Frequent values POJ

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
------------------------详解请看http://blog.sina.com.cn/s/blog_6635898a0100kzpx.html

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1e8
#define lson l , m , rt << 1
#define rson m+1,r,  rt << 1 | 1
using namespace std;

const int maxn=100050;

struct mnode{ int l,r,max; }node[4*maxn];
struct mnseg{ int sta,end; }seg[maxn]; 

int n,q;
int num[maxn],hash[maxn];

void Inite(){                                        //将序列处理成点的集合
    for(int i=1;i<=n;i++) scanf("%d",&num[i]);
    int k=0,pre=999999;
    for(int i=1;i<=n;i++){
        if(num[i]!=pre){
            pre=num[i];
            k++;
            seg[k].sta=i;
            seg[k].end=i;
        }
        else seg[k].end=i;
        hash[i]=k;
    } 
}

void Build(int l,int r,int rt){
    node[rt].l=l;
    node[rt].r=r;
    if(l==r){ node[rt].max=seg[l].end-seg[l].sta+1; return; }
    int m=(l+r)>>1;
    Build(lson);
    Build(rson);
    node[rt].max=max(node[rt<<1].max,node[(rt<<1)|1].max); 
}

int Query(int l,int r,int rt){                                //key point！！！！
    if(node[rt].l==l&&node[rt].r==r) return node[rt].max;
    if(r<=node[rt<<1].r) return Query(l,r,rt<<1);
    if(l>=node[(rt<<1)|1].l) return Query(l,r,(rt<<1)|1);
    int a=Query(l,node[rt<<1].r,rt<<1);
    int b=Query(node[(rt<<1)|1].l,r,(rt<<1)|1);
    return max(a,b);
}

int main()
{   while(~scanf("%d",&n)){
        if(n==0) break;
        scanf("%d",&q);
        
        Inite();
        Build(1,n,1);
        while(q--){
            int a,b,pos1,pos2;
            scanf("%d%d",&a,&b);
            pos1=hash[a],pos2=hash[b];
            if(pos1==pos2) cout<<b-a+1<<endl;
            else{
                int n1,n2,n3;
                n1=seg[pos1].end-a+1;
                n2=0;
                n3=b-seg[pos2].sta+1;
                if(pos2-pos1>1) n2=Query(pos1+1,pos2-1,1);
                cout<<max(n1,max(n2,n3))<<endl;
            }
        }
    }
    return 0;
}