Codeforces 535C Tavas and Karafs

题目链接：CF - 535C

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.

题目描述：给出一个等差数列，操作严格要求从最左边不为零的连续m个数减去1，最多执行t次后问离最左边最远的位置在哪里。

算法分析：我们可以二分位置，仔细想想这两个条件：max(hl，hl+1，hl+2，，，，hr)<=t && hl+(hl+1)+(hl+2)，，，，hr<=m*t

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<cstdlib>

 5 #include<cmath>

 6 #include<algorithm>

 7 #define inf 0x7fffffff

 8 using namespace std;

 9 typedef long long LL;

10 const LL maxn=1000000+10;

11 

12 LL A,B,n;

13 LL l,t,m;

14 

15 LL getValue(LL u) {return A+(u-1)*B; }

16 LL getSeg(LL u,LL v)

17 {

18     return (getValue(u)+getValue(v))*(v-u+1)/2;

19 }

20 

21 int main()

22 {

23     while (scanf("%I64d%I64d%I64d",&A,&B,&n)!=EOF)

24     {

25         while (n--)

26         {

27             scanf("%I64d%I64d%I64d",&l,&t,&m);

28             LL left=A+(l-1)*B;

29             if (left>t) printf("-1\n");

30             else

31             {

32                 LL L=l,R=(t-A)/B+1;

33                 while (L<=R)

34                 {

35                     LL mid=(L+R)>>1;

36                     if (getSeg(l,mid)<=m*t) L=mid+1;

37                     else R=mid-1;

38                 }

39                 printf("%I64d\n",L-1);

40             }

41         }

42     }

43     return 0;

44 }