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…

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];…

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…

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 <…

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…

人生中第一次写RMQ....一看就知道 RMQ+2分但是题目文不对题....不知道到底在问什么东西....各种WA,TLE,,RE...后就过了果然无论错成什么样都可以过的,就是 上层的样例 啊  Interviewe Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3865    Accepted Submission(s): 956…

#include <cstdio> #include <iostream> #include <algorithm> using namespace std; + ; int a[maxn]; ]; int n, k; void RMQ(int num){ ; i <= num; i++){ maxsum[i][] = a[i]; } ; j < ; j++){ ; i <= num; i++){ << j) - <= num){ m…

Problem Description 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…

题意:n个人,每个人的价格a[  i  ] ,求最少分几组,每组取一个人,多出来的人就不考虑,使得这取出人的价格大于k.(每组人数一样) 分析:每组取一个人,那这个人肯定是这组最大的,枚举多少组就可以了. 代码: #include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> using namespace std; typedef long long ll; ; ],a…

HDU 1000 A + B Problem  I/O HDU 1001 Sum Problem  数学 HDU 1002 A + B Problem II  高精度加法 HDU 1003 Maxsum  贪心 HDU 1004 Let the Balloon Rise  字典树,map HDU 1005 Number Sequence  求数列循环节 HDU 1007 Quoit Design  最近点对 HDU 1008 Elevator  模拟 HDU 1010 Tempter of th…