H - Frequent values
Problem F: Frequent values
You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integers ai , ... , aj.
Input Specification
The input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a1 , ... , an (-100000 ≤ ai ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: ai ≤ ai+1. The following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which indicate the boundary indices for the query.
The last test case is followed by a line containing a single 0.
Output Specification
For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.
Sample Input
10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0
Sample Output
1
4
3 RMQ问题
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cstdlib>
#include <sstream>
using namespace std;
typedef long long LL;
const int INF=0x5fffffff;
const double EXP=1e-;
const int MS=; int dp[MS][];
int a[MS],cnt[MS];
int num[MS],l[MS],r[MS];
int n,q; void RMQ_init()
{
for(int i=;i<n;i++)
dp[i][]=cnt[i];
for(int j=;(<<j)<=n;j++)
{
for(int i=;i+(<<j)-<n;i++)
dp[i][j]=max(dp[i][j-],dp[i+(<<(j-))][j-]);
}
} int RMQ(int l,int r)
{
int k=;
while(<<(k+)<=r-l+)
k++;
return max(dp[l][k],dp[r-(<<k)+][k]);
} int main()
{
while(scanf("%d%d",&n,&q)==&&n)
{
for(int i=;i<n;i++)
scanf("%d",&a[i]);
a[n]=a[n-]+;
int start=;
int id=;
for(int i=;i<=n;i++)
{
if(i>&&a[i]>a[i-])
{
for(int j=start;j<i;j++)
r[j]=i-;
cnt[id]=i-start;
id++;
start=i;
}
l[i]=start;
num[i]=id;
}
n=id;
RMQ_init();
int x,y;
while(q--)
{
scanf("%d%d",&x,&y);
x--;
y--;
int ans=;
if(num[x]==num[y])
{
printf("%d\n",y-x+);
continue;
}
ans=max(r[x]-x+,y-l[y]+);
if(num[x]+<num[y])
ans=max(ans,RMQ(num[x]+,num[y]-));
printf("%d\n",ans);
} }
return ;
}
H - Frequent values的更多相关文章
- [POJ] 3368 / [UVA] 11235 - Frequent values [ST算法]
2007/2008 ACM International Collegiate Programming Contest University of Ulm Local Contest Problem F ...
- POJ 3368:Frequent values
Frequent values Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 14764 Accepted: 5361 ...
- UVA - 11235 Frequent values
2007/2008 ACM International Collegiate Programming Contest University of Ulm Local Contest Problem F ...
- poj 3368 Frequent values(RMQ)
/************************************************************ 题目: Frequent values(poj 3368) 链接: http ...
- Frequent values && Ping pong
Frequent values 题意是不同颜色区间首尾相接,询问一个区间内同色区间的最长长度. 网上流行的做法,包括翻出来之前POJ的代码也是RMQ做法,对于序列上的每个数,记录该数向左和向右延续的最 ...
- 【暑假】[实用数据结构]UVa11235 Frequent values
UVa 11235 Frequent values Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 11241 Accep ...
- [HDU 1806] Frequent values
Frequent values Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)T ...
- 数据结构(RMQ):UVAoj 11235 Frequent values
Frequent values You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. I ...
- poj 3368 Frequent values(段树)
Frequent values Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 13516 Accepted: 4971 ...
随机推荐
- 第三百五十五天 how can I 坚持
快一年了,三百五十五天了,等写个程序算算时间,看看日期和天数能不能对的上,哈哈. 计划还是未制定,天气预报还是没有写完,立马行动,发完这个博客,立马行动. 计划:设计模式1个月,三大框架3个月,计算机 ...
- AnnotationSessionFactoryBean用法介绍
http://blog.csdn.net/flyingfalcon/article/details/8273618 —————————————————————————————————————————— ...
- Terrain & Light & Camera
[Terrain Engine] 1.When you press F, wherever your mouse is positioned will be moved to the center o ...
- UVALive 7077 - Song Jiang's rank list(模拟)
https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_probl ...
- hdu 1171 Big Event in HDU(多重背包+二进制优化)
题目链接:hdu1171 思路:将多重背包转为成完全背包和01背包问题,转化为01背包是用二进制思想,即件数amount用分解成若干个件数的集合,这里面数字可以组合成任意小于等于amount的件数 比 ...
- 实现jsp网页设为首页功能
var url = location.href; var browser_name = navigator.userAgent; if(browser_name.indexOf('Chrome')!= ...
- HTML5中script的async属性异步加载JS
HTML5中script的async属性异步加载JS HTML4.01为script标签定义了5个属性: charset 可选.指定src引入代码的字符集,大多数浏览器忽略该值.defer 可 ...
- xshell linux传文件
yum install lrzsz 安装完毕即可使用 rz,sz是便是Linux/Unix同Windows进行ZModem文件传输的命令行工具 windows端需要支持ZModem的telnet/s ...
- 安装配置tomcat环境
安装配置tomcat环境 #所需要软件包 apache-tomcat-7.0.65.tar.gz jdk-7u80-linux-x64.gz #建立 个专用账户 usradd tomcat ...
- C# 反射 通过类名创建类实例
“反射”其实就是利用程序集的元数据信息. 反射可以有很多方法,编写程序时请先导入 System.Reflection 命名空间. 1.假设你要反射一个 DLL 中的类,并且没有引用它(即未知的类型): ...