POJ 3368 Frequent Values(RMQ)

来源：互联网发布：java object方法详解编辑：程序博客网时间：2024/05/05 23:19

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;elsef[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 , ... , 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 10 102 31 105 100

Sample Output

0 0