B. Radio Station time limit per test2 seconds memory limit per test256 megabytes Problem Dsecription As the guys fried the radio station facilities, the school principal gave them tasks as a punishment. Dustin's task was to add comments to nginx conf…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 用map模拟一下映射就好了. [代码] #include <bits/stdc++.h> using namespace std; int n,m; map<string,string> dic; int main(){ #ifdef LOCAL_DEFINE freopen("rush_in.txt", "r", stdin); #endif ios::sync_with_…

D. MADMAX time limit per test1 second memory limit per test256 megabytes Problem Description As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actual game just to prove that…

A. Eleven time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. El…

题目链接:http://codeforces.com/contest/979/problem/D 参考大神博客:https://www.cnblogs.com/kickit/p/9046953.html: 解题心得: 题目给了你很多条件,具体起来就是输入三个数x,k,s,在数列中找到一个数num,要求:1. GCD(x, num)%k == 0: 2. x + num <= s:3. num异或x最大 刚开始一看数据量这么大,条件这么多怎么搞.其实前面两个条件是用来剪枝的.首先可以开很多个set…

题目链接:http://codeforces.com/contest/979/problem/B 解题心得: 这个题题意就是三个人玩游戏,每个人都有一个相同长度的字符串,一共有n轮游戏,每一轮三个人必须改变自己字符串中的一个字母,最后得分就是字符串中出现字符最多的字母的次数. 感觉这个题从题目描述到做法都像一个脑筋急转弯.主要明白一点,如果一个数要变回自己要怎么变.自己->其他->自己.自己->其他->其他->自己,推几个特例很容易就出来了. #include <bit…

F. Ivan and Burgers 题目链接:https://codeforces.com/contest/1100/problem/F 题意: 给出n个数,然后有多个询问,每次回答询问所给出的区间的异或和最大值. 题解: 考虑离线做法,先把所有的询问区间按照右端点进行排序,然后从1开始逐个将ai插入,插入线性基的同时记录一下位置,最后扫一下,看看哪些的位置是不小于li的即可加入答案. 这种做法在时间复杂度上面是可行的,但是需要注意的是,如果在i这个位置插入最高位为x的线性基,同时在j这个位…

C. Enlarge GCD 题目链接:https://codeforces.com/contest/1047/problem/C 题意: 给出n个数,然后你可以移除一些数.现在要求你移除最少的数,让剩下数的gcd变大. 题解: 首先可以先让所有数都除以他们的gcd,让他们互质,好让问题简单化. 由唯一分解定理,题目中的问题可以转化为:找出最多数都共有的质因子,假设其数目为mx,答案就是n-mx. 上面的想法也是基于贪心,具体做法还是有点技巧,就是在筛素数的时候就进行判断,具体见代码吧: #in…

D. Nature Reserve 题目链接:https://codeforces.com/contest/1059/problem/D 题意: 在二维坐标平面上给出n个数的点,现在要求一个圆,能够容纳所有的点,并且与x轴相切的最小半径为多少. 题解: 容易知道圆的纵坐标的绝对值等于其半径,并且半径越大,容纳圆的可能性越大,那么就考虑二分其半径,这样y0值也确定了. 但x值不是很好求.这里我们找到y=y0的那一条线,然后根据半径以及y0,yi值,可以求出当x0在哪一段时,能够包含(xi,yi)这…

D. MADMAX time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actu…