Frequent values（ST）

描述

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.

输入

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 nintegers 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.

输出

For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

样例输入

10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0

样例输出

1
4
3

提示

A naive algorithm may not run in time!

题目大意：

给定一个序列，若干个查询，查询区间内出现最频繁的数的个数。

先预处理当前位置的值前面和它相同的个数，例如样例转化为1,2,1,2,3,4,1,1,2,3。然后ST预处理区间最大值，单纯的查询最大值会出现问题，例如[4,7]ST表得出的答案为4，所以可以将我们要查询的左端点更新一下，更新到一个新的1的位置，这样就不会出现那种问题了。

#include <bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int N=1e5+;
int n,m,a[N],rt[N],pre[N];
int st[N][];
int query(int l,int r)
{
    if(l>r) return ;
    int k=log2(r-l+);
    return max(st[l][k],st[r-(<<k)+][k]);
}
int main()
{
    while(~scanf("%d",&n),n)
    {
        memset(st,,sizeof st);
        scanf("%d",&m);
        a[]=INF;
        for(int i=;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=;i<=n;i++)
        {
            if(a[i]==a[i-]) pre[i]=pre[i-]+;
            else pre[i]=;
        }
        for(int i=;i<=n;i++)
            st[i][]=pre[i];
        for(int j=;j<;j++)
            for(int i=;i+(<<(j-))<=n;i++)
                st[i][j]=max(st[i][j-],st[i+(<<(j-))][j-]);
        rt[n]=n;
        for(int i=n-;i>=;i--)
        {
            if(a[i]==a[i+]) rt[i]=rt[i+];
            else rt[i]=i;
        }
        while(m--)
        {
            int l,r,tmp;
            scanf("%d%d",&l,&r);
            if(rt[l]==rt[l-])  tmp=min(r,rt[l])+;
            else tmp=l;
            int ans=query(tmp,r);
            ans=max(ans,tmp-l);
            printf("%d\n",ans);
        }
    }
    return ;
}