Frequent values(HDU - 1806 )

Frequent values

 

You are given a sequence of n integers a ₁ , a ₂ , ... , 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 . 

InputThe 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 a ₁ , ... , 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. 

OutputFor 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 

Hint

A naive algorithm may not run in time! 

题意:查询区间[l,r]中众数的个数.

因为数列是非递减的，所以可以转化为ST表.

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define N 100005
using namespace std;
int maxn[N][64],a[N],pre[N];
int n,q,i,j;
int main()
{
    while(scanf("%d",&n)!=EOF&&n)
    {
        scanf("%d",&q);
        memset(maxn,0,sizeof maxn);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(i==1)
            {
                pre[i]=1;
                continue;
            }
            if(a[i]==a[i-1])
            pre[i]=pre[i-1]+1;
            else
            pre[i]=1;
        }
        for(i=1;i<=n;i++)
        maxn[i][0]=pre[i];
        int t=log2(n);
        for(j=1;j<=t;j++)
        {
            for(i=1;i+(1<<j)-1<=n;i++)
            {
                maxn[i][j]=max(maxn[i][j-1],maxn[i+(1<<(j-1))][j-1]);
            }
        }
        while(q--)
        {
            int MAX;
            int l,r; 
            scanf("%d%d",&l,&r);
            int m=l;
            while(m<=r&&a[m]==a[m-1])//区间第一个数字要特判. 
            m++;
            if(r<m)
            MAX=0;
            else
            { 
                int k=log(r-m+1)/log(2);
                MAX=max(maxn[m][k],maxn[r-(1<<k)+1][k]);
            } 
            MAX=max(MAX,m-l);
            printf("%d\n",MAX);
        } 
    }
}