zoukankan      html  css  js  c++  java
  • HDU--杭电--3415--Max Sum of Max-K-sub-sequence--暴力或单调队列

    Max Sum of Max-K-sub-sequence

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4913    Accepted Submission(s): 1791


    Problem Description
    Given a circle sequence A[1],A[2],A[3]......A[n]. Circle sequence means the left neighbour of A[1] is A[n] , and the right neighbour of A[n] is A[1].
    Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequence means a continuous non-empty sub-sequence which length not exceed K.
     


     

    Input
    The first line of the input contains an integer T(1<=T<=100) which means the number of test cases.
    Then T lines follow, each line starts with two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).
     


     

    Output
    For each test case, you should output a line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the minimum start position, if still more than one , output the minimum length of them.
     


     

    Sample Input
    4 6 3 6 -1 2 -6 5 -5 6 4 6 -1 2 -6 5 -5 6 3 -1 2 -6 5 -5 6 6 6 -1 -1 -1 -1 -1 -1
     


     

    Sample Output
    7 1 3 7 1 3 7 6 2 -1 1 1

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace
    std;
    int
    a[222222],sum[222222]={0},que[222222];
    int
    main (void)
    {

        int
    t,n,m,s,i,j,k,l,max,aa,ss,qian,hou;
        scanf("%d",&t);
        while
    (
    t--&&scanf("%d%d",&n,&m))
        {

            for
    (
    i=1;i<=n;i++)
                scanf("%d",&a[i]),a[n+i]=a[i];
            for
    (
    i=1,sum[0]=0;i<=n+m;i++)
                sum[i]=sum[i-1]+a[i];  //把总和记录下来
            max=-1000000,aa=ss=qian=hou=0;
            for
    (
    i=1;i<n+m;i++)
            {

                while
    (
    qian<hou&&sum[i-1]<sum[que[hou-1]])hou--;  //保持单调
                que[hou++]=i-1;
                while
    (
    qian<hou&&i-que[qian]>m)qian++;  //保持长度
                if
    (
    sum[i]-sum[que[qian]]>max)
                {

                    max=sum[i]-sum[que[qian]];
                    aa=que[qian]+1;ss=i;
                }
            }

            if
    (
    aa>n)aa-=n;
            if
    (
    ss>n)ss-=n;
            printf("%d %d %d ",max,aa,ss);
        }

        return
    0;
    }

  • 相关阅读:
    spark性能调优 数据倾斜 内存不足 oom解决办法
    python2的中文编码
    spark UDF函数
    spark cache table
    spark 创建稀疏向量和矩阵
    mysql 分组排序
    给pyspark 设置新的环境
    CF662C Binary Table
    bzoj 4310 跳蚤
    3.29省选模拟赛 除法与取模 dp+组合计数
  • 原文地址:https://www.cnblogs.com/jiangu66/p/3223526.html
Copyright © 2011-2022 走看看