Hard Process(二分)】的更多相关文章

A Task Process Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1332    Accepted Submission(s): 656 Problem Description There are two kinds of tasks, namely A and B. There are N workers and the…
Hard Process Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 660C Description You are given an array a with n elements. Each element of a is either 0 or 1. Let's denote the length of the…
C. Hard Process 题目连接: http://www.codeforces.com/contest/660/problem/C Description You are given an array a with n elements. Each element of a is either 0 or 1. Let's denote the length of the longest subsegment of consecutive elements in a, consisting…
题目链接: C. Hard Process time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given an array a with n elements. Each element of a is either 0 or 1. Let's denote the length of the longest su…
题目链接:http://codeforces.com/problemset/problem/660/C 题意: 给你一个长度为 $n$ 的 $01$ 串 $a$,记 $f(a)$ 表示其中最长的一段连续 $1$ 的长度. 现在你最多可以将串中的 $k$ 个 $0$ 变成 $1$,求操作后的 $f(a)$. 题解: (说实话这道题不看tag我不一定能想得出来……) 首先考虑二分枚举答案,对于一个假定的 $f(a)=x$,我们需要判断能不能满足: 用 $dp[i]$ 表示 $a[i-x+1], \c…
链接 二分时间,在时间内dp[i][j]表示截止到第i个人已经做了j个A最多还能做多少个B #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> #include<cmath> #include<queue> #include<set>…
HDU 3433 A Task Process Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1368    Accepted Submission(s): 684 Problem Description There are two kinds of tasks, namely A and B. There are N workers…
题意: 有n个员工,每个员工完成一件A任务和一件B任务的时间给出,问要完成x件A任务y件B任务所需的最短时间是多少 思路: DP + 二分我也是第一次见到,这个我只能说太难想了,根本想不到. dp[i][j]表示在t时间内前i个人完成j件A任务后所能完成B任务的最大数量. 代码中还有一些注释. //#define LOCAL #include <iostream> #include <cstdio> #include <cstring> using namespace…
题目链接: http://codeforces.com/contest/660/problem/C 题意: 将最多k个0变成1,使得连续的1的个数最大 题解: 二分连续的1的个数x.用前缀和判断区间[i,i+x-1]里面0的个数是否小于等于k. 代码: #include<iostream> #include<cstdio> #include<vector> #include<cstring> using namespace std; ; int n,k; i…
题目链接 题意:n个人, 要完成a个x任务, b个y任务. 求,最短的时间 思路:由于时间较大,用 二分来找时间. dp[i][j]表示 i个人完成j个x任务, 最多能完成的y任务个数 这个题 不是很好想, 还参考了一下大神的博客 #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namesp…
题目链接: http://codeforces.com/problemset/problem/660/C 题意: 给定0.1组成的数组,可以改变k个0使其为1,问最终可以得到的连续的1的最大长度. 分析: 很容易想到二分答案的做法, 二分长度,然后找是否存在满足题意的区间. 还可以用尺取法,这样在O(n)时间负责度内就可以完成,但是个人感觉写起来没有二分直观.. 代码: 二分: #include<cstdio> #include<cmath> #include<iostrea…
A. Anton and Danik time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Anton likes to play chess, and so does his friend Danik. Once they have played n games in a row. For each game it's known…
题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5008 Description In this problem, you are given a string s and q queries. For each query, you should answer that when all distinct substrings of string s were sorted lexicographically, which one is…
Description Professor Clumsey is going to give an important talk this afternoon. Unfortunately, he is not a very tidy person and has put all his transparencies on one big heap. Before giving the talk, he has to sort the slides. Being a kind of minima…
C. Anton and Making Potions 题目连接: http://codeforces.com/contest/734/problem/C Description Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next level he has to prepare n potions. Anton has a…
目录 . 引言 . Linux进程 . Linux命名空间 . Linux进程的相关标识 . 进程标识编程示例 . 进程标志在Linux内核中的存储和表现形式 . 后记 0. 引言 在进行Linux主机的系统状态安全监控的过程中,我们常常会涉及到对系统进程信息的收集.聚类.分析等技术,因此,研究Linux进程原理能帮助我们更好的明确以下几个问题 . 针对Linux的进程状态,需要监控捕获哪些维度的信息,哪些信息能够更好地为安全人员描绘出黑客的入侵迹象 . 监控很容易造成的现象就是会有短时间内的大…
Drying Time Limit: 2000MS Memory Limit: 65536K Description It is very hard to wash and especially to dry clothes in winter. But Jane is a very smart girl. She is not afraid of this boring process. Jane has decided to use a radiator to make drying fas…
上学期刷过裸的RMQ模板题,不过那时候一直不理解>_< 其实RMQ很简单: 设f[i][j]表示从i开始的,长度为2^j的一段元素中的最小值or最大值 那么f[i][j]=min/max{d[i][j-1], d[i+2^j-1][j-1]} RMQ的ST算法: void ST() //初始化 { memset(RMQ,,sizeof(RMQ)); ;i<=n;i++) RMQ[i][]=a[i]; ;(<<j)<=n;j++) ;i+(<<j)-<=…
C. Anton and Making Potions time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next…
题目链接:http://poj.org/problem?id=3104                                                                                                      Drying Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11128   Accepted: 2865 Description It is very…
Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now OO want to know ∑i=1n∑j=inf(i,j) mod (109+7).   Input There are m…
D. Max and Bike Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/595/problem/D Description For months Maxim has been coming to work on his favorite bicycle. And quite recently he decided that he is ready to take part in a cy…
Bridging signals Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9234   Accepted: 5037 Description 'Oh no, they've done it again', cries the chief designer at the Waferland chip factory. Once more the routing designers have screwed up co…
描述 http://poj.org/problem?id=3104 n件衣服,第i件衣服里面有水a[i],自然风干每分钟干1个水,用吹风机每分钟干k个水,但是同时只能对一件衣服使用吹风机,求干完所有衣服所需时间的最小值. Drying Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 12639   Accepted: 3256 Description It is very hard to wash and especial…
Yukari's Birthday Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4934    Accepted Submission(s): 1163 Problem Description Today is Yukari's n-th birthday. Ran and Chen hold a celebration party…
F - 二分二分 Crawling in process... Crawling failed Time Limit:6000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Description Given a N × N matrix A, whose element in the i-th row and j-th column Aij is an number that equals i2…
Drying Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 12568   Accepted: 3243 Description It is very hard to wash and especially to dry clothes in winter. But Jane is a very smart girl. She is not afraid of this boring process. Jane has…
Description Data-mining huge data sets can be a painful and long lasting process if we are not aware of tiny patterns existing within those data sets. One reputable company has recently discovered a tiny bug in their hardware video processing solutio…
What a Mess Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice Gym 100947I Description standard input/output  Announcement   Statements Alex is a very clever boy, after all he is the son of the greates…
欢迎参加——每周六晚的BestCoder(有米!) Bridging signals Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 941    Accepted Submission(s): 614 Problem Description 'Oh no, they've done it again', cries the chief…