POJ 3368：Frequent values

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14764		Accepted: 5361

Description

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

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

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

题意是给出n个数，q次询问。每次询问一个区间中出现次数最多的那个数。

这题自己没有做出来，一直TLE，像当年做扩展欧几里德题目似的。看网上代码，才知道RMQ这种算法，总之是要记住了这种有动态规划思想的东西，还有这种询问这么多次的题目要离线处理啊啊啊啊，这都不懂你还搞毛啊光速小子。。。

我自己都不好意思舔这个大脸上这个代码，为了以后自己常常看看，记住它吧。

觉得这个代码写得真的好，很多细节做的真的不错(。。。。。。)：

#include <iostream>
#include <cmath>
#include <algorithm>
#pragma warning(disable:4996) 
using namespace std;

int num[100005];
int fre[100005];
int n,q;

int max_v[100005][20];

void RMQ()
{
	int i,j;
	for(i=1;i<=n;i++)
	{
		max_v[i][0]=fre[i];
	}
	int temp = log((double)n)/log(2.0)+1;
	for(j=1;j<=temp;j++)
	{
		for(i=1;i+(1<<j)-1<=n;i++)
		{
			max_v[i][j]=max(max_v[i][j-1],max_v[i+(1<<(j-1))][j-1]);
		}
	}
}

int cal(int h,int k)
{
	if(h>k)
		return 0;
	int temp=k-h+1;
	int temp2=log((double)temp)/log(2.0);

	return max(max_v[h][temp2],max_v[k-(1<<temp2)+1][temp2]);
}

int main()
{
	//freopen("input.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	while(scanf("%d",&n))
	{
		int i,h,k;
		if(n==0)
			break;
		scanf("%d",&q);

		memset(fre,0,sizeof(fre));
		memset(max_v,0,sizeof(max_v));
		for(i=1;i<=n;i++)
		{
			scanf("%d",&num[i]);
			if(i==1)
			{
				fre[1]=1;
			}
			else
			{
				if(num[i]==num[i-1])
					fre[i]=fre[i-1]+1;
				else
					fre[i]=1;
			}
		}

		RMQ();
		
		for(i=1;i<=q;i++)
		{
			scanf("%d%d",&h,&k);
			int temp;
			int t=0;

			while((h+t<=k)&&(num[h+t]==num[h]))
			{
				t++;
			}
			temp=h+t;
			int temp2=cal(temp,k);
			cout<<max(t,temp2)<<endl;
		}
	}
    return 0;
}