zoukankan      html  css  js  c++  java
  • DZY Loves Sequences

    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    DZY has a sequence a, consisting of n integers.

    We'll call a sequence ai, ai + 1, ..., aj (1 ≤ i ≤ j ≤ n) a subsegment of the sequence a. The value (j - i + 1) denotes the length of the subsegment.

    Your task is to find the longest subsegment of a, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.

    You only need to output the length of the subsegment you find.

    Input

    The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).

    Output

    In a single line print the answer to the problem — the maximum length of the required subsegment.

    Sample test(s)
    input
    6
    7 2 3 1 5 6
    output
    5
    Note

    You can choose subsegment a2, a3, a4, a5, a6 and change its 3rd element (that is a4) to 4.

    思路:预处理出包含第i个元素的左边最长连续递增长度 和 包含第i个元素的右边最长连续递增长度, 枚举可以任意改变的位置 pos, 然后判断该位置的左右两边是否可以连接以满足递增

    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <iostream>
    #include <algorithm>
    #include <map>
    using namespace std;
    const int N = 100005;
    int a[N], lt[N], rt[N], n;
    int main()
    {
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
        lt[1] = 1;
        for(int i = 2; i <= n; ++i) { //预处理i的左边且包含i的连续最长递增数列的长度
            if(a[i] > a[i - 1]) lt[i] = lt[i - 1] + 1;
            else lt[i] = 1;
        }
    
      //  for(int i = 1; i <= n; ++i) cout << lt[i] << ' ' ; cout  << endl;
        rt[n] = 1;
        for(int i = n - 1; i >= 1; --i) { ////预处理i的右边且包含i的连续最长递增数列的长度
            if(a[i] < a[i + 1]) rt[i] = rt[i + 1] + 1;
            else rt[i] = 1;
        }
      //  for(int i = 1; i <= n; ++i) cout << rt[i] << ' ' ; cout  << endl;
        int ans = max(rt[2] + 1, lt[n - 1] + 1);
        for(int i = 2; i < n; ++i) {//枚举可任意更换的位置
            if(a[i - 1] + 1 >= a[i + 1]) ans = max(ans, max(lt[i - 1], rt[i + 1]) + 1); // 若i的两边不能衔接, 取较大的长度
            else ans = max(ans, lt[i - 1] + rt[i + 1] + 1);//否则为答案为 两段和
        }
        printf("%d
    ", ans);
    }
    

      

  • 相关阅读:
    python线程池ThreadPoolExecutor用法
    redis学习笔记
    selenium + chrome 被检测,反反爬小记
    RabbitMQ单机快速安装使用
    NFS配置使用
    通过Rsync实现文件远程备份
    MySQL笔记-高可用方案
    Redis主从同步、哨兵模式、集群模式配置
    Redis维护笔记
    MongoDB笔记03——分片集群
  • 原文地址:https://www.cnblogs.com/orchidzjl/p/4691224.html
Copyright © 2011-2022 走看看