排序+暴力 #include<bits/stdc++.h> using namespace std; #define int long long #define N 1005000 int arr[N]; signed main(){ int _,n; cin>>_; while(_--){ cin>>n; ;i<=n;i++) cin>>arr[i]; sort(arr+,arr++n); ; ;i<=n;i++){ ; ; ;j--){ if…

Codeforces Round #599 (Div. 2) D. 0-1 MST Description Ujan has a lot of useless stuff in his drawers, a considerable part of which are his math notebooks: it is time to sort them out. This time he found an old dusty graph theory notebook with a descr…

B2. Character Swap (Hard Version) This problem is different from the easy version. In this version Ujan makes at most 2…

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…

This problem is different from the easy version. In this version Ujan makes at most 2n2n swaps. In addition, k≤1000,n≤50k≤1000,n≤50 and it is necessary to print swaps themselves. You can hack this problem if you solve it. But you can hack the previou…

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 try to clean up his house again. He…

久违的写篇博客吧 A. Maximum Square 题目链接:https://codeforces.com/contest/1243/problem/A 题意: 给定n个栅栏,对这n个栅栏进行任意排序,问可形成的最大正方形面积是多少 分析: 水题. 先排个序 , 然后暴力枚举正方形边长就可以了 #include<bits/stdc++.h> #define ios std::ios::sync_with_stdio(false) #define sd(n) scanf("%d&qu…

难题不会啊…… 我感觉写这个的原因就是因为……无聊要给大家翻译题面 A. Maximum Square 简单题意: 有$n$条长为$a_i$,宽为1的木板,现在你可以随便抽几个拼在一起,然后你要从这一大块木板中裁出一块最大的正方形. $1 \leq a_i \leq n \leq 1000$ 多测,$T \leq 10$ 给个官网的图: 直接排序然后扫就行了,这数据范围是不是让你想什么神奇东西了? #include <algorithm> #include <iostream> #…

题:https://codeforces.com/contest/1243/problem/D 分析:找全部可以用边权为0的点连起来的全部块 然后这些块之间相连肯定得通过边权为1的边进行连接 所以答案就是这些块的总数-1: #include<bits/stdc++.h> using namespace std; typedef long long ll; #define pb push_back ; set<int>s,g[M]; int vis[M]; void bfs(int…

这题写起来真的有点麻烦,按照官方题解的写法 先建图,然后求强连通分量,然后判断掉不符合条件的换 最后做dp转移即可 虽然看起来复杂度很高,但是n只有15,所以问题不大 #include <iostream> #include <fstream> #include <vector> #include <set> #include <map> #include <bitset> #include <algorithm> #in…

C. Sum Balance Ujan has a lot of numbers in his boxes. He likes order and balance, so he decided to reorder the numbers. There are…

D. 0-1 MST Ujan has a lot of useless stuff in his drawers, a considerable part of which are his math notebooks: it is time to sort them out. This time he found an old dusty graph theory notebook with a description of a graph. It is an undirected weig…

C. Tile Painting Ujan has been lazy lately, but now has decided to bring his yard to good shape. First, he decided to paint the path from his house to the gate. The path consists of…

A. Maximum Square Ujan decided to make a new wooden roof for the house. He has…

Ujan has been lazy lately, but now has decided to bring his yard to good shape. First, he decided to paint the path from his house to the gate. The path consists of nn consecutive tiles, numbered from 11 to nn. Ujan will paint each tile in some color…

Ujan decided to make a new wooden roof for the house. He has nn rectangular planks numbered from 11 to nn. The ii-th plank has size ai×1ai×1 (that is, the width is 11 and the height is aiai). Now, Ujan wants to make a square roof. He will first choos…

题意:就是给你一个n,然后如果  n mod | i - j | == 0  并且 | i - j |>1 的话,那么i 和 j 就是同一种颜色,问你最大有多少种颜色? 思路: 比赛的时候,看到直接手推,发现有点东西,直接打表找出了规律 —— 如果 n的质因子只有一个,那么总数就是 那个 质因子.其它都为 1. 今天上课的时候无聊,还是试着推了一下原理. 1.如果一个数只有一个质因子 x ,那么  n-x .n-2x.n-3x ……等等全为一种颜色,也就是说每隔 x个就是同种颜色,这样的话就是有…

https://codeforces.com/contest/1234/problem/A A. Equalize Prices Again #include<bits/stdc++.h> using namespace std; typedef long long ll; int main(){ int n,a; int t; cin>>t; ll sum = ,ans; while(t--){ cin>>n;sum = ; ;i < n;++i){ cin&g…

Codeforces Round #529(Div.3)个人题解 前言: 闲来无事补了前天的cf,想着最近刷题有点点怠惰,就直接一场cf一场cf的刷算了,以后的题解也都会以每场的形式写出来 A. Repeating Cipher 传送门 题意:第一个字母写一次,第二个字母写两次,依次递推,求原字符串是什么 题解:1.2.3.4,非常明显的d=1的等差数列,所以预处理一个等差数列直接取等差数列的每一项即可 代码: #include<bits/stdc++.h> using namespace s…

半个月没看cf 手生了很多(手动大哭) Problem - A - Codeforces 题意 给定数字n, 求出最大数字k, 使得  n & (n−1) & (n−2) & (n−3) & ... (k) = 0 题解 &:有0则0 假如n转为二进制共有x位 , 要使&后=0, k的最高位需=0 我们使最高位=0,后面都为1; 那么此数+1=什么 =100000(x-1个0), 这样就能&后=0了 -------------------------…

题目链接:Codeforces Round #298 (Div. 2) A. Exam An exam for n students will take place in a long and narrow room, so the students will sit in a line in some order. The teacher suspects that students with adjacent numbers (i and i + 1) always studied side…

Codeforces Round #272 (Div. 2) A. Dreamoon and Stairs time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Dreamoon wants to climb up a stair of n steps. He can climb 1 or 2 steps at each move.…

Codeforces Round #578 (Div. 2) 传送门 A. Hotelier 暴力即可. Code #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 2e5 + 5; int n; char s[N]; int res[10]; int main() { ios::sync_with_stdio(false); cin.tie(0); cin >> n;…

Codeforces Round #590 (Div. 3) Editorial 题目链接 官方题解 不要因为走得太远,就忘记为什么出发! Problem A 题目大意:商店有n件商品,每件商品有不同的价格,找出一个最小的可能值price,使得price * n >= sum,sum指的是原来商品价格的总和. 知识点:模拟 思路:求出sum/n向上取整即可,有两种方法.一是使用ceil()函数,但注意ceil()返回的是向上取整后的浮点数,所以要进行强制类型转换:二是直接向下取整,然后用if语句…

Codeforces Round #572 (Div. 1) A2 题意:给一棵树,带边权,互不相同且为偶数.每次操作是选两个叶子然后在路径上同时加一个数.初始边权全是0,求一个操作序列使得最终边权与给定边权相同. \(n \le 1000\) ,构造的序列长度不超过 \(10^5\) key:构造 首先,如果有二度点,那么一定无解. 所以现在每个点要不是叶子,要不就至少是三度.考虑一条边,我们只把它的边权改变,而不改变其他边.此时需要讨论这条边两段的点的度数. 如果两个点都是大于等于三度的,那…

Codeforces Round #789 (Div. 2) A-C A 题目 https://codeforces.com/problemset/problem/1677/A 题解 思路 知识点:模拟. (比较显然,不写了) 时间复杂度 \(O(nlogn)\) 空间复杂度 \(O(n)\) 代码 #include <bits/stdc++.h> using namespace std; int a[100]; int main(){ std::ios::sync_with_stdio(0)…

Codeforces Round #734 (Div. 3) 20210920.网址:https://codeforces.com/contest/1551. 编程细节:下标定义不要一会[1,n]一会[0,n)啊- A 用价值为1的硬币a枚+价值为2的硬币b枚,买价值为n的东西,希望ab之差的绝对值尽量小.大水题. B1 给我们一个字符串s,用红色和绿色给s上色,规则如下: 一个字母或者不上色,或者被涂成一种颜色,即不能又红又绿. 被涂成一种颜色的所有字母,必须两两不相同. 红色字母数量=绿色字…

Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/output 1 s, 256 MB  x2377 B Queue standard input/output 2 s, 256 MB  x1250 C Hacking Cypher standard input/output 1 s, 256 MB  x740 D Chocolate standard in…

Codeforces Round #116 (Div. 2, ACM-ICPC Rules) 代码 Codeforces Round #116 (Div. 2, ACM-ICPC Rules) A. Defragmentation 按颜色分配\(1\)到\(\sum_{i=1}^{m}{n_i}\)位置,那么可以得到\(t_i\)表示位置\(i\)最后占据位置\(t_i\),-1表示该位置为空. 那么会出现两种情况: \(t_1 \to t_2 \to \cdots \to \ {-1}\),即…

Codeforces Round #531 (Div. 3) 题目总链接:https://codeforces.com/contest/1102 A. Integer Sequence Dividing 题意: 给一个数n,然后要求你把1,2.....n分为两个集合,使得两个集合里面元素的和的差的绝对值最小. 题解: 分析可以发现,当n%4==0 或者 n%3==0,答案为0:其余答案为1.之后输出一下就好了. 代码如下: #include <bits/stdc++.h> using name…