POJ 3368

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17409		Accepted: 6287

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

Source

Ulm Local 2007

一.题意分析

裸的ST表题目

二.思路过程

写一个构建ST表的函数和一个查询函数

三.代码

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

int num[100010], f[100010], MAX[100010][20];
int n;

void ST()
{
    int i, j, k;
    for(i=1;i<=n;i++)
        MAX[i][0]=f[i];
    k=log((double)(n+1))/log(2.0);
    for(j=1;j<=k;j++)
        for(i=1;i+(1<<j)-1<=n;i++)
            MAX[i][j]=max(MAX[i][j-1],MAX[i+(1<<(j-1))][j-1]);
}

int Query(int l,int r)
{
    if(l>r)
        return 0;
    int k=log((double)(r-l+1))/log(2.0);
    return max(MAX[l][k],MAX[r-(1<<k)+1][k]);
}

int main()
{
    int q,i,a,b;
    while(scanf("%d",&n)&&n)
    {
        scanf("%d",&q);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
            if(i==1)
            {
                f[i]=1;
                continue;
            }
            if(num[i]==num[i-1])
                f[i]=f[i-1]+1;
            else
                f[i]=1;
        }
        ST();
        for(i=1;i<=q;i++)
        {
            scanf("%d%d",&a,&b);
            int t=a;
            while(t<=b&&num[t]==num[t-1])
                t++;
            int cnt=Query(t,b);
            int ans=max(t-a,cnt);
            printf("%d\n",ans);
        }
    }
    return 0;
}
/*
10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0
*/