B - Frequent values

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

题意：一组升序的数字，问你某一区间同一数字的最大连续和

解法：RMQ,但同时得注意，每次区间求值的时候得注意更新区间右端点的值，详情看代码

#include
#include
#include
#include
#include
using namespace std;
int a[100010],dp[100010][20],b[100010];
int main()
{
    int n,m,i,j,x,y;
    while(scanf("%d",&n),n)
    {
        scanf("%d",&m);
        for(i=0;i=0;i--)
        {
            if(i==n-1) tmp=1;
            else
            {
                if(a[i]==a[i+1])tmp++;
                else tmp=1;
            }
            b[i]=tmp;//倒着存
        }
        for( i=0;i