E - Frequent values

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

[Submit]   [Go Back]   [Status]

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

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>

using namespace std;

#define MAXN 100100

int a[MAXN];


struct s 
{
	int l;
	int r;
	int value;
}tt[MAXN];

struct node 
{
	int l;
	int r;
	int ma;
}t[MAXN*4];

void make(int st,int ed,int num)
{
	t[num].l=st;
	t[num].r=ed;
	if(st==ed)
	{
		t[num].ma=tt[st].r-tt[st].l+1;
		return ;
	}
	int mid=(t[num].l+t[num].r)/2;
	make(st,mid,num*2);
	make (mid+1,ed,2*num+1);
	t[num].ma=max(t[num*2+1].ma,t[num*2].ma);
}

int query(int st,int ed,int num)
{
	if(st==t[num].l&&ed==t[num].r)
		return	t[num].ma;
	int mid=(t[num].l+t[num].r)/2;
	if(st>mid)
		return query(st,ed,num*2+1);
	else if(ed<=mid)
		return query(st,ed,num*2);
	else 
	{
		return	max(query(st,mid,num*2),query(mid+1,ed,num*2+1));
	}
}

int search(int num,int value)
{
	int st=1,ed=num;
	while(st<ed)
	{
		int mid=(st+ed)/2;
		if(tt[mid].value>value)
			ed=mid-1;
		else if(tt[mid].value<value)
			st=mid+1;
		else return mid;	
	}
	return st;
}

int main()
{
	int n,q;
	int e,i;
	int st,ed;
	int x,y;
	int sum;
	while(scanf("%d%d",&n,&q)!=EOF&&n)
	{
		e=0;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);		//用tt【】去存储想通值的起点和终点//变相的离散化了///
			if(tt[e].value!=a[i])
			{
				e++;
				tt[e].l=i;
				tt[e].r=i;
				tt[e].value=a[i];			
			}
			else 
				tt[e].r++;
		}
	//	printf("%d\n",e);
		make(1,e,1);
		for(i=1;i<=q;i++)
		{
			scanf("%d%d",&x,&y);
			st=search(e,a[x]);
			ed=search(e,a[y]);

			if(st==ed)
				sum=y-x+1;
			else if(st+1==ed)
				sum=max(tt[st].r-x+1,y-tt[ed].l+1);
			else 
				sum=max(query(st+1,ed-1,1),
					max(tt[st].r-x+1,y-tt[ed].l+1));
			printf("%d\n",sum);
		}		
	}
	return 0;
}