hdu 1806 Frequent values（RMQ）

Frequent values

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1443    Accepted Submission(s): 526

Problem Description

You are given a sequence of n integers a₁ , a₂ , ... , a_n 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 a_i , ... , a_j . 

 

Input

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 , ... , a_n(-100000 ≤ a_i ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: a_i ≤ a_i+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

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 10 102 31 105 100

 

Sample Output

143
Hint
A naive algorithm may not run in time! 
 

 

n个数 非降序给出 求任意区间内出现次数最多的数在这个区间出现的次数 

首先从数据最后开始初始化一个数连续出现的次数  然后 在一个区间内求值时分为两种情况 

一种是右端点到中间某个值 这是一段 可以求出坐标 然后相减得到长度 

一种是借助RMQ的想法求出区间的最大值 因为之前是从数据末尾开始初始化那个此数值的 所有从区间中段往左都是可以通过求最值得到答案的

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#include <vector>#include <queue>#define MEM(a,x) memset(a,x,sizeof a)#define eps 1e-8#define MOD 10009#define MAXN 100010#define MAXM 100010#define INF 99999999#define ll __int64#define bug cout<<"here"<<endl#define fread freopen("ceshi.txt","r",stdin)#define fwrite freopen("out.txt","w",stdout)using namespace std;int a[MAXN],b[MAXN],dp[MAXN][30];int n;void RMQ_init(){    for(int i=0;i<n;i++)        dp[i][0]=b[i];    for(int j=1;(1<<j)<=n;j++)        for(int i=0;i+(1<<j)-1<n;i++)            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}int RMQ(int l,int r){    int k=0;    while((1<<(k+1))<=r-l+1) k++;    return max(dp[l][k],dp[r-(1<<k)+1][k]);}int bina(int s,int t)//找到区间中 值和右端点相同的位置{    int temp=a[t];    while(s<t)    {        int mid=(s+t)/2;        if(a[mid]>=temp) t=mid;        else s=mid+1;    }    return t;}int main(){//    fread;    while(scanf("%d",&n)&&n)    {        int q;        scanf("%d",&q);        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        int temp;        for(int i=n-1;i>=0;i--)        {            if(i==n-1) temp=1;            else            {                if(a[i]==a[i+1]) temp++;                else temp=1;            }            b[i]=temp;        }        RMQ_init();        while(q--)        {            int l,r;            scanf("%d%d",&l,&r);            l--; r--;            int temp=bina(l,r);            int num=r-temp+1;            r=temp-1;            if(l>r) printf("%d\n",num);            else printf("%d\n",max(num,RMQ(l,r)));        }    }    return 0;}