[HDU 1806] Frequent values

Frequent values

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1146    Accepted Submission(s): 415

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 10 2 3 1 10 5 10 0

 

Sample Output

1 4 3

 

Hint

A naive algorithm may not run in time!

 

分成三段、中间RMQ、然后求最大值即可

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
#define N 100010

int n,m;
int a[N];
int dp[N][20];
int id[N];
int len[N],l[N],r[N];

void init()
{
    int i,j;
    for(i=1;i<=n;i++)
    {
        dp[i][0]=len[i];
    }
    int k=(int)(log((double)n)/log(2.0));
    for(j=1;j<=k;j++)
    {
        for(i=1;i+(1<<j)-1<=n;i++)
        {
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
        }
    }
}
int query(int i,int j)
{
    int k=(int)(log((double)(j-i+1))/log(2.0));
    int res=max(dp[i][k],dp[j-(1<<k)+1][k]);
    return res;
}
int main()
{
    int i,pos;
    while(scanf("%d",&n),n)
    {
        pos=0;
        scanf("%d",&m);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]!=a[i-1])
            {
                pos++;
                l[pos]=i;
            }
            id[i]=pos;
            r[pos]=i;
            len[pos]=r[pos]-l[pos]+1;
        }
        n=pos;
        init();
        int x,y,xx,yy,ans1,ans2,ans3;
        while(m--)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            xx=id[x];
            yy=id[y];
            if(xx==yy) ans1=ans2=y-x+1; //特殊情况、当x和y在同一个区间、答案是y-x+1
            else
            {
                ans1=r[xx]-x+1;
                ans2=y-l[yy]+1;
            }
            ans3=0;
            if(xx+1<=yy-1)ans3=query(xx+1,yy-1);
            printf("%d\n",max(max(ans1,ans2),ans3));
        }
    }
    return 0;
}