Code Forces 645C Enduring Exodus】的更多相关文章

C. Enduring Exodus time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other…
题目链接 我们将所有为0的位置的下标存起来. 然后我们枚举左端点i, 那么i+k就是右端点. 然后我们三分John的位置, 找到下标为i时的最小值. 复杂度 $ O(nlogn) $ #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <m…
枚举,三分. 首先,这$n+1$个人一定是连续的放在一起的.可以枚举每一个起点$L$,然后就是在$[L,R]$中找到一个位置$p$,使得$p4最优,因为越往两边靠,距离就越大,在中间某位置取到最优解,所以三分一下就可以了. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include…
题目链接: http://codeforces.com/contest/645/problem/C 题意: 给定01串,将k头牛和农夫放进, 0表示可以放进,1表示不可放进,求农夫距离其牛的最大距离的最小值. 分析: 第一遍读题没看清,直接写成dp...然后样例都不过,我开始怀疑人生怀疑自己..... 后来发现是要求中间的到两边的最大距离的最小值,而对于某个距离是否满足条件很好判断啊~~所以直接二分最大距离即可~ 代码: #include<iostream> #include<cstri…
思维题--code forces round# 551 div.2 题目 D. Serval and Rooted Tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Now Serval is a junior high school student in Japari Middle School, and he is…
C. Enduring Exodus 题目连接: http://www.codeforces.com/contest/655/problem/C Description In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he…
题目链接: C. Enduring Exodus time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to…
Code Forces 796C Bank Hacking 题目大意 给一棵树,有\(n\)个点,\(n-1\)条边,现在让你决策出一个点作为起点,去掉这个点,然后这个点连接的所有点权值+=1,然后再将这些刚刚加过一的点的相邻的点的权值+=1 也就是说,除了与根节点相邻的点+=1,其余点+=2 然后求最大集合的最小点权 solution 一看是要求最大的的最小值,首先想到的就是二分,显然想到这个目前没有什么卵用,二分是用来卡最佳答案的,所以使用二分的前提是要给它一个范围去选择,那么现在的任务就是…
Code Forces 833 A The Meaningless Game 题目大意 有两个人玩游戏,每轮给出一个自然数k,赢得人乘k^2,输得人乘k,给出最后两个人的分数,问两个人能否达到这个分数 不得不吐槽一下那么长的英文题面翻译完只有一句话-- solution 也很好想叭 乘积开立方判断是否为两个数的因数 如果是的话,显然不成立 否则输出Yes即可 #include <iostream> #include <cstring> #include <cstdio>…
题目描述 Programmers working on a large project have just received a task to write exactly mm lines of code. There are nn programmers working on a project, the ii -th of them makes exactly a_{i}ai​ bugs in every line of code that he writes. Let's call a…