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

    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
    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    using namespace std;
    #define N 100005
    #define INF 0x3f3f3f3f
    int a[N],b[N],c[N];
    int main()
    {
        int n,i;
        while(~scanf("%d",&n))
        {
            memset(c,0,sizeof(c));
            memset(b,0,sizeof(b));
            for(i = 1; i<= n ; i++) scanf("%d",&a[i]);
            b[1] = 1;
            for(i = 2 ; i <= n ; i++){
                if(a[i]>a[i-1]) b[i] = b[i-1]+1;
                else b[i] = 1;
            }
            c[n] = 1;
            for(i = n-1 ; i>=1 ; i--){
                if(a[i]<a[i+1]) c[i] = c[i+1] +1;
                else c[i] = 1;
            }
            int ans = max(c[2]+1,b[n-1]+1);
            for(i = 2 ; i < n ; i++){
                if(a[i+1]>a[i-1]+1) ans = max(ans,b[i-1]+c[i+1]+1);
                else ans = max(ans,max(b[i-1]+1,c[i+1]+1));
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
  • 相关阅读:
    Codeforces Round #256 (Div. 2/B)/Codeforces448B_Suffix Structures(字符串处理)
    【android】优秀的UI资源站点集合
    升级iOS8系统后,保险箱Pro、私人保险箱、私密相冊打开就闪退的官方解决方式
    js产生随机数
    java实现各种数据统计图(柱形图,饼图,折线图)
    Matlab画图-非常具体,非常全面
    Lucene教程具体解释
    NAND FLASH
    Jenkins(二)
    iOS 本地通知
  • 原文地址:https://www.cnblogs.com/llei1573/p/3852959.html
Copyright © 2011-2022 走看看