POJ 3368 Frequent Values(RMQ)

题意：给出一个非降序排列的整数数组a1,a2,...an，你的任务是对于一系列询问(i, j)，回答ai,ai+1,...aj中出现最多次数的值所出现的次数？

思路：将原始序列处理一下，令f[i]为记录当前i连续的个数，对于每个询问区间（l,r）将它分为两个部分，一个部分为l到它第一个不相同的数，之后的部分直接RMQ查询，然后取最大值即可

Trick；在区间查询的时候没有留意有L>R的情况，WA无数次

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

const int maxn = 200000+100;

int dmax[maxn][20];
int f[maxn];
int a[maxn];
void initmax(int n,int f[])
{
for (int i = 1;i<=n;i++)
dmax[i][0]=f[i];
for (int j = 1;(1<<j)<=n;j++)
for (int i = 1;i+(1<<j)-1<=n;i++)
dmax[i][j]=max(dmax[i][j-1],dmax[i+(1<<(j-1))][j-1]);
}
int getmax(int L,int R)
{
if (L>R)
return 0;
int k = 0;
while ((1<<(k+1))<=R-L+1)
k++;
return max(dmax[L][k],dmax[R-(1<<k)+1][k]);
}
int main()
{
int n,q;
while (scanf("%d",&n) && n)
{
scanf("%d",&q);
f[1]=1;
for (int i = 1;i<=n;i++)
{
scanf("%d",&a[i]);
if (i==1)
continue;
if (a[i]!=a[i-1])
f[i]=1;
else
f[i]=f[i-1]+1;
}
initmax(n,f);
while (q--)
{
int L,R;
scanf("%d%d",&L,&R);
if (L==R)
{
printf("1\n");
}
else
{
int temp = L;
while (temp<=R && a[temp]==a[temp-1])
temp++;
int ans = getmax(temp,R);
ans = max(ans,temp-L);
printf("%d\n",ans);

}
}
}
}

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