E.Boxers time limit per test2 seconds memory limit per test256 megabytes inputstdin outputstdout There are…

E. Boxers 给定N个数字,每个数字可以加一或者减一 使得结果集合中不同数字个数最多 贪心 用桶装数 假如相同的数字$i$超过三个,则上面$i+1$,下面$i-1$都可以分一个 如果相同数字$i$只有两个,优先$i-1$ 如果只有一个也要优先$i-1$ #include<bits/stdc++.h> using namespace std; typedef long long ll; #define sc(x) scanf("%I64d",&x); #defi…

题意:给你一组数,每个数都可以进行一次加一减一,问最后最多能有多少不同的数. 题解:我们可以用桶存每个数的次数,然后枚举\([1,150001]\)来求对答案的贡献,然后贪心,这里我们不用担心其他乱七八糟的东西,直接根据桶中的个数来求贡献即可,但是要先选\(i-1\)的情况,因为后面的数取不到\(i-1\),假如我们不选\(i-1\),而是选了\(i\)和\(i+1\),后面的数可能会冲突不能选,而\(i\)和\(i+1\)就没必要考虑选择顺序了,因为无论他们和后面的数冲不冲突,对答案的贡献都是…

bootstrap-validator是一款与bootstrap相结合的表单前端验证模块,官方网址:http://1000hz.github.io/bootstrap-validator/ 下面内容大部分是从该官方网站翻译过来的. 1.要包含的js文件 <script type="text/javascript" src="js/jquery-1.9.1.min.js"></script> <script type="text…

1.text translation Two hundred years ago, boxing matches were very popular in England. At that time/In those days, boxers fought for prize with bare fists. Because of /By reason of/On account of this, they were known as 'prizefighter'.However, boxing…

Lesson1  A puma at large Pumas are large, cat-like animals which are found in America. When reports came into London Zoo that a wild puma had been spotted forty-five miles south of London, they were not taken seriously. However, as the evidence began…

27W 59 Boxing matches were very popular in England two hundred years ago. In those days, boxers fought with bare fists for Prize money. Because of this, they were known as 'prizefighters'. However, boxing was very crude, for there were no rules and a…

lesson 21 Daniel Mendoza bare 赤裸的 :boxers fought with bare fists crude 天然的:crude sugar, crude oil 粗俗的:A crude manner figure: someone who is important or famous in someway draw up 起草 draw up a contract/will/list/plan 表示喜欢的词语 like/be interested in be f…

codeforces #579(div3) A. Circle of Students 题意: 给定一个n个学生的编号,学生编号1~n,如果他们能够在不改变顺序的情况下按编号(无论是正序还是逆序,但不能既有逆序又有正序,例如12354)围成一个圆圈输出YES,否则输出NO,多组输入. 思路:以5为例有三种情况,12345:54321:32154(这种5在中间的可以归为一类):但他们都满足逐个数组中相邻的元素(首尾元素视为相邻)的差的绝对值最多只有1个是大于1的.也可以通过逆推题意假设他们已经围成…

Codeforces Round #579 (Div. 3) 传送门 A. Circle of Students 这题我是直接把正序.逆序的两种放在数组里面直接判断. Code #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 205; int q, n; int a[N], b[N], c[N]; int main() { ios::sync_with_stdio(false);…

思考之后再看题解,是与别人灵魂之间的沟通与碰撞 A. Circle of Students 题意 给出n个数,问它们向左或者向右是否都能成一个环.比如样例5是从1开始向左绕了一圈 [3, 2, 1, 5, 4] 变成 [1, 2, 3, 4, 5]; 思路 我的方法是差分,假如成立,相邻两个数的差的绝对值要么是1要么是n-1. #include<iostream> #include<cstdio> #include<algorithm> #include<cmat…

A. Circle of Students      题目:https://codeforces.com/contest/1203/problem/A 题意:一堆人坐成一个环,问能否按逆时针或者顺时针正好是 1-n的顺序 思路:水题,把数组开两倍,或者标记当前位置都可以 #include<bits/stdc++.h> #define maxn 100005 #define mod 1000000007 using namespace std; typedef long long ll; int…

比赛链接:https://codeforc.es/contest/1203/ A. Circle of Students 题意:\(T\)组询问,每组询问给出\(n\)个数字,问这\(n\)个数字能否围成圆环.(围成圆环指,从某一位开始顺时针或逆时针遍历,数组为\(1, 2, 3, ..., n\)) 分析:把数组复制一份,两个数组首尾相接,正反判定两次即可. AC代码: #include<bits/stdc++.h> #define SIZE 200010 #define rep(i, a,…

什么是函数式编程 函数式编程是java8的一大特色,也就是将函数作为一个参数传递给指定方法.别人传的要么是基本数据类型,要么就是地址引用 ,我们要穿一个“动作”. Stream 说到函数式编程,就不得不提及Stream,Stream跟我们熟知的io流可不是同一个东西,泛指可以顺序执行或者并行执行的元素序列,主要是针对集合,可以将多个函数通过“.”串起来执行,其特点如下: stream不会存储数据,只是将集合流化,比如说 声明一个stream之后,往集合里面扔东西,stream可以取到新扔到集合里…