Codeforces Round #182 (Div. 1 + Div. 2)
A. Eugeny and Array
- \(r-l+1\)是奇数时,和显然无法为0。
- 奇数的情况需要判断-1和1的个数是否大于等于长度的一半。
B. Eugeny and Play List
- 模拟。
C. Yaroslav and Sequence
- 当负数个数和\(n\)同奇偶时,所有数都可以转化成整数。(随便拿n-1个数和其中一个负数操作,然后再拿另一个负数和这n-1个数再操作即可)。同奇偶,可以转变负数个数。
- 当负数为偶数时,最后都会变成整数。
- 否则,最后剩下一个负数时,与整数的最小值对换,对比看是否会得到更大的和。
D. Yaroslav and Time
- 二分
E. Yaroslav and Algorithm
D. Yaroslav and Divisors
- 序列\(p\)是\(1-n\)的一个排列,则pair对\((q,w)\)满足\(p_q|p_w\)的数量为\(nlogn\)个。
- \((q,w)\)对询问\([l,r]\)产生贡献时,\([l,r]\)包含\((q,w)\),则问题转化成区间覆盖问题,离线后用树状数组维护即可。
Codeforces Round #182 (Div. 1 + Div. 2)的更多相关文章
- Codeforces Round #182 (Div. 1)题解【ABCD】
Codeforces Round #182 (Div. 1)题解 A题:Yaroslav and Sequence1 题意: 给你\(2*n+1\)个元素,你每次可以进行无数种操作,每次操作必须选择其 ...
- Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship
Problem Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec P ...
- Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems(动态规划+矩阵快速幂)
Problem Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems Time Limit: 3000 mSec P ...
- Educational Codeforces Round 43 (Rated for Div. 2)
Educational Codeforces Round 43 (Rated for Div. 2) https://codeforces.com/contest/976 A #include< ...
- Educational Codeforces Round 35 (Rated for Div. 2)
Educational Codeforces Round 35 (Rated for Div. 2) https://codeforces.com/contest/911 A 模拟 #include& ...
- Codeforces Educational Codeforces Round 44 (Rated for Div. 2) F. Isomorphic Strings
Codeforces Educational Codeforces Round 44 (Rated for Div. 2) F. Isomorphic Strings 题目连接: http://cod ...
- Codeforces Educational Codeforces Round 44 (Rated for Div. 2) E. Pencils and Boxes
Codeforces Educational Codeforces Round 44 (Rated for Div. 2) E. Pencils and Boxes 题目连接: http://code ...
- Educational Codeforces Round 63 (Rated for Div. 2) 题解
Educational Codeforces Round 63 (Rated for Div. 2)题解 题目链接 A. Reverse a Substring 给出一个字符串,现在可以对这个字符串进 ...
- Educational Codeforces Round 39 (Rated for Div. 2) G
Educational Codeforces Round 39 (Rated for Div. 2) G 题意: 给一个序列\(a_i(1 <= a_i <= 10^{9}),2 < ...
- Educational Codeforces Round 48 (Rated for Div. 2) CD题解
Educational Codeforces Round 48 (Rated for Div. 2) C. Vasya And The Mushrooms 题目链接:https://codeforce ...
随机推荐
- JS中浏览器兼容问题
body的子节点IE 火狐 显示3个 其他显示7个 float浮动中的兼容 link 操作外联的兼容
- git命令入门
http://www.cocoachina.com/ios/20160629/16855.html 译者序:这是一篇给像我这样的新手或者是熟悉图形工具的老鸟看的.仅作为快速入门的教程. git 现在的 ...
- VLSM(可变长子网掩码)
http://blog.sina.com.cn/s/blog_635e1a9e0100yk51.html(转载) VLSM的介绍: VLSM(VLSM(Variable Length Subnetwo ...
- 【JZOJ4814】【NOIP2016提高A组五校联考2】tree
题目描述 给一棵n 个结点的有根树,结点由1 到n 标号,根结点的标号为1.每个结点上有一个物品,第i 个结点上的物品价值为vi. 你需要从所有结点中选出若干个结点,使得对于任意一个被选中的结点,其到 ...
- SiteMesh:一个优于Apache Tiles的Web页面布局、装饰框架
一.SiteMesh项目简介 OS(OpenSymphony)的SiteMesh是一个用来在JSP中实现页面布局和装饰(layout and decoration)的框架组件,能够帮助网站开发人员较容 ...
- Django-2.2.1版本关于无法使用makemigrations的错误
报错 django.core.exceptions.ImproperlyConfigured: mysqlclient 1.3.13 or newer is required; you have 0. ...
- 安装vagrant&virtualBox
https://blog.csdn.net/dream_188810/article/details/78218073 VirtualBox是一款开源免费的虚拟机软件(之前一直使用vm,vm功能较多, ...
- 【Django入坑之路】基础操作(过滤,继承,跳转)
1:自定过滤器 1创建templatetags文件夹 2在里面创建自定义py文件:固定格式: from django import template from django.utils.safestr ...
- hdu4310 贪心
考虑每次血口的要少 就按照一滴血多少伤害来计算.由于直接相除有小数.考虑x/y > a/b => x*b >y*a; #include<stdio.h> #inclu ...
- sql —— top
用于规定要返回的记录的数目 原表: 我们如果只想看前三个学生信息的话: