()
14 {
15     while(~scanf("%lld%lld%lld",&a,&b,&n))
16     {
17         for(int i=0;i<n;i++)
18         {
19             scanf("%lld%lld%lld",&l,&t,&m);
20             if(a+(l-1)*b>t)
21             {
22                 puts("-1");
23                 continue;
24             }
25             long long ll=l,r=(t-a)/b+1,mid;
26
27             while(ll<=r)
28             {
29                 mid=(ll+r)/2;
30                 if( (2*a+(mid+l-2)*b)*(mid-l+1)/2  <=t*m)
31                 {
32                     ll=mid+1;
33                 }
34                 else
35                 {
36                     r=mid-1;
37                 }
38             }
39             cout<<ll-1<<endl;
40         }
41     }
42
43     return 0;
44 }

我的代码

 1 #include<queue>
 2 #include<math.h>
 3 #include<stdio.h>
 4 #include<string.h>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;
 8 #define N 12345678
 9 #define M 1234
10
11 long long a,b,n,l,t,m;
12
13 int main()
14 {
15     while(~scanf("%lld%lld%lld",&a,&b,&n))
16     {
17         for(int i=0;i<n;i++)
18         {
19             scanf("%lld%lld%lld",&l,&t,&m);
20             if(a+(l-1)*b>t)
21             {
22                 puts("-1");
23                 continue;
24             }
25             long long ll=l,r=(t-a)/b+1,mid;
26
27             while(ll<=r)
28             {
29                 mid=(ll+r)/2;
30                 if( (2*a+(mid+l-2)*b)*(mid-l+1)/2  <=t*m
31                     &&  (2*a+(mid+1+l-2)*b)*(mid+1-l+1)/2  >t*m)
32                                         //这里wa了一次，  把 > 写成了 >= ,
33                 {
34                     break;
35                 }
36                 else if( (2*a+(mid+l-2)*b)*(mid-l+1)/2  <=t*m)
37                 {
38                     ll=mid+1;
39                 }
40                 else
41                 {
42                     r=mid-1;
43                 }
44             }
45             cout<<mid<<endl;
46         }
47     }
48
49     return 0;
50 }