UVA11235:Frequent values(RMQ)

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.

Input Specification

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 , ... , 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.

Output Specification

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

A naive algorithm may not run in time!


题意:

给出一个非降序的整数数组,你的任务是对于一系列询问,回答区间内出现最多的值的次数


思路:

因为这个数组是非降序的,所以可以把所有相等的元素组合起来用二元组表示,例如-1,1,1,2,2,2,4就可以表示为(-1,1)(1,2)(2,3)(4,1),其中(a,b)代表有b个连续的a。

val[i]代表第i段的数值

cnt[i]代表第i段连续的长度

num[i]表示第i个位置的数在那一段

lef[i],righ[i]分别表示第i段的左右端点位置

所求的最大值就是

1.从L到L所在的段的结束处的元素个数:righ[L]-L+1

2.从R到R所在的段的开始处的元素个数:R-lef[R]+1

3.中间第num[L]+1段到第num[R]-1段的cnt的最大值(RMQ)

答案就是1,2,3中的最大值


#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 100005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)

int n,q;
int a[N];
int val[N],cnt[N],num[N],lef[N],righ[N];
int d[N][50],len;

void RMQ_init()
{
    int i,j,k;
    for(i = 1; i<=len; i++)
        d[i][0] = cnt[i];
    for(j = 1; (1<


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