zoukankan      html  css  js  c++  java
  • Maximum Subsequence Sum

    Given a sequence of K integers { N1​​, N2​​, ..., NK​​ }. A continuous subsequence is defined to be { Ni​​, Ni+1​​, ..., Nj​​ } where 1. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

    Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

    Input Specification:

    Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤). The second line contains K numbers, separated by a space.

    Output Specification:

    For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

    Sample Input:

    10
    -10 1 2 3 4 -5 -23 3 7 -21
    

    Sample Output:

    10 1 4

    题目描述略坑

     1 #include<iostream>
     2 #include<cstdio>
     3 using namespace std;
     4 
     5 const int MAXN=10005;
     6 const int INF=0x7f7f7f7f;
     7 int k,x,y,maxn=-INF;
     8 int a[MAXN],f[MAXN];
     9 
    10 int main()
    11 {
    12     scanf("%d",&k);
    13     for(int i=1;i<=k;i++)
    14         scanf("%d",&a[i]);
    15     f[0]=-1;
    16     for(int i=1;i<=k;i++)
    17     {
    18         f[i]=max(f[i-1]+a[i],a[i]);
    19         if(f[i]>maxn)
    20             y=i,maxn=f[i];
    21     }
    22     for(x=y;x>=1;x--)
    23         if(f[x-1]<0) break;
    24     if(maxn<0)
    25         printf("0 %d %d",a[1],a[k]);
    26     else
    27         printf("%d %d %d",maxn,a[x],a[y]);
    28     return 0;
    29 }
  • 相关阅读:
    docker在Linux环境下的安装
    docker在Windows环境下的安装
    tcpdump和windump
    Centos7下安装Elasticsearch 5.6.6
    使用concurrent.futures模块并发,实现进程池、线程池
    Nginx配置Gzip
    linux常用命令
    Linux下文档与目录结构
    快速读取大文件的几种方式
    linux 将大文件分解为多个小文件
  • 原文地址:https://www.cnblogs.com/InWILL/p/10524736.html
Copyright © 2011-2022 走看看