easy版本可以用O(n2)的算法,我们可以枚举左右端点 假设一段的最优排序方法如图 任意长度的一段序列排序可能是排序多个子序列 所以我们需要讨论加上一个点的情况 一号点比之前所有点大,所以不需要排序 二号点在第二段中间,所以需要和第二段放在一起排序,代价+1 三号点和第二段一起排序,代价+1,同上一情况 观察...
解:首先a最大值大于b[1]时无解。一开始想的是贪心,对于每一段a[i...j],如果其max(a[j...n])>=b[k+1]且sum(a[j...n])<=b[k],那么算一次贡献,否则寻找最合适的k。然后WA test3了,想了一下是因为可以选择更大的b[k],使得一次删除的数更多,从而贡献更少。 那么只能dp了。令dp[i][j]...
个数已经在set里了,因此如果k很小,那么插入操作次数就会很多,最后遍历的复杂度不会很高 ,因为 但是存在 AI检测代码解析 #include <bits/stdc++.h> using namespace std; typedef long long ll; ll i,j,k,n,m,t; int main(){ ios::sync_with_stdio(0); cin.tie(0); map<ll,ll> f,mp; mp[0]...
Speed & PTO: Set the vehicle’s highest speed and power.Clutch Reset, Trans Adaptation, Airbag Reset, VIN, FRM Reset, Transport Mode, Rain/Light Sensor, ABSSRS, Cylinder, VGT Learn, A/F Setting, Automatic Start/Stop, Electronic Water Pump Start, HV Battery, ACC, AC... Several Car ...
Codeforces Round #738 (Div. 2) D1. Mocha and Diana (Easy Version),传送门题意:给你两个相同数量点的森林,问你最多能在两个森林中加多少相同的e>#include&
Ideal for pressure or vacuum Easy setpoint adjustment NEMA 4 (Housed Models) Up to 3 setpoints available in one switch Applications Pump & compressor m o nit o ring Engine monitoring Machine tools Hydraulic power units Medical equipment Accuracy: ± 0.5% of the adjustable range Switch:...
Whether you're building a tiny arduino board or looking to replace an existing node, this module is a reliable choice. **Ease of Use and Expansion** Getting started with the NodeMcu Mini D1 Module is a breeze. The module comes with a CH340G driver, which is easy to set up. For ...
work. But for patients — including me — who suffer from diseases that aren’t among the major league targets, it can be hard to find much interest among a set of companies that have to think big in order to survive. These days, with IPOs still out of favor, the blockbuster strategy...
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include#include<set>#include<vector>#include<iomanip>#defineALL(x) (x).begin(), (x).end()#definesz(a) int(a.size())#defineall(a) a.begin(), a.end()#definerep(i,...
D1. Prefix-Suffix Palindrome (Easy version) D2. Prefix-Suffix Palindrome (Hard version) 题意: 对于给出的字符串,可截取其前缀和后缀,求能组成的最长回文串。 思路: 正常来说暴力的思路是先匹配前缀pre和后缀suf,找到第一个不匹配的l和r,然后在由l开始从左向右求最长的回文串palindrome,以及由r开始从右...