• Codeforces Round #FF (Div. 2):C. DZY Loves Sequences


    C. 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.


    题意就是, 你能够改变字符串中的一个字符。 就出其最长的连续字串

    如案列, 7 2 3 1 5 6 —————7 2 3  4 5 6

    输出即为5.




    
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #include<vector>
    #include<queue>
    #include<sstream>
    #include<cmath>
    
    using namespace std;
    
    #define f1(i, n) for(int i=0; i<n; i++)
    #define f2(i, m) for(int i=1; i<=m; i++)
    #define f3(i, n) for(int i=n; i>=1; i--)
    #define f4(i, n) for(int i=1; i<=n; i++)
    #define f5(i, n) for(int i=2; i<=n; i++)
    #define M 1005
    
    const int INF = 0x3f3f3f3f;
    int n, a[100005], b[100005];
    
    int main()
    {
        cin>>n;
        f4(i, n)
        cin>>a[i];
        b[1]=1;
        f5(i, n)
        {
            b[i]=1;
            if (a[i]>a[i-1])
                b[i]=b[i-1]+1;
        }
        int ans=-INF;
        f3(i, n)
        {
            if (b[i]==n)
                ans=max(b[i], ans);
            else
                ans=max(ans, b[i]+1);
            if (a[i-b[i]+1]-1>a[i-b[i]-1])
                ans=max(ans,b[i]+b[i-b[i]]);
            if (a[i-b[i]+2]-1>a[i-b[i]])
                ans=max(ans,b[i]+b[i-b[i]]);
        }
        cout<<ans<<endl;
        return 0;
    }
    


  • 相关阅读:
    CentOS7设置tomcat开机自启动
    MySQL复制表结构和表数据
    SpEL表达式总结
    Flink快速入门
    Flink架构、原理与部署测试
    第十一篇:Spark SQL 源码分析之 External DataSource外部数据源
    第十篇:Spark SQL 源码分析之 In-Memory Columnar Storage源码分析之 query
    第九篇:Spark SQL 源码分析之 In-Memory Columnar Storage源码分析之 cache table
    第八篇:Spark SQL Catalyst源码分析之UDF
    第七篇:Spark SQL 源码分析之Physical Plan 到 RDD的具体实现
  • 原文地址:https://www.cnblogs.com/yxwkf/p/5170048.html
Copyright © 2020-2023  润新知