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
#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 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?≤?rand
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.
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