poj3368（RMQ——ST）

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16543		Accepted: 5985

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

Source

Ulm Local 2007

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分析

因为题上说是有序的，就可以求一样的数的数量，再来使用RMQ

开始我也不会做，看了别人的写的，思路还放不开

还记得以前求众数只会用桶排序，现在想起来实在可笑，但是路还很长

AC代码

#include<cstdio>
#include<iostream>
#include<cmath>
#define MAX 100010
using namespace std;
int f[MAX][20];
int num[MAX];
int sum[MAX];
int n,Q,ans,a,b;
int _max(int a,int b)
{
    return a>b?a:b;
}
void ST()
{
    for(int i=1;i<=n;i++)
        f[i][0]=sum[i];
    int k=log((double)(n+1))/log(2.0);
    for(int i=1;i<=k;i++)
      for(int j=1;j+(1<<i)<=n+1;j++)//n+1
        f[j][i]=_max(f[j][i-1],f[j+(1<<(i-1))][i-1]);    
}
int RMQ_Query(int l,int r)
{
    if(l>r)return 0;
    int k=log((double)(r-l+1))/log(2.0);
    return _max(f[l][k],f[r-(1<<k)+1][k]);//r-(1<<k)一定要+1 
}
int main()
{
    while(scanf("%d",&n)!=EOF,n)
    {
        scanf("%d",&Q);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",num+i);
            if(i==1)
            {
                sum[i]=1;
                continue;
            }
            if(num[i]==num[i-1])
            sum[i]=sum[i-1]+1;
            else sum[i]=1;
        }
        ST();
        while(Q--)
        {
            scanf("%d%d",&a,&b);
            if(a==b||num[a]==num[b])
            {
                cout<<b-a+1<<endl;
                continue;
            }
            int t=a;
            while(num[t]==num[t-1]&&t<=b)//t 必须小于 b 
                t++;
            int cnt=RMQ_Query(t,b);
            ans=_max(cnt,t-a);
            printf("%d\n",ans);
        }
    }
    return 0;
}