hdu-1806 Frequent values（RMQ，求区间最大频率）

Frequent values

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1475    Accepted Submission(s): 540

Problem 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

14
3

题意：给一个长度为N的序列，求s~t之间出现最多的数字出现的次数
思路：用a数组存题目给的序列，b数组存后面序列有多少个数字和本身一样，再加上1（本身）
最后求的时候比一下求最大值就好。用RMQ求区间（b数组上）最大值。
代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define N 100010int a[N];int b[N];int dp[N][20];void makermq(int n){    for(int i=0; i<n; i++)        dp[i][0]=b[i];    for(int j=1; (1<<j)<=n; j++)    {        for(int i=0; i+(1<<j)-1<n; i++)        {            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);        }    }}int getmax(int s,int t){    int k=(int)(log(t-s+1.0)/log(2.0));    return max(dp[s][k],dp[t-(1<<k)+1][k]);}int finds(int s,int t){    int i=s;   while(1)   {       if(i>=t||a[i]!=a[i+1])        break;       i++;   }   return i;}int main(){    int n,q;    int k,l;    while(scanf("%d",&n)&&n)    {        scanf("%d",&q);        for(int i=0; i<n; i++)            scanf("%d",&a[i]);        int t=1;        b[0]=1;        for(int i=1; i<n; i++)        {            if(a[i]==a[i-1])                t++;            else                t=1;            b[i]=t;        }        makermq(n);        while(q--)        {            scanf("%d %d",&k,&l);            k--;            l--;            int tm=finds(k,l);            int num=tm-k+1;            if(tm+1>=l)                printf("%d\n",num);            else            printf("%d\n",max(num,getmax(tm+1,l)));        }    }    return 0;}