23333,在扒了一天题解之后发现我竟然还能秒题,虽然这是个pj的sb题... (排个序,然后upper_bound和lower_bound一用就行了(是不是有O(1)的查询方法啊??貌似要离散啊,一样蛋疼..)) #include <bits/stdc++.h> #define LL long long #define inf 1e60 using namespace std; inline int ra() { ,f=; char ch=getchar(); ; ch=getchar();…

原题链接https://www.lydsy.com/JudgeOnline/problem.php?id=4747 先将原数组排序,然后二分查找即可.时间复杂度\(O((N+Q)logN)\). #include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #define maxn 100010 using namespace std; inline int read(){…

[题目链接] https://www.lydsy.com/JudgeOnline/problem.php?id=4392 [算法] 线段树 时间复杂度 : O(MlogN) [代码] #include<bits/stdc++.h> using namespace std; #define MAXN 200010 typedef long long LL; int n , m; LL a[MAXN]; struct SegmentTree { struct Node { int l , r; L…

4742: [Usaco2016 Dec]Team Building Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 21  Solved: 16[Submit][Status][Discuss] Description Every year, Farmer John brings his NN cows to compete for "best in show" at the state fair. His arch -rival, F…

Counting Haybales 题目描述 Farmer John is trying to hire contractors to help rearrange his farm, but so far all of them have quit when they saw the complicated sequence of instructions FJ wanted them to follow. Left to complete the project by himself, he…

Counting Haybales 时间限制: 50 Sec  内存限制: 256 MB提交: 52  解决: 18[提交][状态][讨论版] 题目描述 Farmer John is trying to hire contractors to help rearrange his farm, but so far all of them have quit when they saw the complicated sequence of instructions FJ wanted them…

P3184 [USACO16DEC]Counting Haybales数草垛 题目描述 Farmer John has just arranged his NN haybales (1 \leq N \leq 100,0001≤N≤100,000 ) at various points along the one-dimensional road running across his farm. To make sure they are spaced out appropriately, pl…

如果我们将两个人拥有的牛混在一起,并按照战斗力从小到大排序,同时把第一个人选的牛看成$)$,第二个人选的牛看成$($的话,那么我们会发现一个合法的方案对应了一个长度为$2k$的括号序列. 于是DP即可,$f[i][j][k]$表示考虑了前$i$头牛,目前选了$j$个左括号,括号序列的前缀和为$k$的方案数. 时间复杂度$O(nk^2)$. #include<cstdio> #include<algorithm> const int N=2010,M=13,P=1000000009;…

原题链接https://www.lydsy.com/JudgeOnline/problem.php?id=4749 可以对于每个点\(i\),往跟\(i\)距离小于等于\(p[i]\)的点\(j\)都连一条边,表示\(i\)可以传输到\(j\).然后由于图的规模最多\(N^2=40000\),我们可以直接从每个点出发遍历一遍算出可以到达多少点. 建图复杂度\(O(n^2)\),搜索复杂度O(N^3). #include<iostream> #include<cstring> #in…

Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem 10983 18765 Y 1036 [ZJOI2008]树的统计Count 5293 13132 Y 1588 [HNOI2002]营业额统计 5056 13607 1001 [BeiJing2006]狼抓兔子 4526 18386 Y 2002 [Hnoi2010]Bounce 弹飞绵羊 43…