3486 ( Interviewe )RMQ
YaoYao decides to make the interview as follows. First he queues the interviewees according to their coming order. Then he cuts the queue into m segments. The length of each segment is , which means he ignores the rest interviewees (poor guys because they comes late). Then, each segment is assigned to an interviewer and the interviewer chooses the best one from them as the employee.
YaoYao’s idea seems to be wonderful, but he meets another problem. He values the ability of the ith arrived interviewee as a number from 0 to 1000. Of course, the better one is, the higher ability value one has. He wants his employees good enough, so the sum of the ability values of his employees must exceed his target k (exceed means strictly large than). On the other hand, he wants to employ as less people as possible because of the high salary nowadays. Could you help him to find the smallest m?
In the first line of each case, there are two numbers n and k, indicating the number of the original people and the sum of the ability values of employees YaoYao wants to hire (n≤200000, k≤1000000000). In the second line, there are n numbers v1, v2, …, vn (each number is between 0 and 1000), indicating the ability value of each arrived interviewee respectively.
The input ends up with two negative numbers, which should not be processed as a case.
7 100 7 101 100 100 9 100 100 110 110
-1 -1
We need 3 interviewers to help YaoYao. The first one interviews people from 1 to 3, the second interviews people from 4 to 6,
and the third interviews people from 7 to 9. And the people left will be ignored. And the total value you can get is 100+101+100=301>300.
题解 : RMQ
C++代码:
#include<iostream>
#include<cstdio>
#include<string>
#include<cmath>
#include<cstring>
using namespace std;
#define MAXN 200000 +9
#define MAXE 22
int h[MAXN],mmax[MAXN][MAXE];
int N,Q;
int L,R;
void RMQ_ST(){
for(int i=;i<=N;i++){
mmax[i][]=h[i]; }
int end_j=log(N+0.0)/log(2.0);
int end_i;
for(int j=;j<=end_j;j++){
end_i=N+-(<<j);
for(int i=;i<=end_i;i++){
mmax[i][j]=max(mmax[i][j-],mmax[i+(<<(j-))][j-]);
// mmin[i][j]=min(mmin[i][j-1],mmin[i+(1<<(j-1))][j-1]);
}
}
}
int QueryMax(int L,int R){ int k=log(R-L+1.0)/log(2.0);
return max(mmax[L][k],mmax[R-(<<k)+][k]);
} int main(){
while(~scanf("%d%d",&N,&Q)&&(N > || Q > )){
int maxx = ;
for(int i=;i<=N;i++){
scanf("%d",&h[i]);
maxx = max(h[i],maxx);
}
int m = ; RMQ_ST();
m = Q / maxx; int flag = ;
if(m == ) {
m = ;
flag = ;
goto out ;
}
for(;m <= N; m ++){
int res = ;
int sss = ; int s = N / m;
int i = ;
int j = m;
while(j--){
res += QueryMax(i ,i + s- );
i += s;
}
if(res > Q) {
flag = ;
break;
}
}
out :
if(!flag) printf("-1\n");
else
cout << m << endl; } return ;
}
3486 ( Interviewe )RMQ的更多相关文章
- hdu 3486 Interviewe (RMQ+二分)
Interviewe Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ...
- 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 ...
- HDOJ 3486 Interviewe
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 I ...
- 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. S ...
- hdu 3484 Interviewe RMQ+二分
#include <cstdio> #include <iostream> #include <algorithm> using namespace std; + ...
- 【转载】图论 500题——主要为hdu/poj/zoj
转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并 ...
- hdu图论题目分类
=============================以下是最小生成树+并查集====================================== [HDU] 1213 How Many ...
- HDU图论题单
=============================以下是最小生成树+并查集====================================== [HDU] 1213 How Many ...
- [数据结构]RMQ问题小结
RMQ问题小结 by Wine93 2014.1.14 1.算法简介 RMQ问题可分成以下2种 (1)静态RMQ:ST算法 一旦给定序列确定后就不在更新,只查询区间最大(小)值!这类问题可以用倍增 ...
随机推荐
- 【leetcode】1144. Decrease Elements To Make Array Zigzag
题目如下: Given an array nums of integers, a move consists of choosing any element and decreasing it by ...
- js 文件下载 兼容ie
前置条件:后台接口返回二进制流文件 一.设置前端请求的的 responseType: 'blob' 二.接收请求数据并调用下载 var content = res.data // 接口返回的二进制流v ...
- ionic slide组件使用
ionic学习使用笔记 slide 组件的使用 开始做的时候,遇到了个要用ionic实现 有一系列的序列需要展示,但是当前页面上只能展示一小部分,剩余的在没有出现时是隐藏的,还得能滑动出现,但是又 ...
- SpringBoot:使用IDEA快速构建项目
西部开源-秦疆老师:基于SpringBoot 2.1.6 的博客教程 秦老师交流Q群号: 664386224 未授权禁止转载!编辑不易 , 转发请注明出处!防君子不防小人,共勉! SpringBoot ...
- 高级软件测试技术(测试管理工具实践day2)
今天在紧张的学习之余,我们小组选定了bugzilla,并且打算在今天晚上刚进行下载安装. 在安装bugzilla需要的软件有MySQL数据库软件,activeperl软件,bugzilla安装包,II ...
- Window7系统安装Ubuntu16双系统
在电脑上插入ubuntu系统启动盘,之前做好的u盘启动盘,重启计算机,进入BIOS设置界面,设置系统启动为u盘启动,保存后退出.之后进入ubuntu系统安装界面. 在安装界面中选择系统语言,选择安装u ...
- oracle 表连接 - sort merge joins 排序合并连接
https://blog.csdn.net/dataminer_2007/article/details/41907581一. sort merge joins连接(排序合并连接) 原理 指的是两个表 ...
- 通用 C# DLL 注入器injector(注入dll不限)
为了方便那些不懂或者不想用C++的同志,我把C++的dll注入器源码转换成了C#的,这是一个很简单实用的注入器,用到了CreateRemoteThread,WriteProcessMemory ,Vi ...
- LinkedTransferQueue 源码分析
LinkedTransferQueue LinkedTransferQueue 能解决什么问题?什么时候使用 LinkedTransferQueue? 1)LinkedTransferQueue 是基 ...
- (一)Maven之使用入门
目录 今天是端午节哦,昨天大学同学举个了会.鱼头泡饼贼拉香,嗯哼,有点跑题了:之后去了同学家里坐了坐:发现同我有一样的书,即:<maven实战>:记得是从二手网店淘到的,已经买了有小半年, ...