Frequent values-------最多的连续的个数

给定一个非递减顺序的整数序列,你需要处理一些查询,每个查询包含两个索引i和j。对于每个查询,找出范围[i, j]内的最常出现的数值。" 133106443,20036795,Linux进程内存布局与单片机内存管理,"['Linux', '单片机', '内存管理']

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Frequent values

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 58   Accepted Submission(s) : 22
Problem 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
<p>The input consists of several test cases. Each test case starts with a line containing two integers <strong>n</strong> and <strong>q</strong> (<i>1 ≤ n, q ≤ 100000</i>). The next line contains <strong>n</strong> integers <strong>a<sub>1</sub> , ... , a<sub>n</sub></strong> (<i>-100000 ≤ a<sub>i</sub> ≤ 100000</i>, for each <i>i ∈ {1, ..., n}</i>) separated by spaces. You can assume that for each <i>i ∈ {1, ..., n-1}: a<sub>i</sub> ≤ a<sub>i+1</sub></i>. The following <strong>q</strong> lines contain one query each, consisting of two integers <strong>i</strong> and <strong>j</strong> (<i>1 ≤ i ≤ j ≤ n</i>), which indicate the boundary indices for the< br>query.</p><p>The last test case is followed by a line containing a single <i>0</i>.</p>
 

Output
<p>For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.</p>
 

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
 
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<cstring>
#include<cmath>
#define M 100001
using namespace std;
struct node
{
    int l,r;
    int lmax,rmax,amax;//记lmax为最左边的个数,同理
}tree[M*4];
int a[M];
void build(int id,int l,int r)
{
    tree[id].l=l;
    tree[id].r=r;
    if(l==r)
    {
        tree[id].lmax=tree[id].rmax=tree[id].amax=1;
        return;
    }
    int mid=(l+r)/2;
     build(id*2,l,mid);
    build(id*2+1,mid+1,r);


    int count;
    if(a[tree[id*2].r]==a[tree[id*2+1].l])
        count=tree[id*2].rmax+tree[id*2+1].lmax;
    else
        count=0;
    tree[id].amax=max(max(tree[id*2].amax,tree[id*2+1].amax),count);
    //此时还没有更新lmax最左边数字的连续的个数,还有最右边的数字的连续的个数,所以接下来要判断
    tree[id].lmax=tree[id*2].lmax;//但是此时有bug,假如左子树左右的数相同,并且与右子树左边的相等,则tree[id]就不准确了
    if(tree[id*2].lmax==mid-l+1&&a[tree[id*2].r]==a[tree[id*2+1].l])
        tree[id].lmax=mid-l+1+tree[id*2+1].lmax;
    tree[id].rmax=tree[id*2+1].rmax;
    if(tree[id*2+1].rmax==r-mid&&a[tree[id*2+1].l]==a[tree[id*2].r])//之前一直有疑问,后来想到当范围内的数全相等时,他的左右max相等
    tree[id].rmax=r-mid+tree[id*2].rmax;
    //build(id*2,l,mid);
    //build(id*2+1,mid+1,r);


}




int query(int id,int l,int r)
{
    if(tree[id].l==l&&tree[id].r==r)
        return tree[id].amax;
    int mid=(tree[id].l+tree[id].r)/2;
    if(r<=mid)
        return query(id*2,l,r);
    else if(l>mid)
        return query(id*2+1,l,r);
    else
    {
        int a1=query(id*2,l,mid);
        int a2=query(id*2+1,mid+1,r);
        int a3=0;
        if(a[tree[id*2].r]==a[tree[id*2+1].l])
        {
            a3=min(tree[id*2].rmax,mid-l+1)+min(tree[id*2+1].lmax,r-mid);
        }
        return max(max(a1,a2),a3);
    }
}
int main()
{
    int n;
    int p;
    int x,y;
    while(scanf("%d",&n)!=0)
    {
        if(n==0)
            break;
        scanf("%d",&p);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        build(1,1,n);
        for(int i=0;i<p;i++)
        {
            scanf("%d%d",&x,&y);
            cout<<query(1,x,y)<<endl;
        }


    }
    return 0;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值