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 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 , ... , 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 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

RMQ经典题,代码如下:

#include 
  
   
#include 
   
     #include 
    
      #define MAXN 100010 using namespace std; typedef long long LL; int cot[MAXN],A[MAXN],num[MAXN],LEFT[MAXN],RIGHT[MAXN],dp[MAXN][17]; void RMQ_init(int n) { //cout<<"RMQ"<
     
      r)return 0; int k=0; while(1<<(k+1)<=r-l+1)k++; return max(dp[l][k],dp[r-(1<
      
       ql){ RIGHT[ti--]=i-1; } ql=LEFT[i]=i; RIGHT[i]=i; } else { num[i]=p; LEFT[i]=ql; cot[p]++; RIGHT[i]=i; } } RMQ_init(p); int l,r,ans; while(q--) { scanf("%d%d",&l,&r); ans=0; if(num[l]!=num[r]) { ans=max(RIGHT[l]-l+1,r-LEFT[r]+1); ans=max(ans,RMQ(num[RIGHT[l]+1],num[LEFT[r]-1])); } else ans=r-l+1; printf("%d\n",ans); } } return 0; }