思路 观察到答案一定是连续的一段下凸函数或者上凸函数 直接模拟找出即可 时间复杂度为\(O(n)\) 代码 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int n,a[200200],times[200200],minval=0x3f3f3f3f,maxval=0,ansbegin,anslast,ansnum,beginx,last,num,belong;…

F. Maximum Balanced Circle 题目链接 题意 给出\(n\)个数,现在要从中选出最多的数\(b_i,b_{i+1},\cdots,b_k\),将这些数连成一个环,要求两两相邻的数相差不超过1. 最后要求输出具体的方案. 题解 一开始想了一个dp,似乎也可以做 这个题也不用这么复杂,因为相差绝对值不超过1,直接统计一下每个数的个数就行了. 因为如果将最后的环给展开,以每个数的值为高,呈现出来的图形一定是先上升后下降的.那么中间部分的数的个数一定大于等于2,最左边和最右边的两…

题目大意:给定一个长度为 N 的序列,求是否能够从序列中选出一个集合,使得这个集合按照特定的顺序排成一个环后,环上相邻的点之间的权值差的绝对值不超过 1. 题解:集合问题与序列顺序无关,因此可以先将序列排序. 可以发现,题目中描述的环,拆成序列之后应该满足 \(a_l,a_{l+1},...,a_{r},a_{r-1},...,a_{l+1}\) 的形态,即:除了 \(a_l,a_r\) 之外的其他所有值应该都有至少两个.因此,开一个桶记录一下每个元素出现的次数,并对原序列进行去重.可知,对于满…

题目: https://codeforces.com/contest/1157/problem/F 给出一个序列 , 我们要从序列里面挑出一些数构造成一个相邻元素之间绝对值为小于1的最大环 , 挑选的数不要求连续 分析: 不要求连续 , 我们可以先排个小序 对于一个满足条件的环我们可以这样的构造 : al , al+1 , al+2 .... ar , ar-1 , ar-2 ,...al+1 因为这样肯定是一个满足条件的结构 , 即:除了 al,aral,ar 之外的其他所有值应该都有至少两个…

不得不说这场div3是真的出的好,算得上是从我开始打开始最有趣的一场div3.因为自己的号全都蓝了,然后就把不经常打比赛的dreagonm的号借来打这场,然后...比赛结束rank11(帮dreagonm上蓝果然没有食言qwq). (震惊...HA省A队CF青名...) upd:system test之后的最终排名是rank9,dreagonmTQL!!! CF1157A Reachable Numbers 水题,分析一下不难发现不超过\(10\)次就会少一位,然后\(10^9\)范围内的数可以…

对没错下面的代码全部是python 3(除了E的那个multiset) 题目链接:https://codeforces.com/contest/1157 A. Reachable Numbers 按位算贡献,一位数的贡献直接算即可 n=int(input()) ans=0 while (n>=10): tmp=n%10 tmp=10-tmp ans+=tmp n+=tmp while (n>0) and (n%10==0): n//=10 ans+=9 print(ans) B. Long N…

c2:Increasing Subsequence (hard version) 那边小取那边,然后相等比较后面的长度 #include<bits/stdc++.h> using namespace std; #define maxn 500005 int a[maxn]; int main(){ ,in; scanf("%d",&n); ;j<n;j++){ scanf("%d",&a[j]); if(mx<a[j]){ m…

Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 比较左子树和右子树的深度,取小的那个返回上来,并+1. 需要注意的是如果没有左子树或右子树.那么无条件取存在的那一边子树深…

Minimum Depth of Binary Tree OJ: https://oj.leetcode.com/problems/minimum-depth-of-binary-tree/ Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf…

Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. 求二叉树的最大深度问题用到深度优先搜索DFS,递归的完美应用,跟求二叉树的最小深度问题原理相同.代码如下: C++ 解法一: class Solution { public: in…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Circle and Points Time Limit: 5000MS   Memory Limit: 30000K Total Submissions: 6850   Accepted: 2443 Case Time Limit: 2000MS Description You are given N points in the xy-plane. You have a cir…

题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1338 1338: Pku1981 Circle and Points单位圆覆盖 Time Limit: 3 Sec  Memory Limit: 162 MBSubmit: 190  Solved: 79[Submit][Status][Discuss] Description You are given N points in the xy-plane. You have a circ…

题目: You are a coach at your local university. There are nn students under your supervision, the programming skill of the ii-th student is aiai. You have to form kk teams for yet another new programming competition. As you know, the more students are…

104. Maximum Depth of Binary Tree -- Easy 方法 使用递归 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int…

124. Binary Tree Maximum Path Sum https://www.cnblogs.com/grandyang/p/4280120.html 如果你要计算加上当前节点的最大path和,这个节点的左右子树必定是纯左右树(即没有拐点), 用另一个参数保留整个二叉树的最大path和,然后计算每一个以当前节点为拐点的路径和并且不断更新整个二叉树的最大值 函数的返回值是纯左右子树的最大path,没有拐点 这个题目给root定位为非空,所以直接这样写可以.如果root为空,这样写就会…

Balanced Sequence Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of this type is said to be balanced:…

地址:http://poj.org/problem?id=1981 题目: Circle and Points Time Limit: 5000MS   Memory Limit: 30000K Total Submissions: 8198   Accepted: 2924 Case Time Limit: 2000MS Description You are given N points in the xy-plane. You have a circle of radius one and…

The improved Quicksort method of the present invention utilizes two pointers initialized at opposite ends of the array or partition to be sorted and an initial partition value Pvalue located at the center of the array or partition. The value at each…

RMQ (Range Minimum/Maximum Query)问题是指:对于长度为n的数列A,回答若干询问RMQ(A,i,j)(i,j<=n),返回数列A中下标在i,j里的最小(大)值,也就是说.RMQ问题是指求区间最值的问题. id=10244" target="_blank" style="color:blue; text-decoration:none">Balanced Lineup Time Limit: 5000MS   Mem…

Problem Description Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of this type is said to be balanced: + if it is the empty string+ if A and B are balanced, AB is balanced,+ if A is balanced, (A) is balanced. Chiaki can reorder…

A balanced binary tree is something that is used very commonly in analysis of computer science algorithms. In this lesson we cover how to determine the maximum number of items it can accommodate. We follow this with a discussion on the maximum height…

Balanced Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1320    Accepted Submission(s): 316 Problem Description Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of t…

Balanced Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6207    Accepted Submission(s): 1616 Problem Description Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of…

题目传送门 题目大意:给出n个字符串,定义了平衡字符串,问这些字符串组合之后,最长的平衡字符子序列的长度. 思路: 首先肯定要把所有字符串先处理成全是不合法的,记录右括号的数量为a,左括号的数量为b,考虑两个字符串,这两个如果min(a1,b2)小于min(a2,b1)时,第一个字符串是不是应该排在前面呢?如果两个相同的话,当然应该吧右括号比较多的放在前面,左括号比较多的放在后面. #include<iostream> #include<cstdio> #include<cm…

先写一道水题的博客,为后面要写的博客做一个铺垫. ヾ(◍°∇°◍)ﾉﾞ RMQ(Range Minimum/Maximum Query),即区间最值查询,对于长度为n的数列A,回答若干询问RMQ(A,i,j)(i,j<=n),返回数列A中下标在i,j之间的最小/大值. 时间复杂度: 1.朴素(即搜索),O(n)-O(qn) online. 2.线段树,O(n)-O(qlogn) online. 3.ST(实质是动态规划),O(nlogn)-O(q) online. ST算法(Sparse Tab…

Circle and Points Time Limit: 5000MS   Memory Limit: 30000K Total Submissions: 8131   Accepted: 2899 Case Time Limit: 2000MS Description You are given N points in the xy-plane. You have a circle of radius one and move it on the xy-plane, so as to enc…

题目链接: Poj 3264 Balanced Lineup 题目描述: 给出一个n个数的序列,有q个查询,每次查询区间[l, r]内的最大值与最小值的绝对值. 解题思路: 很模板的RMQ模板题,在这里总结一下RMQ:RMQ(Range Minimum/Maximum Query) 即区间最值查询,是指这样一个问题:对于长度为n的数列A,回答若干询问RMQ(A,i,j)(i,j<=n),返回数列A中下标在i,j之间的最小/大值. RMQ有三种求法:1:直接遍历查找,炒鸡暴力: 2:线段树也可以解…

http://codeforces.com/contest/1133/problem/Ctime limit per test 2 secondsmemory limit per test 256 megabytesinputstandard inputoutputstandard output You are a coach at your local university. There are $n$ students under your supervision, the programm…

不想用treap和Splay,那就用SB树把,哈哈,其实它一点也SB,厉害着呢. 先膜拜一下作者陈启峰.Orz 以下内容由我搜集整理得来. 一.BST及其局限性 二叉查找树(Binary Search Tree),也称有序二叉树(ordered binary tree),排序二叉树(sorted binary tree),是指一棵空树或者具有下列性质的二叉树: 1.若任意节点的左子树不空,则左子树上所有结点的值均小于它的根结点的值: 2.任意节点的右子树不空,则右子树上所有结点的值均大于它的根结…

Balanced Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6311    Accepted Submission(s): 1648 Problem Description Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of…