G1 - Into Blocks (easy version) 参考:Codeforces Round #584 - Dasha Code Championship - Elimination Round (rated, open for everyone, Div. 1 + Div. 2) G1. Into Blocks (easy version) 思路:先把数据预处理一遍,找到每一种数的最右端的位置,和每一种数的出现的次数,然后,从第一个数字开始遍历,用r保存当前这一块的最大右端,用MAX…

题目:https://codeforc.es/contest/1209/problem/G1 题意:给你一个序列,要你进行一些操作后把他变成一个好序列,好序列的定义是,两个相同的数中间的数都要与他相同,可以把某一种数统一变成另一个数,问最少变得个数 思路:我们可以考虑贪心,对于一个互相牵扯的区间,我肯定是用总长减掉里面出现次数最多的元素才是最划得来的,我们直接求出这些牵扯区间的长度即可,方法就是我用数组记录每个元素出现最右的位置,然后从左到右遍历,我保留当前区间往右延伸的最长位置,然后减去出现最…

http://codeforces.com/problemset/problem/1216/E1 E1. Numerical Sequence (easy version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The only difference between the easy and the hard ver…

B. Ping-Pong (Easy Version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to in…

3868 - Earthstone: Easy Version Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3867 Description Earthstone is a famous online card game created by Lizard Entertainment. It is a collectible card g…

题目链接:Pictures with Kittens (easy version) 题意:给定n长度的数字序列ai,求从中选出x个满足任意k长度区间都至少有一个被选到的最大和. 题解:$dp[i][j]$:以i为结尾选择j个数字的最大和. $dp[i][j]=max(dp[i][j],dp[s][j-1]+a[i])$,$s为区间[i-k,i)$. 以i为结尾的最大和可以由i之前k个位置中的其中一个位置选择j-1个,再加上当前位置的ai得到. #include <cstdio> #includ…

Problem UVA12569-Planning mobile robot on Tree (EASY Version) Accept:138  Submit:686 Time Limit: 3000 mSec  Problem Description  Input The first line contains the number of test cases T (T ≤ 340). Each test case begins with four integers n, m, s, t (…

Coffee and Coursework (Easy version) time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The only difference between easy and hard versions is the constraints. Polycarp has to write a coursewor…

1114 ModricWang's FFT EASY VERSION 思路 利用FFT做大整数乘法,实际上是把大整数变成多项式,然后做多项式乘法. 例如,对于\(1234\),改写成\(f(x)=1*x^3+2*x^2+3*x+4\),那么\(x=10\)处的值就是原数.类似的,对于输入的两个大整数,转换为\(f(x)\) 和\(g(x)\) ,利用FFT求出\(h(x)=f(x)*g(x)\) ,此时\(h(10)\) 就是乘积. 代码 #include <cstdio> #include…

06-图2 Saving James Bond - Easy Version (25 分) This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land…

任意门:http://codeforces.com/contest/1118/problem/F1 F1. Tree Cutting (Easy Version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given an undirected tree of nn vertices. Some vert…

任意门:http://codeforces.com/contest/1118/problem/D1 D1. Coffee and Coursework (Easy version) time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The only difference between easy and hard versions…

F1. Pictures with Kittens (easy version) 题目链接:https://codeforces.com/contest/1077/problem/F1 题意: 给出n个数,以及k,x,k即长度为k的区间至少选一个,x的意思是一共要选x个,少一个或者多一个都不行. 选一个会得到一定的奖励,问在满足条件的前提下,最多得到多少的奖励. 题解: 简单版本数据量比较小,考虑比较暴力的动态规划. dp[i,j]表示前i个数,要选第i个数,目前选了j个所得到的最大奖励,那么当…

05-图2. Saving James Bond - Easy Version (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was capture…

06-图2 Saving James Bond - Easy Version(25 分) This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land…

D1.Remove the Substring (easy version) time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output The only difference between easy and hard versions is the length of the string. You are given a string…

B1. Character Swap (Easy Version) This problem is different from the hard version. In this version Ujan makes exactly one exchange. You can hack this problem only if you solve both problems. After struggling and failing many times, Ujan decided to tr…

CF1225B1 TV Subscriptions (Easy Version) 洛谷评测传送门 题目描述 The only difference between easy and hard versions is constraints. The BerTV channel every day broadcasts one episode of one of the kk TV shows. You know the schedule for the next nn days: a seque…

D1. RGB Substring (easy version) time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output The only difference between easy and hard versions is the size of the input. You are given a string s consistin…

D1. Kirk and a Binary String (easy version) 01串找最长不降子序列 给定字符串s,要求生成一个等长字符串t,使得任意l到r位置的最长不降子序列长度一致 从后往前暴力枚举,枚举每个一替换成0后是否改变了l到r位置的最长不降子序列长度 01串的最长不降子序列,可以通过线性dp求解 dp i表示以i结尾的最长不降子序列长度 dp[0]=dp[0]+s[i]=='0'; dp[1]=max(dp[0],dp[1])+s[i]=='1'; #include<bi…

题目链接: C1. Skyscrapers (easy version) 题目描述: 有一行数,使得整个序列满足 先递增在递减(或者只递增,或者只递减) ,每个位置上的数可以改变,但是最大不能超过原来的值. 最后找到满足这样的序列并且满足 这种方案 所有数加起来 和 是最大的. 考察点 : 贪心,对数据范围的掌握程度,计算每次加数时有可能会 爆 int 析题得侃: 比赛的时候看到这道题直接找了 最大值,然后以最大值为中心向两侧递减,交了一发, WA 后来想到可能会有重复的最大值,因为每个值并不是…

E1. String Coloring (easy version) time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output This is an easy version of the problem. The actual problems are different, but the easy version is almost…

Codeforce 1420 C1. Pokémon Army (easy version) 解析(DP) 今天我們來看看CF1420C1 題目連結 題目 對於一個數列\(a\),選若干個數字,求alternating-series的最大值. 前言 C2真的想不到 @copyright petjelinux 版權所有 觀看更多正版原始文章請至petjelinux的blog 想法 \(dp[i][0]\)代表:考慮到第i個數字為止,最後一個數字是負的的最大值 \(dp[i][1]\)代表:考慮到第…

Saving James Bond - Easy Version This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the cente…

给定一个数组,找出最长的子序列,满足 a,a,..a,b,b,..b,a,a,..a 前面的a和后面的a都要是x个,中间的b是y个. 其中,x>=0且y>=0. \(\color{Red}{---------------------华丽分割线w(ﾟДﾟ)w------------------------}\) 看到这数据,就觉得暴力无疑. \(三种情况\) \(Ⅰ.当x=0,也就是只有中间部分,答案就是出现次数最多的那个数字.\) \(Ⅱ.当y=0,也就是只有两边的部分,那就要枚举前半部分区间…

D2. Kirk and a Binary String (hard version) time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output The only difference between easy and hard versions is the length of the string. You can hack this pro…

The only difference between easy and hard versions is a number of elements in the array. You are given an array aa consisting of nn integers. The value of the ii-th element of the array is aiai. You are also given a set of mm segments. The jj-th segm…

传送门easy 传送门hard 切水题的感觉真好 看到数据范围这么小,所以暴力枚举所有的可能,然后用map+vector存下每种值的区间,然后贪心去选 代码: #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<map> using namespace std; void read(int…

The only difference between the easy and the hard versions is constraints. A subsequence is a string that can be derived from another string by deleting some or no symbols without changing the order of the remaining symbols. Characters to be deleted…

The only difference between easy and hard versions is constraints. There are nn kids, each of them is reading a unique book. At the end of any day, the ii-th kid will give his book to the pipi-th kid (in case of i=pii=pi the kid will give his book to…