显然将扩张按从大到小排序之后,只有不超过前34个有效. d[i][j]表示使用前i个扩张,当length为j时,所能得到的最大的width是多少. 然后用二重循环更新即可, d[i][j*A[i]]=max(d[i][j*A[i]],d[i-1][j]); d[i][j]=max(d[i][j],d[i-1][j]*A[i]); 当某次更新时,检验其符合了答案的条件,就输出. 显然可以用滚动数组优化到空间为线性. 注意爆int的问题. 此外,瞎几把搜+花式剪枝也能过. #include<cstd…

D. Field expansion time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output In one of the games Arkady is fond of the game process happens on a rectangular field. In the game process Arkady can buy…

题目传送门:http://codeforces.com/contest/799/problem/C C. Fountains time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output   Arkady plays Gardenscapes a lot. Arkady wants to build two new fountains. T…

C. Fountains time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Arkady plays Gardenscapes a lot. Arkady wants to build two new fountains. There are n available fountains, for each fountain it…

A. Carrot Cakes time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output In some game by Playrix it takes t minutes for an oven to bake k carrot cakes, all cakes are ready at the same moment t minut…

题目链接:http://codeforces.com/contest/799/problem/C 题目: 题意: 给你n种喷泉的价格和漂亮值,这n种喷泉题目指定用钻石或现金支付(分别用D和C表示),C和D之间不能相互转换.你现在需要修建两个喷泉,给你硬币数和现金数,问你怎样才能使修建的两个喷泉的总漂亮值最大. 思路: 易知,要修建的两个喷泉如果一个是用钻石支付,另一个用现金支付,那么只需找到小于给的钻石和现金的上限的漂亮值最大的两个温泉相加,此处遍历一边即可.对于这两个温泉都用同一种支付方式的情…

我从来没想过自己可以被支配的这么惨,大神讲这个场不容易掉分的啊 A. Carrot Cakes time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output In some game by Playrix it takes t minutes for an oven to bake k carrot cakes, all cakes are…

E - Aquarium decoration 枚举两个人都喜欢的个数,就能得到单个喜欢的个数,然后用平衡树维护前k大的和. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define PLI pair<LL, int> #define ull unsigned…

分几种情况讨论: (1)仅用C或D买两个 ①买两个代价相同的(实际不同)(排个序) ②买两个代价不同的(因为买两个代价相同的情况已经考虑过了,所以此时对于同一个代价,只需要保存美丽度最高的喷泉即可)(预处理b[i],表示代价小于等于i的物品中,美丽度最大的是多少.为了防止重复购买,枚举其中一个,然后另一个只买代价小于其代价的物品.) (2)用C和D各买一个 按这几种情况分类,可以比较好地避免买到同一个喷泉的情况. #include<cstdio> #include<algorithm&g…

http://codeforces.com/contest/799/problem/C 题意: 有n做花园,有人有c个硬币,d个钻石 (2 ≤ n ≤ 100 000, 0 ≤ c, d ≤ 100 000) ,每一个花园用三个维度描述(a,b,c),分别是美丽度,所花钱币个数,钱币种类,当然,钱币之间不能兑换,该人必须要建筑两座花园,如果可以,输出两座花园总的美丽度,否则输出0: 思路: 首先想到分三种情况讨论,两个花园都用硬币,两个花园都用钻石,一个用钻石一个用硬币. 大神的代码真的是很厉害…

2021.12.10 P2516 [HAOI2010]最长公共子序列(动态规划+滚动数组) https://www.luogu.com.cn/problem/P2516 题意: 给定字符串 \(S\) . \(T\) ,都以 \(.\) 结尾,求 \(S\) . \(T\) 最长公共子序列的长度及个数. 分析: 一顿操作猛如虎,一看分数250--爆零了.原本就没准备拿几分,结果令人心塞. 第一问就是求最长公共子序列长度,数据范围比较小, \(O(n^2)\) 就行,上来就是一顿树状数组+LIS,…

Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) 全场题解 菜鸡只会A+B+C,呈上题解: A. Bear and Big Brother 题意:我也没看太清,就是给你两个10以内的数a,b.a每天乘以3,b每天乘以2,求多少天后a大于b. 思路:应该是有公式的,不过看到数据这么小直接暴力乘求解.官方题解貌似就是这样,数据小就是水题. const int N=1e3+10; int main() { int a…

VK Cup 2015 - Round 1 (unofficial online mirror, Div. 1 only)E. The Art of Dealing with ATM Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  Solved: 2xx 题目连接 http://codeforces.com/contest/529/problem/E Description ATMs of a well-known bank of a sm…

Codeforces Round #504 (rated, Div. 1 + Div. 2, based on VK Cup 2018 Final) A. Single Wildcard Pattern Matching 题意就是匹配字符的题目,打比赛的时候没有看到只有一个" * ",然后就写挫了,被hack了,被hack的点就是判一下只有一个" * ". 代码: //A #include<iostream> #include<cstdio>…

Problem   Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems Time Limit: 3000 mSec Problem Description Input The input contains a single line consisting of 2 integers N and M (1≤N≤10^18, 2≤M≤100). Output Print one integer, the total n…

Educational Codeforces Round 129 (Rated for Div. 2) A-D A 题目 https://codeforces.com/contest/1681/problem/A 题解 思路 知识点:贪心. 先手的一方拥有大于等于对方最大牌的牌即可获胜,所以考虑取两组牌各自的最大值进行比较. 时间复杂度 \(O(n)\) 空间复杂度 \(O(1)\) 代码 #include <bits/stdc++.h> #define ll long long using…

Educational Codeforces Round 43 (Rated for Div. 2) https://codeforces.com/contest/976 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define IT set<ll>::iterator #define sqr(x)…

Educational Codeforces Round 35 (Rated for Div. 2) https://codeforces.com/contest/911 A 模拟 #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define IT set<ll>::iterator #define sqr(…

Codeforces Round #539 (Div. 1) A. Sasha and a Bit of Relax description 给一个序列\(a_i\),求有多少长度为偶数的区间\([l,r]\)满足\([l,mid]\)的异或和等于\([mid+1,r]\)的异或和. solution 等价于询问有多少长度为偶数的区间异或和为\(0\). 只需要两个位置的异或前缀和与下标奇偶性相同即可组成一个合法区间. #include<cstdio> #include<algorith…

Educational Codeforces Round 59 (Rated for Div. 2) D. Compression 题目链接:https://codeforces.com/contest/1107/problem/D 题意: 给出一个n*(n/4)的矩阵,这个矩阵原本是一些01矩阵,但是现在四个四个储存进二进制里面,现在给出的矩阵为0~9以及A~F,表示0~15. 然后问这个矩阵能否压缩为一个(n/x)*(n/x)的矩阵,满足原矩阵中大小为x*x的子矩阵所有数都相等(所有子矩阵构…

Educational Codeforces Round 65 (Rated for Div. 2)题解 题目链接 A. Telephone Number 水题,代码如下: Code #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 2e5 + 5; int a[N] ; int n, T; char s[N] ; int main() { cin >> T; whil…

Educational Codeforces Round 64 (Rated for Div. 2)题解 题目链接 A. Inscribed Figures 水题,但是坑了很多人.需要注意以下就是正方形.圆以及三角形的情况,它们在上面的顶点是重合的. 其余的参照样例判断一下就好了了.具体证明我也不会 代码如下: Code #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 2e5 +…

Educational Codeforces Round 37 (Rated for Div. 2)C. Swap Adjacent Elements time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have an array a consisting of n integers. Each integer from 1…

[Educational Codeforces Round 81 (Rated for Div. 2)]E. Permutation Separation(线段树,思维,前缀和) E. Permutation Separation time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given a permutat…

\(Educational\ Codeforces\ Round\ 85\ (Rated\ for\ Div.2)\) \(A. Level Statistics\) 每天都可能会有人玩游戏,同时一部分人会过关,玩游戏的人数和过关的人数会每天更新,问记录的数据是否没有矛盾 //#pragma GCC optimize("O3") //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<…

Problem   Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec Problem Description Input Output The only line should contain the minimal number of days required for the ship to reach the point (x2,y2)(x2,y2). If it…

Codeforces Educational Codeforces Round 44 (Rated for Div. 2) F. Isomorphic Strings 题目连接: http://codeforces.com/contest/985/problem/F Description You are given a string s of length n consisting of lowercase English letters. For two given strings s an…

Codeforces Educational Codeforces Round 44 (Rated for Div. 2) E. Pencils and Boxes 题目连接: http://codeforces.com/contest/985/problem/E Description Mishka received a gift of multicolored pencils for his birthday! Unfortunately he lives in a monochrome w…

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行一次翻转,问是否存在一种方案,可以使得翻转后字符串的字典序可以变小.   这个很简单,贪心下就行了. 代码如下: Code #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 3e5…

Educational Codeforces Round 39 (Rated for Div. 2) G 题意: 给一个序列\(a_i(1 <= a_i <= 10^{9}),2 <= n <= 200000\), 如果至多删除其中的一个数之后该序列为严格上升序列,那么称原序列为几乎严格上升序列. 现在每次将序列中的任意数字变成任意数字,问最少要操作几次才能将序列变成几乎严格上升子序列. 思路: 如果不考虑删除,求让整个序列都变成严格上升子序列的次数 求出\(序列a_i - i\)…