Frequent values（倍增RMQ）

Posted 2022-09-09 背着代码的蜗牛

Frequent values

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
题目大意：
给出n个数和m个询问（l,r)，对于每个询问求出(l,r)之间连续出现次数最多的次数。
思路：
RMQ算法
预处理一个数组：
if(num[i]==num[i-1])
a[i]=a[i-1]+1;
else
a[i]=1;
对于每个询问(l,r)，分为两个部分，前半部分求与l之前相同的数的个数直到t，后半部分从t开始直接用RMQ求解最大值就行了。
最后结果为max(前半部分,后半部分)。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=100010;
int n,m,num[maxn],a[maxn],f[maxn][20];
int init()

    int f=1,p=0;char c=getchar();
    while(c>9||c<0)if(c==-)f=-1;c=getchar();
    while(c>=0&&c<=9)p=p*10+c-0;c=getchar();
    return f*p;

void prepare()

    memset(f,-1,sizeof(-1));
    for(int i=1;i<=n;i++)
    f[i][0]=a[i];
    int k=log((double)(n+1))/log(2.0);
    for(int j=1;j<=k;j++)
      for(int i=1;i+(1<<j)-1<=n;i++)
      f[i][j]=max(f[i][j-1],f[i+(1<<j-1)][j-1]);

int rmq(int l,int r)

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

int main()

    int l,r;
    while(cin>>n&&n)
    
        m=init();
        for(int i=1;i<=n;i++)
        
            num[i]=init();
            if(i==1)
            
                a[i]=1;
                continue;
            
            if(num[i]==num[i-1])
            a[i]=a[i-1]+1;
            else a[i]=1;
        
        prepare();
        for(int i=1;i<=m;i++)
        
            l=init();r=init();
            int t=l;
            while(t<=r&&num[t]==num[t-1])
            t++;
            int ans=rmq(t,r);
            ans=max(ans,t-l);
            printf("%d\\n",ans);
        
    
    return 0;