UVA 11235 Frequent values

Frequent values

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

#include<cstdio>
#include<iostream>
#include<vector>
#include<cmath>
#include<queue>
#include<string>
#include<map>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn=1e5+5;
struct code
{
    int l,r,id;
}p[maxn];
int val[maxn],cnt[maxn],d[maxn][32],no,n,q;
void rmq_init()
{
    for(int i=1;i<=no;i++)d[i][0]=cnt[i];
    for(int j=1;(1<<j)<=no;j++)
        for(int i=1;i+(1<<j)<=no;i++)
            d[i][j]=max(d[i][j-1],d[i+(1<<(j-1))][j-1]);
}
int rmq(int l,int r)
{
    if(l>r)return 0;
    int k=0;
    while(1<<(1+k)<=r-l+1)k++;
    return max(d[l][k],d[r-(1<<k)+1][k]);
}
int main()
{
    while(scanf("%d",&n),n)
    {
        scanf("%d",&q);
        memset(cnt,0,sizeof(cnt));
        memset(d,0,sizeof(d));
        no=0;
        for(int i=1;i<=n;i++)
        {
            int tt;
            scanf("%d",&tt);
            if(tt!=val[no-1]||i==1)
            {
                val[no++]=tt;
                cnt[no]=1;
            }
            else
                cnt[no]++;
        }
        int i=1,num=1;
        for(int j=1;j<=no;j++)
        {
            for(int t=0;t<cnt[j];t++)
            {
                p[i].id=j;
                p[i].l=num;
                p[i].r=num+cnt[j]-1;
                i++;
            }
            num+=cnt[j];
        }
        rmq_init();
        while(q--)
        {
            int l,r,ans;
            scanf("%d%d",&l,&r);
            if(p[l].id==p[r].id)
                ans=r-l+1;
            else
            {
                ans=rmq(p[l].id+1,p[r].id-1);
                ans=max(ans,p[l].r-l+1);
                ans=max(ans,r-p[r].l+1);
            }
            printf("%d
",ans);
        }
    }
    return 0;
}