Frequent values_poj3368_rmq

Description

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You are given a sequence of n integers a1 , a2 , … , an 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 ai , … , aj.

Input

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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 , … , an (-100000 ≤ ai ≤ 100000, for each i ∈ {1, …, n}) separated by spaces. You can assume that for each i ∈ {1, …, n-1}: ai ≤ ai+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

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For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

Analysis

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In order to ZHUANGBI I have decided to use English to write my analysis from now ;-P

Just kidding.

又是线段树例题强行转rmq

题目给出的是不降序，所以相邻的相同项个数可以看成是这个数字的数量(语文水平低原谅措辞)
对于一段询问可以拆成左右两边，右边我们用ST直接求，左边枚统计一下相同的数，这两个的最大值就是答案

Code

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#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;
int f[100001][17],t[100001],c[100001];
int max(int x,int y)
{
    return x>y?x:y;
}
int main()
{
    int n,m;
    while (scanf("%d",&n)&&n)
    {
        scanf("%d",&m);
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&t[i]);
            c[i]=1;
            if (t[i]==t[i-1])
                c[i]=c[i-1]+1;
            f[i][0]=c[i];
        }
        for (int j=1;j<=16;j++)
            for (int i=1;i+(1<<j)-1<=n;i++)
                f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);

        for (int i=1;i<=m;i++)
        {
            int x,y,cnt=0;
            scanf("%d%d",&x,&y);
            for (int j=x;j<=y&&t[j]==t[x];j++)
                cnt++;
            x+=cnt;
            if (x>=y)
                printf("%d\n",cnt);
            else
            {
                int v=floor(log10(y-x+1)/log10(2));
                int ans=max(cnt,max(f[x][v],f[y-(1<<v)+1][v]));
                printf("%d\n",ans);
            }
        }
    }
    return 0;
}