Interviewe HDU - 3486（暴力rmq）

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Interviewe HDU - 3486（暴力rmq）

面试n个人,可以分任意组数,每组选一个,得分总和严格大于k,问最少分几组就是暴力嘛...想到就去写吧.. #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include <vector> #include <stack> #include <…

Interviewe HDU - 3486 （ST表+枚举）（非二分，看下这个数据、）

YaoYao has a company and he wants to employ m people recently. Since his company is so famous, there are n people coming for the interview. However, YaoYao is so busy that he has no time to interview them by himself. So he decides to select exact m i…

hdu 3486 Interviewe (RMQ+二分)

Interviewe Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4543 Accepted Submission(s): 1108 Problem Description YaoYao has a company and he wants to employ m people recently. Since his compa…

HDU 3486 Interviewe RMQ

题意: 将\(n\)个数分成\(m\)段相邻区间,每段区间的长度为\(\left \lfloor \frac{n}{m} \right \rfloor\),从每段区间选一个最大值,要让所有的最大值之和大于\(k\).求最小的\(m\). 分析: 预处理RMQ,维护区间最大值. 然后二分\(m\),将每段区间最大值加起来判断即可. #include <cstdio> #include <cstring> #include <algorithm> using namespa…

HDU 3486 Interviewe

题目大意:给定n个数的序列,让我们找前面k个区间的最大值之和,每个区间长度为n/k,如果有剩余的区间长度不足n/k则无视之.现在让我们找最小的k使得和严格大于m. 题解:二分k,然后求RMQ检验. ST算法: #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int maxn=200010; int d[maxn][30]; int a[maxn];…

HDU 5289 Assignment rmq

Assignment 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5289 Description Tom owns a company and he is the boss. There are n staffs which are numbered from 1 to n in this company, and every staff has a ability. Now, Tom is going to assign a special…

HDU 5726 GCD (RMQ + 二分)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5726 给你n个数,q个询问,每个询问问你有多少对l r的gcd(a[l] , ... , a[r]) 等于的gcd(a[l'] ,..., a[r']). 先用RMQ预处理gcd,dp[i][j] 表示从i开始2^j个数的gcd. 然后用map存取某个gcd所对应的l r的数量. 我们可以在询问前进行预处理,先枚举i,以i为左端点的gcd(a[i],..., a[r])的种类数不会超过log2(n)…