Codeforces 535C Tavas and Karafs(二分)

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题意 有一个等差数列 从A开始 公差为B然后n个询问 每个询问给定l,t,m然后要求如果每次可以最多选择m个数 使这m个数-1 那么在t次操作中可以使l为左端点的最长序列中使所有数为0 输出这个最长序列的右端序号

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

那么就可以二分右端点来解决了 下限为l 上限为hi不超过t的最大i

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#include <bits/stdc++.h>
typedef long long LL;
LL a, b, n, l, t, m;


LL getv(LL p)
{
    return a + (p - 1) * b;
}


LL getsum(LL r)
{
    return (getv(r) + getv(l)) * (r - l + 1) / 2;
}


int main()
{
    scanf("%lld%lld%lld", &a, &b, &n);
    while(n--)
    {
        scanf("%lld%lld%lld", &l, &t, &m);
        if(getv(l) > t) puts("-1");
        else
        {
            LL le = l, ri = (t - a) / b + 1, mid;
            while(le <= ri)
            {
                mid = (ri + le) / 2;
                if(getsum(mid) <= t * m) le = mid + 1;
                else ri = mid - 1;
            }
            printf("%lld\n", le - 1);
        }
    }
    return 0;
}

C. Tavas and Karafs

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 ≤ rand 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 test(s)

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