codeforces535C:Tavas and Karafs(二分)

Karafs is some kind of vegetable in shape of an1 × 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 infinite1-based sequence of Karafses. The height of thei-th Karafs issi = 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 ≤ rand sequence sl, sl + 1, ..., sr can be eaten by performingm-bite no more thant times or print -1 if there is no such numberr.

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 test(s)

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

 

这题比较坑，题意真的很难看懂，但是题意 理解了就会觉得其实挺水的~~

一个递增的等差数列，a为首项，b为公差，n个测试组

输入l，t，m

这段数列从第l个开始，m个数一起每次减一，若m个数中首项为0，则向后推移，t次以后输出最后一项是0的是第几项，若不存在为0的项，则输出-1

 

假设最后一项为0的是第r项，通过观察我们可以发现s[ r ] >= t && ( s[ l ] + s[ l+1 ] + s[ l+2 ] +.....+s[ r ] )<=t*m

接下来二分一下就能解决问题了

<span style="font-size:18px;">#include <iostream>using namespace std;int main(){int a,b,n;cin>>a>>b>>n;while(n--){__int64 l,t,m,sl,smid,sll;cin>>l>>t>>m;sl = a+(l-1)*b;if(sl>t) cout<<"-1"<<endl;else{__int64 ll=l,lr=(t-a)/b+1;while(ll<=lr){__int64 mid = (ll+lr)/2;smid = a+(mid-1)*b;if((sl+smid)*(mid-l+1)/2<=t*m) ll=mid+1;else lr=mid-1;}cout<<ll-1<<endl;}}return 0;}</span>